Pushforward of structure sheaf along a torsor for a finite group

Question

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.

[3] https://arxiv.org/pdf/2107.03127.pdf

Answer 1

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.

Answer 2

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.