4
$\begingroup$

I've been trying to find the cohomology for the trivial module for $\operatorname{PSL}_2(r^n)$ over $\mathbb{F}_p$ for $2 \neq p \neq r$ and have managed to reduce this to the cohomology of a maximal torus $D_{r \pm 1}$ (where $|D_{2n}| = 2n$), dependent on which is divisible by $p$, but am struggling to find a reference for $\operatorname{H}^i(D_{2n}, \mathbb{F}_p)$ for $i > 2$ though this must almost certainly be known.

I can see by computations in magma that the answer should be $0$ for $i \equiv 1, \, 2 \mod 4$ and $\mathbb{F}_p$ otherwise.

$\endgroup$
  • 2
    $\begingroup$ Is it not sufficient to apply the UCT to $H^\ast(D_{2n},\mathbb{Z})$? The latter groups are known. $\endgroup$ – Chris Gerig Feb 4 '19 at 20:17
5
$\begingroup$

EDIT: Thanks a lot to Mike Miller for pointing out in the comments significant simplifications to the proof I wrote

First suppose that $p$ does not divide $n$. Then we have $$H^*(D_{2n};\mathbb{F}_p)=\begin{cases}\mathbb{F}_p & \textrm{ if }n=0\\ 0 &\textrm{ otherwise}\end{cases}\,.$$ Otherwise, let us suppose that $p$ divides $n$.

Let $C_n$ be the subgroup of $D_{2n}$ consisting of rotations. This is a normal cyclic subgroup of index 2. Hence, it has cohomology of the form $$H^*(C_n;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/C_n;\,H^*(C_n;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/C_n$ has order prime to $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(C_n;\mathbb{F}_p)^{D_{2n}/C_n}\,.$$ Note that $D_{2n}/C_n\cong \mathbb{Z}/2$ acts on $C_n$ by sending $z$ to $z^{-1}$. To conclude then it's enough to determine the action of $\mathbb{Z}/2$ on $a$ and $b$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_n→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $C_n$ in $S^1$, and the action of $\mathbb{Z}/2$ extends to $S^1$. By the Hurewicz theorem we have an isomorphism $$\mathrm{Hom}(\pi_2BS^1;\mathbb{F}_p)\cong H^2(BS^1;\mathbb{F}_p)\cong H^2(BC_n;\mathbb{F}_p)$$ compatible with the action of $\mathbb{Z}/2$. In particular the action is nontrivial (since the action on $\pi_2BS^1\cong \pi_1S^1\cong\mathbb{Z}$ sends $1$ to $-1$), so $\sigma b= - b$.

Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I am oscillating between being convinced that this answer is correct and being convinced that it is complete bunk. Right now I think it's correct and I've got to leave now, so I'll offer it to you clemence.. $\endgroup$ – Denis Nardin Feb 4 '19 at 17:27
  • $\begingroup$ Everything is correct. Your argument can be streamlined by working with the LHSS for $\Bbb Z/n \to D_{2n} \to \Bbb Z/2$ directly instead of passing to a $p$-Sylow, then reducing to the case of $S^1$, which has an automorphism extending negation on $\Bbb Z/n$. The induced map of this automorphism (complex conjugation) on $S^1$ is complex conjugation on $\Bbb{CP}^\infty$, so $\sigma b = -b$. $\endgroup$ – Mike Miller Feb 4 '19 at 17:50
  • $\begingroup$ @MikeMiller Thanks! I was trying to avoid using the fiber sequence $C_p\to S^1\to S^1$ to give a more "algebraic" proof, but you're right, the argument flows much better if we embrace it. $\endgroup$ – Denis Nardin Feb 4 '19 at 22:32
  • $\begingroup$ I thought your answer was fine as it was, I just wanted to add something for anybody passing by :) Some small points: there is a surviving $C_p$ somewhere. Because $S^1$ is a non-discrete space as well as a group, can I suggest writing $H^2(BS^1; \Bbb F_p)$ on the second-to-last displayed line (and maybe similarly for $BC_n$) to avoid confusion? $\endgroup$ – Mike Miller Feb 4 '19 at 22:48
  • $\begingroup$ @MikeMiller Aaand there was still a residue mistake (I assumed that the Sylow subgroup wouldn't be trivial). Fixed now :) $\endgroup$ – Denis Nardin Feb 4 '19 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.