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Let $X$ be a separated scheme of finite type over a field $k$, and let $G$ be a smooth linear algebraic group over $k$ acting on $X$.

Let $K_G(X)$ denote the Grothendieck group of coherent $(G,\mathcal{O}_X)$-modules on $X$. That is, $K_G(X)$ is generated by the classes $[F]$, where $F$ is a coherent $(G,\mathcal{O}_X)$-module on $X$, subject to the relations $[F] = [F'] + [F'']$ for all exact sequences $$0\rightarrow F'\rightarrow F\rightarrow F''\rightarrow 0$$

If $\pi : X\rightarrow Y$ is a finite $G$-equivariant map, then since $\pi$ is affine, higher direct images of $\pi$ are trivial, so $\pi_*$ is an exact functor from $(G,\mathcal{O}_X)$-modules to $(G,\mathcal{O}_Y)$-modules, and thus we get a natural map $\pi_* : K_G(X)\rightarrow K_G(Y)$.

However, according to the last few lines of page 8 of

https://arxiv.org/pdf/1301.0425.pdf

and the middle of page 255 of "Equivariant Lefschetz Formula for Finite Reductive Groups" by Ellingsrud-Lonsted, they claim that the same is true when $\pi$ is an arbitrary proper $G$-map.

I don't see why this is true. For example, if $G$ is trivial and $\pi$ is say the structure morphism of a smooth proper curve, then you definitely get nontrivial higher direct images, and so I don't see why $\pi_*$ needs to be exact. Does the connecting homomorphism of the long exact sequence in cohomology just happen to always be zero?

Also, in the Ellingsrud-Lonsted paper they say "Let $X$ be an algebraic $k$-scheme". In today's language do they just mean a $k$-scheme?

EDIT: I have never encountered $K$-theory before, so I'm somewhat unfamiliar with the language.

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    $\begingroup$ Nobody claims that $\pi _*$ is exact. You use $R\pi _*$, as you do for ordinary K-theory. $\endgroup$
    – abx
    Commented Nov 16, 2016 at 19:28
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    $\begingroup$ You can define $K_G(X)$ using perfect complexes instead of vector bundles. That allows you to get a lot more functoriality (for example you get $f_*[F]=\sum_{i\ge0} (-1)^i [R^if_*(F)]$) $\endgroup$
    – Denis Nardin
    Commented Nov 16, 2016 at 19:29
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    $\begingroup$ To be completely explicit, the short exact sequence $0 \to A \to B \to C \to 0$ gives a long exact sequence $0 \to f_{\ast} A \to f_{\ast} B \to f_{\ast} C \to R^1 f_{\ast} A \to R^1 f_{\ast} B \to \cdots$ from which we deduce $\sum (-1)^i [R^i f_{\ast} A] + \sum (-1)^i [R^i f_{\ast} C] = \sum (-1)^i [R^i f_{\ast} B]$. This is the linearity in question. $\endgroup$
    – David E Speyer
    Commented Nov 16, 2016 at 20:55
  • $\begingroup$ @DavidSpeyer AAH! I see! Thank you! Any thoughts on the terminology "algebraic $k$-scheme?" Someone should post these comments as an answer so this can be marked as answered. $\endgroup$
    – stupid_question_bot
    Commented Nov 16, 2016 at 21:02
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    $\begingroup$ Algebraic k-scheme = k-scheme of finite type $\endgroup$
    – ACL
    Commented Nov 17, 2016 at 0:33

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