# GIT lecture 12

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

### Excerpts

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 $h$$M$ 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 $d$th 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.