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Let $G$ be a (split) reductive group over $k$, $T$ a split maximal torus, and W its Weyl group. I sometimes see the Chevalley restriction theorem stated as

(1) $k[G]^G \xrightarrow{\sim} k[T]^W$

and sometimes as

(2) $k[\mathfrak{g}]^G \xrightarrow{\sim} k[\mathfrak{t}]^W$.

I think both can be proved in essentially the same way, but is there a way to directly deduce them from each other?

I think (1) implies (2) directly: equip $k[G]$ and $k[T]$ with the filtrations by the powers of the maximal ideals at the origins. Then the map in (2) is the associated graded of the map in (1). Since the map in (1) is a strict map of filtered rings, it being an isomorphism implies its associated graded is an isomorphism.

It's also clear that the injectivity of (2) implies the injectivity of (1). However I'm not quite seeing what to do about surjectivity -- is there an adjective one can put in front of "filtered" to guarantee that a map is surjective if its associated graded is? (This is the case when the filtration is finite, but that is not satisfied here...)

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    $\begingroup$ There cannot be a completely formal way to prove equivalence of these two, because (1) is always true but (2) can fail in positive characteristic. (But I like this question, and hope that there is a good answer that avoids this in some sense shallow pitfall! It had never occurred to me to try to connect these two facts, rather than just proving both.) $\endgroup$
    – LSpice
    Commented Aug 5, 2020 at 2:14
  • $\begingroup$ Interesting, thanks for pointing that out. Although it seems that your objection applies to the implication (1) => (2) whereas I thought I was stuck on (2) => (1). Does my argument for (1) => (2) sound right in char 0? Certainly it breaks down in positive characteristic because formation of invariants doesn't commute with formation of associated graded. $\endgroup$
    – user125639
    Commented Aug 5, 2020 at 4:38
  • $\begingroup$ It sounds plausible, but I wouldn't want to claim I've thought about it enough that I'd catch a subtle error if there were one. $\endgroup$
    – LSpice
    Commented Aug 5, 2020 at 4:44

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