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.

- (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.