Proof. Since the action of $F$ on $E$ is fiberwise, there is a homotopy commutative diagram

Since rows of this diagram are homotopy fibrations, there is a map between associated homology Serre spectral sequences. Since the homology Serre spectral sequences of the top row is isomorphic with $(E^{r}\otimes H_*(F),\,d^{r}\otimes 1)$, we obtain the map $\mu$. By definition, $\mu$ is a coalgebra map. The second statement holds because the differential on $H_*(F)$ in $E^{r}\otimes H_*(F)$ is trivial. Let $p\colon E\to B$ denote the projection. For each $y\in E$, $\pi ^{-1}(\pi (y))$ is homotopy equivalent to the orbit space $y\cdot F$ including $y$. Therefore, by the construction of the Serre spectral sequence, the first statement holds.

Proof. Let $i\colon \overline {E}^{2}\to E^{2}$ and $q\colon E^{2}\to \overline {E}^{2}$ denote the inclusion and the retraction, respectively. Since

\[ d^{2}(x)=d^{2}(i_*(x))=i_*(\overline{d}^{2}(x))=0 \]

the hypothesis that $\overline {d}^{2}=0$ implies that $d^{2}=0$. Therefore $\overline {E}^{3}=\overline {E}^{2}$ is a retract of $E^{3}=E^{2}$. Iterating this argument, since $\overline {d}_{3}=\cdots =\overline {d}^{r-1}=0$, we also obtain $d^{2}=\cdots =d^{r-1}=0$ and therefore $\overline {E}^{r}$ is a retract of $E^{r}$. The inclusion and retraction at the $r^{th}$-stage may still be denoted by $i\colon \overline {E}^{2}\to E^{2}$ and $q\colon E^{2}\to \overline {E}^{2}$. Suppose that $\overline {d}^{r}(x)=0$ for all $x\in H_p(B)=\widehat {E}^{r}_{p,0}$. Then

\[ d^{r}(x)=d^{r}(i_*(x))=i_*(\overline{d}^{r}(x))=0, \]

implying $d^{r}=0$ by lemma 2.1 and corollary 2.2. This is a contradiction, so the first statement is proved.

Let $y$ be an element of least degree element in $H_*(B)=\overline {E}^{r}_{*,0}$ such that $\overline {d}^{r}y\ne 0$. Let $\Delta$ denote comultiplication, and set

\[ \Delta(y)=y\otimes 1+1\otimes y+\sum_iy_i'\otimes y_i'' \]

where $|y_i'|<|y|$ and $|y_i''|<|y|$ for each $i$. Since $y$ is an element of least degree in $H_*(B)=E^{r}_{*,0}$ with $d^{r}y\ne 0$, we have $d^{r}y_i'=0$ and $d^{r}y_i''=0$ for each $i$. Therefore

\[ \Delta(d^{r}(y))=d^{r}(\Delta(y))=d^{r}\left(y\otimes 1+1\otimes y+\sum_iy_i'\otimes y_i''\right)=d^{r}(y)\otimes 1+1\otimes d^{r}(y) \]

so $d^{r}(y)$ is primitive.

If $r<|y|$, then by corollary 2.2, $d^{r}(y)=\mu (a\otimes b)$ for some non-trivial $a\in H_{|y|-r}(B)$ and $b\in H_{r-1}(F)$. This is impossible because $\mu (a\otimes b)$ is not primitive by lemma 2.1. Thus $r\ge |y|$. Clearly, $r\le |y|$ since $r>|y|$ implies $d^{r}(y)$ lands in the second quadrant of the spectral sequence, which is zero. Therefore $r=|y|$, implying that $y$ is transgressive. Moreover, since

\[ \overline{d}^{r}(y)=\overline{d}^{r}(q_*(y))=q_*(d^{r}(y)) \]

and $d^{r}y$ is primitive, $\overline {d}^{r}y$ is also primitive. Therefore the second statement is proved.

Proof. Since there are isomorphisms

\[ (E^{2}_n)_{p,q}\cong H_p(B)\otimes H_q(F_n)\quad\text{and}\quad\widehat{E}^{2}_{p,q}\cong H_p(B)\otimes\left(\bigoplus_{n\in\mathbb{Z}}H_q(F_n)\right), \]

the inclusion $E_n\to E$ induces an injection $E^{2}_n\to \widehat {E}^{2}$. Then the statement is proved by induction on $r$.

Proof. Let $\epsilon _i\colon S^{2n_i-1}\to G$ be the inclusion for $i=1,\,\ldots,\,l$, where $G\simeq \prod _{i=1}^{l} S^{2n_i-1}$. By [Reference Lang15] $\partial _k$ corresponds to the Samelson product $\langle k\epsilon _1,\,1_G\rangle$ through the adjoint congruence $[G,\,\Omega ^{3}G\langle 3\rangle ]\cong [S^{3}\wedge G,\,G]$, where $\Omega ^{3}G\langle 3\rangle$ is homotopy equivalent to the component of $\Omega ^{3}G$ containing the basepoint. By the linearity of Samelson products, $\langle k\epsilon _1,\,1_G\rangle =k\langle \epsilon _1,\,1_G\rangle$. Thus we aim to get a condition that guarantees the triviality of $\langle \epsilon _1,\,1_G\rangle$. Arguing as in [Reference Kaji and Kishimoto10], we can see that $\langle \epsilon _1,\,1_G\rangle$ is trivial if and only if $\langle \epsilon _1,\,\epsilon _i\rangle$ is trivial for all $i$. It is shown that $\langle \epsilon _1,\,\epsilon _i\rangle$ is trivial for all $i$ if and only if $n_l< p-1$ in [Reference Kishimoto and Tsutaya13] when $G$ is a classical group and in [Reference Hasui, Kishimoto and Ohsita9] when $G$ is an exceptional group. Thus the proof is complete.

Proof. First, for $n\geq 2$, consider the homotopy fibration $\Omega ^{4} S^{2n+1}\to \ast \to \Omega ^{3} S^{2n+1}$. We claim that the element $a_{2np-3}\in H_*(\Omega ^{3} S^{2n+1})$ transgresses to a nonzero element in $H_*(\Omega ^{4} S^{2n+1})$. To see this, let $E^{2}\colon S^{2n-1}\to \Omega ^{2} S^{2n+1}$ be the double suspension. Let $W_{n}$ be the homotopy fibre of $E^{2}$. Then there is a homotopy fibration $\Omega S^{2n+1}\xrightarrow {\Omega E^{2}}\Omega ^{3} S^{2n+1}\to W_{n}$. By [Reference Cohen, Lada and May4] this fibration induces an isomorphism of Hopf algebras

\[ H_*(\Omega^{3} S^{2n+1})\cong H_*(\Omega S^{2n+1})\otimes H_*(W_{n}). \]

In particular, since $W_{n}$ is $(2np-4)$-connected, the element $a_{2np-3}\in H_*(\Omega ^{3} S^{2n+1})$ corresponds to an element $c\in H_{2np-3}(W_{n})$. On the other hand, $W_{n}$ has a single cell in dimension $2np-3$, so $c$ represents the inclusion of the bottom cell. Now consider the homotopy fibration diagram

As $c$ represents the bottom cell in $H_{2np-3}(W_{n})$, it transgresses nontrivially to a class $d\in H_{2np-4}(\Omega W_{n})$. Since $a_{2np-3}$ maps to $c$, the naturality of the transgression implies that $a_{2np-3}$ must transgress to a nontrivial class in $H_{2np-4}(\Omega ^{4} S^{2n+1})$.

Next, consider the homotopy fibration diagram

(4.1)

By corollary 4.2, the map $\Omega g_k$ has a left homotopy inverse. In particular, $(\Omega g_k)_{\ast }$ is an injection. Since $G$ is $p$-regular for $n_l< p$, the left column in the fibration diagram is a product of the homotopy fibrations $\Omega ^{4} S^{3}\langle 3\rangle \to \ast \to \Omega ^{3} S^{3}\langle 3\rangle$ and, for $2\leq i\leq n_{l}$, $\Omega ^{4} S^{2n_{i}-1}\to \ast \to \Omega ^{3} S^{2n_{i}-1}$. Therefore, by the argument in the first paragraph of the proof, the element $a_{2n_{i}-1}\in MH_*(\Omega ^{3} G\langle 3\rangle )$ transgresses to a nontrivial element in $H_{2n_{i}p-4}(\Omega ^{4} G\langle 3\rangle )$. As this element injects into $H_{2n_{i}p-4}(\mathcal {G}_k)$, the naturality of the transgression in (4.1) implies that $(g_k)_*(a_{2n_{i}p-3})$ must be nontrivial.

Finally, consider the element $\bar {a}_{2p-3}\in H_*(\Omega ^{3} G\langle 3\rangle )$. It comes from an element $x\in H_{2p-3}(\Omega ^{3} S^{3}\langle 3\rangle )$. The description of $H_*(\Omega ^{3} S^{3}\langle 3\rangle )$ in (3.4) implies that $\Omega ^{3} S^{3}\langle 3\rangle$ is $(2p-4)$-connected and has a single cell in dimension $2p-3$. Thus $x$ represents the inclusion of the bottom cell. Therefore, $x$ transgresses to a nontrivial element in $H_{2p-4}(\Omega ^{4} S^{3}\langle 3\rangle )$, and so $\bar {a}_{2p-3}$ transgresses to a nontrivial element in $H_{2p-4}(\Omega ^{4} G\langle 3\rangle )$. Since $(\Omega g_k)_*$ is an injection, the naturality of the transgression in (4.1) implies that $(g_k)_*(a_{2p-3})$ is nontrivial.

Proof. Let $(E^{r},\,d^{r})$ denote the mod-$p$ homology Serre spectral sequence for (4.2). Let $z\in H_{m}(\operatorname {Map}^{*}(S^{4},\,BG))$ be an element in the kernel of $g_*$ of least dimension, where $g\colon \operatorname {Map}^{*}(S^{4},\,BG)\to \operatorname {Map}(S^{4},\,BG)$ is the fibre inclusion of (4.2). We begin by establishing some properties of $z$.

Property 1: $z$ is primitive. If not, then $\overline {\Delta }(z)=\Sigma _{\alpha \in A} z'_{\alpha }\otimes z''_{\alpha }$ for some elements $z'_{\alpha },\,z''_{\alpha }$ of degrees $< m$ such that $\{z'_\alpha \otimes z''_\alpha \}_{\alpha \in A}$ is linearly independent, where $\overline {\Delta }$ is the reduced diagonal. The reduced diagonal is natural for any map of spaces, so $(g_*\otimes g_*)\circ \overline {\Delta }=\overline {\Delta }\circ g_*$. Now

\begin{align*} (g_*\otimes g_*)\circ\overline{\Delta}(z)=(g_*\otimes g_*)(\Sigma_{\alpha} z''_{\alpha}\otimes z''_{\alpha})=\Sigma_{\alpha}g_*(z'_{\alpha})\otimes g_*(z''_{\alpha}) \end{align*}

is a sum of linearly independent elements since $z'_{\alpha },\,z''_{\alpha }$ are of degrees $< m$ while the element of least degree in $\mathrm {Ker}\,g_*$ is of degree $m$. On the other hand,

\begin{align*} \overline{\Delta}\circ g_*(z)=\overline{\Delta}(0)=0. \end{align*}

This contradiction implies that $\overline {\Delta }(z)$ must be $0$; that is, $z$ is primitive.

Property 2: $\mathcal {P}^{n}_*(z)=0$ for every $n\ge 1$. Suppose $\mathcal {P}^{n}_{\ast }(z)=y$ for some $n\ge 1$, where $y$ is nonzero. The (dual) Steenrod operations are natural for any map of spaces, so $g_*\mathcal {P}^{n}_*=\mathcal {P}^{n}_*g_*$. Now

\begin{align*} g_*\mathcal{P}^{n}_{{\ast}}(z)=g_*(y) \end{align*}

is nonzero, since $|y|< m$ while the element of least degree in $\mathrm {Ker}\,g_*$ is of degree $m$. On the other hand,

\[ \mathcal{P}^{n}_{{\ast}}g_*(z)=\mathcal{P}^{n}_*(0)=0. \]

This contradiction implies that $\mathcal {P}^{n}_{\ast }(z)=0$ for every $n\ge 1$.

Property 3: $\beta ^{r}(z)=0$ for every $r\geq 1$. The reasoning is exactly as in the proof of property 2.

Property 4: $z$ is in the image of the transgression for the mod- $p$ homology Serre spectral sequence for the homotopy fibration (4.2). By assumption, the first nontrivial differential $d^{r}$ in the mod-$p$ homology Serre spectral sequence for (4.2) satisfies the properties of lemma 2.3. In particular, $d^{r}$ is determined by how it acts on the elements in $H_*(BG)$. The $E^{2}$-term of the spectral sequence is $H_*(BG)\otimes H_*(\operatorname {Map}^{*}(S^{4},\,BG))$. Since $z$ is an element of least degree in the kernel of $(g_k)_*$ and is of degree $m$, the map $(g_k)_*$ is an injection in degrees $< m$. Thus the spectral sequence is totally non-homologous to zero in this degree range, implying that it collapses at $E^{2}$. In particular, the differential $d^{r}$ on an element of $H_*(BG)$ of degree $\leq m$ is zero, and so as $d^{r}$ satisfies the properties of lemma 2.3, $d^{r}(x\otimes y)=0$ for $x\in H_t(BG)$ with $t\leq m$ and $y\in H_*(\operatorname {Map}^{*}(S^{4},\,BG))$. Hence, for degree reasons, if $z$ is in the image of $d^{r}$ then the only possibility is that $z=d^{r}(x)$ and $r=m+1$ for some $x\in H_{m+1}(BG)$. That is, $z$ is in the image of the transgression. On the other hand, as $z\in \mathrm {Ker}\,g_*$, it cannot survive the spectral sequence and so must be hit by some differential.

Property 5: $z$ has odd degree. By property 4, in the mod-$p$ homology Serre spectral sequence for (4.2), we have $z=d^{m+1}(x)$ for some $x\in H_{m+1}(BG)$. As $H_*(BG)$ is concentrated in even degrees, the degree of $z$ must be odd.

Let's examine the consequences of properties 1 to 5. Collectively, Properties 1 to 3 imply that $z\in MH_*(\operatorname {Map}^{*}(S^{4},\,BG))$. Property 5 then implies that $z\in MH_{odd}(\operatorname {Map}^{*}(S^{4},\,BG))$. But $g_*$ is an injection on $MH_{odd}(\operatorname {Map}^{*}(S^{4},\,BG))$ as we saw above, implying in turn that $z\notin \mathrm {Ker}\,g_*$, a contradiction. Thus $g_*$ is an injection, and this completes the proof.

Proof of theorem 1.1. By proposition 4.4, the mod-$p$ homology Serre spectral sequence for (4.2) collapses at the $E^{2}$-term. Thus, as (4.2) is the fiberwise coproduct of the fibrations $\operatorname {Map}^{*}_k(S^{4},\,BG)\to \operatorname {Map}_k(S^{4},\,BG)\xrightarrow {ev} BG$ in (1.1) for all $k\in \mathbb {Z}$, it follows from lemma 2.4 that the mod-$p$ homology Serre spectral sequence for each fibration collapses at the $E^{2}$-term. Equivalently, the mod-$p$ homology Serre spectral sequence for $\Omega ^{3} G\langle 3\rangle \to B\mathcal {G}_k\xrightarrow {ev} BG$ collapses at the $E^{2}$-term for all $k\in \mathbb {Z}$.

Proof. By theorem 1.1, $P_t(B\mathcal {G}_k)=P_t(BG)P_t(\Omega ^{3}G\langle 3\rangle )$. Since $n_l< p-1$, the mod-$p$ cohomology of $BG$ is a polynomial algebra with generators in dimension $2n_1,\,\ldots,\,2n_l$, and so $P_t(BG)=\prod _{i=1}^{l}\frac {1}{1-t^{2n_i}}$. By (3.1), $P_t(\Omega ^{3}G\langle 3\rangle )=P_t(\Omega ^{3}S^{3}\langle 3\rangle )\prod _{i=2}^{l} P_t(\Omega ^{3}S^{2n_i-1})$. By (3.3), (3.4) and (3.5), $P_t(\Omega ^{3}S^{3}\langle 3\rangle )=P(t)Q_{p-1}(t)$ and $P_t(\Omega ^{3}S^{2n_i-1})=P_{n_i-1}(t)$, completing the proof.

Proof. By hypothesis, $(k,\,3)=1$ so the above discussion implies that $\partial _{k}$ is nontrivial. Thus it represents a nontrivial element in $\pi _{3}(\Omega ^{3} S^{3}\langle 3\rangle )\cong \pi _{6}(S^{3}\langle 3\rangle )\cong \pi _{6}(S^{3})$. On the other hand, by [Reference Toda20], the $3$-component of $\pi _{6}(S^{3})$ is isomorphic to $\mathbb {Z}/3\mathbb {Z}$. Therefore $\partial _{k}$ represents a generator of the $3$-component of $\pi _{3}(\Omega ^{3} S^{3}\langle 3\rangle )$. Next, since $\Omega ^{3}S^{3}\langle 3\rangle$ is $2$-connected, the Hurewicz map $\pi _3(\Omega ^{3}S^{3}\langle 3\rangle )\otimes \mathbb {Z}/3\mathbb {Z}\to H_*(\Omega ^{3}S^{3}\langle 3\rangle )$ is an isomorphism, where we take homology with mod-$3$ coefficients. By the naturality of Hurewicz maps there is a commutative diagram

where the vertical maps are the Hurewicz maps. We saw that the top and the right maps are isomorphisms. The left arrow is an isomorphism by the Hurewicz theorem. Thus the bottom arrow is an isomorphism, completing the proof.

Proof of theorem 1.2 Consider the homotopy fibration $S^{3}\xrightarrow {\partial _k}\Omega ^{3} S^{3}\langle 3\rangle \xrightarrow {g_k}B\mathcal {G}_k$. By lemma 5.2, $(\partial _k)_*$ is an injection. Therefore the homology Serre spectral sequence for this homotopy fibration is totally non-homologous to zero, implying that it collapses at the $E^{2}$-term, so $H_*(\Omega ^{3} S^{3}\langle 3\rangle )\cong H_*(S^{3})\otimes H_*(B\mathcal {G}_k)$. Since $H_*(S^{3})\cong \Lambda (y_{3})$ for $|y_3|=3$ such that $y_3$ corresponds to $x_3$ and as observed above, $\Lambda (x_{3})$ is a tensor product factor of $H_*(\Omega ^{3} S^{3}\langle 3\rangle )$, we obtain the isomorphism asserted in the statement of the theorem.