GIT lecture 12

This lecture ( GIT11-12) was given on 9/23/09


In the notes \mathscr{H}_{n,d} = \mathbb{P}(\mathbb{V}_{n,d}), where \mathbb{V}_{n,d} is the vector space of all homogeneous forms in n+1 variables of degree d.  There is a smooth and singular part: \mathscr{H}^{sm}, \mathscr{H}^{sing}.  The former is characterized when all partials of the given form only vanish at the origin.

Recall M = \{(x,h)\} where h is a hypersurface and x is a singular point of hM sits inside a product so there are two projections.

Now \dim \pi_1^{-1}(x) \ge \dim \mathscr{H}_{n,d} - (n+1) because (h,x) \in \pi^{-1}_1(x) iff all partials of h vanish at x which is n+1 conditions.

There is also a claim that \dim M \ge \dim \mathscr{H}_{n,d} -1.  For a generic point x \in \mathbb{P}^n should have \dim M = \dim \mathbb{P}^n + \dim \pi^{-1}_1(x) so \dim M \ge \dim H_{n,d} - 1.  The point is we want to say the fibers of \pi_2 are generically 0-dimensional so \dim M = \dim \mathscr{H}_{n,d}^{sing} and from the previous inequality conclude that codim \mathscr{H}_{n,d}^{sing} = 1 and is in fact a hypersurface cut out by a polynomial, the descriminant D. By why is \dim M = \dim \mathscr{H}_{n,d}^{sing}?

Essentially because powers of the determinant are the only nontrivial characters on GL(n), and because the action of GL(n) preserves the smooth and singular locus, it follows that D is a semi invariant which, in particular, will be fixed by SL(n).

Lemma 12.2 d>2 implies the stabilizer of any h \in H^{sm}_{n,d} in SL(n+1) is finite.

the proof uses a proposition in lecture 8 about reducing to lie algebras (among other things).

Proposition 12.3 Every h \in H^{sm}_{n,d} is properly stable with respect to the action of SL(n+1).

Classical Binary Invariants

This next bit is about finding invariants.  An important tool is the resultant; its a good thing to look up on wikipedia.  The important point is that the resultant of two polynomials vanishes when they have common zeros.  So the resultant applied to f_x, f_y will determine if f is singular or not.

Note about SL(2).  Say it acts on a vector space V, then so their is an action by elements diag(q, q^{-1}); let’s represent this as m(q) \in GL(V).  Then tr_V(m(q)) gives a character on V.  This is a function on \mathbb{G}_m.

Taking dth symmetric powers we also get an inducted action m^d(q) \colon S^d(V) \to S^d(V).  It can be checked

Ch V_d := tr(m^d(q)) = q^d + q^{d-2} + \cdots + q^{-d}

Now (q - q^{-1})(Ch V_d) = q^{d+1} - q^{-d-1}.  In general, Ch V is a linear combination of Ch V_d (why is this so?) This can be used to determine \dim V^G. An element of V^G determines a trivial 1 dimensional representation of G, so the dimension can be determined by counting the number of trivial representations that appear in Ch V, that is the number of Ch V_0.  This is exactly

Res_{q = 0} (q - q^{-1})Ch V

Note there are some error in the posted notes, basically a lot of places things like 1 - det_v(m(q)) are written where really it should be det(1 - m(g)).  This makes a HUGE difference, be cautious.


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