20

Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a ...

15

The plethysm $\mathrm{Sym}^k \rho$ contains the irreducible representation with highest weight $(2,\ldots,2,0,\ldots,0)$ exactly once. It looks like a tricky problem to say much about its other irreducible constituents.
Let $\Delta^\lambda$ denote the Schur functor corresponding to the partition $\lambda$, and let $E$ be an $n$-dimensional complex vector ...

invariant-theory classical-invariant-theory dg.differential-geometry rt.representation-theory plethysm

15

No, there is no reason that $X$ should be a complete intersection. For instance, begin with $Z=\mathbb{A}^2_k$ with coordinates $(z_0,z_1)$. Consider the action on $Z$ of the group of $n^{\text{th}}$ roots of unity, $\mu_n$, by $\zeta\cdot(z_0,z_1) = (\zeta z_0,\zeta z_1)$. The ring of invariants, $k[z_0,z_1]^{\mu_n}$, is generated by the monomials $y_{a,...

15

In general, the only definition I know of GIT quotient is $Proj$ of the invariant ring. The obvious statements one can make about the rational map $Proj\ R\to Proj\ R^G$ are that it collapses $G$-orbits, and if one semistable orbit is in the closure of several others, they all collapse together.
Your example is of a very special type, where the action of $...

13

By my understanding, your question is not "Why is Mumford's construction better than the affine quotient". As you note, Proj is better than Spec of invariants for taking quotients by G_m. Instead, your question is "Why can't we get an even better quotient by going further, involving more characters somehow?"
I would say an answer is that any construction ...

13

This is a quick answer to explain the statement that the hard direction of Schur-Weyl duality is the same thing the First Fundamental Theorem of invariant theory.
Let $V$ be a finite dimensional vector space and $V^{\ast}$ the dual space. The FFT (or a special case there of) says that the $GL(V)$ invariant multilinear functions $V^n \times (V^{\ast})^n$ ...

answered Nov 29 '16 at 1:47

David E Speyer

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13

Dolgachev (2012, p. 57; pdf) observes that your circulant (with determinant $\operatorname{Cat} f$) is the matrix of a symmetric bilinear form $\Omega_{\,f}$ on $\smash{\operatorname{Sym}^n(\mathbf k^{2})}$ in a certain basis, then states as obvious that $f\mapsto\Omega_{\,f}$ is a $\smash{\operatorname{GL}(\mathbf k^2)}$-equivariant map
$$
\operatorname{...

answered Sep 15 '17 at 14:42

Francois Ziegler

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12

This order of terms in the determinant is the most mysterious point for me. Is there any reason for it? What happens if one chooses some other ordering of terms: will the Capelli identity be modified somehow or it will not work at all?
I think the answers to these questions can be found in the paper http://arxiv.org/abs/0809.3516 (Noncommutative ...

12

To the best of my knowledge this is an open problem. In fact, there is strong evidence that the problem is very hard indeed: Consider the action of $S_n$ on the "two-sets," i.e., on the subsets of $\{1,\ldots,n\}$ of two elements. It is easy to see that this is a subrepresentation of the regular representation. So if generating invariants for the regular ...

ag.algebraic-geometry reference-request rt.representation-theory ac.commutative-algebra invariant-theory

11

Even without knowing an explicit set of generators, you can compute the Hilbert series with very little work as follows. In general, suppose a finite group $G$ acts on a vector space $V$ over a field $k$ of characteristic not divisible by $|G|$ via an action map $\rho : G \to \text{GL}(V)$. Then $G$ acts on the symmetric algebra $S(V^{\ast})$ (a coordinate-...

answered Feb 23 '16 at 8:26

Qiaochu Yuan

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11

Here is a complete answer to this question: the map $\varphi$ is an embedding if and only if the group $H$ contains a maximal torus of $G$. I'm assuming (as in the question) that all groups are complexified but originate from a compact group.
The map $\varphi$ is injective if and only if no nonzero $G$-invariant function on $\mathfrak g$ vanishes on $X:=G\...

11

The following construction reduces your problem it to a classical and well-studied problem in invariant theory. First, I claim that there is a natural way to interpret an $n$-tuple of points in $S^2$ as a point of $\mathbb{CP}^n$ and vice-versa. This depends on interpreting $S^2$ as $\mathbb{CP}^1$ and the classical fact that the symmetric product of $n$ ...

answered Nov 15 '12 at 22:44

Robert Bryant

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11

The map $\mathbb C[\mathfrak g]^G\to\mathbb C[\mathfrak g]^P$ is an isomorphism for trivial reasons: In any quasi-affine $G$-variety, $P$ and $G$ have the same fixed points. Just look at the orbit map of a $P$-fixed point which factors through the complete variety $G/P$. Applied to representations, this means $V^G=V^P$ for any rational $G$-module. This holds ...

11

Let $G_0$ be the image of $A_5$ under one of the $3$-dimensional
representations, and $G = \pm G_0$. Then $G$ is the group of
symmetries of the icosahedron, which is a Euclidean reflection group
(type $H_3$, Shephard-Todd #23). Thus $G$ has a polynomial invariant group,
and in this case the generator degrees are $2, 6, 10$. For invariants
$\phi_2, \phi_6,...

answered Jul 5 '16 at 15:18

Noam D. Elkies

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11

Standard monomial theory has been extended to all classical groups by Lakshmibai, Seshadri and others in the series of papers "Geometry of $G/P$ I-IX".
A very concise description of standard tableaux in this setting can be found in the appendix of "Littelmann, Peter: A generalization of the Littlewood-Richardson rule. J. Algebra 130 (1990), no. 2, 328–368".
...

ag.algebraic-geometry rt.representation-theory algebraic-groups invariant-theory geometric-invariant-theory

10

If $G$ is reductive, try looking at Fogarty, Kirwan, Mumford, Geometric Invariant Theory, p. 27

10

If by $V/G$ you mean the space of orbits, this is not true. Consider $\mathbb C^*$ acting on the affine space $\mathbb A^1$ by multiplication; the space of orbits has two points, but the only variety with two points is disconnected, while $\mathbb A^1$ is connected.

10

The invariants are generated by the quadratic polynomials $(u,u)$, $(u,v)$, and $(v,v)$ where $(.,.)$ is the scalar product defining $O(n)$. This pattern generalizes to arbitrary many copies of $\mathbb R^n$. This is called the first fundamental theorem for the orthogonal group.

10

It is a theorem of Brion and, independently, of Vinberg that varieties with an open $B$-orbit (a.k.a. spherical varieties) have in fact only finitely many orbits. A shorter argument is due to Matsuki (see his ICM talk) and independently (using the same idea) by me (On the set of orbits $\ldots$). Thus multiplicity free spaces are visible. Kac implies that ...

10

The principal isotropy group is $H=SL(3)\times SL(3)$: it has the right dimension (namely 16) and occurs as an isotropy group (namely of a general element of $W$). Now it is a general result of Luna-Richardson that the restriction map $\mathbb C[V]^G\to\mathbb C[W]^N$ is an isomorphism where $W=V^H$ and $N=N_G(H)/H$.
For a concrete construction of $\alpha$ ...

10

Let $d=2n$ be the degree of your binary form $f$.
Let me introduce $n+1$ pairs of formal variables $\alpha^{(1)}=(\alpha^{(1)}_{1},\alpha^{(1)}_{2}),\ldots, \alpha^{(n+1)}=(\alpha^{(n+1)}_{1},\alpha^{(n+1)}_{2})$.
Let $\mathcal{S}$ be the polynomial
$$
\mathcal{S}=\prod_{1\le i<j\le n+1} (\alpha^{(i)}\alpha^{(j)})^2
$$
where I used the classical bracket ...

answered Sep 15 '17 at 15:55

Abdelmalek Abdesselam

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10

Yes. By Chevalley-Shepard-Todd, $S(V)^G$ and $S(V)^H$ are polynomial rings. Let $S(V)^G=\mathbb{R}[g_1, \ldots, g_n]$ and $S(V)^H = \mathbb{R}[h_1,\ldots, h_n]$ where the $g_i$ and $h_i$ are homogenous. Then
$$S(V)_G^H = \mathbb{R}[h_1,\ldots,h_n]/\langle g_1,\ldots, g_n \rangle.$$
Here the denominator is the ideal of $\mathbb{R}[h_1,\ldots,h_n]$ generated ...

answered Nov 22 '17 at 21:19

David E Speyer

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10

Here's another way to do it that you might find useful:
Recall that $\mathrm{SL}(2,\mathbb{C})$
acts on the polynomial ring $\mathbb{C}[x,y]$
by linear substitution in $x$ and $y$,
making the subspace $V_d\subset \mathbb{C}[x,y]$, consisting
of polynomials homogeneous of degree $d$ in $x$ and $y$, into
an irreducible $\mathrm{SL}(2,\mathbb{C})$-...

answered Dec 17 '17 at 21:24

Robert Bryant

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9

Yes, this is correct. Let me hit it with a more general statement in case that becomes useful for further generalizations.
There is a general formula (Cauchy identity) for the action of $GL(V) \times GL(W)$ on the exterior power $\bigwedge^n(V \otimes W)$. This is written as
$\bigoplus_{|\lambda|=n} S_\lambda(V) \otimes S_{\lambda^\dagger}(W)$
where $S_\...

9

Let me start with some remarks about the classical symbolic method (without which one cannot understand 19th century invariant theory)
and multisymmetric functions.
I will use an example first. Take four series of three variables $a=(a_1,a_2,a_3)$, $b=(b_1,b_2,b_3)$,
$c=(c_1,c_2,c_3)$ and $d=(d_1,d_2,d_3)$.
Now define the polynomial $\mathcal{A}$ in these 12 ...

ag.algebraic-geometry co.combinatorics rt.representation-theory computational-complexity invariant-theory

answered Jan 9 '17 at 22:14

Abdelmalek Abdesselam

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9

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.
For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is
$$
H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,.
$$
The determinant of this Hessian form is ...

answered Nov 25 '18 at 18:40

Robert Bryant

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8

Reposting VA's wonderful answer (with a trivial correction), since someone else just asked me this question:
This is just to add 1% to Dmitri's 99% complete answer. Change the
coordinates to $w_0,\dots, w_{n-1}$ defined by the formula
$$ w_i = x_0 + \mu^i x_1 + \mu^{2i} x_2 + \dots, $$
where $\mu$ is a primitive $n$-th root of identity. Then ...

8

I believe the answer is yes. Since $X$ is an affine $G$-variety, $G$ acts on $\mathbb{C}[X]$ by $\mathbb{C}$-algebra automorphisms. This yields an action of $G$ on the fraction field of $\mathbb{C}[X]$ by algebra automorphisms. This restricts to an action on the integral closure of $\mathbb{C}[X]$, so that inclusion into the integral closure is $G$-...

8

I am somewhat late for a year, sorry, but since the wiki-article is mainly written by me and I somehow worked on the subject, it is difficult to resist writing an answer.
Currently there is some understanding about non-commutative determinants which makes identity less mysterious:
Point 1. There are determinants (and whole linear algebra) for very ...

answered Feb 14 '16 at 8:53

Alexander Chervov

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8

Yes, and regularity isn't needed (assuming noetherian). By the Eakin-Nagata Theorem (3.7, Matsumura CRT), it is enough that $R$ is $R^G$-finite. For the Cohen ring $W$ of the perfect residue field, the unique local map $W\rightarrow R$ lifting the identity on residue fields is $G$-invariant. Pick a surjection $W[\![x_1, \dots, x_n]\!]\rightarrow R$. Let $t_{...

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