# GIT First Lecture

Question: if $G$ is an alg. group acting on an affine variety $X$ say over an algebraically closed field $k$, then is there a good geometric object to call $X/G$?  Considering just the topological structure, there is a topological space $(X/G)_{top}$ (constructed by modding out by an equivalence relation determined by G-orbits), but the usual quotienting business for a top. sp might not produce an alg. variety.

Let $R$ be the affine coordinate ring of $X$.  The action of $G$ on $X$ produces ring automorphisms $R \to R$.  A good thing to do is to consider the ring of invariants $R^G$, and consider $\mbox{Spec } R^G$ as a candidate for a geometric quotient. Now

1. In general $R^G$ is not finitely generated, but it is fin. gen. in the case of reductive groups ( meaning G is a reductive group?)
2. In general, as topological spaces, $(X/G)_{top} \ne \mbox{Spec } R^G$ (example 1)
3. In general, $Frac(R^G) \ne Frac(R)^G$ (example 1); but it is true if there is a stable orbit (more later)

### Examples

Example 1: $X = \mathbb{A}^n = k^n$, so $R = k[x_1, ...., x_n]$ and $G = k^*$ with action

$t.(x_1, ..., x_n) = (tx_1, ..., tx_n)$

In this case its not hard to see that the only regular function that are invariants are quotients of homogeneous polynomials of the same degree, but there are no quotients in $R$, so $R^G = k$.  Thus $\mbox{Spec } R^G$ is a point $\ne (X/G)_{top}$.  Also

$Frac(R)^G = \{ \frac{f}{g} | f,g \mbox{ homogeneous of the same deg.} \}$

But $Frac(R^G) = R^G = k$.

An action $G$ on $X$ is a closed action if the orbit of any point is a closed set.

Example 2: In the affine plan, $t.(x,y) = (x+t, y)$ is a closed action of $G = \mathbb{A}^1$.

Example 3: $\mathscr{X} = \mbox{Mat }_{n \times n}(k)$, $G = GL_n(k)$ with action given by $g.x = gxg^{-1}$.  So in particular the orbits are the conjugacy classes which are classified by the Jordan normal form.

Now $R = k[x_{11}, x_{12},..., x_{nn}]$, so what is $R^G$ and what are the invariant matrices?

Recall the $\mbox{char}_X(t):= \mbox{det } (X - tI_n)$ is invariant under conjugation, so writing

$\mbox{char}_X(t) = (-1)^nt^n + \mathfrak{S}_1t^{n-1} + ... + \mathfrak{S}_n$

where, for example,

$\mathfrak{S}_1 = (-1)^{n-1} \mbox{trace }(X)$

$\mathfrak{S}_n = \det X$

It follows that $\mathfrak{S}_i \in R^G$; in fact (claim!) $R^G = k[\mathfrak{S}_1, ..., \mathfrak{S}_n]$.

Let $\mathscr{L} \subset \mathscr{X}$ be the diagonal matrices.  Then it is readily verified that $\mathfrak{S}_i|_{\mathscr{L}}$ are the elementary symmetric polynomials in n variables, hence alg. independent.

In this case $Frac(R^G) = Frac(R)^G$.

Example 4:  Here a moduli space is constructed for degree 2 curves in the affine plane considered up to roation and translation.  A general element $C(x,y) = ax^2 + 2bxy + cy^2 + 2dx + 2ey + f$ can be equivalently given by the matrix

$C_v = \left(\begin{array}{ccc} a & b & d \\ b & c & e \\ d & e & f \end{array}\right)$

Now the group we want to quotient out by is the semidirect product of the rotation group with $G_0 = \mathbb{A}^2$ the group of rigid translations: $G = SO(2) \rtimes G_0$.  In view of $C_v$, there is representation of G ( suppressing full sine and cosine notation):

$G = \{g(\theta, l,m)\} = \left\{\left(\begin{array}{ccc} c(\theta) & -s(\theta) & l \\ s(\theta) & c(\theta) & m \\ 0 & 0 & 1 \end{array}\right) \mbox{ with } l,m,\theta \in k\right\}$

My normal definition of rotations and translations would say that positive rotation are counterclockwise and translating by (l,m) takes the origin to (l,m), meaning the equations in original coordinates changes from $C(x,y)$ to $C(x-l,y-m)$.  But if instead the curve always remains fixed, and the ambient plane moves then all of the above get switched.  If the space in which the curve sits rotates counterclockwise by some amount, with the curve fixed, it would appear to from the ambient space’s perspective that the curve has rotated clockwise; now translating by (l,m) takes $C(x,y)$ to $C(x+l,y+m)$.  Then there seems to be the following interpretation:

$g(\theta, 0, 0).C_v = g(\theta,0,0)^\intercal\cdot C_v \cdot g(\theta, 0, 0)$

$g(0, l, m).C_v = g(0,l,m)^\intercal\cdot C_v \cdot g(0,l,m)$

$g(\theta, l,m).C_v = g(0,l,m).(g(\theta,0,0).C_v) = g(\theta,l,m)^\intercal C_v g(\theta,l,m)$

This interpretation is consistent (I think) with matrix multiplication for the representation given above.

As $R = k[a,b,c,d,e,f,g]$ is 6 dimensional and the representation above is 3 dimensional we expect the ring of invariants to be 6-3 = 3 dimensional.  Here are some invariants:

1. $D = \det f_v$
2. $E = ac - b^2$
3. $T = a + c$

Proposition: $R^G = k[D,E,T]$

proof: It can be checked that $k[a,b,c]^{SO(2)} \cong k[E,T]$.  Also, $a,b,c \in R$ are invariant under the induced automorphisms of $G_0$ on $R$.  Thus, it suffices to prove

$k[a,b,c,d,e,f]^{G_0} = k[a,b,c, D]$

Calculating the determinant gives $D = fE + 2bde - cd^2 -ae^2 =: fE + v(a,b,c,d,e)$.  But the determinant is invariant (since $\det g(\theta, l,m) = 1$) so after applying a general element of $latex G_0$, get $D = f'E + v(a,b,c,d+al+bm, e+bl+cm)$.  Using these observations, conclude that $f$ does not depend on $d,e$.  (apparently) this is enough info to conclude that $a,b,c, D$ generate all other invariants.  Q.E.D

Now consider the group $\tilde G$ generated by $G$ and $k^*$ (acting by multiplication of scalar matrices).  Also assume $D \ne 0$ meaning deal only with nondegenerate quadratics.  Then can get something interesting when considering $k[D,1/D,E,T]^{\tilde G}$.  Here we’re looking for ratios of homogeneous polynomials where the denominator is some power of D.

In fact for $A = E^3/D^2, B = T^3/D, C = ET/D$ can show (exercise!) that $A,B,C$ generate all the invariants of $k[D\pm, E,T]^{k^*} = k[A,B,C]/(AB - C^3)$.  So it would appear that a moduli space for nondegenerate degree 2 curves in the plane is given by the surface defined by  $AB - C^3$.