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Consider the action of $\mathbb{Z}/p^\times$ the units of $\mathbb{Z}/p$ on the classifying space $B\mathbb{Z}/p$ by left multiplication on the $n$-simplices $$ \alpha\cdot (g_1,...,g_n)=(\alpha g_1,...,\alpha g_n). $$ Here $B\mathbb{Z}/p$ denotes the usual model as the realization of the simplicial set whose $n$-simplices are $(\mathbb{Z}/p)^n$. Note that the action is free on the $n$-tuples that are different than the one consisting of only zeros.

Then what can be said about the quotient space $(B\mathbb{Z}/p)/\mathbb{Z}/p^\times$? Such as its homotopy type, homology groups...

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Shouldn't this be $B$ of the semidirect product of $\mathbb Z/p$ and $\mathbb Z/p^\times$? –  Will Sawin Apr 27 '13 at 4:53
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The Borel construction should give the $B$ of the semidirect product. –  user19409 Apr 27 '13 at 5:43
    
I think the action is compatible, for $d_i$ we have $\alpha(g_i+g_{i+1})=\alpha g_i+\alpha g_{i+1}$ and similarly $d_0$ and $d_n$ are also compatible . –  user19409 Apr 27 '13 at 5:47
    
@unknown (google): My bad. I misinterpreted what you meant. Sorry about that. [I removed my earlier incorrect comment, so as to not clutter the discussion.] –  Ricardo Andrade Apr 27 '13 at 6:06
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2 Answers

up vote 5 down vote accepted

$\newcommand{\Ext}{\operatorname{Ext}}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\dash}{\text{-}}$$\newcommand{\sSet}{\mathrm{sSet}}$$\newcommand{\ZZ}{\mathbb{Z}}$For convenience, I will denote by $G$ the group $(\ZZ/p)^\times$, and by $X$ the simplicial set $B(\ZZ/p)$ for $p$ a prime. I will prove later in this answer that there are natural weak equivalences $$ X/G \simeq (X\times_G EG)/BG \simeq B(\ZZ/p \rtimes G)/BG \rlap{\qquad\qquad\text{(I)}} $$

The fundamental group, cohomology, and homology of $X/G$

Before I prove formula (I), let me describe what the fundamental group, homology, and cohomology of the space $X/G$ look like. Assume that $p>2$, otherwise $X/G = B(\ZZ/p)$.

The fundamental group: By the van Kampen theorem, the fundamental group of $X/G \simeq B(\ZZ/p \rtimes G)/BG$ is the quotient of $\ZZ/p \rtimes G$ by the normal subgroup generated by $G$. This is easily seen to be trivial, as it is a quotient of $\ZZ/p$ with non-trivial kernel. So $X/G$ is simply connected.

The cohomology: We will make use of the Leray–Serre spectral sequence described by Mariano Suárez-Alvarez in his answer. Here is what we need to apply the conclusion from his answer:

  • The action of $\alpha\in G$ on $H_1 X = \pi_1 X = \ZZ/p$ is given by multiplication by $\alpha$.

  • So $\alpha\in G$ acts on $H^2 X = \Ext^1(H_1 X,\ZZ)$ by multiplication by $\alpha$. We have used the universal coefficient theorem and the fact that $H_2 X = H_2 B(\ZZ/p) = 0$.

  • Therefore, the action of $\alpha\in G$ on $H^\ast X = \ZZ[x]/\langle px \rangle$ (where $x$ is a generator in degree $2$) is by multiplication by $\alpha^n$ on $H^{2n} X$.

  • Since $G$ is cyclic of order $p-1$, the action of $G$ on $H^{2n} X$ is trivial if and only if $p-1$ divides $n$. In particular, the invariant subalgebra is: $$ (H^\ast X)^G = \ZZ[x^{p-1}]/\langle p x^{p-1} \rangle $$

Now we use the single row, single column spectral sequence for $H^\ast(X\times_G EG)$ that Mariano Suárez-Alvarez obtains in his answer.

  • Note that $G$ is cyclic of order $p-1$. So the single non-zero row of the spectral sequence is $H^\ast(BG)=\ZZ[y]/\langle (p-1)y \rangle$ for a generator $y$ in degree $2$.

  • Therefore, the spectral sequence collapses at $E_2$ for degree reasons. The end result of the spectral sequence is then a canonical short exact sequence of graded abelian groups $$ 0 \To \widetilde{H}^\ast BG \To H^\ast(X\times_G EG) \overset{f}{\To} (H^\ast X)^G \To 0 $$ where $f$ is actually an algebra map.

Finally, the long exact sequence for the cohomology of the pair $(X\times_G EG, BG)$ — whose quotient is equivalent to $X/G$ — gives that $H^\ast(X\times_G EG, BG)$ is the kernel of the map $H^\ast(X\times_G EG) \to H^\ast BG$ (which is surjective because it is induced by a section). The previous exact sequence then shows that the composition $$ H^\ast(X/G) = H^\ast((X\times_G EG)/BG) \To H^\ast(X\times_G EG) \To (H^\ast X)^G $$ is an isomorphism. In conclusion, we have an isomorphism of graded algebras: $$ H^\ast(X/G) = H^\ast(X)^G = \ZZ[x^{p-1}]/\langle p x^{p-1} \rangle= \ZZ[u]/\langle pu \rangle $$ where $u=x^{p-1}$ sits in degree $2p-2$. In particular, $H^0(X/G)=\ZZ$, $H^{n(2p-2)}(X/G)\simeq\ZZ/p$ for $n>0$, and all other integral cohomology groups of $X/G$ are zero.

The homology: Observe that $X/G$ is a simplicial set which is finite in each degree. Applying the universal coefficient theorem in reverse, we conclude that $H_0(X/G)=\ZZ$, $H_{n(2p-2)-1}(X/G)\simeq\ZZ/p$ for $n>0$, and all other integral homology groups of $X/G$ are zero.

Proof of the equivalences in formula (I)

As observed in the question, $X$ is a pointed $G$-simplicial set, and the action of $G$ on $X$ is free on every simplex which is not in the basepoint of $X$. This implies that the inclusion of the basepoint $1\to X$ is a cofibration in the projective model structure on $G$-simplicial sets. In other words, $X$ is a cofibrant object in the projective model structure on pointed $G$-simplicial sets.

Now observe that quotienting out by the action of $G$ is a left Quillen adjoint from pointed $G$-simplicial sets to pointed simplicial sets. I will denote this left Quillen adjoint by $F$: $$ F : G\dash\sSet_\ast \To \sSet_\ast $$ Since $X$ is a cofibrant object in the domain of $F$, the quotient $X/G = F(X)$ is weakly equivalent to $LF(X)$, the left derived functor of $F$ applied to $X$: $$ X/G \simeq LF(X) $$ Importantly, $LF(X)$ is weakly equivalent to the quotient of the Borel construction on $X$ by its subspace $BG$: $$ LF(X) \simeq (X\times_G EG)/BG \rlap{\qquad\qquad\text{(#)}} $$ Here, $BG$ includes into the Borel construction via the basepoint of $X$: $$ BG = EG/G = 1\times_G EG \To X\times_G EG $$ At the end of this answer, I will give a proof of formula (#), which simply states the well-known result that the homotopy quotient by $G$ in $\sSet_\ast$ (pointed simplicial sets) is equivalent to the homotopy quotient by $G$ in $\sSet$ further modded out by $BG$.

Finally, as observed in the comments, the Borel construction $X\times_G EG = B(\ZZ/p)\times_G EG$ is actually equivalent to $B(\ZZ/p \rtimes G)$, the classifying space of the semi-direct product. In conclusion: $$ X/G \simeq B(\ZZ/p \rtimes G)/BG $$


Proof of equivalence (#): Consider the sequence of left Quillen adjoints: $$ EG \downarrow G\dash\sSet \overset{T}{\To} 1 \downarrow G\dash\sSet \overset{F}{\To} 1\downarrow\sSet \ (= \sSet_\ast) $$ where the first functor $T$ is left adjoint to pre-composing with $EG\to 1$, i.e. $T$ quotients out by $EG$. Since $EG\to 1$ is a weak equivalence of $G$-simplicial sets, and $G$-simplicial sets form a left proper model category, the functor $T$ is actually a Quillen equivalence. So $X \simeq LT(X)$, where $X$ is seen as an object under $EG$ via $EG\to 1\to X$, and $LT$ is the left derived functor of $T$. Consequently: $$ LF(X) \simeq LF\circ LT(X) \simeq L(F\circ T)(X) \simeq (X\times_G EG)/BG $$ The last weak equivalence is obtained by expressing $F\circ T$ as the composite of the left Quillen adjoints $$ EG\downarrow G\dash\sSet \overset{(-)/G}{\To} BG\downarrow\sSet \overset{(-)/BG}{\To} 1\downarrow\sSet $$ and noting that since $EG$ is cofibrant, the left derived functor of $(-)/G$ is weakly equivalent to the Borel construction.

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Great! I appreciate your detailed proof. –  user19409 Apr 28 '13 at 5:22
    
@unknown (google): You are welcome. It was a fun problem to work on. –  Ricardo Andrade Apr 28 '13 at 5:56
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See the comments: the action is not properly discontinuous so the spectral sequence converges to the Borel construction rather than to the quotient, so this does not really compute what is wanted.

Let $\ell$ be a prime number.

Let $G=(\mathbb Z/\ell)^\times$ and $C_p=\mathbb Z/\ell$, and suppose $G$ acts on $BC_\ell$ in some way properly discontinuously. There is a spectral sequence with $E_2=H^\bullet(G,H^\bullet(BC_\ell,\mathbb Z))$ converging to $H^\bullet(BC_\ell/G,\mathbb Z)$.

Let $p>0$. If $q$ is odd, then $H^q(BC_\ell,\mathbb Z)=0$. If $q$ is even and positive, then $H^q(BC_\ell,\mathbb Z)$ is cyclic of order $\ell$, which is coprime to the order of $G$, and then it follows that $H^p(G,H^q(BC_\ell,\mathbb Z))=0$.

It follows that the spectral sequence has exactly one row and one column.

The $0$th row is the group cohomology of $G$. The $0$th column is $H^0(G,H^q(BC_\ell,\mathbb Z))$, which is just the invariant subalgebra $H^q(BC_\ell,\mathbb Z)^G$, and there is an isomorphism of graded algebras $H^\ast(BC_\ell,\mathbb Z)\simeq\mathbb Z[\zeta]/(\ell\zeta)$, with $\zeta$ a class of degree two. To see what the action of $G$ is on the whole thing it is enough to see what it is on $H^2(BC_\ell,\mathbb Z)$, which is the group cohomology group $H^2(C_\ell,\mathbb Z)$. That should be doable.

Convergence gives a long exact sequence involving all differentials in the spectral sequence. Since the spectral sequence has an algebra structure, this should not be impossible to make completely explicit.

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The spectral sequence seems to converge to the cohomology of the Borel construction $EG\times_G BC_l$? –  user19409 Apr 27 '13 at 5:50
    
I was assuming that the action is properly discontinuous, actually, but I did not say so. This seems to be the case in the question, I think. (The spectral sequence I have in mind is the one coming from the last chapter of Grothendieck's Tōhoku; I hope the one coming from the fibration attached to the Borel construction is the same one) –  Mariano Suárez-Alvarez Apr 27 '13 at 6:13
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Yes it is the spectral sequence of that fibration. The action is not free the simplex $(0,...,0)$ is fixed by the group, otherwise it acts freely. –  user19409 Apr 27 '13 at 6:17
    
Ah, right! ${}{}{}$ –  Mariano Suárez-Alvarez Apr 27 '13 at 6:18
    
Thanks for your answer my first approach was the same. –  user19409 Apr 27 '13 at 6:23
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