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Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$. $S(\mathfrak{g})^G$ is a polynomial algebra with rank $\mathfrak{g}$ generators. Call them $c_i(x)$, where $x\in \mathfrak{g}$ and $i=1,\ldots,\mathrm{rank}\ g$.

Now, consider $G$ acting diagonally on $\mathfrak{g}\oplus\mathfrak{g}$. What is the ring of invariants $R=[S(\mathfrak{g})\otimes S(\mathfrak{g})]^G$ ? (I guess this is a textbook material; sorry for not doing the homework of looking up the books, but it's a Saturday today and it's easier to ask here than to go to the library at my institute in a cold weather...)

For $x\oplus y \in \mathfrak{g}\oplus\mathfrak{g}$, I can easily see $R\ni c_i(x + t y)$ for a scalar $t$. Do they generate $R$ (polynomially)?

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It seems $c_i(x+ty)$ is not enough to generate the ring since its dimension is something like n^2 (may be n^2+-n). and you have only 2n generators from $c_i(x+ty)$. –  Alexander Chervov Jan 14 '12 at 15:09
    
If you consider (gl(n), gl(n),..., gl(n)) - many copies, then invariants here are subject of "Weyl’s fundamental theorem of invariant theory". The theorem says that Tr( P(X1, X2,..XN)) taking all possible polynoms "P" will generate the ring of invariants. However choosing appropriate number of generators and what are relations might be not that much simple. –  Alexander Chervov Jan 14 '12 at 15:11
    
@Alexander: by taking the derivatives w.r.t. t, I guess $c_i(x+ty)$ give us something like $n^2/2$ generators, but yes, they're not enough. Hmm. –  Yuji Tachikawa Jan 15 '12 at 4:38
    
@Yuji Hmm, derivates does not give new invariants, if i understand correctly, cause derivatives of polynom can be expressed via the coeffiecients of polynom - we already counted $n$ coefficients of polynom as invariants, so i think you will not get more from it. –  Alexander Chervov Jan 15 '12 at 13:32
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Yuji, what you are trying to do is to generate all invariants for the $G$-action on $V\oplus V$ (where $V=\mathfrak{g}$) using the polarization operators corresponding to the commuting $GL_2$-action. If $V$ is a vector representation of $G=GL_n$ or its contragredient (and in a few other cases), because of the Howe duality, the commuting algebra is indeed generated by the polarization operators. However, this is very far from being true when $V$ is the "matrix" representation. –  Victor Protsak Jan 15 '12 at 17:57
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4 Answers

When $G=GL(n),$ this is the "invariants of matrices", as Alexander Chervov has pointed out. The full description by generators and relations is only known for small $n.$ So it's not "textbook material" in the same sense as the description of $S(\mathfrak{g})^G.$ For a general $\mathfrak{g},$ there are papers of Bezrukavnikov-Ginzburg and Kostant, among others, but it's Saturday and cold here as well....

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Most of what is known in the case of $GL(n,K)$ acting diagonally by conjugation on $r$-tuples of $n \times n$ matrices can be found in Chapters 8-10 of my CBMS notes "The Polynomial Identities and Invariants of $n \times n$ matrices". There $K$ is a field of characteristic zero, the fixed ring is called the ring of invariants of $n \times n$ matrices, and it is denoted $C(n,r)$.

First Fundamental theorem (H. Weyl): $C(n,r)$ is generated by traces of monomials.

Second Fundamental Theorem (B. Konstant, C. Procesi, Y. P. Razmyslov): Let $S_r$ be the symmetric group on $r$ letters, and let $J(n,r)$ be the two-sided ideal of the group ring $KS_r$ corresponding to Young diagrams with $\geq n+1$ rows. Then there is an easily defined map from $KS_r$ onto the multilinear elements of degree $r$ in $C(n,r)$ whose kernel is $J(n,r)$.

Nagata-Higman Theorem: If an algebra $A$ (noncommutative, without a unit) satisfies $x^n = 0$ for every $x \in A$, then there is an integer $N = N(n)$ such that $x_1 \cdots x_N = 0$ for all $x_1, \dots ,x_N \in A$. Procesi made the brilliant observation that the least such $N$ valid for all $A$ is also the least integer such that $C(n,r)$ is generated by traces of monomials of length $\leq N$. Using the Second Fundamental Theorem Procesi and Razmyslov showed that $N(n) \leq n^2$. Studying the Nagata-Higman Theorem, E. N. Kuzmin showed that $N(n) \geq n(n+1)/2$ and conjectured that $N(n) = n(n+1)/2$. His conjecture has only been verified for $n \leq 4$.

General theorems of M. Hochster - J. L. Roberts and M. P. Murthy imply that $C(n,r)$ is a unique factorization domain and Gorenstein, and M. van den Bergh proved that it is Cohen-Macaulay.

Results of several authors show that $C(n,r)$ is a polynomial ring if and only if $n = 1$, or $r = 1$, or $(n,r) = (2,2)$, and is a complete intersection if and only it is a polynomial ring or $(n,r) = (2,3)$, or $(n,r) = (3,2)$.

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Great answer !!! –  Alexander Chervov Jan 15 '12 at 18:12
    
For $G=SL(n,C)$ the papers by Teranishi are I believe the best published results. I think this is where you can find $N(n) = n(n+1)/2$ for $n=3,4$. Teranishi, Yasuo The ring of invariants of matrices. Nagoya Math. J. 104 (1986), 149–161. The following paper shows that the ring of invariants is polynomial only for G=SL(2,C) and r=2 (for r>1). Teranishi, Yasuo A theorem on invariants of semi-simple Lie algebras. Perspectives in ring theory (Antwerp, 1987), 37–40, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 233, Kluwer Acad. Publ., Dordrecht, 1988. –  David Wehlau Jan 28 '12 at 16:27
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An extended comment, not a full answer: It may help to view the direct sum of $\mathfrak{g}$ with itself as another semisimple Lie algebra. From the context, $\mathfrak{g}$ is the Lie algebra of $G$, so this construction yields the Lie algebra of $G \times G$, in which the semisimple group $G$ is embedded diagonally as a subgroup. In particular, the $G \times G$-invariants on polynomials over its own Lie algebra are clear, so you want a concrete description of the larger algebra of $G$-invariants. By classical invariant theory this algebra should still be finitely generated. It's probably a polynomial algebra, though I'm not sure how far the literature deals with this set-up.

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No, it's not a polynomial algebra (the cases when the algebra of invariants is polynomial have all been classified, by the way). –  Victor Protsak Jan 14 '12 at 20:20
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Shamely I cannot answer by heart, later I'll ask some colleagues doing invariant theory.

But let me give some example and some comments (partly extending Victor's).

Example: consider g=sl(2), then if I understand correctly the factor will be C^3 and invariant polynoms can be taken like this:

Let me denote by $(\Lambda_1, \Lambda_2)$ points in $sl(2) \oplus sl(2)$

$t_1 = Tr (\Lambda_1^2), ~~ t_2=Tr(\Lambda_1 \Lambda_2), ~~ t_3=Tr(\Lambda_2^2)$

Clearly these invariant polynoms and clearly they are algebraically independent.

(Probably??????) They generate the coordinate ring in invariants.


Some comment on this coordinates. See our paper with Dmitry Talalaev

http://arxiv.org/abs/hep-th/0303069 Hitchin system on singular curves I page 30.

The the space $(Mat_n^n)^{gl(n)}$ can be identified as a moduli space of vector bundles on the very singular curve which is $P^1$ with n-cusps. In particular if n=2 we get the curve with 2 cusps: $y^2=(x-a)^3 (x-b)^3$. This is degeneration of the curve of genus 2. For smooth curves of genus 2 there is well-known result by Narasimhan and Ramanan that moduli space of rank 2 bundles is $CP^3$

Our curve is degeneration of the smooth curve, nevertheless we may try to find analogs of Narasimhan-Ramanan coordinates. So in this paper we argued that coordinates above are in some sense analogs of their coordinates. There is description of Hitchin system in this coordinates in that paper.


Next two comments related to the following questions $g\oplus g$ can be endowed with Poisson and symplectic structure. You may see $g\oplus g$ as a Lie algebra, and also as $T^*g$ (the first is Poisson, the second is symplectic). As vector spaces they are the same.

So having poisson manifolds you may want to study the SYMPLECTIC reduction.

The case $T^*g$ and its quantum version $Diff(g)$ is subject of famous paper by Etingof and Ginzburg

http://arxiv.org/abs/math/0011114 Symplectic reflection algebras, Calogero-Moser space, and deformed Harish-Chandra homomorphism

Where the moral is roughly speaking the rational Cherednik algebra in A_n case is quite related to $Diff(g)//g$.


The case of reduction $U(g)\oplus U(g) // g$ has been studied quite recently by Sergey Khoroshkin, Oleg Ogievetsky

http://arxiv.org/abs/0912.4055 Diagonal reduction algebras of $gl$ type

They called these algebras Mickelson-Zhlobenko algebras (as far as I understand) and they were used in several subsequent works by different authors

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