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Let $G$ be an algebraic group over an algebraically closed field of characteristic zero $K$ and let $L$ be another algebraically closed field, together with an embedding $K \hookrightarrow L$.

Why is it true that the extension of scalars is an equivalence of categories from finite dimensional $K$-representations of $G$ to finite-dimensional $L$ representations of $G \times_K L$?

Are all the assumptions on the fields (algebraically closed, characteristic zero) needed?

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  • $\begingroup$ Why do you think this is true? What happens if you take the one-dimensional trivial $K$-representation of $G$? $\endgroup$ – Jeremy Rickard Dec 22 '17 at 20:08
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    $\begingroup$ The trivial $K$-representation has endomorphism ring $K$, but the trivial $L$-representation has endomorphism ring $L$, and an equivalence of categories would preserve endomorphism rings. $\endgroup$ – Jeremy Rickard Dec 22 '17 at 22:28
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    $\begingroup$ @Jeremy: But the question posed here changes the group along with the field, which makes the question less straightforward.. $\endgroup$ – Jim Humphreys Dec 23 '17 at 2:10
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    $\begingroup$ @JimHumphreys I don't understand your comment. Equivalence of categories implies that the functor $F$ is full and faithful, so for any objects $v,v'$, $Hom(v,v')\to Hom(F(v),F(v'))$ is bijective. This is not the case. On the other hand, one can consider the category of $K$-defined $L$-reps with $K$-defined morphisms, which if I'm correct, is obtained from the category of $K$-reps by tensoring all morphism spaces with $L$. Then we have a functor to $L$-reps, and this one is more likely to be an equivalence of categories. $\endgroup$ – YCor Dec 23 '17 at 3:44
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    $\begingroup$ Anyway if $L\neq K$ this functor is not essentially surjective (whether or not we consider the category of $K$-reps with morphisms tensored with $L$). Indeed, consider a 1-dimensional rep of $L^2$ (viewed as 2-dimensional abelian unipotent) whose kernel is not defined over $K$. Then this rep is not isomorphic to any rep in the image of the functor (since isomorphisms preserves kernels). $\endgroup$ – YCor Dec 23 '17 at 3:48
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First of all, even setting aside the issue with scalar automorphisms noted in comments, at the level of objects there is a "problem": the functor is not essentially surjective for unipotent groups (by a consideration of Ext$^1$-groups with their natural structure of vector space over the ground field). Probably nobody cares, so I won't get into the details (but if someone does care then hopefully someone else will summon the energy to write out an actual argument, since I don't feel like going down that road here).

Now let's focus on positive statements that someone might care about. Let $K'/K$ be an extension of fields (any characteristic, but characteristic 0 is especially nice when $G^0$ is reductive because of the Remark below), and $G$ a smooth affine group over $K$. Assume $K$ is algebraically closed (but $K'$ is not assumed to be algebraically closed). Then we claim:

Theorem: For any semisimple linear representation $\rho':G_{K'} \to {\rm{GL}}(V')$, there exists a semisimple linear representation $\rho:G \to {\rm{GL}}(V)$ unique up to isomorphism such that $\rho' \simeq \rho_{K'}$.

Remark: In characteristic 0 the semisimplicity hypothesis on the representations automatically holds if $G$ has reductive identity component. There is a version of the Theorem in characteristic 0 that doesn't require $K$ to be algebraically closed if $G$ is split connected reductive, but that involves an entirely different ingredient than is used below, so I'll pass over it in silence.

To prove the Theorem, we begin with:

Lemma The coefficients of the characteristic polynomial of $\rho'$ come from $K[G] \subset K'[G_{K'}]$.

Proof. An element $f' \in K'[G]$ comes from $K[G]$ if and only if its restriction to a Zariski-dense subset of $G(K)$ belongs to $K$. (Indeed, if we apply "spread out and specialize" to $f’$ then we get some $f \in K[G]$ such that $f_{K'}$ agrees with $f'$ on a Zariski-dense subset of $G(K)$, but such a subset is also Zariski-dense in $G_{K'}$ (exercise!), so $f' = f_{K'}$ as desired.)

Let $B$ be a Borel $K$-subgroup of $G$, so the $G(K)$-conjugates of $B(K)$ cover $G(K)$ (since $K$ is algebraically closed). Hence, it suffices to show that the coefficients on all such $G(K)$-conjugates have values in $K$. Let $T$ be a maximal $K$-torus in $B$, so $B = T \ltimes U$ for $U := \mathscr{R}_u(B)$. Applying Lie-Kolchin to $B_{K'}$ acting on $V'$, those coefficients on any $b \in B(K)$ only depend on the $T$-component of $b$. Thus, since the characteristic polynomial is conjugation-invariant, we're reduced to studying these coefficients on points of $T(K)$. All weights of $T_{K'}$ are "$K$-rational" (since $T$ is a split $K$-torus, as $K$ is algebraically closed), so we win.

QED Lemma

Now if we apply "spreading out and specialization" followed by semisimplification, from $\rho'$ we get a semisimple $$\rho: G \to {\rm{GL}}(V)$$ such that $\rho_{K'}$ has the same characteristic polynomial as $\rho'$ due to the Lemma. But $\rho_{K'}$ is semisimple because $\rho$ is semisimple with $K$ algebraically closed (i.e., $V$ is a semisimple representation of the abstract group $G(K)$ with $K$ algebraically closed, so it is "absolutely semisimple" and hence — by consideration of the endomorphism algebra — $V_{K'}$ is semisimple over $K'$ as a representation of the abstract group $G(K)$ and thus as a representation of the algebraic group $G_{K'}$ by Zariski-density of $G(K)$ in $G_{K'}$). Hence, $\rho_{K'}$ and $\rho'$ are semisimple representations of $G(K')$ with the same characteristic polynomial, so by Brauer-Nesbitt (which applies to semisimple representations of finite dimension for any abstract group at all) these representations are isomorphic. That isomorphism amounts to a single conjugation in the language of matrices, so it says that as algebraic representations they’re $K'$-isomorphic. This gives that $\rho'$ descends to a semisimple representations of $G$ over $K$ as desired.

QED Theorem

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    $\begingroup$ An issue with Ext groups is that two non-isomorphic extensions can be isomorphic as representations. As an example, I believe the functor is essentially surjective for the unipotent group $\mathbb{G}_a$, even though the map on Ext groups is not surjective. $\endgroup$ – Julian Rosen Dec 23 '17 at 5:28
  • $\begingroup$ So $(-)_{K'}$ is bijective on isomorphism classes of objects? Is it faithful? (Clearly it is not full in all nontrivial cases) $\endgroup$ – მამუკა ჯიბლაძე Dec 23 '17 at 5:48
  • $\begingroup$ @მამუკაჯიბლაძე: Your first question is affirmative under the hypotheses of the Theorem above, and the second is "trivially yes" since field extensions are faithfully flat (though of course one doesn't need the notion of faithful flatness to see this, just some bases). $\endgroup$ – nfdc23 Dec 23 '17 at 5:58
  • $\begingroup$ @JulianRosen: Oh, good point. I'm still pretty sure that for general unipotent $K$-groups (say in characteristic 0) the essential surjectivity does fail, but I'd need to think it through some more, on another day. $\endgroup$ – nfdc23 Dec 23 '17 at 6:00
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This actually strikes me as a question about the foundations chosen for algebraic geometry, which in turn affect formlulations about linear/affine algebraic groups and their representations. It's useful to recall that the foundations were in flux for a long time, while people like Weil experimented with working over a very big "universal domain" with an unlimited supply of indeterminates. Eventually Grothendieck's school developed a more flexible language for talking about "schemes" (including algebraic groups) over arbitrary commutative base rings including fields which may or may not be algebraically closed.

In the 1960s Borel adapted the scheme theory in an ad hoc way to reach the structure theory of semisimple (or reductive) groups more quickly: Bruhat decomposition, parabolic subgroups, etc. Anyway, there is general agreement now that the choice of an algebraically closed field won't affect significantly the properties of a scheme/variety or in particular of an algebraic group. This applies equally well to the (finite dimensional) representations which are "rational" (algebraic). So an extension involving just algebraically closed fields leaves all the basic data intact. See for example Milne's recently published book on algebraic groups here, though it's still quite a long story when told in the "correct" language of algebraic geometry.

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