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 ...


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 $V^n \times (V^{\ast})^n$ ...


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{...


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 ...


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-...


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 $n$ ...


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,...


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". ...


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 ...


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 ...


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 $\mathrm{SL}(2,\mathbb{C})$-...


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 ...


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 ...


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 ...


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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