6
$\begingroup$

Let $G$ be a reductive group over a field $k$ with maximal torus $H$. Let $\mathfrak{g}$ and $\mathfrak{h}$ denote the corresponding Lie algebra. If $k$ is algebraically closed, we have a theorem of Chevalley which says that $k[\mathfrak{g}]^G\simeq k[\mathfrak{h}]^W$. A theorem of Chevalley–Shephard–Todd then states that this is a polynomial algebra.

I'm wondering to what extent these theorems remain true if $k$ is not algebraically closed. The case I'm actually interested in is when $k=\mathbb{C}((t))$ in which case, the conjugacy class of Cartan subalgebras are in bijection with conjugacy classes in the absolute Weyl group (i.e., the Weyl group over the algebraic closure).

Most references I know assume $k$ is algebraically closed or that $\mathfrak{h}$ is split. References dealing with non-split Cartan would be appreciated.

$\endgroup$
  • 2
    $\begingroup$ We lose nothing by restricting to $G$ semisimple and simply connected, so $G = {\rm{R}}_{k'/k}(G')$ for a finite etale $k$-algebra $k'$ and $k'$-group $G'$ with connected semisimple fibers that are absolutely simple and simply connected. One can thereby reduce to the case of absolutely simple $G$. Since $k$ is perfect with cohomological dimension 1, by a theorem of Steinberg $G$ is quasi-split (so only types A, D, and E$_6$ can be non-split). A preferred $G(k)$-conjugacy class of maximal $k$-tori is those in Borel $k$-subgroups, which are "induced tori"; maybe using these simplifies things? $\endgroup$ – nfdc23 Jul 23 '16 at 3:28
4
$\begingroup$

Taking invariants commutes with flat base change, so, in particular, with extensions of the base field. This implies that Chevalley's restriction theorem works over any field of characteristic zero. Also Shephard-Todd works over any field since a polynomial ring with a positive grading has only trivial forms. To see this take any $k$-basis of $\mathfrak m/\mathfrak m^2$ (with $\mathfrak m$ being the maximal homogeneous ideal) and lift it to homogeneous ring elements which are defined over $k$. These will freely generate the ring.

$\endgroup$
  • $\begingroup$ Can you explain your argument with the grading in some more detail? That is, how do you know that this $k$-algebra of invariants has a grading that recovers the usual one on the polynomial ring after scalar extension to the split case? I suppose this is using that $G$ is quasi-split and hence can be described concretely as a twisting of the split form via "diagram automorphisms", but that involves making a careful choice of maximal $k$-torus (namely, from a Borel $k$-subgroup), among other things. $\endgroup$ – nfdc23 Jul 23 '16 at 18:31
  • $\begingroup$ The statement is that a graded $k$-algebra which becomes a graded polynomial ring after extension of scalars is a graded polynomial ring itself. This has nothing to do with groups. $\endgroup$ – Friedrich Knop Jul 23 '16 at 19:41
  • $\begingroup$ I agree, but my question was: how are you defining the asserted graded $k$-algebra structure on these $k$-algebras of invariants beyond the split case? It seemed that perhaps one is using that $G$ is quasi-split and a Cartan chosen inside of a Borel subalgebra (rather than an arbitrary Cartan $k$-subalgebra) in order to define the desired grading, but I'm not seeing the details of such a procedure offhand (and am guessing that it rests on how one makes a quasi-split form by a specific type of Galois-twisting of the split form). $\endgroup$ – nfdc23 Jul 23 '16 at 20:11
  • $\begingroup$ I am lost. For the argument, a Cartan subalgebra $\mathfrak h$ (split or not) is just a $k$-vector space. Then $k[\mathfrak h]$ is a polynomal ring with the obvious grading by degree. Its subalgebra $R:=k[\mathfrak h]^W$ then inherits this grading. There is a slight complication in that the Galois group is acting on $W$ and therefore doesn't really act on $\mathfrak h$. Maybe that is what is bothering you. I interpreted $R$ as $k[\mathfrak h]\cap \overline k[\mathfrak h]^W$, or, what is the same, the Galois invariants in $\overline k[\mathfrak h]^W$. $\endgroup$ – Friedrich Knop Jul 24 '16 at 7:45
  • $\begingroup$ Since the finite etale $k$-group $W$ acts via a linear representation on $\mathfrak{h}$, I agree that $k[\mathfrak{h}]^W$ inherits a grading from $k[\mathfrak{h}]$. But to be useful, it must agree with the grading on $k[\mathfrak{h}]^W$ as a polynomial ring in the split case (so applying the compatibility over $\overline{k}$ gives that $k[\mathfrak{h}]^W$ has a grading recovering a polynomial-ring grading over $\overline{k}$, and so is a polynomial-ring grading by your argument). But that compatibility usually fails: $k[x_1,\dots,x_n]^{S_n} = k[s_1,\dots,s_n]$ with ${\rm{deg}}_x(s_i) = i$. $\endgroup$ – nfdc23 Jul 24 '16 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.