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20 votes

How to constructively/combinatorially prove Schur-Weyl duality?

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 ...
David E Speyer's user avatar
16 votes
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Why is Mumford's GIT-quotient so effective?

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 $\mathbb{G}_m$. ...
Will Sawin's user avatar
  • 141k
14 votes

Why is the catalecticant invariant under coordinate changes?

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 ...
Francois Ziegler's user avatar
14 votes

Has anyone researched additive analogues of toric geometry in characteristic zero?

The theory of Luna and Vust (Plongements d'espaces homogènes. Comment. Math. Helv. 58 (1983), 186–245.) on equivariant compactifications of homogeneous varieties works actually for any connected ...
Friedrich Knop's user avatar
13 votes

What generalizes symmetric polynomials to other finite groups?

The name of the body of theory you are asking for is "invariant theory of permutation groups." You will also find relevant papers by searching for "polynomial permutation invariants.&...
benblumsmith's user avatar
  • 2,831
13 votes
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Continuous version of the fundamental theorem of invariant theory for the orthogonal group

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 ...
Terry Tao's user avatar
  • 112k
12 votes
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Invariant polynomials under diagonal action of the orthogonal group

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 $\...
Friedrich Knop's user avatar
12 votes
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Standard Monomial basis for other types

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 ...
Friedrich Knop's user avatar
12 votes
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To describe an invariant trivector in dimension 8 geometrically

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}(...
Robert Bryant's user avatar
11 votes
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Are multiplicity-free representations weight multiplicity free?

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 (...
Friedrich Knop's user avatar
11 votes
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Three dimensional representations of Alternating group

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 (...
Noam D. Elkies's user avatar
11 votes

Geometric interpretation of characteristic polynomial

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 ...
Sean Lawton's user avatar
  • 8,434
11 votes
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Invariant theory for parabolics

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 ...
Friedrich Knop's user avatar
11 votes
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Separating closed $SO(p,q)$ orbits by invariant polynomials

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 ...
Friedrich Knop's user avatar
11 votes

Why is the catalecticant invariant under coordinate changes?

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^{(...
Abdelmalek Abdesselam's user avatar
11 votes
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Explicit formulas for invariants of binary quintic forms

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 ...
Robert Bryant's user avatar
11 votes
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Moduli of smooth curves in $|\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(2,2)| $ and their invariants

[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 ...
Noam D. Elkies's user avatar
11 votes

To describe an invariant trivector in dimension 8 geometrically

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 ...
Abdelmalek Abdesselam's user avatar
11 votes

Has anyone researched additive analogues of toric geometry in characteristic zero?

Firstly $\mathbb{G}_a$ and $\mathbb{G}_m$ are definitely not isomorphic as group schemes even in characterstic $0$, as the exponential is not an algebraic map. But there is a foundational paper on the ...
Daniel Loughran's user avatar
10 votes

Invariant polynomials under diagonal action of the orthogonal group

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) ...
Sean Lawton's user avatar
  • 8,434
10 votes

Chevalley–Shephard–Todd theorem

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. (...
David E Speyer's user avatar
10 votes
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Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$

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-...
Friedrich Knop's user avatar
10 votes
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A duality result for Coxeter groups

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 ...
David E Speyer's user avatar
10 votes
Accepted

Ring of invariants of some special type of subgroups of $GL_3(\mathbb C)$

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+{...
Robert Bryant's user avatar
10 votes
Accepted

Basis of invariant tensors of rank n in three dimensions

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 ...
Noah Snyder's user avatar
  • 27.9k
9 votes
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Explicit invariant of tensors nonvanishing on the diagonal

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 ...
Abdelmalek Abdesselam's user avatar
9 votes
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If an equivariant map is smooth on diagonal matrices, is it smooth everywhere?

I think, one can argue as follows. Let $D\subseteq\text{Sym}$ be the diagonal matrices. Since $\exp:D\to D^+$ and $\exp:\text{Sym}\to\text{Sym}^+$ are compatible diffeomorphisms it suffices to answer ...
Friedrich Knop's user avatar
9 votes
Accepted

Invariant theory over $\mathbb R$

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{...
Friedrich Knop's user avatar
9 votes

Invariants of $\mathrm{GL}_n$ representations

When $k=1$ the ring of invariants of degree $d$ separately in each of $W = \mathrm{Sym}^2 V$ and $W^\star = \mathrm{Sym^2} V^\star$ has dimension equal to the number of partitions of $d$ with at most $...
Mark Wildon's user avatar
8 votes

How to constructively/combinatorially prove Schur-Weyl duality?

This a continuation of my first answer. I was trying to edit the previous one but the MathJax processing was freezing my computer. I suppose that answer was getting too long. @Darij: There has been ...
Abdelmalek Abdesselam's user avatar

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