Proof. Since $\operatorname{Sq}^1$ is a derivation,

\begin{equation*}\operatorname{Sq}^1\circ w_1=w_1^2+w_1\circ \operatorname{Sq}^1=w_1\circ \operatorname{Sq}^1_\mathscr{L},\end{equation*}

which is also equal to $\operatorname{Sq}^1\circ \operatorname{Sq}^1_\mathscr{L}$ using the vanishing relations $\operatorname{Sq}^1\circ \operatorname{Sq}^1=0$ and

\begin{equation*}\operatorname{Sq}^1_\mathscr{L}\circ\operatorname{Sq}^1_\mathscr{L}=(w_1+\operatorname{Sq}^1)\circ \operatorname{Sq}^1_\mathscr{L}=w_1\circ \operatorname{Sq}^1_\mathscr{L}+\operatorname{Sq}^1\circ \operatorname{Sq}^1_\mathscr{L}=0.\end{equation*}

The proof of the other equality is entirely analogous.

Proof. For the first half of the statement, it is enough to note that the ideal $(w_1)$ is a Steenrod submodule:

\begin{equation*}\operatorname{Sq}^k(x\cdot w_1)=\operatorname{Sq}^{k}(x)\cdot w_1+\operatorname{Sq}^{k-1}(x)\cdot w_1^2.\end{equation*}

For the second statement, the first map is a cohomological pullback and therefore compatible with Steenrod operations. The second morphism (projection to K) factors as a composition

\begin{equation*} {\rm H}^i(\widetilde{\operatorname{Gr}}_k(n);\mathbb{F}_2)\xrightarrow{\partial}{\rm H}^{i+1}(\mathscr{L},\mathscr{L}\setminus 0;\mathbb{F}_2)\xrightarrow{{\rm Th}}{\rm H}^i(\operatorname{Gr}_k(n);\mathbb{F}_2). \end{equation*}

Here, the first map is the boundary map in the long exact sequence for the pair $(\mathscr{L},\mathscr{L}\setminus 0)$ of the total space of the determinant line bundle $\mathscr{L}$ and the complement $\mathscr{L}\setminus 0$ of the zero section. The second map is the Thom isomorphism for the line bundle $\mathscr{L}$. Stability of the Steenrod operations means that they commute with the boundary map $\partial$. In general, Steenrod operations don’t commute with the Thom isomorphism. Rather ${\rm Th}\circ{\rm Sq}\circ{\rm Th}^{-1}(1)=w$ is the total Stiefel–Whitney class and therefore ${\rm Th}\circ{\rm Sq}\circ{\rm Th}^{-1}=w\cdot{\rm Sq}$. The key point in our case is that the total Stiefel–Whitney class is $w=1+w_1$, and any element in the image of δ is in $K=\ker(w_1)\subseteq {\rm H}^i(\operatorname{Gr}_k(n);\mathbb{F}_2)$. This implies that the Thom isomorphism commutes with Steenrod squares for elements in the image of $\partial$, and consequently the projection ${\rm H}^*(\widetilde{\operatorname{Gr}}_k(n);\mathbb{F}_2)\to K$ is compatible with Steenrod operations.

Proof. We first note that the tautological sub-bundle $S_{n+j}$ over $\operatorname{Gr}_k(n+j)$ pulls back via i to the tautological bundle S_n over $\operatorname{Gr}_k(n)$. Since ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)$ is generated by $w_l(S_{n})=i^*w_l(S_{n+j})$, for $l=1,\ldots, k$, we find that $i^*$ is surjective. On the other hand, $I=(Q_{n-k+1},\ldots, Q_{n-k+j})\subseteq \ker i^*$, since the pull-back of the quotient bundle splits off a rank j trivial bundle, so its top j Stiefel–Whitney classes $Q_{n-k+1},\ldots, Q_{n-k+j}$ are zero. Since by (2),

\begin{align*} &\underbrace{\mathbb{F}_2[w_1,\ldots, w_k]/(Q_{n-k+1},\ldots, Q_{n})}_{{\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)}\\ &\quad\cong \underbrace{\mathbb{F}_2[w_1,\ldots, w_k]/(Q_{n+j-k+1},\ldots, Q_{n+j})}_{{\rm H}^*(\operatorname{Gr}_k(n+j);\mathbb{F}_2)}\big/(Q_{n-k+1},\ldots, Q_{n-k+j}) \end{align*}

we can conclude that $\ker i^*\subseteq I$ and therefore $\ker i^*=I$.

For the second part of the statement, choose flags $F_\bullet$ in $\mathbb{R}^n$ and $E_\bullet$ in $\mathbb{R}^{n+j}$ compatible with ι, in the sense that $\iota(F_l)=E_l$ for $l\leq n$. Then by the definition of Schubert varieties (e.g. [Reference Munkholm16, 9.4]), we have $i(\sigma_\lambda(F_\bullet))=\sigma_{\lambda'}(E_\bullet)$, where $\lambda'=(\lambda_1+j,\lambda_2+j,\ldots, \lambda_k+j)$, so

\begin{equation*}i_![\sigma_\lambda]=[\sigma_{\lambda'}].\end{equation*}

Since the cohomology has a basis of Schubert classes, this implies that the pushforward $i_!$ is injective. Finally, the normal bundle of i is $\operatorname{Hom}(S,\mathbb{R}^j)$, whose Euler class is $w_k^j$, which implies by the adjunction formula that $i_!w^a=i_!i^*w^a=w^a\cdot i_!1=w^a\cdot w_k^j$.

Proof. By a theorem of Atiyah and Hirzebruch (a Grothendieck–Riemann–Roch-type statement for Steenrod operations), cf. [Reference Atiyah and Hirzebruch1, Satz 3.2], we have

\begin{equation*}\operatorname{Sq}(i_! x)=i_!\bigl(\operatorname{Sq}(x)\cdot w(\nu_i)\bigr),\end{equation*}

where $w(\nu_i)$ denotes the total Stiefel–Whitney class of the normal bundle of the inclusion i. After taking the appropriate degree part, we get:

\begin{equation*}\operatorname{Sq}^1i_!x=i_!(x\cdot w_1(\nu_i))+i_!\operatorname{Sq}^1x=i_!\operatorname{Sq}^1_\mathscr{L} x,\end{equation*}

where we use that $w_1(\nu_i)=w_1(S_{n-1})=w_1(\mathscr{L})$. Similarly,

\begin{equation*}\operatorname{Sq}^1_\mathscr{L} i_!x= w_1\cdot i_!x+\operatorname{Sq}^1 i_!x=i_! (w_1\cdot x)+i_!\operatorname{Sq}^1_\mathscr{L} x=i_!\operatorname{Sq}^1 x. \end{equation*}

Proof. Using $\operatorname{rk} (\operatorname{O}(m))= m$, $\operatorname{rk} (\operatorname{SO}(m))=m-1$, we get $\operatorname{rk}(G)=N$, and in (i) $\operatorname{rk}(K)=N$, and in (ii) $\operatorname{rk}(K)=N-r$, and we can apply (12).

Proof. We have an exact sequence

If there is a relation of the form (13), then q_j is a redundant generator expressible in terms of the other q_i. In this case, the exact sequence fails to be a presentation, but we obtain a presentation upon removing q_j. The number of remaining q_i equals the number of w_i, hence the deficiency is 0. On the other hand, if there is no relation of the form (13), none of the q_i is redundant, and the exact sequence is already a presentation. Consequently, the deficiency is 1 in this case.

Proof. A non-trivial W ₂-relation between the q_j’s implies a non-trivial class in the Koszul homology $H_1(Q,W_2)$, for this statement and relevant notation, see [Reference Matszangosz27, Section 5.2]. A non-zero class in Koszul homology $H_1(Q,W_2)$ in degree d implies that $\mathrm{crk}(S\to\widetilde{\operatorname{Gr}}_k(n))\leq d-2$, see the beginning of Section 5.3 of loc. cit. Conversely, if $\mathrm{crk}(S) \gt t$, then non-zero Koszul homology classes have to have degree $ \gt t+2$.

Proof. These follow by applying proposition 4.10 to results from [Reference Fukaya15], [Reference Jovanović24], [Reference Basu and Chakraborty3], and [Reference Matszangosz27].

First, we cover the cases δ = 0. For k = 2, $q_{\text{odd}}=0$, which is a relation of the form (13) for all n.

For k = 3 and k = 4, we have $q_{2^t-3}=0$ by [Reference Fukaya15], [Reference Jovanović24]. This is a relation of the form (13) for k = 3 and $n=2^{t}-3,2^{t}-2,2^{t}-1$, and for k = 4 and $n=2^{t}-3, 2^{t}-2,2^{t}-1, 2^t$.

For k odd and $n=2^t$, (14) is a relation of the form (13).

This covers all the cases when δ = 0. Now let us turn to the cases when δ = 1. In a number of cases, there are no relations in degrees $\leq n$ at all.

For $\widetilde{\operatorname{Gr}}_3(n)$ with $2^{t-1} \lt n\leq 2^t-4$, there are no relations in degrees $\leq n$ by [Reference Basu and Chakraborty3] or [Reference Matszangosz27, theorem 1.1], showing that the degrees of anomalous generators are $\geq n$.

For $\widetilde{\operatorname{Gr}}_4(n)$ with $2^{t-1} \lt n\leq 2^t-4$, there are no relations in degrees $\leq n$ since the characteristic rank of $\widetilde{\operatorname{Gr}}_4(n)$ is $ \gt n-2$, cf. [Reference Baum31, theorem 6.6].

Proof. For the proof, we will use the following general observation: Since every torsion element is of order 2, the reduction morphism $\rho_\mathscr{L}\colon{\rm H}^*(X;\mathbb{Z}(\mathscr{L}))\to {\rm H}^*(X;\mathbb{F}_2)$ is injective on 2-torsion. Thus, if both sides of an equality are 2-torsion, it is enough to show the mod 2 reduction of the statement.

(1) Since the Bocksteins and $e(\mathscr{L})$ are 2-torsion, both sides of the first equality are 2-torsion. The mod 2 reduction follows from the classical Wu formula:

\begin{equation*}\operatorname{Sq}^1_{\mathscr{L}}(w_1x)=\operatorname{Sq}^1(w_1x)+w_1^2x=w_1\operatorname{Sq}^1(x)+w_1^2x+w_1^2x= w_1\operatorname{Sq}^1(x).\end{equation*}

(2) For the second equality, the right-hand side is a Bockstein, and the left hand-side is the $\delta_{\mathscr{L}}$-image of a 2-torsion element, so again both sides are 2-torsion. The mod 2 reduction is the statement that, in our case, $\operatorname{Sq}^1$ commutes with the coboundary, cf. proposition 2.5:

\begin{equation*} \delta\operatorname{Sq}^1=\operatorname{Sq}^1\delta.\end{equation*}

As an alternative formulation, we are showing the commutativity of the upper-right square in (15) by noting that the lower-right square commutes from naturality of boundary maps and the commutativity of the composition of the squares, which is $\delta\operatorname{Sq}^1=\operatorname{Sq}^1\delta$ from proposition 2.5.

Proof. First, we show that the inclusion of (2) always holds. The inclusion $\operatorname{Im} (w_1\circ \rho)\subseteq \operatorname{Im} w_1$ is immediate, so we only have to show $\operatorname{Im}(w_1\circ \rho)\subseteq \operatorname{Im} \operatorname{Sq}^1_\mathscr{L}$. For all elements, $z\in \operatorname{Im} \rho$, $\operatorname{Sq}^1(z)=0$. Therefore, $w_1\cdot z=(\operatorname{Sq}^1_\mathscr{L}+\operatorname{Sq}^1)(z)=\operatorname{Sq}^1_\mathscr{L}(z)$. The inclusion in point (3) is entirely analogous.

(2) $\Rightarrow$(1): Assume that there is an element in the difference, i.e. there exists an element

\begin{equation*}z=\operatorname{Sq}^1_\mathscr{L}(a)=w_1\cdot z'\in {\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2),\end{equation*}

which is not in the image of $w_1\circ \rho$.

We claim that $\pi^*_\mathscr{L}\beta_\mathscr{L}(a)$ is a non-zero, 2-divisible, 2-torsion element, i.e. it is two times a 4-torsion class. First, it is non-zero, otherwise $\beta_\mathscr{L}(a)$ would be in the image of $e(\mathscr{L})$, which is impossible since then z would be in the image of $w_1\circ \rho$, contradicting the base assumption. It is clearly 2-torsion, since it is in the image of the Bockstein map. Finally, $z=w_1\cdot z'$ implies $\pi^*(z)=\rho\circ \pi^*_\mathscr{L}\circ \beta_\mathscr{L}(a)=0$. Therefore, the mod 2 reduction of $\pi^*_\mathscr{L}\beta_\mathscr{L}(a)$ is zero, hence it is 2-divisible.

(1) $\Rightarrow$(2): By theorem 3.1, we can have at worst 4-torsion, so assume ${\rm H}^*(\widetilde{\operatorname{Gr}}_k(n);\mathbb{Z})$ contains non-trivial 4-torsion. Then it arises as y in Diagram (1), together with some $a,c,b,x,x'$ as in the diagram, see remark 5.3. Then $\rho_\mathscr{L}(x)$ is an element in $\operatorname{Im} w_1\cap \operatorname{Im} \operatorname{Sq}^1$. We claim that it is not in $\operatorname{Im}(w_1\circ \rho)$. Indeed, if $\rho_\mathscr{L}(x)=w_1\cdot \rho(v)$ for some $v\in {\rm H}^*(\operatorname{Gr}_k(n);\mathbb{Z})$, then $e(\mathscr{L})\cdot v-x\in \ker \rho_\mathscr{L}$, i.e. it is 2-divisible. Since x is 2-torsion, and $e(\mathscr{L})\cdot v$ is also 2-torsion, this means that $e(\mathscr{L})\cdot v=x$. But that would mean $y=\pi_\mathscr{L}^*x=0$, which is impossible, so $\rho_\mathscr{L}(x)$ is not contained in $\operatorname{Im}(w_1\circ \rho)$.

The equivalence (1) $\Leftrightarrow$(3) is obtained by repeating the proof for the other Gysin sequence, respectively, the diagram obtained from (15) by switching the role of the coefficients $\mathbb{Z}$ and $\mathscr{L}$.

Proof. The statement on Q_j follows from the recursion $Q_j=\sum_{i=1}^k w_iQ_{j-i}$ since $\binom{|a|}{a}$ counts the number of ways the monomial w^a appears in the recursion. Since q_j is the reduction of Q_j modulo w ₁,

\begin{equation*}q_j=\sum_{j=2a_2+\ldots +ka_k}\binom{|a|}{a}w^a,\end{equation*}

the statement for $P_j=Q_j+\tilde{q}_j$ follows by taking the sum of the two expressions.

Proof. Using the definitions (17) and (18), Eq. (20) states that the following equality holds in ${\rm H}^*(\operatorname{Gr}_k(n-1);\mathbb{F}_2)$:

\begin{equation*} c_1p_{n-1}+\sum_{j=2}^{k-1}(c_0w_{j}+c_{j})p_{n-j}+c_0w_kp_{n-k}=i_{n-1}^*\left( \sum_{j=0}^{k-1}c_jp_{n-j} \right). \end{equation*}

Using the recursion (16), $p_n=Q_{n-1}+\sum_{l=2}^k w_l p_{n-l}$ since $i^*w_j=w_j$, the right hand side is equal to

\begin{align*} \sum_{j=1}^{k-1}c_jp_{n-j}+c_0\left(Q_{n-1}+\sum_{l=2}^k w_l p_{n-l}\right)&=c_1p_{n-1}+\sum_{j=2}^{k-1}(c_0w_j+c_j)p_{n-j}\\ &\quad +c_0w_kp_{n-k}+c_0Q_{n-1}, \end{align*}

and noting that $Q_{n-1}=0$ in ${\rm H}^*(\operatorname{Gr}_k(n-1);\mathbb{F}_2)$, this proves the first equality.

The second equality (21)—using the definitions (17) and (19)—states that

\begin{equation*} c_{k-1}p_{n+1}+c_0w_kp_n+\sum_{j=1}^{k-2}(c_jw_k+c_{k-1}w_{j+1})p_{n-j}=(i_n)_!\left(\sum_{j=0}^{k-1}c_jp_{n-j}\right). \end{equation*}

Using the recursion $p_{n+1}=Q_{n}+\sum_{l=2}^k w_l p_{n+1-l}$, we can write the left-hand side as

\begin{equation*} \begin{split} &c_{k-1}Q_n+c_{k-1}\sum_{l=2}^kw_lp_{n+1-l}+c_0w_k p_n+\sum_{j=1}^{k-2}(c_jw_k+c_{k-1}w_{j+1})p_{n-j}\\ =&\,c_0w_kp_n+c_{k-1}Q_n+\sum_{l=1}^{k-2}(c_{k-1}w_{l+1}+c_lw_k+c_{k-1}w_{l+1})p_{n-l}+c_{k-1}w_{k}p_{n+1-k}\\ =&\,w_k\left( \sum_{l=0}^{k-1}c_lp_{n-l}\right)+c_{k-1}Q_n \end{split} \end{equation*}

which by lemma 2.6 is equal to the right-hand side, modulo the ideal $(Q_n)$. Since $k\geq 2$ by assumption, Q_n is one of the relations defining ${\rm H}^*(\operatorname{Gr}_k(n+1);\mathbb{F}_2)=\mathbb{F}_2[w_1,\ldots, w_k]/(Q_{n-k+2},\ldots, Q_{n+1})$, and this concludes the proof.

Proof. The claim is equivalent to the following equality:

\begin{equation*} \sum_{i\geq 0}w_{2i}P_{2^t-2i}=w_1^{2^t}. \end{equation*}

This will be easier to establish, thanks to lemma 6.1. The proof will proceed along the lines of a similar result for the q_j, cf. [Reference Matszangosz27, Sections 4.1 and 4.2].

We will show the following claim, which is an analogue of [Reference Matszangosz27, proposition 4.5] (but including w ₁ in the monomials): for a sequence $a=(a_1,\dots,a_k)$ with $\sum_{i=1}^kia_i=2^t$ and $a_1 \gt 0$, we have

(25)\begin{equation} \binom{|a|}{a}=\sum_{j=1}^{\lfloor k/2\rfloor}\binom{|a|-1}{\hat{a}_{2j}}, \end{equation}

unless $a=(2^t,0,\dots,0)$. Here, for $1\leq i\leq k$, $\hat a_{i}$ denotes the sequence $(a_1,\ldots, a_{i-1},a_i-1,a_{i+1},\ldots, a_k)$.

The key thing to note is that [Reference Matszangosz27, lemma 4.3] still applies, with the same proof. Consequently, we can still make the same arguments as in the proof of [Reference Matszangosz27, proposition 4.5]. If $\binom{|a|}{a}\equiv 1\bmod 2$, there is a unique index l such that $\binom{|a|-1}{\hat{a}_l}\equiv 1\bmod 2$. By the same argument as in loc.cit., we find that if $\sum_{i=1}^kia_i=2^t$ then l must be even unless we are in the exceptional case $a=(2^t,0,\dots,0)$. Similarly, by the same argument as in [Reference Matszangosz27, proposition 4.5], we find that for $\binom{|a|}{a}\equiv 0\bmod 2$, then for any $\binom{|a|-1}{\hat{a}_{2j}}\equiv 1\bmod 2$, there is a unique $l\neq 2j$ with $\binom{|a|-1}{\hat{a}_l}\equiv 1\bmod 2$, and l is even. This establishes (25) and thus the claim of the proposition.

Proof. This follows from proposition 6.3 and [Reference Matszangosz27, theorem 4.6]. Alternatively, it can be established with the same proof as used for these results.

Proof. First, we consider the case k = 3. We first note that by lemma 6.1,

\begin{equation*} Q_{2^t-1}=\sum_{a_1+2a_2+3a_3=2^t-1}\binom{a_1+a_2+a_3}{a_1,a_2,a_3} w_1^{a_1}w_2^{a_2}w_3^{a_3}, \end{equation*}

i.e. it consists of all the monomials $w_1^{a_1}w_2^{a_2}w_3^{a_3}$ of degree $2^t-1$ for which the exponents $a_1,a_2,a_3$ have pairwise disjoint binary expansions. By lemma 6.1 again, $P_{2^t-3}$ consists of all the monomials $w_1^{a_1}w_2^{a_2}w_3^{a_3}$ of degree $2^t-3$ where the exponents $a_1,a_2,a_3$ have disjoint binary expansions, and where $a_1\geq 1$.

We show that $Q_{2^t-1}-w_1^{2^t-1}$ is divisible by w ₃. Assume a monomial $w_1^{a_1}w_2^{a_2}$ of degree $2^t-1$ has non-zero coefficient in $Q_{2^t-1}$. Then $2^t-1=a_1+2a_2$, which means that the binary expansions of a ₁ and $2a_2$ form a disjoint decomposition of a string of 1s of length t. Since the binary expansion of a ₂ removes one trailing 0, this will introduce overlap between the expansions of a ₁ and a ₂, unless $a_2=0$. In particular, there is no monomial of the form $w_1^{a_1}w_2^{a_2}$ with $a_2 \gt 0$ in $Q_{2^t-1}$, so the only monomial in $Q_{2^t-1}$ with $a_3=0$ is $w_1^{2^t-1}$.

Now we want to show that a monomial $w_1^{a_1}w_2^{a_2}w_3^{a_3}$ appears in $Q_{2^t-1}$ if and only if $w_1^{a_1+1}w_2^{a_2}w_3^{a_3-1}$ appears in $P_{2^t-3}$.

Assume that $w_1^{a_1}w_2^{a_2}w_3^{a_3}$ appears in $Q_{2^t-1}$. Then $a_1,a_2,a_3$ have disjoint binary expansions, and so we must have a ₂ even, otherwise a ₁ and a ₃ are both even, which is impossible because $a_1+2a_2+3a_3=2^t-1$. So exactly one of a ₁ and a ₃ is odd, and we can make a case distinction.

If a ₁ is even and a ₃ is odd, then in the binary expansions of $(a_1,a_2,a_3)$ and $(a_1+1,a_2,a_3-1)$ we exchange the last bit of a ₁ and a ₃, which results in disjoint binary expansions.

In case a ₁ is odd and a ₃ is even, let q denote the number of trailing ones in a ₁ and r denote the number of trailing zeros in a ₃; by disjointness $q\leq r$. Note that by disjointness, a ₂ must have at least q many trailing zeros. Then for the q + 1’st digit in

\begin{equation*}a_1+2a_2+a_3+2a_3=2^{t}-1\end{equation*}

to be 1, we must also have $r\leq q$. This implies that r = q, and therefore $(a_1+1,a_2,a_3-1)$ are also disjoint because we are switching the last q + 1 bits of a ₁ and a ₃ to get $a_1+1$ and $a_3-1$, respectively.

Assume that $w_1^{a_1+1}w_2^{a_2}w_3^{a_3-1}$ appears in $P_{2^t-3}$. Then as before, exactly one of $a_1+1$ and $a_3-1$ has to be odd. In the case where $a_1+1$ is odd and $a_3-1$ is even, we see that $(a_1,a_2,a_3)$ have disjoint binary expansions and so $w_1^{a_1}w_2^{a_2}w_3^{a_3}$ appears in $Q_{2^t-1}$. In case $a_1+1$ is even and $a_3-1$ is odd, we can make the same argument counting trailing zeros and ones to see that $(a_1,a_2,a_3)$ also have disjoint binary expansions because we swapped the last q + 1 bits of a ₁ and a ₃.

We have thus shown that we get a bijection between monomials of $Q_{2^t-1}-w_1^{2^t-1}$ and $P_{2^t-3}$, sending a monomial $w_1^{a_1}w_2^{a_2}w_3^{a_3}$ to $w_1^{a_1+1}w_2^{a_2}w_3^{a_3-1}$. This shows that

\begin{equation*}Q_{2^t-1}=w_1^{2^t-1}+w_3P_{2^t-3}/w_1=w_1^{2^t-1}+w_3p_{2^t-3},\end{equation*}

which shows the claim since $Q_{2^t-1}=0$ in ${\rm H}^*(\operatorname{Gr}_3(2^t);\mathbb{F}_2)$.

The argument still works for k = 4 with minor obvious changes. The key points above are that the only odd Stiefel–Whitney classes are w ₁ and w ₃.

Proof. We have $w_1p_{2^t-3}=P_{2^t-3}$. Moreover, $q_{2^t-3}=0$ by [Reference Fukaya15] and [Reference Jovanović24, lemma 2.3], cf. also [Reference Matszangosz27, proposition 4.7]. So we have $P_{2^t-3}=Q_{2^t-3}$. Therefore, whenever $Q_{2^t-3}$ is among the relations defining ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)$, we get $p_{2^t-3}\in\ker w_1$. The claim follows because ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)\cong\mathbb{F}_2[w_1,\dots,w_k]/(Q_{n-k+1},\dots,Q_n)$.

Proof. To see that the kernel is generated by these classes a_n and d_n (defined by the ascending and descending operations described in (18) and (19)) see [Reference Matszangosz27, proposition 6.5]. For $n=2^t-1$, the kernel is generated by $p_{2^t-3}$. Using proposition 6.2, we can identify the classes with the pullbacks and pushforwards of $p_{2^t-3}$, which are respectively $p_{2^t-3}$ and $w_3^j\cdot w_1^{2^{t-1}-1}$, using proposition 6.5 for the latter.

Proof. For hooks λ of length 2^t, the conditions of lemma 6.8 are satisfied, so p_λ is odd for all such λ. More explicitly, if $\lambda=(2^t-i,1^i)$, then $p_\lambda=\binom{2^t-1}{i}$ is odd by Lucas’ theorem.

Any other partition λ of 2^t does not contain a hook of length 2^t, so p_λ is even by lemma 6.8.

Proof. (i) We show this case directly: the hook-lengths are row-by-row

\begin{equation*}(1,\ldots, 2^{t}-2k+2,2^t-2k+4,\ldots, 2^t-k,2^t-k+2),(1,\ldots, k-3,k-1),(1).\end{equation*}

Using the symmetry that the 2-adic valuation of $2^t-i$ is equal to the 2-adic valuation of $2^t \gt i \gt 0$, this is the same set of 2-adic valuations as

\begin{equation*}(1,\ldots, 2^{t}-2k+2,2^t-2k+4,\ldots, 2^t-k,2^t-k+2),(2^t-k+1,2^t-k+3,\ldots, 2^t-1),(1).\end{equation*}

This contains all the even numbers between $1,\ldots, 2^t-1$ exactly once, so its 2-adic valuation is equal to the 2-adic valuation of $(2^t)!$ and therefore p_λ is odd by (27).

(ii) We check this case using lemma 6.8. Indeed, it has a unique hook of length 2^t, so by the lemma it is enough to show that by removing it, we obtain a partition µ, for which p_µ is odd. Removing the hook, we obtain the partition $\mu=(2^t-k,k-2,1)$, which was covered in case (i).

Proof. By proposition 6.9, $w_1^{2^t}$ is equal to the sum $\sum s_\lambda$, where λ runs through hooks of length 2^t. However, the maximal hook that fits into the $k\times (n-k)$ rectangle has length n − 1, so $w_1^{2^t}=0$ in ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)$ for $n\leq 2^t$, which gives the bound $\operatorname{ht}(w_1)\leq 2^t-1$.

It remains to show that $w_1^{2^t-1}$ is non-zero. By the duality $\operatorname{Gr}_k(n)\cong \operatorname{Gr}_{n-k}(n)$, we can assume that $k\leq 2^{t-1}$. Moreover, it is enough to show the proposition for $n=2^{t-1}+1$, since for $i\colon\operatorname{Gr}_k(n)\to \operatorname{Gr}_k(n+1)$ we have $i^*w_1=w_1$.

So it is enough to give a partition $\lambda\subseteq k\times(2^{t-1}+1-k)$ of $|\lambda|=2^{t}-1$, such that p_λ is odd. The partition in part (ii) of lemma 6.10 is such a partition, which allows us to conclude.

Proof. (1) Consider the Gysin sequence associated with F:

\begin{equation*} \cdots \to {\rm H}^{i-1}(X;\mathbb{F}_2)\to {\rm H}^{i-1}(F;\mathbb{F}_2)\xrightarrow{\delta} {\rm H}^{i-n}(X;\mathbb{F}_2) \xrightarrow{w_n(E)} {\rm H}^i(X;\mathbb{F}_2)\to\cdots. \end{equation*}

By assumption, the characteristic classes of E generate ${\rm H}^*(X;\mathbb{F}_2)$, so classes which are not characteristic classes are detected by $\ker w_n(E)$ under the boundary map δ. In particular, the lowest non-zero degree of $\ker w_n(E)$ (plus n − 1, the degree of δ) is equal to $\mathrm{crk}(\pi_F^*E\to F)+1$. Set $h=\operatorname{ht}(w_n(E))$, which is finite since X is a manifold. Then by definition, $w_n(E)^h$ is a non-zero element of degree nh in $\ker w_n(E)$.

(2) The argument is the same, using the Gysin sequence for the determinant bundle instead.

Proof. Let $S_0\to \operatorname{Gr}_k(n)$ be the tautological bundle; it is well-known that its characteristic classes generate ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)$. Since $\widetilde{\operatorname{Gr}}_k(n)$ is the sphere bundle of $\det S_0$ and $w_1(\det S_0)=w_1(S_0)$ and $S\to\widetilde{\operatorname{Gr}}_k(n)$ is the pullback of S ₀ along $\pi\colon\widetilde{\operatorname{Gr}}_k(n)\to \operatorname{Gr}_k(n)$, from the discussion at the beginning of this section (or part (2) of proposition 7.1), we obtain the upper bound

\begin{equation*}\mathrm{crk}(S)\leq \operatorname{ht} w_1(S_0)-1=2^t-2\end{equation*}

where the last equality is the theorem of Stong [Reference Stong33], cf. theorem 6.11.

To obtain the other upper bound, recall from lemma 2.6 that if $i\colon\operatorname{Gr}_k(m)\subseteq \operatorname{Gr}_k(n)$ is the canonical inclusion induced by the inclusion of vector spaces $\mathbb{R}^m\subseteq\mathbb{R}^n$, then $\ker i_!=(0)$. Also, if $x\in \ker w_1\subseteq {\rm H}^*(\operatorname{Gr}_k(m);\mathbb{F}_2)$, then $i_!x\in \ker w_1 \subseteq {\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)$, by the projection formula (since w ₁ on $\operatorname{Gr}_k(m)$ is the restriction of w ₁ on $\operatorname{Gr}_k(n)$). For $m=2^{t-1}$, we can use lemma 2.6 to see that

\begin{equation*} i_! w_1^{2^{t-1}-1}=w_k^{n-2^{t-1}}w_1^{2^{t-1}-1} \end{equation*}

is a non-zero element in $\ker w_1$, giving the other upper bound.

Proof. We already know that $d_{2^t}$ is non-zero from [Reference Matszangosz27, theorem 5.7] or via the identification with $w_1^{2^{t}-1}$ in proposition 6.3. It suffices to show that there are no anomalous classes in degrees below $2^t-1$. In the Koszul homology picture of [Reference Matszangosz27], this means that there are no relations between $q_{2^t-4},\dots,q_{2^t}$ in degrees $\leq 2^t-1$. By the lemma of Korbaš in [Reference Jovanović24], we know that $q_{2^t-4}$ and $q_{2^t-3}$ are non-zero, so there cannot be relations in these degrees. We discuss the two remaining degrees below.

(degree $2^t-2$) We want to show that $q_{2^t-2}$ and $w_2q_{2^t-4}$ are independent. Since $2^t-8$ is divisible by 8, the equation $2a+4=2^t-4$ has an even solution and therefore $w_2^aw_4$ with a even appears as monomial in $q_{2^t-4}$. But the monomial $w_2^{a+1}w_4$ cannot appear in $q_{2^t-2}$ and thus cannot be cancelled, showing the independence.

(degree $2^t-1$) We want to show that $q_{2^t-1}$, $w_2q_{2^t-3}$, and $w_3q_{2^t-4}$ are independent. Since $2^t-8$ is divisible by 8, the equation $4a+5=2^t-3$ has an even solution and therefore $w_4^aw_5$ with a even appears as a monomial in $q_{2^t-3}$. But $w_2w_4^aw_5$ cannot appear in $q_{2^t-1}$ because of two odd exponents and obviously cannot appear in $w_3q_{2^t-4}$ as well. On the other hand, $2^t-4$ is exactly divisible by 4 and therefore $w_4^a$ appears with odd exponent a in $q_{2^t-4}$. Also, $w_3w_4^a$ cannot appear in $q_{2^t-1}$ and thus cannot cancel. So there cannot be any non-trivial relation.

Proof. The characteristic rank is $\leq 2^t-2$ by theorem 7.2.

To prove that the characteristic rank is $\geq 2^t-2$, we need to show that there are no relations between $q_{2^t-5},\dots, q_{2^t-1}$ in degrees $\leq 2^t-1$. Again, by Korbaš’ lemma in [Reference Jovanović24], none of these elements are zero, so there are no relations in degrees $2^t-5$ and $2^t-4$. Showing that there are no relations in degrees $2^t-3,\dots,2^t-1$ is a lengthy case distinction carried out below.

(degree $2^t-3$) We want to show that $w_2 q_{2^t-5}$ and $q_{2^t-3}$ are independent. For this, it suffices to find a monomial in $q_{2^t-3}$ which is not divisible by w ₂. We note that for $t\geq 4$ the number $2^t-8$ is divisible by 8, and therefore the solution to $4a+5=2^t-3$ is even. Then the monomial $w_4^aw_5$ appears in $q_{2^t-3}$.

(degree $2^t-2$) We want to show that $w_3q_{2^t-5}, w_2q_{2^t-4}$, and $q_{2^t-2}$ are independent. We first want to show that $q_{2^t-2}$ doesn’t appear in a relation, i.e. we want to show that $q_{2^t-2}$ contains a monomial $w_4^aw_5^2$ with $a-1\equiv 0\bmod 4$ which consequently cannot be cancelled by monomials from $w_3q_{2^t-5}$ or $w_2q_{2^t-4}$. We first note

\begin{equation*} a=\frac{2^t-12}{4}=2^{t-2}-3, \end{equation*}

and consequently a − 1 is divisible by 4. Therefore, the monomial $w_4^aw_5^2$ appears in $q_{2^t-2}$.

To show that $w_3q_{2^t-5}$ and $w_2q_{2^t-4}$ are independent, we note that there is a monomial $w_2^aw_3$ in $q_{2^t-5}$ because $2^t-8$ is divisible by 4, i.e. $2a+3=2^t-5$ has an even solution. Moreover, $a=2^{t-1}-4$ is exactly divisible by 4. Thus, the binary expansion of a − 1 ends with the two digits 11 and therefore a − 1 and 2 don’t have disjoint binary expansions. This means that the monomial $w_2^{a-1}w_3^2$ cannot appear in $q_{2^t-4}$ and consequently the monomial $w_2^aw_3^2$ in $w_3q_{2^t-5}$ cannot be cancelled, showing the independence.

(degree $2^t-1$) Finally, we want to show the independence of $q_{2^t-1}$, $w_2q_{2^t-3}$, $w_3q_{2^t-4}$, $w_4q_{2^t-5}$, and $w_2^2q_{2^t-5}$.

We first show that $w_2q_{2^t-3}$ cannot appear in any relation. For this, we claim that there is a monomial $w_2^aw_5$ in $q_{2^t-3}$ with a even. Since $w_2^{a+1}w_5$ cannot occur in $q_{2^t-1}$ and $w_2^{a-1}w_5$ cannot occur in $q_{2^t-5}$, such a monomial cannot be cancelled. To see that the monomial actually occurs in $q_{2^t-3}$, we note that $2^t-8$ is divisible by 8, and thus $2a+5=2^t-3$ has an even solution. This means that $w_2q_{2^t-3}$ appears trivially.

Now we want to show that $w_3q_{2^t-4}$ cannot appear in any relation. We claim that there is a monomial $w_2^aw_5^2$ in $q_{2^t-4}$ with $a\equiv 1\bmod 8$. This follows since $2a+10=2^t-4$ has a solution $a\equiv 1\bmod 8$ and thus the binary expansions of a and 2 are disjoint. Because a is odd, the monomial $w_2^aw_3w_5^2$ cannot appear in $q_{2^t-1}$, and the monomial $w_2^{a-2}w_3w_5^2$ cannot appear in $q_{2^t-5}$. So there is no possibility to cancel, and hence $w_3q_{2^t-4}$ cannot appear in any relation.

To exclude $w_4q_{2^t-5}$ from any relation, we note that $3+4a=2^t-5$ has a solution with a even and therefore $w_3w_4^a$ appears as monomial in $q_{2^t-5}$, but $w_3w_4^{a+1}$ cannot appear in $q_{2^t-1}$ and hence cannot cancel. This excludes $w_4q_{2^t-5}$ from any relation.

We are left to show that $q_{2^t-1}+w_2^2q_{2^t-5}$ is non-zero. We claim that a monomial $w_3w_4^a$ with a even is contained in $q_{2^t-5}$. This follows since $2^t-8$ is divisible by 8 and thus $3+4a=2^t-5$ has an even solution, but the monomial $w_2^2w_3w_4^a$ with a even cannot appear in $q_{2^t-1}$. This concludes the proof.

Proof. The statement $d_{2^t-1}\neq 0$ has been proved in [Reference Matszangosz27, theorem 5.7], which implies $\mathrm{crk}\leq 2^t-2$.

The lower bound follows from [Reference Prvulović and Radovanović31, theorem 3.1] and theorem 7.6, i.e. one has

\begin{equation*} \mathrm{crk}(S\to \widetilde{\operatorname{Gr}}_6(2^t)) \geq \mathrm{crk}(S \to \widetilde{\operatorname{Gr}}_5(2^t -1)) = 2^t -2. \end{equation*}

Proof. Since $\operatorname{Sq}^1 w_1^{2^t-1}=w_1^{2^t}$, this is 0 by Stong’s theorem, and $\operatorname{Sq}^1_\mathscr{L} w_1^{2^t-1}=w_1^{2^t}+w_1^{2^t}=0$.

For the second equality, note that if j is odd, $\operatorname{Sq}^1\left(w_k^j\right)=w_k^{j-1}\operatorname{Sq}^1(w_k)=w_1w_k^j$, since $w_{k+1}=0$ in ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{F}_2)$. Then the derivation property implies that

\begin{equation*} \operatorname{Sq}^1(a_n)=\operatorname{Sq}^1(w_1^{2^{t-1}-1}w_k^j)=\begin{cases} w_1a_n\qquad& \text{if } j \,\text{even},\\ w_1a_n+w_1a_n\qquad &\text{if } j \,\text{odd}.\\ \end{cases} \end{equation*}

Since $a_n\in \ker w_1$, this is zero in both cases. From this, we also get

\begin{equation*}\operatorname{Sq}^1_\mathscr{L} w_1^{2^{t-1}-1}w_k^j=w_1 a_n+\operatorname{Sq}^1 a_n=0,\end{equation*}

since $a_n\in \ker w_1$.

Proof. For the proof, we will use the following fact concerning the degrees of non-zero rational classes in ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{Q})$:

1. (*) There are no rational classes in odd degree of either twist unless k is odd and n is even.

This holds, since the rational cohomology algebra is generated by Pontryagin classes of degree 4i, Euler classes of even degree, and the hook class of odd degree, if k is odd and n is even, see [Reference Wendt36].

For the assertion concerning d_n, let $n=2^t-1$. By proposition 8.1, d_n is the reduction of an integral (untwisted) class. Since $n=2^t-1$ is odd, by (*), there are no non-torsion classes in degree n and so d_n is the reduction of a 2-torsion class. Therefore, $d_{2^t-1}$ is in the image of $\operatorname{Sq}^1$, i.e. $d_{2^t-1}=\operatorname{Sq}^1 \delta_{2^t-1}$ for some $\delta_{2^t-1}$. For $2^{t-1} \lt n\leq 2^t-1$, set $\delta_{n}=i^*\delta_{2^t-1}$ for the natural inclusion $i\colon\operatorname{Gr}_k(n)\to\operatorname{Gr}_k(2^t-1)$. By naturality, we have

\begin{equation*}\operatorname{Sq}^1\delta_{n}=\operatorname{Sq}^1i^*\delta_{2^t-1}=i^*\operatorname{Sq}^1\delta_{2^t-1}=i^*d_{2^t-1}=d_{n}\end{equation*}

which proves that $d_n\in \operatorname{Im} \operatorname{Sq}^1$.

Next, assume that k is even. We want to show that depending on the parity of n,

(35)\begin{equation} a_{n}\in \operatorname{Im} \operatorname{Sq}^1\qquad \text{or} \qquad a_{n}\in \operatorname{Im} \operatorname{Sq}^1_\mathscr{L}. \end{equation}

We first show for $n=2^t$ that $a_n\in\operatorname{Im}\operatorname{Sq}^1$. The rest of the cases is then established by induction.

Since k is even, $\deg a_{2^t}=2^t-1$ is odd. By (*), there are no odd-degree cohomology classes in ${\rm H}^*(\operatorname{Gr}_k(n);\mathbb{Q})$; therefore, a_n is the reduction of some 2-torsion class from the trivial twist, i.e. $a_n=\operatorname{Sq}^1 \alpha_n$ for some α_n. This proves (35) for $n=2^t$.

To prove the rest of the cases, set $\alpha_n=i_!\alpha_{n-1}$ for $i\colon\operatorname{Gr}_k(n-1)\to \operatorname{Gr}_k(n)$. There are two cases. If by induction $a_{n-1}=\operatorname{Sq}^1_\mathscr{L}\alpha_{n-1}$, then by lemma 2.7:

\begin{equation*}\operatorname{Sq}^1\alpha_n=\operatorname{Sq}^1i_!\alpha_{n-1}=i_!\operatorname{Sq}^1_\mathscr{L} \alpha_{n-1}=i_!a_{n-1}=a_n.\end{equation*}

If by induction $a_{n-1}=\operatorname{Sq}^1\alpha_{n-1}$, then

\begin{equation*} \operatorname{Sq}^1_\mathscr{L} \alpha_n=\operatorname{Sq}^1_\mathscr{L} i_!\alpha_{n-1}=i_!\operatorname{Sq}^1 \alpha_{n-1}=i_!a_{n-1}=a_n. \end{equation*}

This concludes the proof.

Proof. We claim that p_λ defined in (27) is odd for the partition $\lambda=(2^t-k, 1^{k-1})$. Indeed, $p_\lambda=\binom{2^t-2}{k-1}$, which is even iff k is even, by Lucas’ theorem. So for k odd, the coefficient of s_λ in $w_1^{2^{t}-1}$ is non-zero. For k odd, this class is the reduction of a non-zero rational class; indeed, the Schubert variety $\sigma_{(n-k,1^{k-1})}\subseteq \operatorname{Gr}_k(n)$ is a smaller Grassmannian $\operatorname{Gr}_{k-1}(n-2)$ for an appropriate flag, cf. [Reference Wendt36, proposition 5.8] and [Reference Matszangosz26, lemma 5.4]. Since this class appears with zero coefficient in $\operatorname{Sq}^1\sigma_\mu$ for all µ, $w_1^{2^t-1}\not\in \operatorname{Im} \operatorname{Sq}^1$—this follows, e.g. from the main result of [Reference Korbaš25], or via checkerboard colourings, see [Reference Ehresmann11], [Reference Wendt35], [Reference Matszangosz26, (5.4), proposition 6.3.].

Proof. Assuming the characteristic rank conjecture (36) and the conditions of conjecture 8.6, we will show that the smaller-degree class among a_n and d_n satisfies one of the conditions of proposition 5.4. Explicitly, we show that the smaller-degree element is in $\operatorname{Im} w_1$ and $\operatorname{Im} \operatorname{Sq}^1$ (or $\operatorname{Im} \operatorname{Sq}^1_\mathscr{L}$), but it is not in the image of $w_1\circ \rho_\mathscr{L}$ (or $w_1\circ \rho$).

First, $d_n=w_1^{2^t-1}$ is clearly in the image of w ₁ and it is in the image of $\operatorname{Sq}^1$ by proposition 8.2 in both cases (3) and (4). Note that the case k odd and $n=2^t$ in which proposition 8.2 (1) doesn’t apply is also excluded in point (4) of conjecture 8.6. To show that d_n is not in the image of $w_1\circ \rho_\mathscr{L}$, note that for any z ^ʹ such that $w_1z'=w_1^{2^t-1}$, we have $z'-w_1^{2^t-2}\in \ker w_1$. Assuming $\deg d_n \lt \deg a_n$, by (36), $\ker w_1$ is just (0) in degree $\deg d_n-1$, so $z'=w_1^{2^t-2}$. Therefore, it is enough to show that $w_1^{2^t-2}$ is not in the image of $\rho_\mathscr{L}$. And for this, it is enough to show that $\operatorname{Sq}^1_\mathscr{L}(w_1^{2^t-2})\neq 0$:

\begin{equation*} \operatorname{Sq}^1_\mathscr{L}(w_1^{2^t-2})=\operatorname{Sq}^1(w_1^{2^t-2})+w_1^{2^t-1}, \end{equation*}

where the first term is zero by the derivation property, and the second term is non-zero by Stong’s theorem.

The case when $\deg a_n \lt \deg d_n$ proceeds similarly. Note that this implies that we are in case (3) of conjecture 8.6, i.e. k is even. First, $a_n=w_1^{2^{t-1}-1}w_k^j$ with $j=n-2^{t-1}$ is in the image of w ₁. It is also in the image of $\operatorname{Sq}^1$ if n is even and in the image of $\operatorname{Sq}^1_\mathscr{L}$ if n is odd, by proposition 8.2. Similarly to the first case, it is enough to show that $w_1^{2^{t-1}-2}w_k^j$ is not in the image of $\rho_\mathscr{L}$ if n is even and not in the image of ρ if n is odd. Since $\operatorname{Sq}^1 w_k=w_1w_k$, by the derivation property

\begin{equation*}\operatorname{Sq}^1(w_1^{2^{t-1}-2}w_k^j)=j\cdot w_1^{2^{t-1}-1}w_k^j=j\cdot a_n,\end{equation*}

which is non-zero if n is odd and

\begin{equation*}\operatorname{Sq}^1_\mathscr{L}(w_1^{2^{t-1}-2}w_k^j)=(j+1)\cdot a_n,\end{equation*}

which is non-zero if n is even, which allows us to conclude.