GIT quotients for linear representations of $SL(2,\mathbb C)$
Let $V$ be the standard two-dimensional representation of $SL(2,\mathbb C)$ and let ${\rm Sym}^2V$ be its symmetric square. Let $n$ be a positive integer and consider the following two representations 1) $V^{\oplus n}$ and 2) $({\rm Sym}^2V)^{\oplus n}$ of $SL(2,\mathbb C)$.
Question 1) Is there some explicit description of the GIT quotient $V^{\oplus n}// SL(2,\mathbb C)$? In particular, is it true that the ring of invariant polynomials is generated by $\frac{n(n-1)}{2}$ quadratic polynomials $vol(v_i,v_j)$ $i\ne j$, where $v=(v_1,\cdots ,v_n)=v\in V^{\oplus n}$ and $vol$ is a volume preserved by $SL(2,\mathbb C)$?
Question 2). Is there some explicit description of the GIT quotient $({\rm Sym}^2V)^{\oplus n}// SL(2,\mathbb C)$? In particular, is it true that the ring of invariant polynomials is generated by $\frac{n(n+1)}{2}$ quadratic polynomials $(v_i,v_j)$ , where $v=(v_1,\cdots ,v_n)=v\in ({\rm Sym}^2V)^{\oplus n}$ and $(.,.)$ is a symmetric bilinear form on $({\rm Sym}^2V)$ preserved by $SL(2,\mathbb C)$?
ag.algebraic-geometry rt.representation-theory invariant-theory
Abdelmalek Abdesselam
aglearneraglearner
$\begingroup$ Mumford's description of the geometric quotient of $V^{\oplus n}$ by the standard action of $\text{SL}(V)$ in Chapter 3 of GIT is so very explicit that it does not actually use any of the general theory of "Geometric Invariant Theory"! That is important, because Mumford wanted to construct $\mathcal{M}_g$ as a scheme over $\text{Spec}(\mathbb{Z})$, whereas GIT (at that time) only worked over a fixed field. So Mumford uses his explicit quotient construction of $V^{\oplus n}$ over any base as a step in constructing $\mathcal{M}_g$ as a quasi-projective scheme over $\text{Spec}(\mathbb{Z})$. $\endgroup$ – Jason Starr Apr 4 '17 at 15:06
Both questions are extensively dealt with in Weyl's book "Classical invariant theory" who investigated the invariant of classical groups on multiple copies of their defining representations. Determining a set of generators is called a "First Fundamental Theorem" (FFT) while the relations are given in a "Second Fundamental Theorem".
Question 1: This is best regarded as the action of $Sp(2n)$ on $V=\mathbb C^{2n}$ for $n=1$. Then, indeed, the ring of invariants is generated by all pairings $f_{ij}:=\omega(v_i,v_j)$. These are neatly organized in a $2n\times 2n$ skew-symmetric matrix. The relations are generated by all "principal" Pfaffian minors of size $(2n+2)\times (2n+2)$. For $n=2$ these are quadratic polynomials in 3 variables called the "Plücker relations". The quotient consists of all skew-symmetric matrices of rank $\le 2$ (which is, of course, the affine cone over a Grassmannian).
Question 2: In this case one is dealing with the group $SO(n)$ acting on $\mathbb C^n$ for $n=3$. Here things are a bit more complicated since $SO(n)$ is not strictly a classical group. The better problem is to look at the group $O(n)$ instead. In this case, the invariants are indeed generated by all pairings $p_{ij}=(v_i,v_j)$ which can be organized into a $n\times n$ symmetric matrix. The relations are generated by all $(n+1)\times(n+1)$-principal minors. In your case, the quotient would be the set of symmetric matrices of rank $\le3$. Since you are dealing with $SO(n)$ instead of $O(n)$ things are more complicated. In this case there are additional generating invariants namely all determinants of the form $\det(v_{i_1},\ldots,v_{i_n})$. For $n=3$ the quotient is the subset of $S^2\mathbb C^n\oplus\wedge^3\mathbb C^n$ given be two sets of relations: the rank conditions and the condition that the square of the determinant can be expressend as a Gram matrix $\det((v_{i_\mu},v_{i_\nu}))$.
Friedrich KnopFriedrich Knop
$\begingroup$ Thank you! I imagine this is contained in your answer, but is it possible to spell out a bit more explicitly what is the GIT quotient a cone over, in the second case? (if it is again a cone over something) $\endgroup$ – aglearner Apr 4 '17 at 20:20
$\begingroup$ It is not a space one encounters in a Linear Algebra course. Varieties like that have been studied. Search for "determinantal varieties". It carries a $GL(n,\mathbb C)$-action which makes it into a spherical variety with probably 3 orbits. It is a singular with rational singularities (in particular normal and Cohen-Macaulay). $\endgroup$ – Friedrich Knop Apr 5 '17 at 5:41
$\begingroup$ Friedrich, many thanks! I have not realized that this is a spherical variety. I am not worried too much that this variety is not from a linear algebra course :). But I am not able to understand the last three lines of your answer. What are "the rank conditions" and the condition that "the square of the determinant can be expressed as a Gram matrix" ? $\endgroup$ – aglearner Apr 5 '17 at 9:06
$\begingroup$ In fact, I understand that there is a morphism from $\rm Sym^2(V)^{\oplus n}//SL(2,\mathbb C)$ to the cone over the Grassmanian $G(3,n)$. For a generic point the fiber is $6$-dimensional. But I don't see how to get an action with one open orbit on $\rm Sym^2(V)^{\oplus n}//SL(2,\mathbb C)$... So I don't understand what is the spherical variety in this case... $\endgroup$ – aglearner Apr 5 '17 at 11:38
For question 1, the answer is most easily described by thinking about $V^{\oplus n} $ as $ Hom(\mathbb C^2, \mathbb C^n) $. From this it follows that the projective GIT quotient $$ V^{\oplus n} //_{det} GL_2 $$ is isomorphic to the Grassmannian $\mathbb G(2,n)$. From this, it follows that the invariant ring $ \mathbb C[V^{\oplus n}]^{SL_2} $ is homogeneous coordinate ring of $\mathbb G(2,n) $. Your quadratic polynomials are the generators of this coordinate ring (and the relations are given by the Plucker relations).
Joel KamnitzerJoel Kamnitzer
$\begingroup$ I suppose the FFT is hidden in the isomorphism with the Grassmannian. $\endgroup$ – Abdelmalek Abdesselam Apr 4 '17 at 17:58
The answers are:
Question 1: yes.
Question 2: no.
Explicit linear generators follow from the first fundamental theorem (FFT) for $SL_2$. You can see my two answers to this MO question for an explicit proof of the FFT for $SL_k$. It is best to use a graphical language to represent these generators as in my article "On the volume conjecture for classical spin networks". J. Knot Theory Ramifications 21 (2012), no. 3, 1250022. This type of graphical representation is very old as you can see in this MO answer. Then the fun begins, namely trying to find polynomial rather than linear generators. Essentially this results from the Grassmann-Plücker relation. For $SL_2$ and forms of degree 1 or 2 this is easy to do by hand. For quadratics, the GP relation can be used to break cycles (the only thing produced by the FFT). In fact the article by Kempe in the second MO answer I mentioned does exactly that, with explanatory pictures. The first invariant which is not expressible by the ones you gave is for three quadratics corresponding to a 3-cycle containing each one of them. This is also the Wronskian of the three quadratics.
For quadratics an explicit system of generators which basically adds these Wronskians for each triple of forms is given in Section 256 "The quadratic types" in the book by Grace and Young.
The basic identity from that book needed for breaking cycles of length at least four is in classical symbolic notation: $$ 2(ab)(bc)(cd)(de)=(bc)(cd)(db)(ae)-(cd)^2(ab)(be)-(bc)^2(ad)(de)+(bd)^2(ac)(ae) $$ where $a=(a_1,a_2)$ etc. and $(ab)$ is the determinant of the matrix with first row $a$ and second row $b$ etc.
Using self-duality of $SL_2$ representations, the LHS can be a interpreted as a product of four $2\times 2$ matrices. This is basically the Amitsur-Levitzki Theorem for $2\times 2$ matrices.
I don't know how explicit you want to be but you can look at the article "Defining Relations for the Algebra of Invariants of 2×2 Matrices" by Drensky for more details. He treats the case of generic matrices while you are interested in matrices coming from symmetric bilinear forms by self-dualization. For generic matrices, traces of words of length 1, 2, 3 form a minimal system of $n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}$ algebra generators. Drensky finds the polynomial relations between these generators. Here, with quadratic binary forms, the words of length 1 disappear.
Abdelmalek AbdesselamAbdelmalek Abdesselam
$\begingroup$ Thank you! I guess the article of Drensky indeed addresses the second half of the question. I would be happy if there was some "geometric" description of this GIT quotient akin to the cone over Grassmannian $G(2,n)$ appearing in the part 1) of the question. (But it might be that such a description does not exist...) $\endgroup$ – aglearner Apr 5 '17 at 3:35
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