16

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 $\mathbb{G}_m$. 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 ...

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 $...

15

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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14

Dolgachev (2012, p. 57; pdf) observes that your matrix $\left( a_{i+j-2}\right) _{1\leq i\leq n+1,\ 1\leq j\leq n+1}$ (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{\...

answered Sep 15 '17 at 14:42

Francois Ziegler

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Yes. It suffices to show that if one has a sequence $\vec v^{(n)} = (v^{(n)}_1,\dots,v^{(n)}_m) \in E^m$ whose Gram matrix $(\langle v^{(n)}_i, v^{(n)}_j \rangle)_{i,j=1,\dots,m}$ converges to a Gram matrix $(\langle v_i, v_j \rangle)_{i,j=1,\dots,m}$ of a tuple $\vec v = (v_1,\dots,v_m) \in E^m$, then after applying linear isometries to each of the $\vec v^...

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

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.

12

Here's another very nice (but still algebraic) interpretation that explains some of the geometry: Recall that $\operatorname{SL}(2,\mathbb{C})$ has a $2$-to-$1$ representation into $\operatorname{SL}(3,\mathbb{C})$ so that the Lie algebra splits as
$$
{\frak{sl}}(3,\mathbb{C}) = {\frak{sl}}(2,\mathbb{C})\oplus {\frak{m}}
$$
where ${\frak{m}}$ is the ($5$-...

answered Nov 24 '20 at 19:40

Robert Bryant

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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

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

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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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

11

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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11

For a purely geometric construction, see further below, after the following algebraic considerations.
There is a Wronskian isomorphism which as a particular case says that the second exterior power of $R_4$ is isometric to the second symmetric power of $R_3$. So the invariant in question is $I(Q,C)$, a joint invariant in a binary quadratic $Q$ and a binary ...

answered Nov 24 '20 at 18:57

Abdelmalek Abdesselam

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10

I am reluctant to answer a question this old that already has a very nice answer, however, looking at the title the first thing that comes to my mind is something quite different from the existing answer (and maybe it will be useful to someone who comes across this question).
When $V$ is a finite dimensional vector space over $\mathbb{C}$, then the ...

10

Torsten's argument is of course beautiful, but it might be worth recording that there is also a slick combinatorial argument, in case you need to teach this to students without algebraic geometry. (After rereading, it looks like Josh Swanson is linking to the same argument, so upvote his answer if you like this.) Let $G$ be any finite subgroup of $GL(V)$ for ...

answered Nov 22 '17 at 21:37

David E Speyer

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10

This is covered in Chapter 2, Section 9 (starting on page 52) in Weyl's The Classical Groups, Their Invariants and Representations.
As already answered, there you will find:
Theorem (Theorem 2.9.A) Every orthogonal invariant in vectors $x^1,...,x^m$ in $\mathbb{R}^n$ is expressible in terms of the $m^2$ scalar products $(x^i,x^j)$.

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

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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10

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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10

[EDITED to exhibit $j$ as a rational function of $J_2,J_3,J_4$,
and to fix various local errors etc.]
The action of ${\rm SL_2} \times {\rm SL_2}$ on the $9$-dimensional space of
$(2,2)$ forms has a polynomial ring of invariants, with generators in degrees
$2,3,4$. If we write a general $(2,2)$ form $P(x_1,x_2;y_1,y_2)$ as
$(x_1^2, x_1 x_2, x_2^2) M_3 (y_1^...

answered May 8 '20 at 1:22

Noam D. Elkies

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10

Planar partitions with no singletons works. You need to pick for each $n>1$ some map with certain properties. One way to do this is to just fix a preferred trivalent tree of each size and interpret each vertex as a cross product. For example, one arbitrary choice gives the map $V^{\otimes n} \rightarrow \mathbb{C}$ given by $$v_1 \otimes \ldots \otimes ...

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

The group $SO(p,q)$ is not per se an algebraic group. Rather there is an algebraic group $G$ such that $SO(p,q)=G(\mathbb R)$ is its group of real points. The main point is that also $G(\mathbb C)$ is defined which leads to two different concepts of orbits. First, there are the orbits $G(\mathbb R)v$ which you are probably after.
But then there are also the ...

9

The answer depends on what you mean by "one-to-one correspondence". Is it bijective or just injective? Robert Bryant's (standard) argument shows that $\mathbb R^N/\mathrm{SO}(n)\to \mathrm{AlgHom}_{\mathbb R}(A,\mathbb R)$ is injective. In general, this map is very far from being surjective, though. In other words, not every Homomorphism $A\to\...

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 $G$-...

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