Physicist trying to understand GIT quotient
I am reading Nakajima's textbook on Hilbert Schemes. I am trying to understand some very basic facts about the GIT quotient.
We start with a vector space $V$ over $\mathbb{C}$. Let $G \subset U(V)$ be a Lie group and $G^{\mathbb{C}}$ its complexification so I guess $G^{\mathbb{C}} \subset GL_V$. I will denote by $G$ the complexification from now on.
Apparently $V/G$ is a very badly behaved space. I do not know really why though. I can imagine that there might be some singularities but can they not be resolved e.g. by blowing up? Also, why sometimes this space is not Hausdorff?
Now, let $A(V)$ be the coordinate ring of $V$. Nakajima says something I did not know, that the $A(V)$ is the same as the symmetric power of the dual space $V^*$. $$ A(V) = Sym^n(V^*) $$
- Why is this true? I have to admit that this seems very basic and I did not know about it.
Next I learn that $G$ has a natural action on $V$ i.e. $v \mapsto gv$ for $g \in G$ and $v \in V$. Then, this induces an action on $A(V)$. We define $$ A(V)^G = \{ {\text{polynomials }a | ga =a \text{ for }\forall g\in G } \}$$ the ring of invariant (polynomials). Finally we define the algebro-geometric quotient of $V$ by $G$ as $$ Spec(A(V)^G)=V//G $$ To me this is the space of prime ideals that are invariant under $G$. But I do not see how exactly this is related to the original space we wanted to construct. It seems quite different actually.
- Intuitively what is this space $V//G$ and why is it useful?
P.S. Nakajima says: The underlying space of $V//G$ is the set of closed $G$-orbits modulo the equivalence relation defined by $x \backsim y$ if some specific condition, that I do not mention here, holds.
Apparently $V/G$ is a very badly behaved space. I do not know really why though. I can imagine that there might be some singularities but can they not be resolved e.g. by blowing up? Also, why sometimes this space is not Hausdorff?
The simplest way to think about this is just to consider an example, and the best one is probably the following:
Consider the action of $\mathbb C^* = \mathbb C -0$ on $\mathbb C^2$, where the action is given by $$(x,y) \mapsto (\lambda x, \lambda^{-1} y) $$ What are the orbits for this action? There are the orbits of the form $xy = c \neq 0$ for any complex number $c$. Then there axial orbits $\{ (x,0) : x \neq 0\}$ and $\{ (0,y): y \neq 0\}$. Finally there is the zero orbit, which just contains one point $0$.
The vast majority of the orbits are of the first type, which suggest the quotient $\mathbb C^2 /\mathbb C^*$ should be $\mathbb C$, but then the question remains, what happens to the other orbits? The axial orbits and the zero orbit all lie arbitrarily close to one another (a sequence of points in the axial orbits can converge to zero, but zero is not in the orbit). Therefore the resulting space taking the quotient naively would be $\mathbb C$, but with three copies of the $0$ point, which is a non-Hausdorff space. Since we are quotienting a variety, we would hope to get another variety, and this is clearly not one.
GIT deals with this by declaring any orbit which contains zero in the closure to be unstable, and the quotient is defined only on the (poly)stable points (I don't want to go into what stability means, since it's a bit complicated and there are multiple competing definitions. I'll link some stuff to read if you want to know more at the bottom.)
To me this is the space of prime ideals that are invariant under $G$. But I do not see how exactly this is related to the original space we wanted to construct. It seems quite different actually.
- Intuitively what is this space $V//G$ and why is it useful?
Again, best to think of an example. In the example above, the polynomial ring associated to the variety $\mathbb C^2$ is the whole polynomial ring in two variables $\mathbb C[x,y]$. Then under the action above, a polynomial $f(x,y)$ gets sent to $f(\lambda x, \lambda^{-1} y)$, and we see therefore that the polynomial $f = xy$ is invariant under the action of $\mathbb C^*$. It's not hard to show that all invariant polynomials are generated by this one, so the invariant ring is $$ \mathbb C[x,y]^{\mathbb C^*} = \mathbb C[xy]$$ It's clear then that $$\operatorname{Spec}(\mathbb C[x,y]^{\mathbb C^*}) = \operatorname{Spec}(\mathbb C[xy]) = \mathbb C,$$ and in this we see that since the axial orbits and the zero orbit all lie in the same invariant class ($xy =0$); hence the GIT quotient treats all 3 as equivalent.
(Try and do this example again for yourself but with a different space and a different action. It's a good exercise. The only way to get your head around this stuff is lots of examples in my opinion.)
As for why the GIT quotient is useful: well, it's the correct quotient to use in algebraic geometry. Since taking quotients is so common in geometry, it's no suprise that mathematicians want a good theory about how to do it. Most interesting to me personally is how it relates to the symplectic reduction through the Kempf-Ness theorem. That's projective GIT, and that's the really interesting bit to me. There's also infinite-dimensional analogue of the Kempf-Ness theorem that concerns the theory of connections on principal $U(n)$-bundles upto gauge equivalence.
Anyway some resources. I learnt GIT and symplectic reduction from the notes by Richard Thomas. There's also a book by Dolgachev on Invariant theory that's pretty good for an algebraic perspective, and the original GIT was of course found in the book by Mumford. I believe the latest edition has some stuff on symplectic reduction as well. There's also a book on invariants and moduli by Mukai which is simply brilliant. I'm currently writing some stuff about this for my master's project, I'll link it here when I'm done.