Invariant theory

Invariant theory

In the previous video, we decomposed Sym 2 Sym 2 C 2 as Sym 4 C 2 plus C. If Sym 2 C 2 has a basis alpha, beta, gamma then Sym 2 Sym 2 C 2 has a basis consisting of quadratic monomials like alpha squared, alpha beta, etc, and the trivial summand C is spanned by beta squared minus alpha gamma.

We also mentioned that Sym 2 C 2 is "basically the same as" quadratic forms in x and y, so that Sym 2 Sym 2 C 2 is "basically the same as" quadratic expressions in the coefficients of a quadratic form. We know that there is a distinguished quadratic expression in the coefficients of a quadratic form Q = a x squared + b x y + c y squared which vanishes if and only if the Q has a repeated root, namely b squared minus 4 a c, which looks an awful lot like beta squared minus alpha gamma. The reason why the factor of 4 is missing is that we have not been sufficiently careful about the isomorphism we used to identify quadratic forms with Sym 2 C 2. In this video, we will be more careful, and will make a comment about how to generalise this to higher degree polynomials or polynomials in more variables: what are the invariant quantities in those cases?

Quadratic forms as a representation of SL(2,C)

A quadratic form can be written as follows: row vector x, y times the 2-by-2 matrix a, b over 2; b over 2, c, times the column vector x, y, so we can identify quadratic forms in x and y with 2-by-2 symmetric matrices like the one in the middle of this expression. We will allow the group SL(2,C) to act by coordinate changes on xy-space in the usual way: x, y maps to M x, y The row vector x, y goes to x, y times M transpose and so the quadratic form turns into x , y row times M transpose times a, b over 2, b over 2 c, times M times x, y column. More explicitly, if M is the 2-by-2 matrix s, t; u, v then the new matrix for the quadratic form is s squared a + s u b + u squared c, s t a + (u t + s v) times b over 2 + u v c, t squared a + t v b + v squared c So if we just keep track of the coefficients of the quadratic form, this transformation is a,b,c gets multiplied by the 3-by-3 matrix s squared, s u, u squared; 2 s t, u t + s v, 2 u v; t squared, t v, v squared. So we can think of the space of quadratic forms as the 3-dimensional (a,b,c)-space, and the formula above tells us how to think of that space as a representation of SL(2,C) except that it is not really a representation: it is an "antirepresentation". You can see this by looking at what happens if we act on x, y first using M_1 and then M_2. This gives M_2 M_1 x, y, and the quadratic form becomes x, y row vector times M_1 transpose times M_2 transpose times the matrix of the quadratic form times M_2 times M_1 times x, y You can see that the matrix M_2 hits the matrix of the quadratic form first, followed by M_1! This is back to front: if we write R(M) for the 3-by-3 matrix above, then R is an antirepresentation in the sense that R(M_2 M_1) = R(M_1) R(M_2). We can fix this by taking the transpose: R(M_2 M_1) transpose equals R(M_2) transpose times R(M_1) transpose. It is this tranposed representation to which we will apply the theory we've developed.

If you want to think about what this trick of transposing the matrices means, it is essentially that instead of thinking of a,b,c as entries of a vector, we're thinking of them as linear functions on the space of quadratic forms which return the x squared, x y and y squared coefficients respectively.

Weights and action of the Lie algebra

We now figure out the weights of this representation and how X in little s l (2, C) acts. Let's take M equals exp i theta H, which equals e to the i theta, 0; 0, e to the minus i theta. Substituting this into our 3-by-3 matrix we get R(M) transpose equals the diagonal matrix with entries e to the 2 i theta, 1, e to the minus 2 i theta. So a has weight 2, b has weight 0 and c has weight minus 2.

Now let's take M equals exp of t X, which equals 1, t; 0, 1. We get R of exp t X, transpose = 1, 2 t, t squared; 0, 1, t; 0, 0, 1. To find the action of the Lie algebra element X, we differentiate with respect to t and set t = 0: which gives the 3-by-3 matrix 0, 2, 0; 0, 0, 1; 0, 0, 0. Note that this is not the same as just substituting s = 0, t = 1, u = 0, v = 0: the formula for R of M transpose relies on M being in the Lie group S L (2,C), not in the Lie algebra.

Therefore the action of X is X a = 0, X b = 2 a, X c = b.

We know that our 3-dimensional representation (R of M, transpose, of S L (2,C) on quadratic forms) is isomorphic to Sym 2 C 2 because it has weights minus 2, 0, 2. However, if we look at the action of X on the basis a,b,c versus the action of X on alpha = e_1 squared, beta equals e_1 e_2, gamma = e_2 squared in Sym 2 C 2, we get something a bit different. Namely, we get X of gamma = X of e_2 squared, which equals 2 e_1 e_2, which equals 2 beta, X of beta equals X of e_1 e_2, which equals e_2 squared, which equals alpha, and X of alpha equals X of e_1 squared, which equals 0. So although these representations are isomorphic, the identification a = alpha, b = beta, c = gamma is not an isomorphism of representations! It turns out we need to take a = alpha, b = 2 beta, c = gamma.

For example, then we get X of c = X gamma = 2 beta = b, X of b = X of 2 beta, = 2 alpha = 2 a. Now if we look at the invariant element beta squared minus alpha gamma, this becomes b squared over 4 minus a c, or a quarter of b squared minus 4 a c.

This is the beginning of a subject called invariant theory. You could do the same trick with cubic forms in 2 variables, looking for a trivial 1-dimensional subrepresentation of Sym n of Sym 2 of C 2 for some n which vanishes if and only if the cubic form has a repeated root. It turns out that this lives in Sym 4 of Sym 3 of C 2. Using our methods, you can check that this representation contains a trivial 1-dimensional subrepresentation, and find an element Delta in the zero weight-space with X Delta = 0. However, if you do it the way we did in the previous video, you need to be careful to use the right identification of Sym 3 C 2 with cubic forms (which turns out to be e_1 cubed = a, e_1 squared s_2 = b over 3, e_1 e_2 squared = c over 3, e_2 cubed = d for the cubic form ax cubed + b x squared y + c x y squared + d y cubed).

You can also do this for forms in more variables, using groups like S L (n,C).