# GIT, wedging, and moment map

So this stuff didn’t come from a GIT lecture, but it is stuff I spent time thinking about because of a GIT lecture.

### Wedging

I should say this is just a discussion about finite dimensional vector spaces (maybe even finite rank free modules?)

I was always told that if $W = V^*$ then elements of $\wedge^r W$ can be thought of as alternating multilinear maps $W^r \to k$, where $k$ is the ground field.

So if $a,b \colon V \to k$ are functionals, then $a\wedge b$ should give rise to an alternating map $V\times V \to k$.  The answer is (or should be)

$a\wedge b (x,y) = a(x)b(y) - a(y)b(x)$

the only question is, how do you determine that this is the answer?

Roughly an answer (Due to Theo)

By the universal property of wedging, the vector space of alternating bilinear forms is naturally isomorphic to $(\wedge^2 V)*$.  The nontrivial step is to explain why, at least if $V$ is finite dimensional, then $(\wedge^2 V)^* \cong \wedge^2 V^*$.

Recall: If $V,W$ are finite dim, then $V^*\otimes W^* \cong (V \otimes W)*$

pf: because of fin dim can choose basis $e_1, ..., e_n \in V; f_1, ..., f_m \in W$.  Then $a \in V^*$ and $\psi \in (V \otimes W)^*$ are determined by $a_i := a(e_i)$ and $\psi_{ij} = \psi(e_i\otimes f_j)$

The map in one direction takes $a,b$ to the bilinear map $ab(e_i\otimes f_j) = a_ib_j$.  In the reverse direction, choose $j$ such that $\psi_{kj} \ne 0$ for some $k$.  Set $a_i = \psi_{ij}$; then $a_k \ne 0$.  Then set $b_l = \psi_{kl}/a_k$. QED.

To simplify stuff set

1. $X = V^* \wedge V^*$
2. $Y = V^* \otimes V^*$
3. $Y' = (V \otimes V)^*$
4. $Z$ is the subset of $Y'$ that give rise to alternating bilinear maps, i.e. $Z \cong (\wedge^2 V)^*$.

Note $X$ is defined as a quotient of $Y$, so there is map in that direction.  There is also an averaging map (analogous to the Renolds operator) $ave: Y' \to Z$.  For example, $ave(x \otimes y) = x\otimes y - y\otimes x$

So we have

$X \xleftarrow{p} Y \cong Y' \xrightarrow{ave} Z$

Let $s \colon X \to Y$ be any lift, for example $s(x \wedge y) = x \otimes y$.  Then $ave \circ \cong \circ s$ produces a multilinear function:

$X \xrightarrow{s} Y \cong Y' \xrightarrow{ave} Z$

In the example at hand, this will take $a \wedge b$ to some multiple of $a\otimes b - b \otimes a$.

Its possible that requiring $p \circ s = id$ will pin down $s$ uniquely, but in any case, as long as the characteristic isn’t 2, this should produce an isomorphism which roughly explains that

$a \wedge b(x,y) = a(x)b(y) - b(x)a(y)$

and if the characteristic is 0, then you’ll get similar isomorphism for higher wedge products.

### The Moment Map

This material is taken directly from ‘Cohomology of Quotients in Symplectic and Algebraic Geometry’ by Kirwan.

This works for any symplectic manifold $X$, but at least now I usually just worry about $X$ being a projective variety.  So there is a closed 2-form $\omega$.  Get iso, $\forall x \in X$

$T_x X \to T^*_x X$

$v \mapsto \omega_x(v, - )$

$H$ is a lie group with lie alg. $\mathfrak{h}$; $H$ acts on $X$ and preserves the symplectic form.

Preliminaries.

###1: For $g \in H$ and $\alpha \in \mathfrak{h}^*$, setting $Ad^*_g(\alpha) = \alpha \circ Ad_{g^{-1}}$ gives an action of $H$ on $\mathfrak{h}^*$.  There might also be some transpose business to work out.  Recall $Ad_g(X) = gXg^{-1}$.

###2:Given a map $X \xrightarrow{m} V^*$ where the latter is the dual of a vector space, then for every $x \in X$ we get a map on tangent spaces $T_x X \xrightarrow{dm_x} V^*$ and for every $v \in V$, the composition

$T_xX \xrightarrow{dm_x} V^* \xrightarrow{e_v} k$

where $e_v$ is evaluation at $v$, is a 1-form.

###3: Thinking of the lie algebra as the tangent space at the identity, any $Y \in \mathfrak{h}$ gives a vector field on $Y$.  Let $\gamma_Y(t)$ be a 1-param. subgroup of $H$ with tan. vec. $Y$.  Then over the point $x$, assing the tan. vec. $\frac{d}{dt} \gamma_Y(t).x =: \vec Y_x$.

A moment map is a map $\mu X \to \mathfrak{h}^*$ s.t.

1. ($H$-equivariance) for $z \in X$ and $g \in H$, $\mu(gz) =$ $Ad^*_g \mu(z)$ holds.
2. $\forall Y \in \mathfrak{h}$, $\omega(\vec Y, - )$ is a 1-form and $e_Y \circ d\mu_z( - )$ is another 1-form.  Its required that

$\omega(\vec Y, - ) = e_Y \circ d\mu_z( - )$

Claim: If $U(n+1)$ acts on $\mathbb{P}^n$ by matrix multiplication on column vectors, then a moment map is determined by

$\langle \mu(\vec z), Y\rangle = c \cdot (\vec z, Y \vec z)$

where $z \in \mathbb{P}^n$, $Y \in \mathfrak{h}$, $\langle, \rangle$ is the natural pairing of vector space and its dual, and $(, )$ is the standard Hermitian form on $\mathbb{C}^{n+1}$ (apparently called the Study-Fubini form.  The constant $c = 1/2\pi i |\vec z|^2$.

Now

$\langle \mu(g.z), Y \rangle = (g.z, Yg.z) = (z, g^{-1}Yg.z) = \langle Ad^*_g \mu(z) , Y \rangle$

This shows 1, and transitivity of the action means 2 can be checked just at an easy point.  An easy point is for example $p = [1: 0: ... : 0]$.

###4 Evidently if $Y$ is a lie algebra element, it acts on $\mathbb{P}^n$ as multiplication by a matrix, $Y = (Y_{ij})$, and $\vec Y_p$ is just the first column of this matrix.  Also $Y \in u(n+1)$ which means $\bar Y^\intercal = - Y$ , so in particular $\overline{Y_{ij}} = - Y_{ji}$, this will be used later.

###5 For the point $p$, local coordinate are $z = (z_1, ..., z_n) \leftrightarrow [1:z_1:...: z_n]$.  Now $(z,Yz)$ is a function that depends on $z_i$:

$(z,Yz) = (1, \bar z_1, ..., \bar z_n)\cdot Y \cdot (1, z_1, ...., z_n)^\intercal =: f(z)$

You can multiply it out, differentiate it via $df = \sum_i \partial f/\partial z_i dz_i$ and evaluate at the point $p = (0, ..., 0) = [1:0:...:0]$.  Then the only terms that don’t disappear are exactly the terms of the form $const. \cdot dz_i, const. \cdot d\bar z_j$.  This produces

$d(z, Yz) = frac{1}{2 \pi i}\sum_i Y_{0i}dz_i + Y_{io}d\bar z_i$

$= \frac{i}{2\pi} \sum_i \overline{Y_{io}} dz_i - Y_{io} d \bar z_i$

where the remark in ###4 is used; note this is a 1-form.

### Finally…

There’s another way of getting a 1-form from $Y$.  Namely, look at $\omega( \vec Y_p, - )$.  To compute this its required to understand what the Study-Fubini form will do to the tangent vector $\vec Y_p = (Y_{10}, ...., Y_{n0})$.  Write this out in real coordinates: set $a_i = Re(Y_{io})$ and $b_i = Im(Y_{io})$ then

$\vec Y_p = \sum_i a_i \partial/\partial x_i + b_i \partial/\partial y_i$

###6: Recall $dx(\partial/\partial x) = 1$.  Which is to say $\partial/\partial x .x = 1$.  Using this as motivation, $\partial/\partial z$ should be some linear combination of $\partial/partial x$ and $\partial/\partial y$ such that $\partial/\partial z .(x + iy) = 1$.  This analysis yields

$\partial/\partial z = \frac{1}{2} \bigl( \partial/\partial x - i \partial/\partial y \bigr)$

$\partial/\partial \bar z = \frac{1}{2}\bigl( \partial/\partial x + i/2 \partial/\partial y \bigr)$

So $\partial/\partial x = \partial/\partial z + \partial/\partial \bar z$ and similarly for $\partial/\partial y$.  Doing this change of basis yeilds

$\vec Y_p = \sum_i Y_{i0} \partial/\partial z_i + \overline{Y_{io}} \partial/\partial \bar z_i$.

Then it becomes clear, that when you apply $\omega( - , - )$ the 1-form you get agrees with the previous computation.