Reference requests: Integral cohomology of $G_2$-homogeneous spaces

Do you know a place where the integral cohomology of $ G_2$ -homogeneous spaces is computed?

Great computational efforts using representation theory in order to determine the characteristic classes of homogeneous spaces were done by Borel and Hirzebruch in a series of papers:

  1. Characteristic classes and homogeneous spaces. I. Amer. J. Math. 80 1958, 491–504.
  2. Characteristic classes and homogeneous spaces. II. Amer. J. Math. 81 1959 315–382.
  3. Characteristic classes and homogeneous spaces. III. Amer. J. Math. 82 1960 458–538.

I was unable to find there a systematic answer to the question.

Cohomology of simple finite groups

Suppose $ G$ and $ H$ are finite simple groups. I think it’s obvious that one expects that the cohomology groups with integer coefficients of $ G$ and $ H$ are not isomorphic unless $ G\cong H$ . Is there any proof of this? Notice, I do not want to compute those cohomology groups (and as far as I know it hasn’t been done yet completely), just to show that they are not isomorphic.

Group Cohomology of $S_3$ in terms of its Sylow subgroups

I am trying to understand $ H^*(S_3, M)$ in terms of it’s Sylow $ p$ subgroups. From III.10.2 and III.10.3 in Brown we know that \begin{equation}H^n(G,M) = \bigoplus_p H^n(H,M)^G\end{equation} where $ p$ ranges over primes dividing $ |G|$ and $ H^n(H,M)^G$ is the $ G$ -invariant submodule. We say $ z \in H^n(H,M)$ is $ G$ -invariant if $ \text{res}_{H \cap gHg^{-1}}^{H} z= \text{res}_{H\cap gHg^{-1}}^{gHg^{-1}}gz$ for all $ g \in G$ .

For $ G = S_3$ , the Sylow subgroups we have to consider are isomorphic to $ C_2$ and $ C_3$ . The cohomology of the cyclic groups are well as $ $ H^i(C_n, M) = \begin{cases}M/nM \text{ for } i = 2,4,6,\ldots \\text{Ann}_M(n) \text{ for } i = 1,3,5,\ldots\ M \text{ for } i = 0\end{cases}$ $ where $ \text{Ann}_M(n)$ is the $ n$ -torsion submodule of $ M$ .

Then, from https://groupprops.subwiki.org/wiki/Group_cohomology_of_symmetric_group:S3 we know $ $ H^i(S_3,M) = \begin{cases}M \text{ for } i = 0\ \text{Ann}_M(2) \text{ for } i \equiv 1 \mod 4\ M/2M \text{ for } i \equiv 2 \mod 4\\text{Ann}_M(6) \text{ for } i \equiv 3 \mod 4\M/6M \text{ for } i > 0, i \equiv 0 \mod 4\end{cases}$ $

My question is how the period 2 cohomology for the cyclic groups can turn into period 4 cohomology for $ S_3$ . More specifically, applying the formula from Brown at each $ i$ , we get $ $ H^1(S_3,M) = \text{Ann}_M(2)^G \oplus \text{Ann}_M(3)^G$ $ So, it seems that $ \text{Ann}_M(3)^G$ should be 0 here? Then $ $ H^2(S_3,M) = M/2M^G \oplus M/3M^G$ $ which also seems to imply $ M/3M^G = 0$ here. Then $ $ H^3(S_3,M) = \text{Ann}_M(2)^G \oplus \text{Ann}_M(3)^G$ $ Assuming that it is true that $ \text{Ann}_M(2) \oplus \text{Ann}_M(3) = \text{Ann}_M(6)$ , then it seems like here, $ \text{Ann}_M(3)^G$ should be $ \text{Ann}_M(3)$ . Finally $ $ H^4(S_3,M) = M/2M^G \oplus M/3M^G$ $ Similarly, here it seems that $ M/3M^G$ should be $ M/3M$ .

It comes down to fact that I don’t understand how the $ G$ -invariant submodules of the same cohomology groups, except at a different indices, could be different. Is there with understanding the formula from Brown is wrong or is my understanding of computing $ G$ -invariants flawed?

Cohomology of the pullback of a coherent sheaf (considered as an abelian sheaf)

Let $ X$ be an affine Noetherian scheme, $ Y$ a separated Noetherian scheme, $ f:X\rightarrow Y$ a morphism of schemes inducing a homeomorphism on the underlying topological spaces.

Let $ F$ be a coherent sheaf on $ Y$ ; treat it as a sheaf of abelian groups and pull it back to $ X$ . It is possible that $ H^1(X, f^{-1}(F))\neq 0$ ?

This question may be confusing because we are essentially computing the sheaf cohomology of the exact same abelian sheaf on the exact same topological space but the non-trivial piece of information here is the existence of the morphism $ f$ (remember, the forgetful functor from schemes to spaces is not full). Thus a random coherent sheaf on a separated scheme with non-vanishing $ H^1$ is not going to do the trick.

A question about the (motivic) integral cohomology of the Eilenberg-MacLane spectrum

Let $ H\mathbb{Z}$ be the Eilenberg-MacLane spectrum. Let $ n\geq 0$ be any integer.

Is it known the structure of the group $ [H\mathbb{Z},\Sigma^{n}H\mathbb{Z}]$ ?

Is there any reference in this direction?

What about of its motivic version, i.e. if $ M\mathbb{Z}$ denotes the motivic Eilenberg-MacLane spectrum,

What $ [M\mathbb{Z},\Sigma^{2n,n}M\mathbb{Z}]$ is?

I would appreciate any help.

Are pullbacks on singular cohomology unique on the nose?

Let $ C$ be the category of even-dimensional connected closed oriented topological manifolds/orientation-preserving continuous maps and $ D$ be the category of finite-dimensional graded $ \mathbb{Q}$ -algebras.

We have a functor $ H:C\rightarrow D$ given by singular cohomology.

Let us call a choice of an isomorphism of graded $ \mathbb{Q}$ -algebras $ H(M)\otimes H(M’)\rightarrow H(M\times M’)$ for every $ M$ , $ M’\in Obj(C)$ a Kunneth system. Projections $ M\times M’\rightarrow M$ , $ M’$ define a Kunneth system which is functorial with respect to $ H$ (Kunneth theorem).

For a contravariant functor $ G:C\rightarrow D$ such that the induced map on objects is equal to one induced by $ H$ , it is meaningful to ask if the Kunneth system above is functorial with respect to $ G$ . The question: does there exist a contravariant functor $ G:C\rightarrow D$ such that

  • $ G$ coincides with $ H$ on the level of objects but not on the level of morphisms;
  • for any non-negative even integer $ i$ , the contravariant functors from $ C$ to $ \mathbb{Q}$ -vector spaces obtained by composing either $ G$ or $ H$ with the “$ i$ -th graded piece” functor are equal;
  • the Kunneth system above is functorial with respect to $ G$ ?

P.S.: this question is inspired by this question.

Pullbacks on $\ell$-adic cohomology

For smooth projective varieties $ X$ over $ F := \overline{\mathbf{F}}_p$ , $ \ell$ -adic cohomology $ $ H^*(X,\mathbf{Q}_{\ell})$ $ is a Weil cohomology theory.

We denote, for a morphism of smooth projective $ F$ -varieties $ f : X\to Y$ , by $ $ f^* : H^*(Y,\mathbf{Q}_{\ell})\to H^*(X,\mathbf{Q}_{\ell})$ $ the usual pullback on $ \ell$ -adic cohomology.

Suppose now that we have another pullback $ \widetilde{f}^*$ on $ \ell$ -adic cohomology, compatible with the $ \ell$ -adic cycle map, and making it into a Weil cohomology again.

That is, $ \widetilde{f}^*$ still makes $ X\mapsto H^*(X,\mathbf{Q}_{\ell})$ into a functor to graded $ \mathbf{Q}_{\ell}$ -algebras that still satisfies the axioms of a Weil cohomology with the usual cycle map on $ \ell$ -adic cohomology.

Do we necessarily have $ \widetilde{f}^* = f^*$ ?

Due to compatibility with cycle maps, $ \widetilde{f}^* = f^*$ in degrees $ 0$ and top.