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I'd like to ask if the following "analogue" of the fact $\mathbb{C}[\mathfrak{g}]^G$ for semisimple $G$ is a polynomial ring is known, or disproved, or anything.

Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$. Let $\mathfrak{g}[[t]]=g\otimes \mathbb{C}[[t]]$, similarly for $\mathfrak{g}((t))$. Pick a nilpotent orbit $O$ of $\mathfrak{g}$, and consider a subset of $\mathfrak{g}((t))$ of the form

$X_O= \{n/t + x \ |\ n\in \overline{O}, \, x\in \mathfrak{g}[[t]]\} $.`

$\mathfrak{g}[[t]]$ acts naturally on $X_O$. I'd like to know if $R=\mathbb{C}[X_O]^{\mathfrak{g}[[t]]}$ is a polynomial ring. More precisely:

Functions on $X_O$ are generated by coordinate functions $a_i:X_O\to \mathbb{C}$ which is defined for $a\in \mathfrak{g}^*$ and $i\in\mathbb{Z}$ via $a_i(y)=$ ($a$ applied on the coefficient of $t^i$ of $y\in X_O$). We assign bidegree $(1,i)$ to $a_i$.

Then, I'm interested in the subring $R_-$ of $R$ whose second degree is negative. I think $R_-$ is a polynomial ring generated by $(\dim O)/2$ variables. I'd like to know the bidegree of the generators, too.

(One might need to require $\mathfrak{g}$ to be simply-laced, and/or $O$ to be special.)

For example, when $O$ is the orbit of zero, $R_-$ is just $\mathbb{C}$.

When $O$ is the regular nilpotent orbit, $R_-$ has $\dim O/2$ generators $c_{k,i}$ where $k=1,\ldots,\mathrm{rank}\, \mathfrak{g}$ and $i=1,\ldots, e_k$ where $e_k$ is the $k$-th exponent of $\mathfrak{g}$. Explicitly, $c_{k,i}$ is given as follows: denote by $c_k\in \mathbb{C}[\mathfrak{g}^*]^G$ a generator of degree $e_k+1$. Then we can consider $c_k(y)\in \mathbb{C}((t))$. for $y\in \mathfrak{g}((t))$. Finally we can define $c_{k,i}:X_O\to \mathbb{C}$ as the coefficient of $t^{-i}$ of $c_k(y)$ where $y\in X_O$.

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are you sure you want to consider invariant of g[[t]] ? I mean, may be just taking Gaudin-Bethe commutative subalgebra is what you want ? Cause if starting with first sentence analogy C[g]^C - polynomial ring - it is center of U(g), but in loop algebra case U(g(t)) has center, not U(g[t]) (modula details like crit. level etc.) - the correct analog of "center" in positive loops U(g[t]) is very simple thing - called Gaudin-Bethe commutative subalgebra in U(g[t]) - it is just image of the projection of Center(U((g (t)) under projection g(t)->g[t]. It is free polynomial and as nice as possible:) –  Alexander Chervov Dec 20 '11 at 6:53
Hmm, maybe what I want is $\mathbb{C}[Ad \mathfrak{g}((t)) X_O ]^{\mathfrak{g}((t))}$. Does it make a difference? I'd like to know how the invariant rings depend on the nilpotent residue chosen. Anyway thank you for the suggestion... I'll study Gaudin-Bethe commutative subalgebras. –  Yuji Tachikawa Dec 20 '11 at 7:46
If I am not making mistake, it makes big difference. I think that properly understood invariants in affine case should behave the same as in non-affine. –  Alexander Chervov Dec 20 '11 at 7:51
Some remarks: on U(g(t)) has two products - standard one and "normal ordered". The center in standard product is commutative subalgebra in normal ordered (Guadin-Bethe). Normal ordering means that positive loops commute with negative loops - means that U(g(t))^{normal product} = U(g[t]) \otimes U(g[t^-1]) , so there is natural projections which are algebra homorphisms U(g(t))^{normal product} -> U(positve loops) or to negative loops. So the center becomes commutative subalgebras in positve (negative) loops. –  Alexander Chervov Dec 20 '11 at 7:59
If one considers "classical limit" e.g. with Poisson brackets, but not commutators, then the explicit formula for the center(and GaudinBethe) is det(L(z) - l ) = C_{ij}z^il^j , C_{ij} will be free generators (it is very simple). The beautiful discovery by Dmitry Talalaev is that that in quantum case the correct formula looks like det(d/dz - L(z)) = C_{ij}z^i (d/dz)^j - so we get differential operator - it is what is called GL_n-oper in the Langlands correspondence theory, see e.g. section 8.2 page 30 in as well as –  Alexander Chervov Dec 20 '11 at 8:10

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