# Pushforward of structure sheaf along a torsor for a finite group

Let $$\pi : P \to X$$ be a torsor for a discrete, finite group $$G$$ of size $$\#G = N$$ on a scheme $$X$$. I want to compare $$\pi_* \mathcal O_P$$ with $$\mathcal{O}_X$$. Locally but not globally, $$\pi_* \mathcal O_P = \mathcal O_X^{N}$$ [1].

Q1: Is the class $$[\pi_* \mathcal O_P] - [\mathcal O_X^N]$$ torsion in (algebraic) $$K$$ theory?

I have two partial arguments in favor, neither of which is quite complete.

1. (from a friend)

The torsor $$P$$ is classified by an element of $$H^1(X, G)$$, which has multiplication coming from that of $$G$$ [2]. As $$G$$ is a finite group, all its elements have order dividing $$N$$ and everything in $$H^1(X, G)$$ is $$N$$-torsion (Lagrange's Theorem). The vector bundle $$\pi_* \mathcal O_P$$ is associated to the regular representation of $$G$$ on itself.

However the multiplication on $$H^1$$ is different from that of $$K$$; I don't think it's comparable to either addition or multiplication on $$K$$ theory. Or is it? $$K$$ theory takes direct sums of bundles, whereas $$H^1$$'s group law is this contracted product of torsors. Is there an easy comparison that I'm not seeing?

If $$\pi_* \mathcal O_P$$ were a line bundle, the group law of $$H^1$$ would correspond to multiplication in $$K$$ theory, but this corresponds to $$G$$ of size $$N = 1$$. The GRR isomorphism should identify multiplication of $$K$$ classes with cup product on $$H^*$$.

2)

In topological $$K$$ theory for a smooth manifold $$X$$, one computes the chern character using connections on the vector bundle. The $$G$$ bundle has a unique flat connection, so its curvature vanishes and all the chern classes except $$ch_0$$ vanish. The topological version of GRR $$ch : K_{\mathbb Q} \simeq H^{even}$$ shows the class $$\pi_* \mathcal O_P$$ is torsion in topological $$K$$ theory.

This is different from algebraic $$K$$ theory. The same argument does not show vector bundles with flat connection have vanishing chern classes in algebraic $$K$$ theory -- in fact, they do not [3]. I could work with topological $$K$$ theory instead, but it's bad for me that this requires smoothness and does not work for stacks. Moreover, it would waste the reader's time and be overkill if there's a proof already in the algebraic $$K$$ setting.

My apologies if this is easy.

Notes:

[1] I've made this erroneous claim on this site before; I'm trying to figure out the right statement so I can edit my past questions appropriately.

[2] This multiplication is only a group operation on cohomology if $$G$$ is abelian, because otherwise multiplication $$G \times G \to G$$ is not a group homomorphism. In any case, one still has contracted products. I would guess the statement about being $$N$$-torsion still holds but I have not checked carefully.

• In the few cases where $\pi_*\mathcal O_P$ *is * $\mathcal O_X^N$, then this rarely is torsion - do you mean in reduced $K$-theory ? Nov 30, 2021 at 8:33
• Oops! Thanks. Yes, either in reduced $K$ theory or just the class that's the difference between those two. I want them to be the same rationally in $K_Q$. Nov 30, 2021 at 16:32

If X is say a smooth projective variety, then yes (I think smoothness is not actually needed). First, note that $$x=[\pi_*O_P]$$ satisfies the equation $$x^2 = |G|x$$ (because $$P\times_XP = G\times P$$ an base-change). Therefore, it would suffice to show that x is invertible in $$K_0(X)\otimes \mathbb{Q}$$, because then $$x-|G|$$ has to be $$0$$. For this, it would suffice to show that $$x-|G|$$ is nilpotent, since $$|G|$$ is invertible rationally.

Now, the claim is that for every vector bundle $$E$$ the class $$[E]-\dim(E)[\mathcal{O}]$$ is nilpotent. To see this, first we can use the splitting principle to reduce to the case of a line bundle (just pull-back to the flag variety of $$E$$, which is injective on $$K$$-theory, and write $$E-\dim(E)$$ as sum of things like $$L_i-1$$-s). Next, since $$X$$ is projective, every line bundle on $$X$$ is the ratio of very ample line bundles, so we may write $$E = U\cdot W^{-1}$$ for $$U$$ and $$W$$ very ample. But then $$E-1 = UW^{-1} - 1 = (UW^{-1} - U) + (U -1) = UW^{-1}(1-W) + (U-1)$$ so it suffices to show the claim for U and W separately, that is, for $$E$$ a very ample line bundle. But such a bundle is pulled back from $$\mathbb{P}^n$$ and there we know the $$K$$-theory is $$K_0(F)[O(1)-1]/(O(1)-1)^{n+1}$$ for $$F$$ the base field, so that $$(O(1)-1)$$ is indeed nilpotent.

• I'm not sure x^2 = Nx holds in K_Q. The multiplication there is tensor product, which is different from contracted product of G-torsors (which is still not quite P \times_X P). Dec 1, 2021 at 0:11
• If it were, I could just use Lagrange's theorem. I like your suggestion to use the splitting principle and the K theory of projective space though. Dec 1, 2021 at 0:25
• I think that $\pi_*\mathcal{O}_P\otimes \pi_*\mathcal{O}_P \simeq \pi_*\mathcal{O}_P^{\oplus G}$. Isn't it follows from the projection formula and flat base-change? I mean, $\pi_*\mathcal{O}_P\otimes \pi_*\mathcal{O}_P \simeq \pi_*\pi^*\pi_*\mathcal{O}_P$ by projection formula and now you can exchange the $\pi^*\pi_*$ with pull-back to $P\times_XP$ followed by push along other projection to $P$. Now use $P\times_XP = |G|\times P$. Dec 1, 2021 at 7:17
• Another way to say this algebraically is that $\pi_*\mathcal{O}_P$ is a co-$G$-torsor in commutative algebras over $\mathcal{O}_X$ (a.k.a. $G$-Galois extension), and this is just the "free transitive axiom". Note that the argument does not involve the contraction of $P$ against itself just the pollback of $P$ along $\pi$. Dec 1, 2021 at 7:20

I just realized Grothendieck-Riemann-Roch solves this problem.

The chern character isomorphism $$ch_X : K_\mathbb Q X \to A_\mathbb Q X$$ is usually only functorial up to Todd class: $$ch_Y(f_!(E)) = f_*(ch_X(E)Td(T_f))$$ for $$f : X \to Y$$ a morphism, $$f_!$$ the pushforward in $$K$$ theory and $$f_*$$ the pushforward in Chow.

In this case, $$\pi : P \to X$$ is 'etale, $$T_\pi = 0$$ and $$Td(T_\pi) = 1$$. I.e., the chern character is functorial on the nose for pushforward along 'etale maps.

Since $$ch_P$$ is a ring isomorphism, it sends 1 to 1, i.e., $$\mathcal O_P$$ to $$[P]$$. Likewise $$ch_X$$ sends $$N$$ to $$N$$, i.e., $$\mathcal O_X^N$$ to $$N[X]$$. The classes $$\pi_* \mathcal O_P$$ and $$\mathcal O_X^N$$ have the same image under $$ch_X$$, so they are the same.

My apologies for not realizing this sooner. I was afraid of the usual lack of multiplicativity.

Edit:

The same argument with Dan Edidin's version of GRR, Theorem 5.4 in Joe Harris's 60th birthday volume, does not work. It shows $$\pi_* \mathcal O_P$$ is not equal to $$\mathcal O_X^N$$ on stacks, just on schemes. A counterexample is given by $$G = \mathbb Z/2$$. The problem is that the chern character isomorphism is to Chow of the inertia stack $$A_*(IBG)$$ and it is not multiplicative.