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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Symplectic geometry\\[15mm]
\small lecture 15: Geometric invariant theory}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small Misha Verbitsky } 
\\[20mm]

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
October 23, 2021
}
\end{center}

\newpage

{\bf \blue Moment maps (reminder)}

\definition
$(M,\omega)$ be a symplectic manifold, and $G$ a 
Lie group acting on $M$ by symplectomorphisms.
{\bf \blue A moment map} $\mu$ of this action is a linear map
${\goth g}\arrow C^\infty M$
associating to each $g\in G$ its Hamiltonian.

\remark
It is more convenient to consider $\mu$
as an element of ${\goth g}^* \otimes_\R C^\infty M$,
or (and this is most standard) {\bf \red as a function
with values in ${\goth g}^*$}. 

\remark
Moment map {\bf \purple always exists} if $M$ 
is simply connected.

\definition 
A moment map $M \arrow  {\goth g}^*$
is called {\bf \blue equivariant}
if it is equivariant with respect to the 
coadjoint action of $G$ on ${\goth g}^*$.

\remark
$M\stackrel\mu \arrow  {\goth g}^*$ is a moment map
iff for all $g\in {\goth g}$, 
$\langle d\mu,g\rangle= i_{\rho_g}(\omega)$.
Therefore, {\bf \purple a moment map is defined up to 
a constant ${\goth g}^*$-valued function.}
An equivariant moment map is is defined up to 
{\bf \purple a constant ${\goth g}^*$-valued function
which is $G$-invariant}, that is, up to addition of a central 
vector $c\in \g^*$. 

\claim
{\bf \red
An equivariant moment map exists whenever $H^1(G, {\goth g}^*)=0$.}
In particular, when $G$ is reductive and $M$ is simply connected,
an equivariant moment map exists.
Further on, all moment maps will be tacitly considered equivariant.


\newpage

{\bf \blue Weinstein-Marsden theorem (reminder)}

%\definition A $G$-invariant $c\in \goth g^*$ is called
%{\bf\blue central}.

\definition
(Weinstein-Marsden)
$(M,\omega)$ be a symplectic manifold, $G$ a 
compact Lie group freely acting on $M$ by symplectomorphisms,
$M\stackrel\mu \arrow  {\goth g}^*$ an equivariant
moment map, and $c\in {\goth g}^*$ a central element.
The quotient $\mu^{-1}(c)/G$ is called {\bf \blue
symplectic reduction} of $M$, denoted by $M\2 G$.

\claim
{\bf \red The symplectic quotient $M\2 G$ is a symplectic
manifold of dimension $\dim M - 2 \dim G$.}\\
\pstep
$T_x (\mu^{-1}(c))= d\mu^{-1}(0)$.
However, the space 
$\langle d\mu, g\rangle \subset \Lambda^1 M$ is $\omega$-dual to the space 
$\tau(\goth g)$ of vector fields
tangent to the $G$-action, {\bf \purple hence 
$d\mu^{-1}(c)= \tau(\goth g)^\bot$.}

{\bf \green Step 2:}
Since $\mu$ is $G$-equivariant, $G$ preserves
$\mu^{-1}(c)$, hence $\tau(\goth g)\subset d\mu^{-1}(0)$.
This implies that {\bf \purple $\tau(\goth g)\subset TM$ is isotropic}
(that is, $\omega\restrict{\tau(\goth g)}=0$).
Its $\omega$-orthogonal complement in $T_x M$ 
is $T_x (\mu^{-1}(c))$ (Step 1).

{\bf \green Step 3:}
Consider the {\bf \blue characteristic foliation}
${\cal F}$ on $\mu^{-1}(c)$. It is 
a bundle because $\mu^{-1}(c)\subset M$
is coisotropic. From Step 2 {\bf \purple we obtain that ${\cal F}=\tau(\goth g)$.}

{\bf \green Step 4:}
Since $\omega\restrict {\mu^{-1}(c)}$ is closed, it
satisfies  $\Lie_v (\omega)=0$ for all $v\in {\cal F}$.
This implies that it is {\em \green basic}, that is, 
lifted from the leaf space of characteristic foliation, 
identified with $M\2 G$.
\endproof

\newpage

{\bf \blue Holomorphic vector fields}

\definition
{\bf\blue A holomorphic vector field} is a vector field
$X\in TM$ 
satisfying $\Lie_X I=0$, that is, such that the
corresponding diffeomorphism flow $e^{tX}$ is holomorphic.

\remark Let $(M,I)$ be a compact complex manifold. 
Then the group of biholomorphisms of $M$
is a Lie group whose Lie algebra is the space of 
(real) holomorphic vector fields. 

\remark
It is not hard to see that {\bf \purple $I(X)$ is holomorphic
whenever $X$ is holomorphic} ({\bf \red prove this}).
Then $X\arrow I(X)$ {\bf \purple defines the complex structure on
the Lie algebra of holomorphic vector fields.}

\newpage

{\bf \blue Symplectic reduction and a K\"ahler potential (reminder)}

\definition
Let $d^c:= I dI^{-1}$.
{\bf \blue K\"ahler potential} on a K\"ahler manifold $(M, \omega)$
is a function $\psi$ such that $dd^c \psi=\omega$.

\proposition
Let $G$ be a real Lie group acting
on a K\"ahler manifold $M$ by holomorphic isometries,
and $\psi$ a $G$-invariant K\"ahler potential.
{\bf \red Then the moment map $\g \times M 
\stackrel {\mu_g} \arrow \R$
can be written as $g, m \arrow \Lie_{Iv}\psi$,}
where $v=\tau(g)\in TM$ is the tangent vector field
associated with $g\in \goth g$.

\proof
Since $\psi$ is $G$-invariant, and $I$ is $G$-invariant,
we  have $0 = \Lie_v d^c \psi = i_v (dd^c\psi) +
d(i_v d^c \psi)$. Using $\omega=dd^c \psi$,
we rewrite this equation as
$i_v\omega = -d(\langle d^c \psi, v\rangle)$,
giving an equation for the moment map 
$\mu_g= -\langle d^c \psi, v\rangle.$
Acting by $I$ on both sides, we obtain
$\mu_g= \langle d \psi, Iv\rangle = \Lie_{Iv}\psi$.
\endproof

\corollary Let $V$ be a Hermitian representation of
a compact Lie group $G$. {\bf \purple Then the corresponding moment
map can be written as $\mu_g(v)= \Lie_{Ig}|v|^2=
\frac 1 4 \langle v, Ig(v)\rangle.$} \endproof

\newpage

{\bf \blue Moment map and extrema on 1-parametric orbit}

\remark
Let $G\subset U(n)$ be a Lie group acting 
on a complex vector space $V=\C^n$, equipped
with the standard Hermitian structure $h$, and 
$G_\C \subset GL(n, \C)$ its complexification.
Denote by $\mu:\; V \arrow \g^*$ the moment map.
Since $\mu_g(v)= \Lie_{Ig}|v|^2$,
a vector $z\in V$ {\bf \purple belongs to $\mu^{-1}(0)$
if and only if the function $l:\; G_\C \cdot z\arrow \R$, $l(z)=|z|^2$
on the orbit $G_\C$ has extremum in $z$.}
It turns own that $z\arrow l(z)$ {\bf \green is convex},
in the following sense.

\claim Let
$V=\C^n$, $A\in {\goth u}(V)$ be an anti-Hermitian endomorphism,
and $G_A:=e^{t A}\subset GL(V)$, $t\in \C$ the corresponding 1-parametric
subgroup. {\bf \red Then 
$\frac{d^2}{d{u}^2} |e^{(t + u) A}(z)|^2= 4
|A(e^{tA}(z))|^2$,} when $u\in \1 \R$.

\proof
Let $A$ be diagonalized in an orthonormal basis 
$x_1, ..., x_n\in \R$, such that $A(x_i) = \1 w_i x_i$,
$w_i \in \R$, and $h$ the Hermitian form.
The operator $e^{tA}(x_i) = e^{\1 tw_i} x_i$ is an isometry when $t$ is real,
and is Hermitian self-adjoint when $t$ is imaginary. 
This gives 
$\frac{d}{d{u}} |e^{(t + u) A}(z)|^2= 
2 h(Ae^{(t) A}(z), e^{(t) A}(z)).$ 
Taking the derivative in $u$ once again, we
obtain 
$\frac{d^2}{d{u}^2} |e^{(t + u) A}(z)|^2=2 h(Ae^{(t) A}(z),
 A e^{(t) A}(z))$.
\endproof

\corollary The function $l$ {\bf \red is convex on $G_A$,} 
and {\bf \red strictly convex
in imaginary direction,} unless $A=0$.
\endproof

\newpage

{\bf \blue Moment map and extrema on 1-parametric orbit}

\corollary
Let $A$ be diagonalized in an orthonormal basis 
$x_1, ..., x_n\in \R$, such that $A(x_i) = \1 w_i x_i$, $w_i \in \R$,
and $z= \sum \alpha_i x_i$, with all $\alpha_i\neq 0$. 
Then the function
$l$ {\bf \red has a minimum on the line $e^{\1 \R A}(z)$
if and only if  there are two basis vectors $x_l, x_k$
such that $w_l <0$ and $w_k >0$.} Moreover, {\bf \red $l$ has no
other extrema on $e^{\1 \R A}(z)$, unless $A=0$.}

\proof
The function $u\arrow |e^{(t + u) A}(z)|^2$
is strictly convex along the imaginary axis,
and constant along the real axis. Therefore,
it has at most one extremum, and it is the minimum.
It has minimum if and only if $\lim |e^{(t + u) A}(z)|^2=\infty$
as $t \arrow \pm \infty$, which happens if and only if
$w_l <0$ and $w_k >0$ for some $k,l$.
\endproof


\newpage

{\bf \blue Extrema of the length function on $G_\C$-orbits}

\theorem
Let $G\subset U(n)$ be a Lie group acting 
on a complex vector space $V=\C^n$, equipped
with the standard Hermitian structure $h$, and 
$G_\C \subset GL(n, \C)$ its complexification.
Let $\g_\C = \Lie(G_\C)= \g \otimes_\R \C$.
Consider the function $\phi:\; \g_ \C \arrow \R$
taking $g$ to $l(e^{g}z)$.
{\bf \red Then $\phi$ is constant in real direction,
convex in imaginary direction, and satisfies
$\frac{d^2}{d{u}^2}|\phi(u)(z)|^2= 4 h(u(z), u(z))$
where $u\in \im \g_\C$.}

\proof 
Write $u\in I({\goth u}(n))$ in an appropriate
orthonormal basis $x_1,...., x_n$ as
$u(x_i) = w_i x_i$, $w_i\in \R$. 
Let $z= \sum \alpha_i x_i$.
Then
$\frac{d}{d{u}}|\phi(u)(z)|^2= 2\sum w_i |x_i|^2 = 2 h(u(z), z)$,
and $\frac{d^2}{d{u}^2}|\phi(u)(z)|^2= 4 h(u(z), u(z))$.
\endproof

{\bf \green COROLLARY 1:}
Either the function $l$ {\bf \red has no extremal points
on the orbit $G_\C \cdot z$, or $l$ has a minimum.}
In the second case, {\bf \red the set of points on $G_\C$ where $l$ has
a minimum is an orbit of $G$.} 
\endproof


\newpage

{\bf \blue Stable and unstable orbits}


\definition
Let $G\subset U(n)$ be a Lie group acting 
on a complex vector space $V=\C^n$, equipped
with the standard Hermitian structure, and 
$G_\C \subset GL(n, \C)$ its complexification.
An orbit $G_\C \cdot z$, $z\neq 0$ is called {\bf\blue stable}
if $l$ reaches minimum on $G_\C \cdot z$,
{\bf\blue unstable} if $0$ belongs to the closure
of $G_\C \cdot z$, and {\bf\blue (strictly) semistable} 
if if it is not stable and not unstable.

\theorem
Let $G\subset U(V)$ be a group acting on
a complex Hermitian vector space $V= \C^n$, and 
$z\in V \backslash 0$.
{\bf \red Then an orbit $G_\C\cdot z$ is stable if and only if
it intersects the zero set of the moment map
$\mu(v, z)= \Lie_{Iv}(l)(z)$. } Moreover, {\bf \red $G$
acts on $G_\C\cdot z\cap \mu^{-1}(0)$ transitively.}

\proof Extrema of $l$ on $G_\C\cdot z$ are its minima
because $l$ is convex. The extrema of $l$ are 
zero set of $\mu$ because $\mu(v, z)= \Lie_{Iv}(l)(z)$.
The set of extrema is one $G$-orbit by Corollary 1.
\endproof


\newpage

{\bf \blue Hilbert-Mumford criterion of stability}

\exercise
Let $\rho:\; U(1) \arrow \End(V)$, $V=\C^n$ be a complex Hermitian
representation of $U(1)$.
Then {\bf \red there exists an orthonormal basis $x_1, ..., x_n$
in $V$ such that $g(x_i)= \1 2\pi w_i x_i$,}
where $w_i\in \Z$ are integer numbers called
{\bf\blue the weights} of the action.

\proof Since the action is unitary, $\rho(t)$ is diagonalizable.
The numbers $w_i$ are integer because $\rho\restrict{\1 \R}$ factorizes through $U(1)$.
\endproof


\theorem
Let $G=U(1)$ act on a complex 
Hermitian vector space $(V=\C^n, h)$, $G_\C= \C^*$
the corresponding complex Lie group, and $z\in V$ 
a non-zero vector, $z = \sum \alpha_{l_k} x_{l_k}$, where $\alpha_{l_k}$
are all non-zero. Consider the weight decomposition of the
generator of this action: $A(x_i)= \1 2\pi w_i x_i$.
Then 

$\bullet$ {\bf \red $G_\C\cdot z$ is unstable if and only if all
$w_{l_k}$ are positive or negative,}

$\bullet$  {\bf \red  $G_\C\cdot z$ is stable if and only if some
$w_{l_k}$ are positive while others are negative.}

$\bullet$  {\bf \red $G_\C\cdot z$ is strictly semistable if some
$w_{l_k}$ vanish and all others are  positive or negative.}

\proof
If some
$w_{l_k}$ are positive while others are negative,
one has $\lim_{\im t \to \pm \infty} |e^{t\1 A}(z)| =\infty$,
hence $l$ reaches the minimum somewhere on the imaginary axis.

If all weights are positive or negative, one has
$\lim_{\im t \to \infty} |e^{t\1 A}(z)| =\infty$
and $\lim_{\im t \to -\infty} e^{t\1 A}(z) =0$
or vice versa, and the orbit is unstable.

If all weights are $\geq 0$ or $\leq 0$,
with one of the weights equal to 0,
one has
$\lim_{\im t \to \infty} |e^{t\1 A}(z)| =\infty$
and $\lim_{\im t \to -\infty} e^{t\1 A}(z) =z$,
and the orbit is strictly semistable.
\endproof

\newpage

{\bf \blue Set of stable orbits}

{\bf \purple For $G_\C=\C^*$, the following theorem immediately
follows from the Hilbert-Mumford criterion.}

\proposition
Let $G\subset U(n)$ be a Lie group acting 
on a complex Hermitian vector space $(V=\C^n, h)$, and 
$G_\C \subset GL(n, \C)$ its complexification,
and ${V_s}$ the set of all $z\in V$ such that
the orbit $G_\C \cdot z$ is stable.
{\bf \red Then ${V_s}\subset V$ is open.}

\pstep  Let $\bar B_R\subset V$ be a closed ball of radius $R$.
The orbit $z\in {V_s}$ is stable if and only if
for each $R\in \R^{>0}$, the intersection
$\bar B_R\cap G_\C \cdot z$ is compact for all $R\in \R^{>0}$.
Indeed, if it is compact, $l$ reaches minimum somewhere on
$G_\C \cdot z$. On the other hand, of $l$ reaches its
minimum, one has $\lim_{\im t \to \pm \infty} |e^{t\1 A}(z)| =\infty$
for all $\1 A \in \1 \g$, hence $\bar B_R\cap G_\C \cdot
z$ is compact.

{\bf \green Step 2:} If
$\lim_{\im t \to \pm \infty} |e^{t\1 A}(z)| =\infty$
for all $\1 A \in \1 \g$, one has
$\lim_{\im t \to \pm \infty} |e^{t\1 A}(z+u)| =\infty$
for $u$ sufficiently small, hence this condition is
open in $z\in V$.
\endproof

\newpage

{\bf \blue Geometric Invariant Theory}

\remark
The following theorem is identifies 
``the GIT reduction'' (taking a $G_\C$-quotient
of the union of all stable orbits) and the 
symplectic reduction.

\theorem
Let $G\subset U(n)$ be a Lie group acting 
on a complex Hermitian vector space $(V=\C^n, h)$, and 
$G_\C \subset GL(n, \C)$ its complexification.
Denote by $\mu:\; V \arrow \g^*$ the moment map,
$\mu(g, z):= (\Lie_{Ig}l)(z)$. {\bf \red Then
an orbit $G_\C \cdot z$ is stable if and only if
$G_\C \cdot z\cap\mu^{-1}(0)\neq 0$.} Moreover,
{\bf \red  $G_\C \cdot z\cap\mu^{-1}(0)$ is
precisely one $G$-orbit, and  $\mu^{-1}(0)/G = V_s/G_\C$,}
where $V_s\subset V$ is the union of all stable orbits.

\proof
$G_\C \cdot z$ is stable if and only if
$G_\C \cdot z\cap\mu^{-1}(0)\neq 0$ because $\mu^{-1}(0)$
intersects the orbit in the points where the length 
$l$ is minimal. This intersection set is precisely
one $G$orbit, which gives $\mu^{-1}(0)/G = V_s/G_\C$.
\endproof







\end{document}
