2.1 The Artin–Tate pairing

We briefly summarize the definition of the generalized pairing $\langle \cdot \,,\,\cdot \rangle _{\text{AT}}$ from [Reference JahnJah15, §§2 and 3]. Let $X$ be a geometrically connected, smooth, projective variety of even dimension $2d$ over $\mathbf{F}_{q}$. Jahn defines the higher Brauer group

$$\begin{eqnarray}\operatorname{Br}^{d}(X):=H_{L}^{2d+1}(X;\mathbf{Z}(d)),\end{eqnarray}$$

where $H_{L}$ denotes Lichtenbaum cohomology. By [Reference JahnJah15, Lemma 1] we have the following interpretation of its non-divisible quotient for $\ell \neq p$:

$$\begin{eqnarray}\operatorname{Br}^{d}(X)_{\operatorname{ nd}}[\ell ^{\infty }]\cong H_{\acute{\text{e}}\text{t}}^{2d+1}(X;\mathbf{Z}_{\ell }(d))_{\operatorname{tors}}.\end{eqnarray}$$

The pairing $\langle \cdot \,,\,\cdot \rangle _{\text{AT}}$ on $\operatorname{Br}^{d}(X)_{\operatorname{nd}}[\ell ^{\infty }]$ is defined as follows. For any abelian group $G$, let $G_{\operatorname{nd}}$ denote its non-divisible quotient (i.e. the quotient by the maximal divisible subgroup). Let

$$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D6FF}}:H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }(d))_{\operatorname{nd}}\rightarrow H_{\acute{\text{e}}\text{t}}^{2d+1}(X;\mathbf{Z}_{\ell }(d))_{\operatorname{tors}}\end{eqnarray}$$

be the boundary map induced by the short exact sequence

$$\begin{eqnarray}0\rightarrow \mathbf{Z}_{\ell }(d)\rightarrow \mathbf{Q}_{\ell }(d)\rightarrow \mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }(d)\rightarrow 0.\end{eqnarray}$$

The map $\widetilde{\unicode[STIX]{x1D6FF}}$ is an isomorphism, so it suffices to define a pairing on $H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }(d))_{\operatorname{nd}}$. Now the key point is that $X$ has a Poincaré duality of dimension $4d+1$, since $X_{\overline{\mathbf{F}}_{q}}$ has a Poincaré duality of dimension $4d$ and $\operatorname{Spec}\mathbf{F}_{q}$ has a Poincaré duality of dimension $1$. (This may be deduced directly from the Hochschild–Serre spectral sequence and the usual Poincaré duality for $X_{\overline{\mathbf{F}}_{q}}$.) In particular, there is a canonical isomorphism

$$\begin{eqnarray}\int :H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }(2d))\xrightarrow[{}]{{\sim}}\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }.\end{eqnarray}$$

Definition 2.1. For $x,y\in H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }(d))_{\operatorname{nd}}$, we define

$$\begin{eqnarray}\langle x,y\rangle _{\text{AT}}:=\int (x\smallsmile \widetilde{\unicode[STIX]{x1D6FF}}y).\end{eqnarray}$$

From Poincaré duality and the fact that $\widetilde{\unicode[STIX]{x1D6FF}}$ is an isomorphism, it is evident that this pairing is non-degenerate. It is also skew-symmetric: this is proved in [Reference JahnJah15, §3], and it also follows from combining Propositions 2.3 and 2.5 below.

2.2 The analogy to the linking form

An analogous pairing exists on any orientable manifold $M$ of odd dimension $4d+1$, and is called the linking form. Our approach was inspired by ideas of Browder used to study a variant of the linking form in [Reference BrowderBro62].

Actually, to make the analogy sharper it is better to work in a slightly more general setup. We do not assume that $M$ is orientable, but we do assume that the orientation sheaf of $M$ is given by the tensor square of a $\mathbf{Z}_{\ell }$-local system ${\mathcal{L}}$. Then there is a pairing on $H^{2d}(M;\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }\otimes {\mathcal{L}})_{\operatorname{nd}}\cong H^{2d+1}(M;\mathbf{Z}_{\ell }\otimes {\mathcal{L}})_{\operatorname{tors}}$ given by

$$\begin{eqnarray}\langle x,y\rangle :=\int x\smallsmile \widetilde{\unicode[STIX]{x1D6FF}}y,\end{eqnarray}$$

where $\widetilde{\unicode[STIX]{x1D6FF}}$ is the analogous boundary map to that in §2.1 and

$$\begin{eqnarray}\int :H^{4d+1}(M;\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }\otimes {\mathcal{L}}^{\otimes 2})_{\operatorname{ nd}}\xrightarrow[{}]{{\sim}}\mathbf{Q}_{\ell }/\mathbf{Z}_{\ell }\end{eqnarray}$$

is the isomorphism furnished by Poincaré duality. The same argument shows that this pairing is skew-symmetric (this is why we require the dimension to be $1\text{ mod }4$). This pairing is called the linking form.Footnote ⁴

We were informed by an anonymous referee that the linking form on an orientable smooth 5-manifold should be alternating if and only if the manifold admits a spin$^{c}$-structure. We were not previously aware of this fact, nor have we been able to locate a proof in the literature, but after hearing it we realized that our method yields a necessary and sufficient criterion for the linking form on any odd-dimensional topological manifold (with orientation sheaf of the above form) to be alternating, which recovers the aforementioned result for orientable smooth 5-manifolds. This will be explained in §7.3. Although our paper is phrased for étale cohomology, the reader can check that every one of the results has a corresponding statement for the singular cohomology of manifolds, which is either easier to prove or already a known theorem.

2.3 An auxiliary pairing

We define an auxiliary pairing on the group $H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))$. As in §2.1 there is a Poincaré duality for $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2^{n}\mathbf{Z}(\ast ))$, which means in particular that there is a fundamental class inducing an isomorphism

$$\begin{eqnarray}\int :H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2d))\xrightarrow[{}]{{\sim}}\mathbf{Z}/2^{n}\mathbf{Z}.\end{eqnarray}$$

Definition 2.2. We have the short exact sequence of sheaves on $X$:

$$\begin{eqnarray}0\rightarrow \mathbf{Z}/2^{n}\mathbf{Z}(d)\rightarrow \mathbf{Z}/2^{2n}\mathbf{Z}(d)\rightarrow \mathbf{Z}/2^{n}\mathbf{Z}(d)\rightarrow 0\end{eqnarray}$$

inducing a boundary map

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}:H_{\acute{\text{e}}\text{t}}^{i}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))\rightarrow H_{\acute{\text{e}}\text{t}}^{i+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d)).\end{eqnarray}$$

We define the pairing

$$\begin{eqnarray}\langle \cdot ,\cdot \rangle _{n}:H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))\times H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))\rightarrow \mathbf{Z}/2^{n}\mathbf{Z}\end{eqnarray}$$

by

$$\begin{eqnarray}\langle x,y\rangle _{n}:=\int x\smallsmile \unicode[STIX]{x1D6FD}y.\end{eqnarray}$$

Proof. The assertion is equivalent to

$$\begin{eqnarray}x\smallsmile \unicode[STIX]{x1D6FD}y+y\smallsmile \unicode[STIX]{x1D6FD}x=0.\end{eqnarray}$$

Since $\unicode[STIX]{x1D6FD}$ is a derivation, we have $x\smallsmile \unicode[STIX]{x1D6FD}y+y\smallsmile \unicode[STIX]{x1D6FD}x=\unicode[STIX]{x1D6FD}(x\smallsmile y)$. Then the result follows from the next lemma.◻

Proof. By the obvious long exact sequence, the image is the kernel of

$$\begin{eqnarray}[2^{n}]:H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2d))\rightarrow H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{2n}\mathbf{Z}(2d)),\end{eqnarray}$$

which is identified with the inclusion $2^{n}\mathbf{Z}/2^{2n}\mathbf{Z}{\hookrightarrow}\mathbf{Z}/2^{2n}\mathbf{Z}$ by Poincaré duality.◻

Proposition 2.5. The boundary map $H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))\rightarrow H_{\acute{\text{e}}\text{t}}^{2d+1}(X;\mathbf{Z}_{2}(d))$ induced by the short exact sequence

$$\begin{eqnarray}0\rightarrow \mathbf{Z}_{2}(d)\xrightarrow[{}]{2^{n}}\mathbf{Z}_{2}(d)\rightarrow \mathbf{Z}/2^{n}\mathbf{Z}(d)\rightarrow 0\end{eqnarray}$$

surjects onto $H_{\acute{\text{e}}\text{t}}^{2d+1}(X,\mathbf{Z}_{2}(d))[2^{n}]$. Moreover, it is compatible for the pairings $\langle \cdot ,\cdot \rangle _{n}$ and $\langle \cdot ,\cdot \rangle _{\text{AT}}$ in the sense that the following diagram commutes.

Proof. The first claim is immediate from the long exact sequence. For the second claim, we will apply the following observation, which is an immediate consequence of naturality for the cup product: given a map of short exact sequences of sheaves

and multiplications fitting into a commutative diagram

then for $a\in H^{\ast }(X;A),c\in H^{\ast }(X;C)$, we have $h(a\smallsmile c)=f(a)\smallsmile h(c)$.

We apply this observation to each of the maps of short exact sequences in the following commutative diagram of sheaves.

Denote by $\widetilde{\unicode[STIX]{x1D6FD}}$ the boundary map in cohomology corresponding to the middle horizontal sequence, recalling that $\unicode[STIX]{x1D6FD}$ and $\widetilde{\unicode[STIX]{x1D6FF}}$ denote the boundary maps for the top and bottom horizontal sequences, respectively. The observation applied to the upper map of sequences shows that for $x,y\in H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))$ we have

$$\begin{eqnarray}x\smallsmile \unicode[STIX]{x1D6FD}(y)=x\smallsmile \widetilde{\unicode[STIX]{x1D6FD}}(y).\end{eqnarray}$$

The observation applied to the lower map of sequences shows that

$$\begin{eqnarray}x\smallsmile \widetilde{\unicode[STIX]{x1D6FD}}(y)\mapsto \biggl[\frac{1}{2^{n}}\biggr]_{\ast }(x)\smallsmile \widetilde{\unicode[STIX]{x1D6FF}}(y).\end{eqnarray}$$

Combining these equations yields the desired conclusion. ◻

Proposition 2.5 immediately implies the following.

Hence to prove Theorem 1.3 we are reduced to proving the following.

The proof of Theorem 2.7 will be the focus of the rest of the paper.

3.1 The Steenrod algebra in topology

We begin with a motivational pitch about Steenrod operations in algebraic topology. An old and fundamental observation in algebraic topology is that the singular cohomology of any space with $\mathbf{Z}/2\mathbf{Z}$ coefficients carries a natural module structure over a ring called the Steenrod algebra, and that it is fruitful to understand this additional structure. The Steenrod algebra may be characterized abstractly as the algebra of stable cohomology operations on $H^{\ast }(-;\mathbf{Z}/2\mathbf{Z})$, i.e. all natural transformations $H^{j}(-;\mathbf{Z}/2\mathbf{Z})\rightarrow H^{k}(-;\mathbf{Z}/2\mathbf{Z})$ commuting with the suspension isomorphisms.

More concretely, one can exhibit a set of cohomology operations $\operatorname{Sq}^{i}$, which generate the Steenrod algebra and which admit an explicit description in terms of homotopies defined on the cochain complex of a topological space, whose existence has to do with the failure of the cup product to be commutative at the level of cochains. This will be explained in §3.3.

A key point in this paper is that we can and should ask about the analogous structure for $H^{\ast }(-;\mathbf{Z}/2^{n}\mathbf{Z})$ for every $n$. In particular, we need analogs of the $\operatorname{Sq}^{i}$ for $\mathbf{Z}/2^{n}\mathbf{Z}$-coefficients. This leads to a construction of operations that we call $\widetilde{\operatorname{Sq}}^{i}$. These turn out to all be induced by the $\operatorname{Sq}^{i}$, so they are not fundamentally new operations. However, they do come up very directly in our calculations, so it will be useful to spell them out explicitly.

3.2 Étale homotopy theory

Using étale homotopy theory, we will be able to transport our definition of (singular) cohomology operations on topological spaces to étale cohomology of algebraic varieties. Here we just summarize the facts that we need.

To any locally noetherian scheme $X$ there is attached a pro-object in simplicial sets, which is called its étale topological type,Footnote ⁵ and which we denote $\acute{\text{E}}\text{t}(X)$. We refer to [Reference FriedlanderFri82, Definition 4.4] for the definition of $\acute{\text{E}}\text{t}(X)$. Given the awkwardness of the expression ‘pro-(simplicial set)’, we will henceforth use the phrase ‘pro-space’ to denote a pro-object in simplicial sets (however, it will be important at certain points that our ‘spaces’ are really simplicial sets).

Definition 3.4. We define the cochain complex of a pro-space $\{T^{i}\}$ with coefficients in a local coefficient system ${\mathcal{F}}$ to be the direct limit of the levelwise cochain complexes:

$$\begin{eqnarray}C^{\ast }(\{T^{i}\};{\mathcal{F}}):=\mathop{\varinjlim }\nolimits_{i}C^{\ast }(T^{i};{\mathcal{F}}).\end{eqnarray}$$

By the exactness of filtered colimits, we have

$$\begin{eqnarray}H^{\ast }(C^{\ast }(\{T^{i}\};{\mathcal{F}}))\cong \mathop{\varinjlim }\nolimits_{i}H^{\ast }(C^{\ast }(\{T^{i}\};{\mathcal{F}})),\end{eqnarray}$$

so this recovers the definition of the cohomology of a pro-space $\{T^{i}\}$ in [Reference FriedlanderFri82, Definition 5.1] as the direct limit of the levelwise cohomology.

In particular, if $\acute{\text{E}}\text{t}(X)=\{U^{i}:i\in I\}$ then

$$\begin{eqnarray}C^{\ast }(\acute{\text{E}}\text{t}(X);{\mathcal{F}}):=\mathop{\varinjlim }\nolimits_{i}C^{\ast }(U^{i};{\mathcal{F}})\quad \text{and}\quad H^{\ast }(\acute{\text{E}}\text{t}(X);{\mathcal{F}}):=\mathop{\varinjlim }\nolimits_{i}H^{\ast }(U^{i};{\mathcal{F}}).\end{eqnarray}$$

Proposition 3.5 [Reference FriedlanderFri82, Proposition 5.9].

If ${\mathcal{F}}$ is a locally constant sheaf on $X$ and $\acute{\text{E}}\text{t}({\mathcal{F}})$ is the corresponding local coefficient system on $\acute{\text{E}}\text{t}(X)$ under the equivalence of categories in Proposition 3.3, then there is a natural isomorphism

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X;{\mathcal{F}})\cong H^{\ast }(\acute{\text{E}}\text{t}(X);\acute{\text{E}}\text{t}({\mathcal{F}})).\end{eqnarray}$$

3.3 Steenrod’s cup-$i$ product

Let $X$ be a topological space. Let $R$ be a local coefficient system in commutative rings, and $C^{\ast }(X;R)$ the singular cochain complex. Steenrod defined sequence of maps

$$\begin{eqnarray}\displaystyle \text{cup}_{i}:C^{r}(X;R)\otimes C^{s}(X;R) & \rightarrow & \displaystyle C^{r+s-i}(X;R)\nonumber\\ \displaystyle u\otimes v & \mapsto & \displaystyle u\smallsmile _{i}v\nonumber\end{eqnarray}$$

called the ‘cup-$i$ products’. We will give a high-level exposition; a reference for this standard (in topology) material is [Reference Mosher and TangoraMT68, ch. 2].

The cup product for $X$ is induced at the level of chain complexes by the composition of the Alexander–Whitney map and the restriction to the diagonal:

(3.3.1)$$\begin{eqnarray}C^{\ast }(X;R)\leftarrow C^{\ast }(X\times X;R)\xleftarrow[{}]{{\sim}}C^{\ast }(X;R)\otimes C^{\ast }(X;R).\end{eqnarray}$$

This composition is not $S_{2}$-equivariant because the Alexander–Whitney map is not $S_{2}$-equivariant; it is only $S_{2}$-equivariant up to homotopy. However, there is a way to rectify it to be an $S_{2}$-equivariant quasi-isomorphism, which we now describe.

Let $\text{E}S_{2}$ be a contractible space with a free $S_{2}$-action; in fact, let us take the explicit model $\text{E}S_{2}=S^{\infty }$. We view $C_{\ast }(\text{E}S_{2};\text{}\underline{R})$ as a cochain complex in non-positive degrees, which provides a free resolution as $S_{2}$-modules of the constant sheaf $\text{}\underline{R}$ in degree $0$. Then there is an $S_{2}$-equivariant quasi-isomorphism

(3.3.2)$$\begin{eqnarray}C^{\ast }(X\times X;R)\simeq C^{\ast }(X;R)\otimes C^{\ast }(X;R)\otimes C_{\ast }(\text{E}S_{2};\text{}\underline{R}),\end{eqnarray}$$

where the $S_{2}$ action on the right-hand side is diagonal for the ‘swap’ action on $C^{\ast }(X;R)\otimes C^{\ast }(X;R)$ and the tautological action on $\text{E}S_{2}$. Tensoring (3.3.2) with $C^{\ast }(\text{E}S_{2};\text{}\underline{R})$ and applying the evaluation pairing yields an $S_{2}$-equivariant cochain map

(3.3.3)$$\begin{eqnarray}C^{\ast }(X;R)\otimes C^{\ast }(\text{E}S_{2};\text{}\underline{R})\leftarrow C^{\ast }(X;R)\otimes C^{\ast }(X;R),\end{eqnarray}$$

where the $S_{2}$-action is via ‘swap’ on the right-hand side, and the tautological action on $C^{\ast }(\text{E}S_{2};\text{}\underline{R})$ on the left-hand side.

Now we use the presentation of $S^{\infty }$ as a simplicial complex with two cells $d_{i}$ and $Td_{i}$ in every dimension, which are interchanged under the $S_{2}$-action. In the chain complex $C_{\ast }(\text{E}S_{2};\text{}\underline{R})$ we then have two corresponding generators $e_{i}\otimes 1$ and $Te_{i}\otimes 1\in C_{i}(\text{E}S_{2};\text{}\underline{R})$. Contracting (3.3.3) with $e_{i}\otimes 1$ gives the cup-$i$ product

$$\begin{eqnarray}C^{r+s-i}(X;R)\leftarrow C^{r}(X;R)\otimes C^{s}(X;R):\text{cup}_{i}.\end{eqnarray}$$

We will also use the notation

$$\begin{eqnarray}u\smallsmile _{i}v:=\text{cup}_{i}(u\otimes v).\end{eqnarray}$$

We have the coboundary formula [Reference Mosher and TangoraMT68, ch. 2, p. 16]

(3.3.4)$$\begin{eqnarray}d(u\smallsmile _{i}v)=(-1)^{i}\,du\smallsmile _{i}v+(-1)^{i+r}u\smallsmile _{i}dv-(-1)^{i}u\smallsmile _{i-1}v-(-1)^{rs}v\smallsmile _{i-1}u,\end{eqnarray}$$

where $|u|=r,|v|=s$. We can rewrite (3.3.4) as

(3.3.5)$$\begin{eqnarray}d(u\smallsmile _{i}v)-(-1)^{i}\,\text{cup}_{i}(d(u\otimes v))=(-1)^{i-1}u\smallsmile _{i-1}v-(-1)^{rs}v\smallsmile _{i-1}u.\end{eqnarray}$$

It is the case $i=1$ in (3.3.5) that will be most important for us. For concreteness, let us spell out the informal meaning of (3.3.5). The cup-$0$ product is just the multiplication on cochains. The cup-$1$ product furnishes a chain homotopy between $u\smallsmile _{0}v$ and $\pm v\smallsmile _{0}u$ ‘witnessing’ the graded commutativity of the cup product. The cup-$2$ product furnishes a chain homotopy between $u\smallsmile _{1}v$ and $\pm v\smallsmile _{1}u$, etc.

We now turn to the task of extracting cohomology operations out of the cup-$i$ product. The cup-$i$ product does not preserve cocycles, except in characteristic $2$, so that is the simplest case in which we get cohomology operations, and we discuss it first.

3.4 Classical Steenrod squares

If $2=0$ in $R$, then it is easily checked from (3.3.4) that the operation

$$\begin{eqnarray}u\mapsto u\smallsmile _{i}u\end{eqnarray}$$

sends cocycles to cocycles and coboundaries to coboundaries, and hence descends to a cohomology operation

$$\begin{eqnarray}\operatorname{Sq}_{i}:H^{r}(X;R)\rightarrow H^{2r-i}(X;R).\end{eqnarray}$$

We then define the Steenrod square

$$\begin{eqnarray}\operatorname{Sq}^{i}:=\operatorname{Sq}_{r-i}:H^{r}(X;R)\rightarrow H^{r+i}(X;R).\end{eqnarray}$$

For $R=\mathbf{Z}/2\mathbf{Z}$, which is the case studied in [Reference Mosher and TangoraMT68, §2], this construction recovers the classical Steenrod squares.

3.5 Generalized Steenrod squares

We now drop our assumption that $2=0$ in $R$. If $u$ is a cocycle, we see from (3.3.4) that

(3.5.1)$$\begin{eqnarray}d(u\smallsmile _{i}u)=[(-1)^{i}-(-1)^{r^{2}}]u\smallsmile _{i-1}u.\end{eqnarray}$$

Suppose $2^{n}=0$in $R$. If $r-i$ is even, then (3.5.1) implies that $2^{n-1}u\smallsmile _{i}u$ is a cocycle. Furthermore, one can check that the operation $u\mapsto 2^{n-1}u\smallsmile _{i}u$ also takes coboundaries to coboundaries, and therefore descends to a cohomology operation

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}_{i}:H^{r}(X;R)\rightarrow H^{2r-i}(X;R).\end{eqnarray}$$

Definition 3.7. If $i$ is even, we define

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}:=\widetilde{\operatorname{Sq}}_{r-i}:H^{r}(X;R)\rightarrow H^{r+i}(X;R).\end{eqnarray}$$

Lemma 3.8. Continue to assume that $2^{n}=0$ in $R$. Let $\operatorname{red}_{2}:H^{\ast }(X;R)\rightarrow H^{\ast }(X;R/2R)$ be the reduction mod $2$, and let $[2^{n-1}]:H^{\ast }(X;R/2R)\rightarrow H^{\ast }(X;R)$ be the map induced by $R/2R\xrightarrow[{}]{2^{n-1}}R$. If $i$ is even, then we have

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}=[2^{n-1}]\circ \operatorname{Sq}^{i}\circ \operatorname{red}_{2}\!.\end{eqnarray}$$

Next suppose that $r-i$ is odd. In this case we do not assume a priori that $2^{n}=0$in $R$. (Although we do not need the operations where $r-i$ is odd in this paper, we construct them for the sake of completeness.) From (3.5.1) we see that if $r-i$ is odd, then $u\smallsmile _{i}u$ is a cocycle if $u$ is a cocycle. Similarly one checks that $u\mapsto u\smallsmile _{i}u$ sends coboundaries to coboundaries, hence descends to a cohomology operation

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}_{i}:H^{r}(X;R)\rightarrow H^{2r-i}(X;R).\end{eqnarray}$$

Definition 3.9. If $i$ is odd, we define

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}:=\widetilde{\operatorname{Sq}}_{r-i}:H^{r}(X;R)\rightarrow H^{r+i}(X;R).\end{eqnarray}$$

Let us elucidate the relationship between the Steenrod squares constructed in the two cases. If $u$ is cocycle, then by (3.3.4) we have

$$\begin{eqnarray}u\smallsmile _{i}u=(-1)^{i}2d(u\smallsmile _{i+1}u).\end{eqnarray}$$

Then the analog of Lemma 3.8 is as follows.

Lemma 3.10. Suppose that the complex

$$\begin{eqnarray}0\rightarrow R/2^{n}R\xrightarrow[{}]{2}R/2^{n+1}R\rightarrow R/2R\rightarrow 0\end{eqnarray}$$

is short exact. Let $\unicode[STIX]{x1D6FD}_{2,2^{n}}:H^{\ast }(X;R/2R)\rightarrow H^{\ast +1}(X;R/2^{n}R)$ be the induced boundary map, and let $\operatorname{red}_{2}$ be as in Lemma 3.8. If $i$ is odd, then we have

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}=\unicode[STIX]{x1D6FD}_{2,2^{n}}\circ \operatorname{Sq}^{i-1}\circ \operatorname{red}_{2}\!.\end{eqnarray}$$

3.6 Application to étale cohomology

Let $R:=\bigoplus _{j\in \mathbf{Z}}\mathbf{Z}/2^{n}\mathbf{Z}(j)$, viewed as a locally constant sheaf on $X$ (where ‘$(j)$’ denotes the Tate twist) valued in rings, with multiplicative structure given by the isomorphisms

$$\begin{eqnarray}\mathbf{Z}/2^{n}\mathbf{Z}(j)\otimes \mathbf{Z}/2^{n}\mathbf{Z}(j^{\prime })\xrightarrow[{}]{{\sim}}\mathbf{Z}/2^{n}\mathbf{Z}(j+j^{\prime }).\end{eqnarray}$$

Applying §3.2 and the construction of §3.5, we obtain operations

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}:H_{\acute{\text{e}}\text{t}}^{r}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))\rightarrow H_{\acute{\text{e}}\text{t}}^{r+i}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2j)).\end{eqnarray}$$

For convenience of the reader, we summarize all the facts that we shall need about the $\widetilde{\operatorname{Sq}}^{i}$ below.

Example 3.11. If $x\in H_{\acute{\text{e}}\text{t}}^{i+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))$ then $\widetilde{\operatorname{Sq}}^{i}(x)$ has the following description. Let $C_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))$ be the étale cochain complex for $X$, defined as in §3.2. There is a chain homotopy

$$\begin{eqnarray}\displaystyle \text{cup}_{1}:C_{\acute{\text{e}}\text{t}}^{r}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))\otimes C_{\acute{\text{e}}\text{t}}^{s}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j)) & \rightarrow & \displaystyle C_{\acute{\text{e}}\text{t}}^{r+s-1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2j))\nonumber\\ \displaystyle u\otimes v & \mapsto & \displaystyle u\smallsmile _{1}v\nonumber\end{eqnarray}$$

such that (by the $i=1$ case of (3.3.5))

$$\begin{eqnarray}d(\text{cup}_{1}(u\otimes v))+\text{cup}_{1}(d(u\otimes v))=u\smallsmile v-(-1)^{rs}v\smallsmile u.\end{eqnarray}$$

Let $u\in C_{\acute{\text{e}}\text{t}}^{i+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))$ be a representative for $x$. If $i$ is even, then we have

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}(x)=[2^{n-1}u\smallsmile _{1}u].\end{eqnarray}$$

Lemma 3.12. Let $[2^{n-1}]:H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z}(j))\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))$ be the map induced by the inclusion of sheaves $\mathbf{Z}/2\mathbf{Z}(j)\xrightarrow[{}]{2^{n-1}}\mathbf{Z}/2^{n}\mathbf{Z}(j)$. Let $\operatorname{red}_{2}$ be the reduction mod $2$. If $i$ is even, then

$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{i}=[2^{n-1}]\circ \operatorname{Sq}^{i}\circ \operatorname{red}_{2}\!.\end{eqnarray}$$

4.1 Higher Bockstein operations

The key technical ingredient in the proof is the study of ‘higher Bockstein operations’. These form a family of cohomology operations $\{\unicode[STIX]{x1D6FD}_{r}\}$ ‘growing off’ of the Bockstein $\unicode[STIX]{x1D6FD}$ in the following sense. We have

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{1}:=\unicode[STIX]{x1D6FD}:H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j)).\end{eqnarray}$$

The operation $\unicode[STIX]{x1D6FD}_{r}$ is only defined on the kernel of $\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{r-1}$, and its image is only defined modulo the image of $\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{r-1}$. (These operations $\unicode[STIX]{x1D6FD}_{r}$ arise as differentials in a spectral sequence, which explains this structure.) In fact we only need $\unicode[STIX]{x1D6FD}_{1}$ and $\unicode[STIX]{x1D6FD}_{2}$ for our purposes.

Remark 4.1. We must now confront the technical subtlety that $\mathbf{Z}_{\ell }$-étale cohomology is not, as defined classically, the cohomology of a cochain complex with $\mathbf{Z}_{\ell }$-coefficients, while the state of affairs is much more naturally reasoned about and phrased in terms of ‘integral cochains’. It is straightforward to translate all our statements into ones about compatible systems of $\ell$-adic sheaves; for example the short exact sequence

$$\begin{eqnarray}0\rightarrow \mathbf{Z}_{\ell }(j)\xrightarrow[{}]{\ell ^{n}}\mathbf{Z}_{\ell }(j)\rightarrow \mathbf{Z}/\ell ^{n}\mathbf{Z}(j)\rightarrow 0\end{eqnarray}$$

should be replaced by the system of sequences

$$\begin{eqnarray}0\rightarrow \mathbf{Z}/\ell ^{N}\mathbf{Z}(j)\xrightarrow[{}]{\ell ^{n}}\mathbf{Z}/\ell ^{N+n}\mathbf{Z}(j)\rightarrow \mathbf{Z}/\ell ^{n}\mathbf{Z}(j)\rightarrow 0\end{eqnarray}$$

for all $N\gg 0$. We leave this translation to the reader so as not to complicate our exposition. (It should also be possible to deal with this problem more uniformly using the pro-étale topology of Bhatt–Scholze [Reference Bhatt and ScholzeBS15].)

Recall that the Bockstein $\unicode[STIX]{x1D6FD}$ is induced by the short exact sequence of sheaves

$$\begin{eqnarray}0\rightarrow \mathbf{Z}/\ell ^{n}\mathbf{Z}(j)\xrightarrow[{}]{\ell ^{n}}\mathbf{Z}/\ell ^{2n}\mathbf{Z}(j)\rightarrow \mathbf{Z}/\ell ^{n}\mathbf{Z}(j)\rightarrow 0.\end{eqnarray}$$

Thanks to the commutative diagram

it admits the following alternative description. For $x\in H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))$ we let $\widetilde{x}\in C_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))$ be a representative for $x$, and $a$ a lift of $\widetilde{x}$ in $C_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}_{\ell }(j))$. Since $\widetilde{x}$ is a cocycle, $da$ is divisible by $\ell ^{n}$ within $C_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}_{\ell }(j))$, so we may define

$$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D6FD}}(x):=\biggl[\frac{1}{\ell ^{n}}da\biggr]\in H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}_{\ell }(j)).\end{eqnarray}$$

Then we define

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}(x):=\overline{\widetilde{\unicode[STIX]{x1D6FD}}(x)}\in H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))\end{eqnarray}$$

to be the reduction of $\widetilde{\unicode[STIX]{x1D6FD}}(x)$ mod $\ell ^{n}$. Note that $\widetilde{\unicode[STIX]{x1D6FD}}$ is the boundary map for the short exact sequence of sheaves

$$\begin{eqnarray}0\rightarrow \mathbf{Z}_{\ell }(j)\xrightarrow[{}]{\ell ^{n}}\mathbf{Z}_{\ell }(j)\rightarrow \mathbf{Z}/\ell ^{n}\mathbf{Z}(j)\rightarrow 0.\end{eqnarray}$$

Definition 4.2. We define operations

$$\begin{eqnarray}\displaystyle \widetilde{\unicode[STIX]{x1D6FD}}_{2}:(\ker \unicode[STIX]{x1D6FD}_{1}\subset H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))) & \rightarrow & \displaystyle H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}_{\ell }(j))/\operatorname{Im}\widetilde{\unicode[STIX]{x1D6FD}}_{1},\nonumber\\ \displaystyle \unicode[STIX]{x1D6FD}_{2}=\overline{\widetilde{\unicode[STIX]{x1D6FD}}_{2}}:(\ker \unicode[STIX]{x1D6FD}_{1}\subset H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))) & \rightarrow & \displaystyle H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))/\operatorname{Im}\unicode[STIX]{x1D6FD}_{1}\nonumber\end{eqnarray}$$

as follows. If $x\in \ker \unicode[STIX]{x1D6FD}_{1}$, then (keeping the notation of the preceding paragraph) we have

$$\begin{eqnarray}\overline{\frac{1}{\ell ^{n}}da}=d\widetilde{y}\quad \text{for some }\widetilde{y}\in C_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j)),\end{eqnarray}$$

where the overline means reduction mod $\ell ^{n}$. Hence we may and do choose a lift $b\in C_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}_{\ell }(j))$ of $\widetilde{y}$ such that $(1/\ell ^{n})da\equiv db\text{ mod }\ell ^{n}$, or in other words

$$\begin{eqnarray}da\equiv d(\ell ^{n}b)\hspace{0.6em}{\rm mod}\hspace{0.2em}\ell ^{2n}.\end{eqnarray}$$

Then we can form the cochain $(1/\ell ^{2n})d(a-\ell ^{n}b)\in C_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}_{\ell }(j))$, which is evidently a cocycle. Finally, we define

$$\begin{eqnarray}\displaystyle \widetilde{\unicode[STIX]{x1D6FD}}_{2}(x) & := & \displaystyle \biggl[\frac{1}{\ell ^{2n}}d(a-\ell ^{n}b)\biggr]\in H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}_{\ell }(j))/\operatorname{Im}\widetilde{\unicode[STIX]{x1D6FD}}_{1},\nonumber\\ \displaystyle \unicode[STIX]{x1D6FD}_{2}(x) & := & \displaystyle \overline{\widetilde{\unicode[STIX]{x1D6FD}}_{2}(x)}\in H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}/\ell ^{n}\mathbf{Z}(j))/\operatorname{Im}\unicode[STIX]{x1D6FD}_{1}.\nonumber\end{eqnarray}$$

We leave it to the reader to check that this is indeed well-defined.

It is straightforward to define $\unicode[STIX]{x1D6FD}_{r}$ in a similar way for all $r$. Since we only need $\unicode[STIX]{x1D6FD}_{1}$ and $\unicode[STIX]{x1D6FD}_{2}$, we do not spell out the explicit construction.

In what follows, we will focus on the operations introduced in Definition 4.2 for $\ell =2$.

Proposition 4.3. Let $\ell =2$ in Definition 4.2. For any $x\in H_{\acute{\text{e}}\text{t}}^{2k}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))$, we have the following identity:

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{2}(2^{n-1}x^{2})=[x\cdot \unicode[STIX]{x1D6FD}(x)-\widetilde{\operatorname{Sq}}^{2k}(\unicode[STIX]{x1D6FD}(x))]\in H_{\acute{\text{e}}\text{t}}^{2k+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2j))/\operatorname{Im}\unicode[STIX]{x1D6FD}.\end{eqnarray}$$

Proof. Note that since $\unicode[STIX]{x1D6FD}$ is a derivation, it indeed kills $2^{n-1}x^{2}$, hence $2^{n-1}x^{2}$ indeed lives in $\ker \unicode[STIX]{x1D6FD}$ so that $\unicode[STIX]{x1D6FD}_{2}(2^{n-1}x^{2})$ is defined.

Let $a$ be any integral cochain in $C_{\acute{\text{e}}\text{t}}^{2k}(X;\mathbf{Z}_{2}(j))$ lifting a representative for $x$. Let $y:=\unicode[STIX]{x1D6FD}(x)\in H_{\acute{\text{e}}\text{t}}^{2k+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(j))$. Then by the definition of $\unicode[STIX]{x1D6FD}$, we have $da=2^{n}b$, where $b\in C_{\acute{\text{e}}\text{t}}^{2k+1}(X;\mathbf{Z}_{2}(j))$ lifts a representative for $y$.

According to the discussion in Definition 4.2, $\unicode[STIX]{x1D6FD}_{2}(2^{n-1}x^{2})$ is calculated by finding an integral cochain lifting a representative for $2^{n-1}x^{2}$, whose boundary is divisible by $2^{2n}$. We check that $2^{n-1}a^{2}+2^{2n-1}(a\smallsmile _{1}b)$ is such an integral cochain, using Example 3.11:

$$\begin{eqnarray}\displaystyle d(2^{n-1}a^{2}+2^{2n-1}(a\smallsmile _{1}b)) & = & \displaystyle 2^{n-1}(a\cdot 2^{n}b+2^{n}b\cdot a-2^{2n}(b\smallsmile _{1}b)+2^{n}(ab-ba))\nonumber\\ \displaystyle & = & \displaystyle 2^{n-1}(2^{n+1}ab-2^{2n}(b\smallsmile _{1}b))\nonumber\\ \displaystyle & = & \displaystyle 2^{2n}(ab-2^{n-1}(b\smallsmile _{1}b)).\nonumber\end{eqnarray}$$

Hence by Definition 4.2 we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FD}_{2}(2^{n-1}x^{2}) & = & \displaystyle \biggl[\,\overline{\frac{1}{2^{2n}}d(2^{n-1}a^{2}-2^{2n-1}(a\smallsmile _{1}b))}\,\biggr]\nonumber\\ \displaystyle & = & \displaystyle [\overline{ab-2^{n-1}(b\smallsmile _{1}b)}]\nonumber\\ \displaystyle & = & \displaystyle [xy-[\overline{2^{n-1}(b\smallsmile _{1}b)}]].\nonumber\end{eqnarray}$$

We then conclude by using Example 3.11 again to identify $[\overline{2^{n-1}(b\smallsmile _{1}b)}]=\widetilde{\operatorname{Sq}}^{2k}y$.◻

Theorem 4.4. Let $X$ be a smooth projective variety over $\mathbf{F}_{q}$ of dimension $2d$. For $x\in H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d))$, we have

$$\begin{eqnarray}x\smallsmile \unicode[STIX]{x1D6FD}(x)=\widetilde{\operatorname{Sq}}^{2d}(\unicode[STIX]{x1D6FD}(x))\in H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2d)).\end{eqnarray}$$

Proof. By Proposition 4.3, we have the identity

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{2}(2^{n-1}x^{2})=[x\cdot \unicode[STIX]{x1D6FD}(x)-\widetilde{\operatorname{Sq}}^{2k}(\unicode[STIX]{x1D6FD}(x))]\in H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2d))/\operatorname{Im}\unicode[STIX]{x1D6FD}.\end{eqnarray}$$

Therefore we will be done if we can show that the images of $\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D6FD}_{2}$ in $H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2d))$ are both $0$. Since $\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D6FD}_{2}$ are the reductions of $\widetilde{\unicode[STIX]{x1D6FD}}$ and $\widetilde{\unicode[STIX]{x1D6FD}}_{2}$, it suffices to prove the stronger statement that $\widetilde{\unicode[STIX]{x1D6FD}}$ and $\widetilde{\unicode[STIX]{x1D6FD}}_{2}$ vanish in the appropriate degree, which is what we shall do.

Note that the image of $\widetilde{\unicode[STIX]{x1D6FD}}$ is automatically $2^{n}$-torsion. Similarly, from the definition of $\widetilde{\unicode[STIX]{x1D6FD}}_{2}$ it is immediate that its image is $2^{2n}$-torsion. Indeed, referring to Definition 4.2 we see that $\ell ^{2n}\widetilde{\unicode[STIX]{x1D6FD}}_{2}(x)=[d(a-\ell ^{n}b)]$ is manifestly a coboundary. (In general, $\operatorname{Im}(\widetilde{\unicode[STIX]{x1D6FD}}_{r})$ is $2^{rn}$-torsion.) But by Poincaré duality we have

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}_{2}^{n}(2d))\cong \mathbf{Z}_{2}\end{eqnarray}$$

is torsion-free, so the images of $\widetilde{\unicode[STIX]{x1D6FD}}$ and $\widetilde{\unicode[STIX]{x1D6FD}}_{2}$ in $H_{\acute{\text{e}}\text{t}}^{4d+1}(X;\mathbf{Z}/2^{n}\mathbf{Z}(2d))$ are necessarily $0$.◻

5.1 Cohomology with supports

Let $i:X{\hookrightarrow}Y$ be a closed subscheme. We recall the definition of the ‘cohomology of $Y$ with supports in $X$’ [Reference Freitag and KiehlFK88, § I.10]. Let $j:U{\hookrightarrow}Y$ denote the inclusion of the (open) complement of $X$ in $Y$. Then $H_{X}^{k}(Y;{\mathcal{F}})$ is defined to be the $k$th right derived functor of

$$\begin{eqnarray}{\mathcal{F}}\mapsto \ker ({\mathcal{F}}(Y)\mapsto j_{\ast }j^{\ast }{\mathcal{F}}(Y)).\end{eqnarray}$$

We want to define Steenrod operations on $H_{X}^{k}(Y;{\mathcal{F}})$. Since we have been in the habit of defining Steenrod operations via étale homotopy theory, we need to realize the cohomology with supports in terms of étale homotopy theory, as the cohomology of a certain pro-space. This is explained in [Reference FriedlanderFri82, §14]. The key features of this construction are summarized below.

5.2 Construction of étale Stiefel–Whitney classes

Let $k$ be a field of characteristic not equal to $2$. Let $i:X{\hookrightarrow}Y$ be a codimension $r$ closed embedding of smooth varieties over $k$. Then we have a cycle class $s_{X/Y}\in H_{X}^{2r}(Y;\mathbf{Z}_{\ell }(r))$, which can be described as the image of $1$ under the Gysin isomorphism

(5.2.1)$$\begin{eqnarray}\unicode[STIX]{x1D719}:H^{j}(X;{\mathcal{F}})\xrightarrow[{}]{{\sim}}H_{X}^{j+2r}(Y;{\mathcal{F}}(r)),\end{eqnarray}$$

which holds for any locally constant constructible sheaf ${\mathcal{F}}$ on $Y$ [Reference Freitag and KiehlFK88, § I.10].

We are going to apply this with $Y$ being the total space of a vector bundle $E$ over $X$, and $i:X{\hookrightarrow}E$ being the zero section.

Definition 5.2. Let $E$ be a vector bundle over $X$. We define the jth Stiefel–Whitney class of $E$ by

(5.2.2)$$\begin{eqnarray}w_{j}(E)=\unicode[STIX]{x1D719}^{-1}(\operatorname{Sq}^{j}s_{X/E}).\end{eqnarray}$$

Define the total Stiefel–Whitney class to be $w(E):=\sum w_{i}(E)$. If no vector bundle is mentioned, then by default we set $w_{i}:=w_{i}(TX)$ and $w:=\sum w_{i}$.

There is a possibly more intuitive way to phrase (5.2.2), which we will use later. The Gysin isomorphism (5.2.1) says that $H_{X}^{\ast }(E;\mathbf{Z}/2\mathbf{Z})$ is a free rank-one module over $H_{\acute{\text{e}}\text{t}}^{\ast }(E;\mathbf{Z}/2\mathbf{Z})$, which can be identified with $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z})$ via $\unicode[STIX]{x1D70B}^{\ast }$ since $E$ is a vector bundle over $X$. Under this identification (5.2.2) is equivalent to

(5.2.3)$$\begin{eqnarray}\operatorname{Sq}^{i}(s_{X/E})=w_{i}\cdot s_{X/E}.\end{eqnarray}$$

Remark 5.3. The reason that we call these ‘Stiefel–Whitney classes’ is that Thom observed in [Reference ThomTho52] that an exactly analogous construction for manifolds produces the usual Stiefel–Whitney classes.Footnote ⁶ The construction goes as follows (a reference is [Reference Milnor and StasheffMS74, §8]). Let $M$ be a topological manifold and $E$ be a vector bundle of rank $r$ over $M$. Let $i:M{\hookrightarrow}E$ denote the inclusion of $M$ as the zero section of $E$. Let $E_{0}=E-i(M)$. We have a Thom isomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D719}:H^{i}(M;\mathbf{Z}/2\mathbf{Z})\cong H^{i+2r}(E,E_{0};\mathbf{Z}/2\mathbf{Z}),\end{eqnarray}$$

and $w_{i}(E)=\unicode[STIX]{x1D719}^{-1}(\operatorname{Sq}^{i}\unicode[STIX]{x1D719}(1))$.

5.3 Steenrod squares of Stiefel–Whitney classes

The following technical lemma is needed later in §7.1. The reader may safely skip this subsection for now and refer back to it when necessary.

Proof of Lemma 5.4.

We will use the identities of Steenrod squares from §3.4. Note that we may assume that $i<j$, since for $i>j$ we have $\operatorname{Sq}^{i}(w_{j})=0$ and for $i=j$ we have $\operatorname{Sq}^{i}(w_{j})=w_{j}^{2}$.

We induct on $j$, and then (for fixed $j$) on $i$, with the base case $j=0$ being trivial, and the base cases $i=0$ being trivial for any $j$ since $\operatorname{Sq}^{0}=\operatorname{Id}$. Consider the expression

$$\begin{eqnarray}\operatorname{Sq}^{i}\operatorname{Sq}^{j}(s_{X/TX}).\end{eqnarray}$$

On the one hand, we have by (5.2.3) that

$$\begin{eqnarray}\operatorname{Sq}^{i}\operatorname{Sq}^{j}(s_{X/TX})=\operatorname{Sq}^{i}(w_{j}\cdot s_{X/TX}).\end{eqnarray}$$

By the Cartan formula,

$$\begin{eqnarray}\displaystyle \operatorname{Sq}^{i}(w_{j}\cdot s_{X/TX}) & = & \displaystyle \mathop{\sum }_{k+\ell =i}\operatorname{Sq}^{k}(w_{j})\operatorname{Sq}^{\ell }(s_{X/TX})\nonumber\\ \displaystyle & = & \displaystyle \biggl(\operatorname{Sq}^{i}(w_{j})+\mathop{\sum }_{k<i}\operatorname{Sq}^{k}(w_{j})w_{i-k}\biggr)s_{X/TX}.\nonumber\end{eqnarray}$$

By the induction hypothesis, $\operatorname{Sq}^{k}(w_{j})$ is a polynomial in the $\{w_{l}\}$ for each $k<i$, so the upshot is that

(5.3.1)$$\begin{eqnarray}\operatorname{Sq}^{i}\operatorname{Sq}^{j}(s_{X/TX})=(\operatorname{Sq}^{i}(w_{j})+\text{Poly}(\{w_{l}\}))s_{X/TX}.\end{eqnarray}$$

From (5.3.1) it is clearly sufficient to show that $\operatorname{Sq}^{i}\operatorname{Sq}^{j}(s_{X/TX})$ is a polynomial in the $\{w_{l}\}$ times $s_{X/TX}$. For this we use the Adem relations: for $0<i<2j$ we have

$$\begin{eqnarray}\operatorname{Sq}^{i}\operatorname{Sq}^{j}=\mathop{\sum }_{k=0}^{[i/2]}\binom{j-1-k}{i-2k}\operatorname{Sq}^{i+j-k}\operatorname{Sq}^{k},\end{eqnarray}$$

hence

$$\begin{eqnarray}\operatorname{Sq}^{i}\operatorname{Sq}^{j}(s_{X/TX})=\mathop{\sum }_{k=0}^{[i/2]}\binom{j-1-k}{i-2k}\operatorname{Sq}^{i+j-k}(w_{k}\cdot s_{X/TX}).\end{eqnarray}$$

Every index $k$ in this sum is strictly less than $j$ since we assumed $i<j$, so every summand is a polynomial in the $\{w_{l}\}$ times $s_{X/TX}$ by the induction hypothesis, which is what we wanted.◻

5.4 Properties of the Stiefel–Whitney classes

We now record that the Stiefel–Whitney classes, as constructed in §5.2, enjoy the usual properties of topological Stiefel–Whitney classes.

1. (i) We have $w_{i}(E)\in H^{i}(X;\mathbf{Z}/2\mathbf{Z})$, with $w_{0}=1$ and $w_{i}=0$ for $i>2\operatorname{rank}E$.

2. (ii) (naturality) If $f:X^{\prime }\rightarrow X$ then

   $$\begin{eqnarray}f^{\ast }w_{i}(E)=w_{i}(f^{\ast }E).\end{eqnarray}$$

3. (iii) (Whitney product formula) We have

   $$\begin{eqnarray}w_{i}(E\oplus E^{\prime })=\mathop{\sum }_{k=0}^{i}w_{k}(E)\smallsmile w_{i-k}(E^{\prime }).\end{eqnarray}$$
   If we set $w=\sum w_{i}$ to be the total Stiefel–Whitney class, then this can be written more succinctly as
   $$\begin{eqnarray}w(E\oplus E^{\prime })=w(E)\cdot w(E^{\prime }).\end{eqnarray}$$

Proofs. It is well-known in the topological setting (cf. [Reference Milnor and StasheffMS74, §8]) that the characteristic properties of Stiefel–Whitney classes can be formally derived from those of the Steenrod squares. Since our étale Stiefel–Whitney classes are also based on Steenrod operations, essentially the same proofs go through. Nonetheless, we spell them out because they will be used in the proof of Theorem 5.10 below.

(i) Immediate from the fact that $\operatorname{Sq}^{0}=\operatorname{Id}$ and $\operatorname{Sq}^{i}$ vanishes on $H^{j}$ if $i>j$.

(ii) Immediate from the naturality of the Gysin map and Steenrod squares.

(iii) We begin by considering a general setup. Suppose we have two closed embeddings of smooth proper varieties

$$\begin{eqnarray}\displaystyle i:X{\hookrightarrow}Y & & \displaystyle \quad \text{codimension }r,\nonumber\\ \displaystyle i^{\prime }:X^{\prime }{\hookrightarrow}Y & & \displaystyle \quad \text{codimension }r^{\prime }.\nonumber\end{eqnarray}$$

We consider the two corresponding Gysin maps obtained:

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}:H^{\ast }(X;\mathbf{Z}/2\mathbf{Z}) & \cong & \displaystyle H_{X}^{\ast +r}(Y;\mathbf{Z}/2\mathbf{Z}),\nonumber\\ \displaystyle \unicode[STIX]{x1D719}^{\prime }:H^{\ast }(X^{\prime };\mathbf{Z}/2\mathbf{Z}) & \cong & \displaystyle H_{X^{\prime }}^{\ast +r}(Y^{\prime };\mathbf{Z}/2\mathbf{Z}).\nonumber\end{eqnarray}$$

These send $\unicode[STIX]{x1D719}(1)=s_{X/Y}$ and $\unicode[STIX]{x1D719}^{\prime }(1)=s_{X^{\prime }/Y^{\prime }}$. By the compatibility of the Gysin map for products, we have that for the closed embedding $X\times X^{\prime }{\hookrightarrow}Y\times Y$, the Gysin isomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D719}\smallsmile \unicode[STIX]{x1D719}^{\prime }:H^{\ast }(X\times X^{\prime };\mathbf{Z}/2\mathbf{Z})\cong H_{X\times X^{\prime }}^{\ast +r+r^{\prime }}(Y\times Y^{\prime };\mathbf{Z}/2\mathbf{Z})\end{eqnarray}$$

sends $1\mapsto s_{X/Y}\smallsmile s_{X^{\prime }/Y^{\prime }}$. Now taking $Y$ and $Y^{\prime }$ to be the total spaces of $E$ and $E^{\prime }$, and applying the Cartan formula of $\operatorname{Sq}$ and the definition of Stiefel–Whitney classes, we obtain

$$\begin{eqnarray}\displaystyle w_{X\times X^{\prime }}\smallsmile (s_{X/Y}\smallsmile s_{X^{\prime }/Y^{\prime }}) & = & \displaystyle \operatorname{Sq}(s_{X/Y}\smallsmile s_{X^{\prime }/Y^{\prime }})\nonumber\\ \displaystyle & = & \displaystyle \operatorname{Sq}(s_{X/Y})\smallsmile \operatorname{Sq}(s_{X^{\prime }/Y^{\prime }})\nonumber\\ \displaystyle & = & \displaystyle (w_{X}\smallsmile s_{X/Y})\smallsmile (w_{X^{\prime }}\smallsmile s_{X^{\prime }/Y^{\prime }})\nonumber\\ \displaystyle & = & \displaystyle (w_{X}\smallsmile w_{X^{\prime }})s_{X/Y}\smallsmile s_{X^{\prime }/Y^{\prime }}.\nonumber\end{eqnarray}$$

Finally, pulling back via the diagonal $\unicode[STIX]{x1D6E5}:X{\hookrightarrow}X\times X$ and using naturality gives the result.

It is formal that the Whitney product formula for direct sums implies it for extensions.

Lemma 5.7. If

$$\begin{eqnarray}0\rightarrow E^{\prime }\rightarrow E\rightarrow E^{\prime \prime }\rightarrow 0\end{eqnarray}$$

is a short exact sequence of vector bundles on $X$, then

$$\begin{eqnarray}w(E)=w(E^{\prime })\smallsmile w(E^{\prime \prime }).\end{eqnarray}$$

5.5 Lifting Stiefel–Whitney classes to integral cohomology

We shall see in §7 that it is crucial to know whether our Stiefel–Whitney classes lift to integral cohomology. The goal of this subsection is to prove Theorem 5.10, which answers this question.

Our first task is to address a technical subtlety that will come up in the proof of Theorem 5.10. There are the two short exact sequences

(5.5.1)$$\begin{eqnarray}0\rightarrow \mathbf{Z}/2\mathbf{Z}\rightarrow \mathbf{Z}/4\mathbf{Z}\rightarrow \mathbf{Z}/2\mathbf{Z}\rightarrow 0\end{eqnarray}$$

and

(5.5.2)$$\begin{eqnarray}0\rightarrow \unicode[STIX]{x1D707}_{2}\rightarrow \unicode[STIX]{x1D707}_{4}\rightarrow \unicode[STIX]{x1D707}_{2}\rightarrow 0.\end{eqnarray}$$

Since $\unicode[STIX]{x1D707}_{2}$ is canonically identified with $\mathbf{Z}/2\mathbf{Z}$, as we are not in characteristic $2$, both sequences induce Bockstein operations $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z})\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast +1}(X;\mathbf{Z}/2\mathbf{Z})$, but they are not necessarily the same. In §3.4 we noted that the Bockstein operation for (5.5.1) is $\operatorname{Sq}^{1}$. Let us denote by $\unicode[STIX]{x1D6FD}^{(1)}$ the Bockstein operation for (5.5.2). We need to quantify the difference between these two operations. For this discussion, it will help to maintain a psychological distinction between $\unicode[STIX]{x1D707}_{2}$ and $\mathbf{Z}/2\mathbf{Z}$.

Lemma 5.8. Let $\unicode[STIX]{x1D6FC}$ be the image of $1\in H_{\acute{\text{e}}\text{t}}^{0}(X;\mathbf{Z}/2\mathbf{Z})\xrightarrow[{}]{{\sim}}H_{\acute{\text{e}}\text{t}}^{0}(X;\unicode[STIX]{x1D707}_{2})$ under the boundary map $\unicode[STIX]{x1D6FD}^{(1)}$. Then for all $c\in H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z})\xrightarrow[{}]{{\sim}}H_{\acute{\text{e}}\text{t}}^{\ast }(X;\unicode[STIX]{x1D707}_{2})$ we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}^{(1)}(c)=\operatorname{Sq}^{1}(c)+\unicode[STIX]{x1D6FC}\smallsmile c.\end{eqnarray}$$

Proof. Since $\unicode[STIX]{x1D707}_{4}$ is a module over $\mathbf{Z}/4\mathbf{Z}$, the cohomology $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\unicode[STIX]{x1D707}_{4})$ is a module over $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/4\mathbf{Z})$. We similarly view $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\unicode[STIX]{x1D707}_{2})$ as a module over $H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z})$.

The reduction map $\unicode[STIX]{x1D707}_{4}\rightarrow \unicode[STIX]{x1D707}_{2}$, viewed as part of the short exact sequence (5.5.2), is compatible with the reduction map $\mathbf{Z}/4\mathbf{Z}\rightarrow \mathbf{Z}/2\mathbf{Z}$, viewed as part of (5.5.1), for the respective module structures. Hence the induced maps on cohomology satisfy the same compatibility: the reduction map

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X;\unicode[STIX]{x1D707}_{4})\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X;\unicode[STIX]{x1D707}_{2})\end{eqnarray}$$

is compatible as a map of modules with respect to the map of rings

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/4\mathbf{Z})\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z}).\end{eqnarray}$$

More precisely, let $x\in H_{\acute{\text{e}}\text{t}}^{i}(X;\unicode[STIX]{x1D707}_{2})$ and $r\in H_{\acute{\text{e}}\text{t}}^{j}(X;\mathbf{Z}/2\mathbf{Z})$, so that $x$ is viewed as a module element and $r$ is viewed as a ring element. Then $rx\in H_{\acute{\text{e}}\text{t}}^{i+j}(X;\unicode[STIX]{x1D707}_{2})$, and we are saying that

(5.5.3)$$\begin{eqnarray}\unicode[STIX]{x1D6FD}^{(1)}(rx)=(\operatorname{Sq}^{1}r)x+r\unicode[STIX]{x1D6FD}^{(1)}(x).\end{eqnarray}$$

This is seen immediately upon going back to the definition of the boundary map, using that the coboundary map on cochains is a derivation.

The lemma then follows from taking $r=c$ and $x=1\in H^{0}(X;\unicode[STIX]{x1D707}_{2})$ in (5.5.3).◻

Remark 5.9. The element $\unicode[STIX]{x1D6FC}\in H_{\acute{\text{e}}\text{t}}^{1}(X;\unicode[STIX]{x1D707}_{2})$ is actually the pullback of a universal $\unicode[STIX]{x1D6FC}\in H_{\acute{\text{e}}\text{t}}^{1}(\operatorname{Spec}\mathbf{F}_{q};\unicode[STIX]{x1D707}_{2})$, which vanishes if and only if $q\equiv 1\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$. Indeed, (5.5.1) and (5.5.2) are obviously the same for $q\equiv 1\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$. We also note for later use that $\unicode[STIX]{x1D6FC}$ lifts to $H_{\acute{\text{e}}\text{t}}^{1}(X;\mathbf{Z}_{2}(1))$, because $\unicode[STIX]{x1D6FD}^{(1)}$ is the reduction of the Bockstein for

$$\begin{eqnarray}0\rightarrow \mathbf{Z}_{2}(1)\xrightarrow[{}]{2}\mathbf{Z}_{2}(1)\rightarrow \unicode[STIX]{x1D707}_{2}\rightarrow 0.\end{eqnarray}$$

Theorem 5.10. Let $X$ be a smooth variety over a finite field $\mathbf{F}_{q}$ of characteristic not $2$ and $E$ a vector bundle on $X$ of rank $r$. Let $\unicode[STIX]{x1D6FC}$ be as in Lemma 5.8. Then we have

$$\begin{eqnarray}w(E):=1+w_{1}+w_{2}+\cdots +w_{2r}=\left\{\begin{array}{@{}ll@{}}(1+\unicode[STIX]{x1D6FC})\overline{c}_{\operatorname{even}}+\overline{c}_{\operatorname{odd}}\quad & r\text{ odd},\\ \overline{c}_{\operatorname{even}}+(1+\unicode[STIX]{x1D6FC})\overline{c}_{\operatorname{odd}}\quad & r\text{ even},\end{array}\right.\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle c_{\operatorname{even}} & = & \displaystyle 1+c_{2}+c_{4}+\cdots \in H_{\acute{\text{e}}\text{t}}^{\ast }\biggl(X;\bigoplus _{i\in \mathbf{Z}}\mathbf{Z}_{2}(i)\biggr),\nonumber\\ \displaystyle c_{\operatorname{odd}} & = & \displaystyle c_{1}+c_{3}+\cdots \in H_{\acute{\text{e}}\text{t}}^{\ast }\biggl(X;\bigoplus _{i\in \mathbf{Z}}\mathbf{Z}_{2}(i)\biggr),\nonumber\end{eqnarray}$$

and $\overline{c}$ means the reduction of $c$ modulo $2$.

Proof. Grothendieck showed [Reference GrothendieckGro58] that the definition of all characteristic classes can be obtained from the axioms in §5.4 plus the definition of the characteristic classes for arbitrary line bundles. Therefore, it suffices to check that the formula above satisfies the properties in §5.4 and is correct for all line bundles.

The fact that it satisfies axiom (i) of §5.4 is evident from the definition. The fact that it satisfies (ii) is immediate from the observation that the Chern classes satisfy the Whitney sum and naturality property. The fact that it satisfies (iii) also follows from the analogous property of Chern classes plus a case analysis of the formula. For example, when summing two bundles ${\mathcal{E}}$ and ${\mathcal{E}}^{\prime }$ of odd rank with Chern classes $c$ and $c^{\prime }$, the product of the classes claimed in the formula is

$$\begin{eqnarray}\displaystyle ((1+\unicode[STIX]{x1D6FC})\overline{c}_{\operatorname{even}}+\overline{c}_{\operatorname{odd}})((1+\unicode[STIX]{x1D6FC})\overline{c^{\prime }}_{\operatorname{even}}+\overline{c^{\prime }}_{\operatorname{odd}}) & = & \displaystyle (\overline{c}_{\operatorname{even}}\overline{c^{\prime }}_{\operatorname{even}}+\overline{c}_{\operatorname{odd}}\overline{c^{\prime }}_{\operatorname{odd}})\nonumber\\ \displaystyle & & \displaystyle +\,(1+\unicode[STIX]{x1D6FC})(\overline{c}_{\operatorname{even}}\overline{c^{\prime }}_{\operatorname{odd}}+\overline{c}_{\operatorname{odd}}\overline{c^{\prime }}_{\operatorname{even}})\nonumber\end{eqnarray}$$

because $(1+\unicode[STIX]{x1D6FC})^{2}=1$, and the Whitney sum formula for Chern classes implies that the right-hand side is indeed $c_{\operatorname{even}}({\mathcal{E}}\oplus {\mathcal{E}}^{\prime })+(1+\unicode[STIX]{x1D6FC})c_{\operatorname{odd}}({\mathcal{E}}\oplus {\mathcal{E}}^{\prime })$.

Finally, we must check the formula for line bundles. What makes this possible is that we only have to verify the formula for $w_{1}$ and $w_{2}$, since the higher Stiefel–Whitney classes vanish for degree reasons. Thus we only need to compute $\operatorname{Sq}^{1}$ and $\operatorname{Sq}^{2}$, and we have ‘explicit’ descriptions of the Steenrod operations on degree $2$ elements in these cases (§3.4).

Let $Y$ be the total space of a line bundle ${\mathcal{L}}$ on $X$. We view $X$ as embedded in $Y$ via the zero section, and identify their étale cohomology groups via pullback for the projection map $\unicode[STIX]{x1D70B}:Y\rightarrow X$.

Calculation of $w_{1}$. Recall from (5.2.3) that $w_{1}$ is defined by

$$\begin{eqnarray}\operatorname{Sq}^{1}s_{X/Y}=w_{1}\smallsmile s_{X/Y}.\end{eqnarray}$$

But the cycle class $s_{X/Y}$ lifts compatibly to $H_{X}^{2}(Y;\unicode[STIX]{x1D707}_{2^{j}})$ for all $j$, and hence even to $H_{Y}^{2}(X;\mathbf{Z}_{2}(1))$ (cf. [Reference Freitag and KiehlFK88, § II.2]). In particular, $s_{X/Y}$ lies in the image of the reduction map $H_{X}^{i}(Y;\unicode[STIX]{x1D707}_{4})\rightarrow H_{X}^{i}(Y;\unicode[STIX]{x1D707}_{2})$. The long exact sequence for (5.5.2) then shows that $\unicode[STIX]{x1D6FD}^{(1)}(s_{X/Y})=0$. We are really interested in the other boundary map $\operatorname{Sq}^{1}$, but Lemma 5.8 tells us the difference between them:

$$\begin{eqnarray}\operatorname{Sq}^{1}(s_{X/Y})=\unicode[STIX]{x1D6FD}^{(1)}(s_{X/Y})+\unicode[STIX]{x1D6FC}\smallsmile s_{X/Y}=\unicode[STIX]{x1D6FC}\smallsmile s_{X/Y}.\end{eqnarray}$$

Hence $\unicode[STIX]{x1D6FC}=w_{1}$, as required.

Calculation of $w_{2}$. The argument is essentially the same as in the proof of [Reference UrabeUra96, Lemma 2.6]. Again, (5.2.3) tells us that

$$\begin{eqnarray}\operatorname{Sq}^{2}s_{X/Y}=w_{2}\smallsmile s_{X/Y}\in H_{X}^{4}(Y;\mathbf{Z}/2\mathbf{Z}).\end{eqnarray}$$

Since $s_{X/Y}\in H_{X}^{2}(Y;\mathbf{Z}/2\mathbf{Z})$ we have that $\operatorname{Sq}^{2}s_{X/Y}=s_{X/Y}\smallsmile s_{X/Y}$ (using one of the explicit ‘special cases’ from §3.4).

We now need to recall a property of the cycle class, which is a special case of a more general discussion to come in §6.3.2. If $X{\hookrightarrow}Y$ is a codimension 1 closed embedding of smooth varieties, then we have a cycle class $\text{cl}_{Y}(X)\in H^{2}(Y;\unicode[STIX]{x1D707}_{2})$, which is the image of the line bundle ${\mathcal{O}}_{Y}(X)$ under the Chern class map $H^{1}(Y;\mathbf{G}_{m})\rightarrow H^{2}(Y;\unicode[STIX]{x1D707}_{2})$. This class $\text{cl}_{Y}(X)$ also coincides with the image of $s_{X/Y}$ under the natural map $H_{X}^{\ast }(Y)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(Y)$. (A reference is [Reference Freitag and KiehlFK88, Propositions II.2.2 and II.2.6].Footnote ⁷)

Consider the following commutative diagram.

An elementary calculation shows that the line bundle ${\mathcal{L}}$ on $X$ whose total space is $Y$ pulls back to ${\mathcal{O}}_{Y}(X)$ on $Y$, i.e. the line bundle associated to the divisor of the zero-section in $Y$. The upshot is that in $H_{X}^{4}(Y;\mathbf{Z}/2\mathbf{Z})$, we have

$$\begin{eqnarray}\displaystyle \overline{c_{1}}({\mathcal{O}}_{Y}(X))\smallsmile s_{X/Y}=\text{cl}_{Y}(X)\smallsmile s_{X/Y}=s_{X/Y}\smallsmile s_{X/Y}=\operatorname{Sq}^{2}(s_{X/Y}), & & \displaystyle \nonumber\end{eqnarray}$$

which shows that $w_{2}=\overline{c_{1}}({\mathcal{O}}_{Y}(X))\in H_{\acute{\text{e}}\text{t}}^{2}(Y;\mathbf{Z}/2\mathbf{Z})$. Since we have already established that ${\mathcal{L}}$ pulls back to ${\mathcal{O}}_{Y}(X)$ under the projection map $\unicode[STIX]{x1D70B}:Y\rightarrow X$, naturality for Chern classes and the fact that $\unicode[STIX]{x1D70B}^{\ast }$ induces an isomorphism on cohomology shows that $w_{2}=\overline{c_{1}}({\mathcal{L}})\in H_{\acute{\text{e}}\text{t}}^{2}(X;\mathbf{Z}/2\mathbf{Z})$.◻

6.1 Wu’s theorem for smooth manifolds

We first explain the classical version of Wu’s theorem. Let $M$ be a closed smooth manifold of dimension $n$, so that the cup product induces a perfect duality on $H^{\ast }(M)$. Then for a cohomology class $x\in H^{n-i}(M;\mathbf{Z}/2\mathbf{Z})$ the map $x\mapsto \operatorname{Sq}^{i}x\in H^{n}(M;\mathbf{Z}/2\mathbf{Z})$ must, by Poincaré duality, be represented by a class $v_{i}\in H^{i}(M;\mathbf{Z}/2\mathbf{Z})$, i.e.

$$\begin{eqnarray}v_{i}\smallsmile x=\operatorname{Sq}^{i}x\quad \text{for all }x\in H^{i}(M;\mathbf{Z}/2\mathbf{Z}).\end{eqnarray}$$

This $v_{i}$ is called the $i$th Wu class.

Let $v:=\sum _{i}v_{i}$ be the total Wu class and $w:=\sum w_{i}$ be the total Stiefel–Whitney class of $TM$. Then Wu’s formula relates the two in the following way.

Example 6.3. We use Wu’s theorem to calculate a few small examples. Equating terms of degree $1$, we deduce that

$$\begin{eqnarray}v_{1}=w_{1}.\end{eqnarray}$$

Equating terms of degree $2$, we deduce that $v_{2}+\operatorname{Sq}^{1}v_{1}=w_{2}$, which we can rewrite as

$$\begin{eqnarray}v_{2}=w_{2}+w_{1}^{2}.\end{eqnarray}$$

6.2 A Wu theorem for varieties over finite fields

The aim of this section is to prove a version of Wu’s theorem in the setting of étale cohomology. For varieties over separably closed fields, this is done in [Reference UrabeUra96, Theorem 0.5]. In that case one can more or less transpose the usual proof for manifolds, essentially because the $\ell$-adic cohomology of smooth varieties over separably closed fields behaves very similarly to the singular cohomology of complex manifolds. In particular, for a surface that lifts to characteristic $0$, the classical version of Wu’s theorem implies the version for geometric $\ell$-adic cohomology. The main result of this section (Theorem 6.5) is that the same formula also holds for absolute étale cohomology over finite fields, with our definitions of the $w_{i}$ from §5. Because the ground field is not separably closed there are some significant new difficulties; one indication of this is that the proof requires étale homotopy theory.

Theorem 6.5. Let $X$ be a smooth, proper, geometrically connected variety over $\mathbf{F}_{q}$. Define the Wu class $v\in H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z})$ to be the unique cohomology class such that

$$\begin{eqnarray}\int \operatorname{Sq}x=\int v\smallsmile x\quad \text{for all }x\in H_{\acute{\text{e}}\text{t}}^{\ast }(X;\mathbf{Z}/2\mathbf{Z}).\end{eqnarray}$$

Then we have $w=\operatorname{Sq}v$.

The reader is recommended to skip the proof on the first pass through the paper, as it is quite lengthy and nothing but the statement of Theorem 6.5 will be used in the following.

6.3 Proof of Theorem 6.5

Our proof of Theorem 6.5 proceeds in four steps. Steps 2 and 3 are essentially a translation of the usual (topological) proof to algebraic geometry. Step 1 performs a technical reduction to the case where the topological argument begins, and is necessary because of the lack of ‘tubular neighborhoods’ in algebraic geometry. Finally, step 4 bridges a new technical difficulty, the spirit of which is discussed in Remark 6.4, that arises here because our ground field is not separably closed.

6.3.1 Step 1

Recall from (5.2.3) that we defined the Stiefel–Whitney classes $w_{i}$ by

$$\begin{eqnarray}\operatorname{Sq}^{i}s_{X/TX}=\unicode[STIX]{x1D70B}^{\ast }(w_{i})\smallsmile s_{X/TX},\end{eqnarray}$$

where $\unicode[STIX]{x1D70B}:TX\rightarrow X$ is the projection. Recall also that the normal bundle of $X$ in $X\times X$ is isomorphic to $TX$. The purpose of this step is to prove the following lemma, which is motivated by the preceding considerations.

Lemma 6.6. Let $s_{X/X\times X}\in H_{X}^{\ast +2n}(X\times X)$ be the image of $1\in H_{\acute{\text{e}}\text{t}}^{0}(X)$ under the Gysin isomorphism

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X)\xrightarrow[{}]{{\sim}}H_{X}^{\ast +2n}(X\times X).\end{eqnarray}$$

Then we have

(6.3.1)$$\begin{eqnarray}\operatorname{Sq}^{i}s_{X/X\times X}=\text{pr}_{1}^{\ast }(w_{i})\smallsmile s_{X/X\times X},\end{eqnarray}$$

where $\text{pr}_{1}:X\times X\rightarrow X$ denotes projection to the first factor.

From the definitions Lemma 6.6 is an immediate consequence of the following lemma.

Lemma 6.7. Let $X{\hookrightarrow}Y$ be a codimension $n$ closed embedding of smooth varieties (over any field) and let

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}_{1}:H^{\ast }(X)\xrightarrow[{}]{{\sim}}H_{X}^{\ast +2n}(N_{X/Y})\unicode[STIX]{x1D719}_{2}:H^{\ast }(X)\xrightarrow[{}]{{\sim}}H_{X}^{\ast +2n}(Y) & & \displaystyle \nonumber\end{eqnarray}$$

be the two Gysin isomorphisms. Then

$$\begin{eqnarray}w(N_{X/Y}):=\unicode[STIX]{x1D719}_{1}^{-1}(\operatorname{Sq}s_{X/N_{X/Y}})=\unicode[STIX]{x1D719}_{2}^{-1}(\operatorname{Sq}s_{X/Y}).\end{eqnarray}$$

Remark 6.8. If $X$ were a smooth manifold, we could argue directly since we have an isomorphism

$$\begin{eqnarray}H_{X}^{\ast +2n}(N_{X/Y})\cong H^{\ast +2n}(U,U\setminus X),\end{eqnarray}$$

where $U$ is a tubular neighborhood of the zero-section in $Y$, and we have

$$\begin{eqnarray}H^{\ast +2n}(U,U\setminus X)\cong H_{X}^{\ast +2n}(Y):=H^{\ast +2n}(Y,Y\setminus X)\end{eqnarray}$$

by excision. Since these isomorphisms are pullbacks induced by maps of spaces, they commute with Steenrod squares. A referee has suggested that an analog of this argument may be carried out in our setting using the Morel–Voevodsky purity theorem.

Proof. The key fact is that if $X{\hookrightarrow}Y$ is a closed embedding, then there is a flat family deforming the inclusion $X{\hookrightarrow}Y$ into the zero-section $X{\hookrightarrow}N_{X/Y}$ (‘deformation to the normal cone’). This allows us to carry out the idea of Remark 6.8.

More precisely, there is a flat family ${\mathcal{Y}}\rightarrow \mathbf{A}^{1}$, which restricts to the trivial family away from the origin, ${\mathcal{Y}}|_{\mathbf{A}^{1}-0}\cong Y\times (\mathbf{A}^{1}-0)$, but such that ${\mathcal{Y}}|_{0}\cong N_{X/Y}$. Furthermore, there is a closed embedding $X\times \mathbf{A}^{1}{\hookrightarrow}{\mathcal{Y}}$, which restricts to the given embedding $X{\hookrightarrow}Y$ away from $0$, and $X{\hookrightarrow}N_{X/Y}$ at $0$. For the construction and proofs of the properties, see [Reference FultonFul98, §5]. The situation is depicted in the diagram below.

Applying the Gysin morphism to $X\times \mathbf{A}^{1}{\hookrightarrow}{\mathcal{Y}}$, we have an isomorphism

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times \mathbf{A}^{1})\xrightarrow[{}]{{\sim}}H_{X\times \mathbf{A}^{1}}^{\ast +2n}({\mathcal{Y}})\end{eqnarray}$$

sending $1\mapsto s_{X\times \mathbf{A}^{1}/{\mathcal{Y}}}$. Note that $X\times \mathbf{A}^{1}$ and $Y$ (viewed as the fiber over $t$) intersect transversely in ${\mathcal{Y}}$, and similarly $X\times \mathbf{A}^{1}$ and $N_{X/Y}$. Hence from the diagram above we obtain a diagram of maps in cohomology (where the vertical maps are the respective Gysin isomorphisms).

Under this diagram the Thom classes are mapped as follows, by compatibility with base change (cf. §2 of Deligne’s exposé ‘Cycle’ in [Reference DeligneSGA4½]).

Since the horizontal maps in the bottom row are pullbacks they are compatible with $\operatorname{Sq}$, and hence send

By definition $\operatorname{Sq}^{i}(s_{X/N_{X/Y}})=\unicode[STIX]{x1D70B}^{\ast }(w_{i})\smallsmile s_{X/N_{X/Y}}$, and since the maps

are isomorphisms of $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\cong H_{\acute{\text{e}}\text{t}}^{\ast }(X\times \mathbf{A}^{1})$-modules, they send

as desired. ◻

6.3.2 Step 2

For a regular embedding $X{\hookrightarrow}Y$, there is an attached cycle class $\text{cl}_{Y}(X)\in H^{\ast }(Y)$, which in topology would be the ‘Poincaré dual to the fundamental class of $X$ in homology’. The goal of this step is to prove the following lemma.

Lemma 6.9. Let $\unicode[STIX]{x1D6E5}:=\text{cl}_{X\times X}(X)\in H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)$ be the cycle class for the diagonal embedding of $X$. Then we have $w=(\text{pr}_{1})_{\ast }(\operatorname{Sq}\unicode[STIX]{x1D6E5})\in H_{\acute{\text{e}}\text{t}}^{\ast }(X)$.

We first review the definition of the pushforward in cohomology for a map of smooth proper varieties, and then the definition of the cycle class.

Definition 6.10 (Pushforwards in cohomology).

If $f:X\rightarrow Y$ is a map of smooth proper varieties over $\mathbf{F}_{q}$ of dimensions $m$ and $n$, respectively, then the pullback map

$$\begin{eqnarray}f^{\ast }:H_{\acute{\text{e}}\text{t}}^{\ast }(Y)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X)\end{eqnarray}$$

induces an adjoint map on the $\mathbf{Z}/2\mathbf{Z}$-dual spaces:

(6.3.2)$$\begin{eqnarray}(f^{\ast })^{\vee }:H_{\acute{\text{e}}\text{t}}^{\ast }(X)^{\vee }\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X)^{\vee }.\end{eqnarray}$$

We can identify $H_{\acute{\text{e}}\text{t}}^{\ast }(X)^{\vee }\cong H_{\acute{\text{e}}\text{t}}^{2m+1-\ast }(X)$ by Poincaré duality, obtaining from (6.3.2) a map

$$\begin{eqnarray}f_{\ast }:H_{\acute{\text{e}}\text{t}}^{\ast }(X)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast +2n-2m}(Y).\end{eqnarray}$$

In particular, we define the cycle class of $X$in $Y$ to be $f_{\ast }(1)=:\text{cl}_{Y}(X)$. Unwrapping the definition, the class $\text{cl}_{Y}(X)$ is characterized by the identity

$$\begin{eqnarray}\int _{X}f^{\ast }\unicode[STIX]{x1D6FE}=\int _{Y}\text{cl}_{Y}(X)\smallsmile \unicode[STIX]{x1D6FE}\quad \text{for all }\unicode[STIX]{x1D6FE}\in H^{\ast }(Y).\end{eqnarray}$$

We recall some basic properties of this pushforward. The proofs are all immediate from the definition except the last, which is [Reference Freitag and KiehlFK88, Proposition 2.7].

1. – It is functorial.

2. – We have the product formula

   (6.3.3)$$\begin{eqnarray}f_{\ast }(\unicode[STIX]{x1D6FC}\smallsmile f^{\ast }\unicode[STIX]{x1D6FD})=(f_{\ast }\unicode[STIX]{x1D6FC})\smallsmile \unicode[STIX]{x1D6FD}.\end{eqnarray}$$

3. – If $X{\hookrightarrow}Y$ is a closed embedding, then the map $H_{X}^{\ast }(Y)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(Y)$ sends $s_{X/Y}\mapsto \text{cl}_{Y}(X)$.

Proof of Lemma 6.9.

We now apply the preceding discussion to the case $Y=X\times X$, with $f$ being the diagonal embedding. By Lemma 6.6 we have

$$\begin{eqnarray}\operatorname{Sq}^{i}s_{X/X\times X}=\operatorname{pr}_{1}^{\ast }(w_{i})\smallsmile s_{X/X\times X}\in H_{X}^{2n+i}(X\times X).\end{eqnarray}$$

Since the map $H_{X}^{\ast }(X\times X)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)$ is induced by a pullback (§5.1) it is automatically compatible with Steenrod operations, so it sends as follows.

Hence by Definition 6.10 and (6.3.3) we have

$$\begin{eqnarray}(\text{pr}_{1})_{\ast }(\text{pr}_{1}^{\ast }(w_{i})\smallsmile \text{cl}_{X\times X}(X))=w_{i}\smallsmile (\text{pr}_{1})_{\ast }f_{\ast }1=w_{i}.\square\end{eqnarray}$$

6.3.3 Step 3

At this point, the classical proof of Wu’s theorem proceeds by computing $(\operatorname{pr}_{1})_{\ast }\text{cl}_{X\times X}(X)$ in a second way, which is predicated upon the Künneth formula

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)\cong H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X),\end{eqnarray}$$

which unfortunately breaks down in our situation. To explain how to repair the argument, we need to make some observations. Note that $H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)$ acts by correspondences on $H_{\acute{\text{e}}\text{t}}^{\ast }(X)$, inducing the map

(6.3.4)$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)\rightarrow \operatorname{End}(H_{\acute{\text{e}}\text{t}}^{\ast }(X))\end{eqnarray}$$

given explicitly by sending $x\in H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)$ to the endomorphism

(6.3.5)$$\begin{eqnarray}\unicode[STIX]{x1D6FE}\mapsto (\operatorname{pr}_{1})_{\ast }(x\smallsmile \operatorname{pr}_{2}^{\ast }\unicode[STIX]{x1D6FE}).\end{eqnarray}$$

Lemma 6.11. Let $\unicode[STIX]{x1D6E5}:=\operatorname{cl}_{X\times X}(X)\in H_{\acute{\text{e}}\text{t}}^{2\dim X}(X\times X)$. Then the map (6.3.4) sends $\unicode[STIX]{x1D6E5}\mapsto \operatorname{Id}$.

Proof. Let $f:X{\hookrightarrow}X\times X$ denote the inclusion of the diagonal. Applying (6.3.5) to $x=\unicode[STIX]{x1D6E5}$ yields the endomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}\mapsto (\operatorname{pr}_{1})_{\ast }(f_{\ast }(1)\smallsmile p_{2}^{\ast }\unicode[STIX]{x1D6FE})=(\operatorname{pr}_{1})_{\ast }f_{\ast }(1\smallsmile f^{\ast }\operatorname{pr}_{2}^{\ast }\unicode[STIX]{x1D6FE}).\end{eqnarray}$$

But since $\operatorname{pr}_{1}\circ f=\operatorname{pr}_{2}\circ f=\operatorname{Id}$, this last expression is just $\unicode[STIX]{x1D6FE}$ again.◻

The map (6.3.4) can be interpreted as a ‘pushforward’ in the following way. The projection morphisms $\operatorname{pr}_{1},\operatorname{pr}_{2}:X\times X\rightarrow X$ induce maps $\operatorname{pr}_{1}^{\ast },\operatorname{pr}_{2}^{\ast }:H_{\acute{\text{e}}\text{t}}^{\ast }(X)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)$. From this we get a pullback map

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)\xrightarrow[{}]{\operatorname{pr}_{1}^{\ast }\smallsmile \operatorname{pr}_{2}^{\ast }}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X).\end{eqnarray}$$

Therefore, we get a dual map in the opposite direction

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)^{\vee }\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X)^{\vee }\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)^{\vee }.\end{eqnarray}$$

Each of these groups is canonically self-dual via Poincaré duality, so we can identify this with a map

(6.3.6)$$\begin{eqnarray}\unicode[STIX]{x1D711}_{\ast }:H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X).\end{eqnarray}$$

Note that this map increases the total degree by 1. It is a straightforward exercise in unraveling the definitions to see that this map is the same as (6.3.4), once one makes the appropriate identifications.

Let $(p_{1}^{\prime })_{\ast }$ and $(p_{2}^{\prime })_{\ast }$ denote the ‘pushforward’ maps

which are dual to the obvious ‘pullbacks’

Remark 6.12. The maps $(p_{i}^{\prime })^{\ast }$ and $(p_{i}^{\prime })_{\ast }$ are not induced by maps of varieties; indeed $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ is not the cohomology of a variety over $\mathbf{F}_{q}$. However, $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)$is the cohomology of the pro-space $\acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X)$. This means, for instance, that it is equipped with a natural cup product, which is just the tensor product of the cup products on $H_{\acute{\text{e}}\text{t}}^{\ast }(X)$. Now, $(p_{i}^{\prime })^{\ast }$ and $(p_{i}^{\prime })_{\ast }$are induced by maps of pro-spaces, namely the obvious projection maps

$$\begin{eqnarray}p_{i}^{\prime }:\acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X)\rightarrow \acute{\text{E}}\text{t}(X).\end{eqnarray}$$

This implies that $(p_{i}^{\prime })^{\ast }$ and $(p_{i}^{\prime })_{\ast }$ share the nice formal properties that are enjoyed by all pullbacks and pushforwards: for example, we will use that they satisfy the projection formula, and that $(p_{i}^{\prime })^{\ast }$ commutes with Steenrod operations.

However, it is not really necessary to use étale homotopy theory to see all this. We can just formally define the cup product on $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ to be the tensor product of the cup products on $H_{\acute{\text{e}}\text{t}}^{\ast }(X)$, and formally define $\operatorname{Sq}$ on $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ to be the tensor product of $\operatorname{Sq}\otimes \operatorname{Sq}$. It is an exercise in elementary algebra to check that this induces a well-defined action of the Steenrod algebra, satisfying all the axioms of §3.4. Similarly, the projection formula for $(p_{i})_{\ast }$ boils down to a tautology.

Lemma 6.13. Let $X$ be a smooth proper variety over a finite field. Let $\{e_{i}\}$ be a basis for $H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ and $\{f_{i}\}$ the dual basis under Poincaré duality. Then, letting $\unicode[STIX]{x1D6E5}$ be as in Lemma 6.11, we have

(6.3.7)$$\begin{eqnarray}\unicode[STIX]{x1D711}_{\ast }\unicode[STIX]{x1D6E5}=\mathop{\sum }_{i}e_{i}\otimes f_{i},\end{eqnarray}$$

where $\unicode[STIX]{x1D711}_{\ast }$ is as in (6.3.6).

Proof. Lemma 6.11 says that the action of $\unicode[STIX]{x1D6E5}$ induced on $H^{\ast }(X)$ by (6.3.5) is just the identity map. Therefore, it suffices to show that the right-hand side of (6.3.7) acts as the identity on $H^{\ast }(X)$, but this is just a straightforward linear algebra exercise about dual bases.◻

Since the pullback $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\xrightarrow[{}]{\operatorname{pr}_{1}^{\ast }}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)$ obviously factors through

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X)\xrightarrow[{}]{(p_{1}^{\prime })^{\ast }}H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)\xrightarrow[{}]{\operatorname{pr}_{1}^{\ast }\smallsmile \operatorname{pr}_{2}^{\ast }}H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)\end{eqnarray}$$

(morally, ‘$\operatorname{pr}_{1}=\unicode[STIX]{x1D711}\circ p_{1}^{\prime }$’) we have

(6.3.8)$$\begin{eqnarray}(p_{1})_{\ast }=(p_{1}^{\prime })_{\ast }\unicode[STIX]{x1D711}_{\ast }.\end{eqnarray}$$

Now, let us summarize where we are. Combining Lemma 6.9 and (6.3.8), we know that

(6.3.9)$$\begin{eqnarray}w=(p_{1})_{\ast }\operatorname{Sq}\unicode[STIX]{x1D6E5}=(p_{1}^{\prime })_{\ast }\unicode[STIX]{x1D711}_{\ast }\operatorname{Sq}\unicode[STIX]{x1D6E5}.\end{eqnarray}$$

Lemma 6.13 gives us an expression for $\unicode[STIX]{x1D711}_{\ast }\unicode[STIX]{x1D6E5}$, and hence also $\operatorname{Sq}\unicode[STIX]{x1D711}_{\ast }\unicode[STIX]{x1D6E5}$. If we could commute $\unicode[STIX]{x1D711}_{\ast }$ and $\operatorname{Sq}$, then this would give us a formula for $\unicode[STIX]{x1D711}_{\ast }\operatorname{Sq}\unicode[STIX]{x1D6E5}$. But although Steenrod squares commute with pullbacks, they do not in general commute with pushforwards. This is the key problem (note that the whole issue disappears when one has the Künneth formula, as in classical algebraic topology or in algebraic geometry over separably closed fields). To address this issue, in the last step of the proof, we will establish the following.

Proposition 6.14. Let $X$ be a smooth proper variety over a finite field, and $\unicode[STIX]{x1D711}_{\ast }$ be as in (6.3.6). Then we have $\operatorname{Sq}\unicode[STIX]{x1D711}_{\ast }=\unicode[STIX]{x1D711}_{\ast }\operatorname{Sq}$.

Assuming Proposition 6.14 for now (it will be shown in §6.3.4), we complete the rest of the proof of Theorem 6.5. Let $\{e_{i}\}$ be a basis for $H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ and $\{f_{i}\}$ the dual basis under Poincaré duality, as above. By (6.3.9), Proposition 6.14, and Lemma 6.13 we have

(6.3.10)$$\begin{eqnarray}w=(p_{1}^{\prime })_{\ast }\operatorname{Sq}\biggl(\mathop{\sum }_{i}(p_{1}^{\prime })^{\ast }e_{i}\smallsmile (p_{2}^{\prime })^{\ast }f_{i}\biggr).\end{eqnarray}$$

By the Cartan formula and the projection formula (which hold by Remark 6.12), we have

(6.3.11)$$\begin{eqnarray}\displaystyle (p_{1}^{\prime })_{\ast }\operatorname{Sq}\biggl(\mathop{\sum }_{i}(p_{1}^{\prime })^{\ast }e_{i}\smallsmile (p_{2}^{\prime })^{\ast }f_{i}\biggr) & = & \displaystyle \mathop{\sum }_{i}(p_{1}^{\prime })_{\ast }((p_{1}^{\prime })^{\ast }\operatorname{Sq}e_{i}\smallsmile (p_{2}^{\prime })^{\ast }\operatorname{Sq}f_{i})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{i}\operatorname{Sq}e_{i}\smallsmile (p_{1}^{\prime })_{\ast }(p_{2}^{\prime })^{\ast }\operatorname{Sq}f_{i}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{i}\operatorname{Sq}e_{i}\otimes (p_{1}^{\prime })_{\ast }(p_{2}^{\prime })^{\ast }\operatorname{Sq}f_{i}.\end{eqnarray}$$

Now, unraveling the definitions shows that

$$\begin{eqnarray}(p_{1}^{\prime })_{\ast }(p_{2}^{\prime })^{\ast }\unicode[STIX]{x1D6FE}=\int \unicode[STIX]{x1D6FE}\quad \text{for all }\unicode[STIX]{x1D6FE}\in H_{\acute{\text{e}}\text{t}}^{\ast }(X),\end{eqnarray}$$

where the right-hand side is viewed in $\mathbf{Z}/2\mathbf{Z}\cong H_{\acute{\text{e}}\text{t}}^{0}(X)$. (It is also easy to see directly that this must be the case for degree reasons, since the left side can only be non-zero for $\unicode[STIX]{x1D6FE}$ in top degree.) Combining this with (6.3.10) and (6.3.11), we find that

$$\begin{eqnarray}\displaystyle w=\mathop{\sum }_{i}\operatorname{Sq}e_{i}\cdot \int \operatorname{Sq}f_{i}=\mathop{\sum }_{i}\operatorname{Sq}e_{i}\cdot \int v\smallsmile f_{i}=\operatorname{Sq}\biggl(\mathop{\sum }_{i}e_{i}\cdot \int v\smallsmile f_{i}\biggr)=\operatorname{Sq}v, & & \displaystyle \nonumber\end{eqnarray}$$

with the last equality using that $\{e_{i}\}$ and $\{f_{i}\}$ are dual bases.◻

6.3.4 Step 4

This step is devoted to the proof of Proposition 6.14. As foreshadowed in Remark 6.12, the difficulty stems from the inability to realize $H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ as the cohomology of an actual variety over $\mathbf{F}_{q}$. For this reason it is useful to pass to étale topological type, where we can interpret

$$\begin{eqnarray}H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)=H_{\acute{\text{e}}\text{t}}^{\ast }(\acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X)).\end{eqnarray}$$

The idea of the argument. The basic geometric idea is that the map

(6.3.12)$$\begin{eqnarray}\unicode[STIX]{x1D711}_{\ast }:H_{\acute{\text{e}}\text{t}}^{\ast }(X\times X)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)\end{eqnarray}$$

looks like a pushforward map on cohomology induced by a ‘homotopy quotient by $\widehat{\mathbf{Z}}$’ at the level of geometric objects. Proposition 6.14 is then motivated by the well-known fact (which we prove below in Proposition 6.23) that Steenrod operations commute with pushforward through a homotopy quotient by $\mathbf{Z}$, and that cohomologically (with finite coefficients) homotopy quotients by $\mathbf{Z}$ and by $\widehat{\mathbf{Z}}$ look the same.

To see why the ‘basic geometric idea’ should be true, our heuristic is that for any field $k$ and $G_{k}:=\operatorname{Gal}(\overline{k}/k)$, we should have

(6.3.13)$$\begin{eqnarray}\text{`}\,\acute{\text{E}}\text{t}(X)=\acute{\text{E}}\text{t}(X_{\overline{k}})_{hG_{k}}\text{'}.\end{eqnarray}$$

Here if a group $G$ acts on a space $Y$ then we write $Y_{hG}:=(Y\times EG)/G$ for the ‘homotopy quotient of $Y$ by $G$’, where $EG$ is some contractible space with a free $G$ action, and the quotient is for the (free) diagonal action. The heuristic (6.3.13) then suggests that

$$\begin{eqnarray}\displaystyle \acute{\text{E}}\text{t}(X\times _{k}X)\cong \acute{\text{E}}\text{t}((X\times _{k}X)_{\overline{k}})_{hG_{k}}\cong \acute{\text{E}}\text{t}(X_{\overline{k}}\times _{\overline{k}}X_{\overline{k}})_{hG_{k}}\cong (\acute{\text{E}}\text{t}(X_{\overline{k}})\times \acute{\text{E}}\text{t}(X_{\overline{k}}))_{hG_{k}}, & & \displaystyle \nonumber\end{eqnarray}$$

where the quotient is for the action of the diagonal $G_{k}$ (leaving a residual action of $G_{k}$), while

$$\begin{eqnarray}\displaystyle \acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X)\cong \acute{\text{E}}\text{t}(X_{\overline{k}})_{hG_{k}}\times \acute{\text{E}}\text{t}(X_{\overline{k}})_{hG_{k}}\cong (\acute{\text{E}}\text{t}(X_{\overline{k}})\times \acute{\text{E}}\text{t}(X_{\overline{k}}))_{h(G_{k}\times G_{k})}. & & \displaystyle \nonumber\end{eqnarray}$$

Hence we would have a homotopy quotient

$$\begin{eqnarray}\acute{\text{E}}\text{t}(X\times _{k}X)\rightarrow \acute{\text{E}}\text{t}(X\times _{k}X)_{hG_{k}}\cong \acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X)\end{eqnarray}$$

whose induced pushforward on cohomology agrees recovers (6.3.12).

In a previous version of this paper, we used some complicated gymnastics in cohomology as a substitute for the fact that we did not know how to rigorously formulate (6.3.13). We are very grateful to an anonymous referee for informing us that there already exists a framework to handle this sort of issue, namely the ‘relative étale homotopy theory’ developed in [Reference Harpaz and SchlankHS13, Reference Barnea and SchlankBS16]. This formalism makes the argument much more efficient and transparent, so we review it next.

Relative étale homotopy theory. If $X$ is a variety over a field $k$ and $L/k$ is a finite extension, then $\acute{\text{E}}\text{t}(X_{L})$ is a pro-object in simplicial sets equipped with an action of $\operatorname{Gal}(L/k)$as a pro-object. However, it would be better to work with an object that enjoys a level-wise action, rather than an action as a pro-object. In [Reference Harpaz and SchlankHS13] Harpaz–Schlank defined a refined variant of the étale homotopy type denoted $\acute{\text{E}}\text{t}_{/k}(X)$, called the relative étale homotopy type [Reference Harpaz and SchlankHS13, §9.2.3]Footnote ⁸ of $X$, which is a pro-object in the homotopy category of $G_{k}$-spaces (which means by definition that every simplex has an open stabilizer).Footnote ⁹ The point is that $\acute{\text{E}}\text{t}_{/k}(X)$ is equipped with a level-wise action of $G_{k}$. In [Reference Barnea and SchlankBS16] Barnea–Schlank lifted this construction to the pro-category of $G_{k}$-spaces, and it is this refinement that we will use in this paper. The improvement is analogous to Friedlander’s refinement of the Artin–Mazur étale homotopy type, which is a pro-object in the homotopy category of spaces, to the étale topological type discussed in §3.2, which is a pro-object in spaces, although the methods of [Reference Barnea and SchlankBS16] are very different.

Here is a very brief summary of the difference between $\acute{\text{E}}\text{t}_{/k}(X)$ and $\acute{\text{E}}\text{t}(X)$; see [Reference Harpaz and SchlankHS13, §9.2.3] and [Reference Barnea and SchlankBS16, §8] for the details. The basic idea is that the usual definition of $\acute{\text{E}}\text{t}(X)$ attaches to each hypercovering $U_{.}$ of $X$ the simplicial set $\unicode[STIX]{x1D70B}_{0}(U_{.})$ of its connected components. On the other hand, $\acute{\text{E}}\text{t}_{/k}(X)$ assigns to $U_{.}$ the simplicial set $\unicode[STIX]{x1D70B}_{0}(U_{.}\times _{k}\overline{k})$, which is equipped with an obvious $G_{k}$-action.

Definition 6.15 [Reference Harpaz and SchlankHS13, §9.6.2].

Given a $G_{k}$-space $Y$, we define the homotopy quotient $Y_{hG_{k}}$ to be the pro-space

$$\begin{eqnarray}Y_{hG_{k}}:=\{(Y\times E(G_{k}/H))/{G_{k}\}}_{H},\end{eqnarray}$$

where the index set runs over $i$ and open normal subgroups $H\lhd G_{k}$.

Given a pro-$G_{k}$-space $\{{Y_{i}\}}_{i}$, we define the homotopy quotient $(\{{Y_{i}\}}_{i})_{hG_{k}}$ to be the pro-space

$$\begin{eqnarray}(\{{Y_{i}\}}_{i})_{hG_{k}}:=\{(Y_{i}\times E(G_{k}/H))/{G_{k}\}}_{i,H},\end{eqnarray}$$

where the index set runs over open normal subgroup $H\lhd G_{k}$.

Remark 6.16. Note that $(Y\times E(G_{k}/H))/G_{k}=(Y/H)_{h(G_{k}/H)}$. In particular, if $H$ acts trivially on $Y$ then $(Y\times E(G_{k}/H))/G_{k}=Y_{h(G_{k}/H)}$.

We next begin discussing the key properties of $\acute{\text{E}}\text{t}_{/k}(X)$. Actually, we replace $\acute{\text{E}}\text{t}_{/k}(X)$ by its Postnikov truncation denoted $\acute{\text{E}}\text{t}_{/k}(X)^{\sharp }$ in [Reference Harpaz and SchlankHS13], which does not alter the cohomology. This is a technical device to guarantee certain finiteness conditions levelwise; for simplicity of notation we will omit the $\sharp$.

1. (i) By [Reference Harpaz and SchlankHS13, Proposition 9.82], we have a homotopy equivalence

   $$\begin{eqnarray}\acute{\text{E}}\text{t}_{/k}(X)_{hG_{k}}\xrightarrow[{}]{{\sim}}\acute{\text{E}}\text{t}(X).\end{eqnarray}$$

2. (ii) By the cofinality of the diagonal in the product of a left filtered index category with itself, [Reference Harpaz and SchlankHS13, Proposition 9.82] also implies that we have a homotopy equivalence

   $$\begin{eqnarray}(\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{hG_{k}\times hG_{k}}\xrightarrow[{}]{{\sim}}\acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X).\end{eqnarray}$$

3. (iii) Finally, [Reference Harpaz and SchlankHS13, Proposition 9.19] asserts that the underlying pro-space of $\acute{\text{E}}\text{t}_{/k}(X)$ (obtained by forgetting the $G_{k}$-action) is homotopy equivalent to $\acute{\text{E}}\text{t}(X_{\overline{k}})$. Denote this forgetful functor by $\text{Oblv}$.

Now, $\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X)$ is naturally a pro-$G_{k}\times G_{k}$-space. Taking the homotopy quotient for the diagonal $G_{k}$-action leaves a residual $G_{k}$-action, making $(\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{hG_{k}}$ a pro-$G_{k}$-space, so that we can take the homotopy quotient again. Putting the facts (i)–(iii) together, we realize the homotopy quotient map (6.3.13) as the vertical map between pro-spaces in the diagram below.

(6.3.14)

Comparing homotopy quotients by $\mathbf{Z}$and $\widehat{\mathbf{Z}}$. Let $Y:=\{{Y_{i}\}}_{i}$ be any pro-$\widehat{\mathbf{Z}}$-space. We can restrict the $\widehat{\mathbf{Z}}$-action to a $\mathbf{Z}$-action, and then form $Y_{h\mathbf{Z}}$. By Definition 6.15 and Remark 6.16, we have $Y_{h\mathbf{Z}}=\{{(Y_{i})_{h\mathbf{Z}}\}}_{i}$ while $Y_{h\widehat{\mathbf{Z}}}=\{{(Y_{i}/n\mathbf{Z})_{h(\mathbf{Z}/n\mathbf{Z})}\}}_{i,n}$. Thus there is a canonical map of pro-$\mathbf{G}_{k}$-spaces:

(6.3.15)$$\begin{eqnarray}Y_{h\mathbf{Z}}\rightarrow Y_{h\widehat{\mathbf{Z}}}.\end{eqnarray}$$

Our next goal is to show that if $Y=\acute{\text{E}}\text{t}_{/k}(X)$ arises as the relative homotopy type of a variety $X/\mathbf{F}_{q}$, then (6.3.15) induces an isomorphism on cohomology.

Lemma 6.17. Let $X$ be a variety over $\mathbf{F}_{q}$. Then the natural map

$$\begin{eqnarray}H^{\ast }(\acute{\text{E}}\text{t}_{/k}(X)_{h\mathbf{Z}};\mathbf{Z}/2\mathbf{Z})\leftarrow H^{\ast }(\acute{\text{E}}\text{t}_{/k}(X)_{h\widehat{\mathbf{Z}}};\mathbf{Z}/2\mathbf{Z})\end{eqnarray}$$

is an isomorphism.

Proof. We consider the pro $G_{k}$-space $Y=(Y_{i}):=\acute{\text{E}}\text{t}_{/k}(X)$. The map on cohomology induced by (6.3.15) is

$$\begin{eqnarray}\mathop{\varinjlim }\nolimits_{i}H^{\ast }((Y_{i})_{h\mathbf{Z}})\leftarrow \mathop{\varinjlim }\nolimits_{i}\mathop{\varinjlim }\nolimits_{n}H^{\ast }((Y_{i})_{h(\mathbf{Z}/n\mathbf{Z})}).\end{eqnarray}$$

Therefore, it will certainly suffice to prove that

(6.3.16)$$\begin{eqnarray}H^{\ast }((Y_{i})_{h\mathbf{Z}})\xleftarrow[{}]{{\sim}}\mathop{\varinjlim }\nolimits_{n}H^{\ast }((Y_{i}/n\mathbf{Z})_{h(\mathbf{Z}/n\mathbf{Z})})\end{eqnarray}$$

individually for each $i$. Moreover, since we assume that $Y=\acute{\text{E}}\text{t}_{/k}(X)$ we can take $Y_{i}$ to be excellent, i.e. we can assume that the $\widehat{\mathbf{Z}}$-action on $Y_{i}$ already factors through a finite quotient [Reference Harpaz and SchlankHS13, §9.2.3, pp. 296–297]. Hence, by restricting to a cofinal subcategory of the indexing category, and renaming $Y_{i}$ to $Y$, we may assume that $Y$ is a $\widehat{\mathbf{Z}}$-space (as opposed to pro-space) on which the action factors through $\mathbf{Z}/n\mathbf{Z}$. Then, obviously, for any $n\mid N$ we have

$$\begin{eqnarray}(Y/N\mathbf{Z})_{h(\mathbf{Z}/N\mathbf{Z})}=Y_{h(\mathbf{Z}/N\mathbf{Z})}.\end{eqnarray}$$

The identity map $Y\rightarrow Y$ is obviously equivariant for the group homomorphism $\mathbf{Z}\rightarrow \mathbf{Z}/N\mathbf{Z}$, and induces the following map of spectral sequences.

(6.3.17)

The following map of spectral sequences is not necessarily an isomorphism for any fixed $N$, but after taking the direct limit as $N$ runs over all positive multiples of $n$, it becomes an isomorphism by the well-known comparison of group cohomology for $\widehat{\mathbf{Z}}$ and $\mathbf{Z}$ with finite coefficients.

(6.3.18)

Now (

6.3.18

) implies that the natural map in (6.3.16) induces an isomorphism on associated gradeds, and hence an isomorphism. ◻

More on pushforwards in cohomology. We now turn towards studying the pushforward on cohomology. We begin by formalizing the construction of the pushforward in §6.3.2. To help distinguish varieties from pro-spaces, we will use boldface letters to denote pro-spaces in this part.

Definition 6.18. We say that a pro-space $\mathbf{Y}$ has Poincaré duality if there exists an $n$ and an isomorphism $\int :H^{n}(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})\cong \mathbf{Z}/2\mathbf{Z}$ such that the cup product induces a perfect pairing

$$\begin{eqnarray}H^{i}(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})\times H^{n-i}(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})\xrightarrow[{}]{\smallsmile }H^{n}(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})\xrightarrow[{}]{\int }\mathbf{Z}/2\mathbf{Z}.\end{eqnarray}$$

(This uniquely determines $n$.) We denote the non-zero element in $H^{n}(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})$ by $\unicode[STIX]{x1D707}_{\mathbf{Y}}$, and call it the fundamental class of $\mathbf{Y}$. From $\int$ we obtain a functional

$$\begin{eqnarray}H^{\ast }(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})\xrightarrow[{}]{\text{project}}H^{n}(\mathbf{Y};\mathbf{Z}/2\mathbf{Z})\xrightarrow[{}]{\int }\mathbf{Z}/2\mathbf{Z},\end{eqnarray}$$

which we will also denote by $\int$.

Definition 6.19. Let $f:\mathbf{W}\rightarrow \mathbf{Y}$ be a map of pro-spaces with Poincaré duality, with fundamental classes $\unicode[STIX]{x1D707}_{\mathbf{W}}\in H^{m}(\mathbf{W})$ and $\unicode[STIX]{x1D707}_{\mathbf{Y}}\in H^{n}(\mathbf{Y})$. Then we have a pullback map on cohomology

$$\begin{eqnarray}f^{\ast }:H^{\ast }(\mathbf{Y})\rightarrow H^{\ast }(\mathbf{W}),\end{eqnarray}$$

which induces a dual map

$$\begin{eqnarray}(f^{\ast })^{\vee }:H^{\ast }(\mathbf{W})^{\vee }\rightarrow H^{\ast }(\mathbf{Y})^{\vee }.\end{eqnarray}$$

We define the pushforward on cohomology

$$\begin{eqnarray}f_{\ast }:H^{i}(\mathbf{W})\rightarrow H^{i+n-m}(\mathbf{Y})\end{eqnarray}$$

to be the map $(f^{\ast })^{\vee }$, where the identifications

$$\begin{eqnarray}H^{i}(\mathbf{W})=H^{m-i}(\mathbf{W})^{\vee }\quad \text{and}\quad H^{i+n-m}(\mathbf{Y})=H^{m-i}(\mathbf{Y})^{\vee }\end{eqnarray}$$

are induced by Poincaré duality. It is easily checked that $f_{\ast }$ takes $\unicode[STIX]{x1D707}_{\mathbf{W}}\mapsto \unicode[STIX]{x1D707}_{\mathbf{Y}}$.

In particular, if $\mathbf{W}=\acute{\text{E}}\text{t}(X)$ for some smooth proper variety $X$ over $\mathbf{F}_{q}$ or $k^{\text{sep}}$, then $\mathbf{W}$ inherits Poincaré duality from $X$. If the map $f:\mathbf{W}\rightarrow \mathbf{Y}$ arises from a map of varieties $X\rightarrow V$, then the pushforward $f_{\ast }$ tautologically agrees, under the identifications $H^{\ast }(\mathbf{W})=H_{\acute{\text{e}}\text{t}}^{\ast }(X)$ and $H^{\ast }(\mathbf{Y})=H_{\acute{\text{e}}\text{t}}^{\ast }(V)$, with the pushforward we defined in §6.3.2. In addition, $\acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X)$ has Poincaré duality by the Künneth theorem. Hence, by the horizontal homotopy equivalences in (6.3.14) we find that $(\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{hG_{k}}$ and $(\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{h(G_{k}\times G_{k})}$ have Poincaré duality. It is tautological that the diagram

commutes, and dualizing it shows the following lemma.

Lemma 6.20. Let $X$ be smooth and proper over $\mathbf{F}_{q}$. Then the map

$$\begin{eqnarray}\unicode[STIX]{x1D711}_{\ast }:H_{\acute{\text{e}}\text{t}}^{\ast }(X\otimes _{\boldsymbol{ F}_{q}}X)\rightarrow H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X)\end{eqnarray}$$

defined in §6.3.2 coincides with the pushforward in cohomology (as defined in Definition 6.19) induced by the vertical map in (6.3.14)

$$\begin{eqnarray}H^{\ast }(\text{Oblv}(\acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X)\times \acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X))_{h\widehat{\mathbf{Z}}})\rightarrow H^{\ast }((\acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X)\times \acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X))_{h(\widehat{\mathbf{Z}}\times \widehat{\mathbf{Z}})})\end{eqnarray}$$

under the identifications obtained by §3.2 and the horizontal equivalences in (6.3.14):

$$\begin{eqnarray}\displaystyle H^{\ast }(\text{Oblv}(\acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X)\times \acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X))_{h\widehat{\mathbf{Z}}}) & = & \displaystyle H_{\acute{\text{e}}\text{t}}^{\ast }(X\times _{\mathbf{F}_{q}}X),\nonumber\\ \displaystyle H^{\ast }((\acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X)\times \acute{\text{E}}\text{t}_{/\mathbf{F}_{q}}(X))_{h(\widehat{\mathbf{Z}}\times \widehat{\mathbf{Z}})}) & = & \displaystyle H_{\acute{\text{e}}\text{t}}^{\ast }(X)\otimes H_{\acute{\text{e}}\text{t}}^{\ast }(X).\nonumber\end{eqnarray}$$

There are other constructions of the pushforward in cohomology, which are more useful for studying the interaction with Steenrod operations, so we prove a general criterion for recognizing when a map on cohomology coincides with the pushforward as we have defined it.

Lemma 6.21. Suppose $f:\mathbf{W}\rightarrow \mathbf{Y}$ is a map of (pro-)spaces with Poincaré duality so that $f_{\ast }$ is defined as in Definition 6.19. Keeping the notation of Definition 6.19, suppose $f_{\ast }^{\prime }:H^{\ast }(\mathbf{W})\rightarrow H^{\ast +n-m}(\mathbf{Y})$ is another map satisfying the following:

1. (i) $f_{\ast }^{\prime }(\unicode[STIX]{x1D707}_{\mathbf{W}})=\unicode[STIX]{x1D707}_{\mathbf{Y}}$;

2. (ii) $f_{\ast }^{\prime }(x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE})=f_{\ast }^{\prime }x\smallsmile \unicode[STIX]{x1D6FE}$ for all $x\in H^{\ast }(\mathbf{W})$ and $\unicode[STIX]{x1D6FE}\in H^{\ast }(\mathbf{Y})$.

Then $f_{\ast }^{\prime }=f_{\ast }$.

Remark 6.22. In other words, condition (ii) is saying that $f_{\ast }^{\prime }$ is a module homomorphism for $H^{\ast }(\mathbf{Y})$, for its natural action on $H^{\ast }(\mathbf{Y})$ and its action on $H^{\ast }(\mathbf{W})$ via $f^{\ast }$.

Proof. By Poincaré duality on $\mathbf{Y}$, it suffices to show that

(6.3.19)$$\begin{eqnarray}\int _{\mathbf{Y}}f_{\ast }^{\prime }x\smallsmile \unicode[STIX]{x1D6FE}=\int _{\boldsymbol{ Y}}f_{\ast }x\smallsmile \unicode[STIX]{x1D6FE}\quad \text{for all }x\in H^{\ast }(\mathbf{W}),\unicode[STIX]{x1D6FE}\in H^{\ast }(\mathbf{Y}).\end{eqnarray}$$

By definition (Definition 6.19), we have

(6.3.20)$$\begin{eqnarray}\int _{\mathbf{Y}}f_{\ast }x\smallsmile \unicode[STIX]{x1D6FE}=\int _{\mathbf{W}}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE}.\end{eqnarray}$$

Substituting (6.3.20) into (6.3.19) and using condition (ii), we see that we need to show that

(6.3.21)$$\begin{eqnarray}\int _{\mathbf{Y}}f_{\ast }^{\prime }(x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE})=\int _{\boldsymbol{ W}}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE}.\end{eqnarray}$$

By the definition of the fundamental class, we can write

$$\begin{eqnarray}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE}=\biggl(\int _{\boldsymbol{ W}}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE}\biggr)\unicode[STIX]{x1D707}_{\boldsymbol{ W}}+\text{(lower degree terms)},\end{eqnarray}$$

hence condition (i) of the lemma implies that

(6.3.22)$$\begin{eqnarray}f_{\ast }^{\prime }(x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE})=\biggl(\int _{\boldsymbol{ W}}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE}\biggr)\unicode[STIX]{x1D707}_{\boldsymbol{ Y}}+\text{(lower degree terms)}.\end{eqnarray}$$

Substituting (6.3.22) into (6.3.21), we have

$$\begin{eqnarray}\int _{\mathbf{Y}}f_{\ast }^{\prime }(x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE})=\biggl(\int _{\boldsymbol{ W}}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE}\biggr)\cdot \biggl(\int _{\boldsymbol{ Y}}f_{\ast }^{\prime }\unicode[STIX]{x1D707}_{\mathbf{W}}\biggr)=\int _{\mathbf{W}}x\smallsmile f^{\ast }\unicode[STIX]{x1D6FE},\end{eqnarray}$$

where the last equality follows from condition (i) that $f_{\ast }^{\prime }\unicode[STIX]{x1D707}_{\mathbf{W}}=\unicode[STIX]{x1D707}_{\mathbf{Y}}$ and the fact that the lower degree terms ‘integrate’ to $0$ by definition.◻

We use this discussion to study the pushforward in cohomology induced by a homotopy quotient of spaces $M\rightarrow M_{h\mathbf{Z}}$. The following proposition is presumably well-known, but we have included a proof since we could not find a reference.

Proposition 6.23. Let $M$ be any simplicial set with $\mathbf{Z}$-action. If

$$\begin{eqnarray}f:M\rightarrow M_{h\mathbf{Z}}\end{eqnarray}$$

denotes the homotopy quotient map, then we can define a pushforward map

$$\begin{eqnarray}f_{\ast }:H^{i}(M)\rightarrow H^{i+1}(M_{h\mathbf{Z}}).\end{eqnarray}$$

If $M$ has Poincaré duality, then so does $M_{h\mathbf{Z}}$ and $f_{\ast }$ agrees with the pushforward defined in Definition 6.19.

Moreover, for all $x\in H^{\ast }(M)$ we have

$$\begin{eqnarray}\operatorname{Sq}\circ f_{\ast }(x)=f_{\ast }\circ \operatorname{Sq}(x).\end{eqnarray}$$

Proof. We can take $\mathbf{R}=E\mathbf{Z}$ as a model for $E\mathbf{Z}$, so that a model for $M_{h\mathbf{Z}}$ is $(M\times \mathbf{R})/\mathbf{Z}$. (The map $M\rightarrow (M\times \mathbf{R})/\mathbf{Z}$ is what might classically be called the ‘inclusion of a fiber into the mapping torus’.) With this model there is an evident homeomorphism

$$\begin{eqnarray}((M\times \mathbf{R})/\mathbf{Z},M)\cong (S^{1}\wedge (M_{+}),\operatorname{pt}),\end{eqnarray}$$

where the left-hand side means the pointed space obtained from $(M\times \mathbf{R})/\mathbf{Z}$ by collapsing $M\times \{0\}$ to a point.

We always have a pushforward map $H^{\ast }(M)\rightarrow H^{\ast +1}(M_{h\mathbf{Z}})$ defined as the following composition.

(6.3.23)

If $M$ has Poincaré duality then so does $M_{h\mathbf{Z}}$ by the same reasoning as in §2.1, so we have fundamental classes $\unicode[STIX]{x1D707}_{M}$ and $\unicode[STIX]{x1D707}_{M_{h\mathbf{Z}}}$ and Definition 6.19 supplies another notion of pushforward $H^{\ast }(M)\rightarrow H^{\ast +1}(M_{h\mathbf{Z}})$. In this case, the alternate pushforward (

6.3.23

) evidently takes $\unicode[STIX]{x1D707}_{M}\mapsto \unicode[STIX]{x1D707}_{M_{h\mathbf{Z}}}$, and is an $H^{\ast }(M_{h\mathbf{Z}})$-module homomorphism since $b$ is induced by a map of spaces $M_{h\mathbf{Z}}\rightarrow S^{1}\wedge (M_{+})$. Therefore, Lemma 6.21 shows that the two notions of pushforward coincide.

Finally, the homomorphism $b$ commutes with Steenrod operations because it is induced by the map of spaces, and the map $a$ commutes with Steenrod operations because it is a suspension isomorphism, and Steenrod operations always commute with suspension (§3.4).◻

The punchline. We now finally complete the proof of Proposition 6.14. We transfer the Poincaré duality structure

$$\begin{eqnarray}\text{from }H^{\ast }(\acute{\text{E}}\text{t}(X\times _{\boldsymbol{ F}_{q}}X))\text{ to }H^{\ast }((\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{h\mathbf{Z}})\end{eqnarray}$$

and

$$\begin{eqnarray}\text{from }H^{\ast }(\acute{\text{E}}\text{t}(X)\times \acute{\text{E}}\text{t}(X))\text{ to }H^{\ast }((\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{h(\mathbf{Z}\times \mathbf{Z})})\end{eqnarray}$$

using the horizontal isomorphisms in (6.3.14) plus Lemma 6.17. Then Definition 6.19 furnishes a notion of pushforward

(6.3.24)$$\begin{eqnarray}H^{\ast }((\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{h\mathbf{Z}})\rightarrow H^{\ast }((\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{h(\mathbf{Z}\times \mathbf{Z})}),\end{eqnarray}$$

which by Lemma 6.20 is compatible the map in Proposition 6.14 in the sense that the following diagram commutes.

On the other hand, we may write $(\acute{\text{E}}\text{t}_{/k}(X)\times \acute{\text{E}}\text{t}_{/k}(X))_{h\mathbf{Z}}=\{{U_{i}\}}_{i}$, with each $U_{i}$ being a $\mathbf{Z}$-space. Then by Lemma 6.21 the map (6.3.24) agrees with the colimit of the levelwise pushforwards

$$\begin{eqnarray}H^{\ast }(U_{i})\rightarrow H^{\ast }((U_{i})_{h\mathbf{Z}}),\end{eqnarray}$$

each of which commutes with Steenrod operations by Lemma 6.23. ◻

7.1 Lifting Wu classes to integral cohomology

The goal of this subsection is to prove that Wu classes lift to integral cohomology, which will turn out to be important later.

Proof. We induct on $s$. The base case is $v_{0}=w_{0}=1$. Consider the equation

$$\begin{eqnarray}\operatorname{Sq}v=w\end{eqnarray}$$

from Theorem 6.5. Equating terms in cohomological degree $s$, we have

$$\begin{eqnarray}v_{s}+\operatorname{Sq}^{1}v_{s-1}+\cdots =w_{s}.\end{eqnarray}$$

By the induction hypothesis each term $v_{s-i}$ is a polynomial in the Stiefel–Whitney classes, so by the Cartan formula for Steenrod squares (§3.4) and Lemma 5.4, each $\operatorname{Sq}^{i}v_{s-i}$ is a polynomial in the Stiefel–Whitney classes. Then solving for $v_{s}$ completes the induction.◻

7.2 Proof of the main theorem

We now combine the preceding ingredients to prove Theorem 2.7, which as already noted implies Theorem 1.3. We wish to show that

$$\begin{eqnarray}x\smallsmile \unicode[STIX]{x1D6FD}(x)=0\quad \text{for all }x\in H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n}\mathbf{Z}(d)).\end{eqnarray}$$

Theorem 4.4 tells us that

(7.2.1)$$\begin{eqnarray}x\smallsmile \unicode[STIX]{x1D6FD}(x)=\widetilde{\operatorname{Sq}}^{2d}(\unicode[STIX]{x1D6FD}(x)).\end{eqnarray}$$

Then Lemma 3.12 implies that

(7.2.2)$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{2d}(\unicode[STIX]{x1D6FD}(x))=[2^{n-1}]\circ \operatorname{Sq}^{2d}(\overline{\unicode[STIX]{x1D6FD}(x)}),\end{eqnarray}$$

where $\overline{\unicode[STIX]{x1D6FD}(x)}$ denotes the reduction of $\unicode[STIX]{x1D6FD}(x)$ mod $2$, and the notation $[2^{n-1}]$ is as in Lemma 3.12.

By the definition of the Wu classes in Theorem 6.5, we have

(7.2.3)$$\begin{eqnarray}\widetilde{\operatorname{Sq}}^{2d}(\unicode[STIX]{x1D6FD}(x))=[2^{n-1}](v_{2d}\smallsmile \overline{\unicode[STIX]{x1D6FD}(x)}).\end{eqnarray}$$

It is immediate from the definition of $[2^{n-1}]$ that

(7.2.4)$$\begin{eqnarray}[2^{n-1}](v_{2d}\smallsmile \overline{\unicode[STIX]{x1D6FD}(x)})=([2^{n-1}]v_{2d})\smallsmile \unicode[STIX]{x1D6FD}(x).\end{eqnarray}$$

Next, since $\unicode[STIX]{x1D6FD}$ is a derivation we have

(7.2.5)$$\begin{eqnarray}([2^{n-1}]v_{2d})\smallsmile \unicode[STIX]{x1D6FD}(x)=\unicode[STIX]{x1D6FD}([2^{n-1}]v_{2d}\smallsmile x)-\unicode[STIX]{x1D6FD}([2^{n-1}]v_{2d})\smallsmile x.\end{eqnarray}$$

Stringing together (7.2.1), (7.2.2), (7.2.3), and (7.2.4), it suffices to show that (7.2.5) is $0$ for all $x$.

Lemma 7.3. We have $\unicode[STIX]{x1D6FD}([2^{n-1}]v_{2d})=\unicode[STIX]{x1D6FD}_{2,2^{n}}(v_{2d})$ where $\unicode[STIX]{x1D6FD}_{2,2^{n}}$ is the boundary map for

$$\begin{eqnarray}0\rightarrow \mathbf{Z}/2^{n}\mathbf{Z}(d)\xrightarrow[{}]{2}\mathbf{Z}/2^{n+1}\mathbf{Z}(d)\rightarrow \mathbf{Z}/2\mathbf{Z}(d)\rightarrow 0.\end{eqnarray}$$

Proof. This follows immediately from the below commutative diagram.

Referring to (7.2.5), we have $\unicode[STIX]{x1D6FD}([2^{n-1}]v_{2d}\smallsmile x)=0$ for all $x$ by Lemma 2.4. But we also have $\unicode[STIX]{x1D6FD}_{2,2^{n}}(v_{2d})=0$ as this is the obstruction to lifting $v_{2d}$ to $H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}/2^{n+1}\mathbf{Z}(d))$, and $v$ even lifts to $v_{2d}\in H_{\acute{\text{e}}\text{t}}^{2d}(X;\mathbf{Z}_{2}(d))$ by Corollary 7.2. By Lemma 7.3 we also get $\unicode[STIX]{x1D6FD}([2^{n-1}]v_{2d})=0$, so the expression in (7.2.5) vanishes for all $x$, which completes the proof.◻

7.3 A criterion for the alternation of the linking form

The argument of §7.2, and the ingredients going into it, can be adapted in a straightforward manner to study linking form of an odd-dimensional manifold (§2.2). It yields the following conclusion, which we state in the orientable case for simplicity.

Theorem 7.4. Let $M$ be an orientable closed manifold of dimension $4d+1$. Then the linking form on $M$ is alternating if and only if $v_{2d}$ lifts to $H^{2d}(M;\mathbf{Z})$, i.e. if and only if $\widetilde{\unicode[STIX]{x1D6FD}}(v_{2d})=0$ where $\widetilde{\unicode[STIX]{x1D6FD}}$ is the boundary map for the short exact sequence

$$\begin{eqnarray}0\rightarrow \mathbf{Z}\xrightarrow[{}]{2}\mathbf{Z}\rightarrow \mathbf{Z}/2\mathbf{Z}\rightarrow 0.\end{eqnarray}$$

Proof. The argument in §7.2, with the appropriate adaptations of its ingredients to singular cohomology, shows that the linking form is alternating if and only if $\unicode[STIX]{x1D6FD}_{2,2^{n}}(v_{2d})=0$ for all $n$. Since $H^{2d+1}(M;\mathbf{Z})$ is finitely generated, the reduction maps induce an injection

$$\begin{eqnarray}H^{2d+1}(M;\mathbf{Z}){\hookrightarrow}\mathop{\varprojlim }\nolimits_{n}H^{2d+1}(M;\mathbf{Z}/2^{n}\mathbf{Z}),\end{eqnarray}$$

which implies that $\ker \widetilde{\unicode[STIX]{x1D6FD}}=\bigcap _{n}\ker \unicode[STIX]{x1D6FD}_{2,2^{n}}$.◻