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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Complex geometry\\[15mm]
\small lecture 13: Cohomology and a circle action}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
 November 4, 2020
}
\end{center}


\newpage

{\bf \blue Stone-Weierstrass approximation theorem}

\definition
Let $M$ be a topological space, and
$\|f\|:= \sup_M |f|$ {\bf\blue the sup-norm on functions}.
{\bf \blue $C^0$-topology} on the space $C^0(M)$ of continuous, 
bounded real-valued functions is the topology defined by the sup-norm.

\exercise
Prove that {\bf \purple $C^0(M)$ with sup-norm is a complete metric space.}

\definition
Let  $A\subset C^0 M$ be a subspace in the space of
continuous functions. We say that $A$ {\bf\blue separates the points} of $M$
if for all distinct points $x, y\in M$, there exists $f\in A$ such that
$f(x) \neq f(y)$.


\theorem {\bf\blue  (Stone-Weierstrass theorem)}\\
Let $A\subset  C^0 M$ be a subring separating points,
and $\bar A$ its closure. {\bf \blue Then  $\bar A= C^0M$.}


\newpage

{\bf \blue Hilbert spaces}

\definition
{\bf \blue Hilbert space} over $\C$ 
is a complete, infinite-dimensional
Hermitian space which is second countable
(that is, has a countable dense set).

\definition
    {\bf \blue Orthonormal basis} in a Hilbert space $H$
    is a set of pairwise orthogonal vectors $\{x_\alpha\}$
    which satisfy $|x_\alpha|=1$, and such that $H$ is the
    closure of the subspace generated by the set $\{x_\alpha\}$.

\theorem
{\bf \red Any Hilbert space has a basis, 
and all such bases are countable.}

\proof
A basis is found using Zorn lemma. If it's not countable,
open balls with centers in $x_\alpha$ and radius $\epsilon < 2^{-1/2}$
don't intersect, which means that the second countability axiom is not satisfied.
\endproof

\theorem
    {\bf \red All Hilbert spaces are isometric}.

    \proof Each Hilbert space has a countable orthonormal basis.
    \endproof

\newpage

{\bf \blue Fourier series}

\example
Let $(M,\mu)$ be a space with measure. Consider 
the space $V$ of measurable functions
$f:\; M \arrow \C$ such that $\int_M |f|^2 \mu< \infty$.
For each $f, g \in V$, the integral $\int f\bar g\mu$ is well defined,
by Cauchy inequality: $\int |fg|\mu < \sqrt{\int_M |f|^2 \mu \int_M |g|^2 \mu}$.
This gives a Hermitian form on $V$.
Let $L^2(M)$ denote the completion of $V$ with respect to this metric.
It is called {\bf \blue the space of square-integrable functions
on $M$}. Its elements are called {\bf \red $L^2$-functions.}

\claim {\bf\blue ("Fourier series")}
Functions $e_k(t)= e^{ 2\pi\1 k t}$, $k\in \Z$ on $S^1=\R/\Z$
{\bf \red form an orthonormal basis in the Hilbert space $L^2(S^1)$}.

\pstep 
Orthogonality is clear from
$\int_{S^1} e^{ 2\pi\1 k t} dt =0$ for all $k\neq 0$ (prove it).

{\bf \green Step 2:}
The space of Fourier polynomials $\sum_{i=-n}^n a_k e_k(t)$
is dense in the space of continuous
functions on the circle by the Stone-Weierstrass approximation theorem.
Therefore, the closure of the space of functions 
which admit Fourier series is $L^2(S^1)$.
\endproof

\newpage

{\bf \blue Fourier series on a torus}

\remark
Let $t_1,..., t_n$ be coordinates on $\R^n$.
We can think of $t_i$ as of angle coordinates on the torus
$T^n=\R^n/\Z^n$, considered as a product of $n$ copies of $S^1$.
Consider the {\bf \blue Fourier monomials}
$F_{l_1, ..., l_n}:= \exp(2\pi\1\sum_{i=1}^n l_it_i)$,
where $l_1, ..., l_n$ are integers. Clearly,
\[ 
L^2(T^n)\cong \underbrace{L^2(S^1) 
\hat\otimes L^2(S^1)\hat \otimes ...\hat \otimes  L^2(S^1)}_{\text{$n$ times}}.
\]
where $\hat\otimes$ denotes the completed tensor product.
This implies that the {\bf \purple Fourier monomials form a Hilbert basis
in $L^2(T^n)$. }

\remark This also follows directly 
from the Stone-Weierstrass theorem.

\theorem
Let $V$ be a Hilbert space, $\Map(T^n, V)$ continuous maps,
and $L^2(T^n, V)$ a completion of $\Map(T^n, V)$
with respect to the $L^2$-norm $|v|^2= \int_{T^n} |v(x)|^2 dx$.
Consider an orthonormal basis $u_1, ..., u_n, ...$ in $V$.
Then {\bf \red an orthonormal basis in $\Map(T^n, V)$ 
is given by monomial maps $F_{l_1, ..., l_n} u_j$}
taking $s\in T^n$ to $F_{l_1, ..., l_n}(s)u_j$.

\proof
Orthonormality of the collection
$\{F_{l_1, ..., l_n} u_j\}$  is clear. 
To prove its completeness (that is, the density
of the subspace generated by 
$\{F_{l_1, ..., l_n} u_j\}$), notice that
$\Map(T^n, V)$ is a completion
of $\oplus_i \Map(T^n,V_i)$,
where $V_i=\langle v_i\rangle$.
Now, $\{F_{l_1, ..., l_n} u_i\}$
is an orthonormal basis in $V_i=\Map(T^n, \C)$.
\endproof

\newpage

{\bf \blue Weight decomposition for $U(1)$-representations}

\exercise
Let $\rho:\; U(1) \arrow GL(V)$ be a finite-dimensional 
irreducible complex representation of the Lie group $U(1)$.
{\bf \purple Prove that $\dim \C=1$ and there exists $n\in \Z$
such that $t\in U(1)=\R/\Z$ acts on $V$ as
$\rho(t)(v)= e^{2\pi\1 nt}v$.}

\definition
A representation of $U(1)$ 
with $\rho(t)(v)= e^{2\pi\1 nt}v$ 
is called {\bf \blue weight $n$} representation.

\definition
Let $V$ be a Hermitian space (possibly infinitely-dimensional)
equipped with an action of $U(1)$,
and $V_k\subset V$ weight $k$ representations, $k\in \Z$.
The direct sum $\bigoplus V_k$ is called {\bf \blue
the weight decomposition} for $V$ if it is dense in $V$.


\example
Let $L^2(S^1, W)$ the space of maps from $S^1$ to 
a Hermitian space $W$.
We define $U(1)$-action on $L^2(S^1, W)$
by $\rho(t)(f)= R_t(f)$
where $R_t(f(x))= f(x+t)$ shifts $S^1$ by $t$. Clearly,
this is a Hermitian representation,
and {\bf \purple its weight decomposition is its
Fourier decomposition.}


\newpage

{\bf \blue Weight decomposition for $U(1)$-representations (2)}

\claim
Let $\bigoplus V_k\subset V$ be the weight decomposition
of a Hermitian representation $\rho$ of $U(1)$.
Then {\bf \red any vector $v\in V$ can be decomposed 
onto a converging serie $v=\sum_{i\in \Z} v_i$, with
$v_i\in V_i$}. This decomposition is called
{\bf \blue the weight decomposition} for $v$.

\pstep
Clearly, all $V_i$ are pairwise orthogonal;
indeed, for any $t\in U(1)$ and $x_p\in V_p$, 
$x_q\in V_q$, $i\neq j$, we have
\begin{multline*} e^{2\pi\1 pt} (x_p, x_q)=
(\rho(t)(x_p), x_q)= (x_p, \rho(-t)x_q)= \\=
 (x_p, e^{-2\pi\1 qt}x_q)=e^{2\pi\1 qt}(x_p, x_q)
\end{multline*}
giving $p=q$ whenever $(x_p, x_q)\neq 0$.

{\bf \green Step 2:} 
Let $\pi_i:\; V \arrow V_i$ be the orthogonal projection.
Then $|x|^2 \geq \sum_{i=-p}^{p} |\pi_i(x)|^2$
because orthogonal projection is always distance-decreasing.
Therefore, the serie $\sum_{i\in \Z} \pi_i(x)$
converges. Its limit is a vector $x'$ which 
satisfies $(x, u)=(x', u)$ for any 
$u\in \bigoplus_{k\in \Z} V_k$. Since 
$\bigoplus_{k\in \Z} V_k$ is dense in $V$,
this implies $x=x'$.
\endproof

\newpage

{\bf \blue Weight decomposition and Fourier series}

\lemma
Let $W$ be a Hermitian representation of $U(1)$ admitting a
weight decomposition. Then {\bf \red any subquotient of $W$ also
admits a weight decomposition.}

\proof 
This is clear for quotients. Any closed subspace $V\subset W$
gives a direct sum decomposition $W=V\oplus V^\bot$, hence it
also can be realized as a quotient.
\endproof


\lemma
Let $\rho:\; U(1)\arrow U(W)$ be a Hermitian representation of $U(1)$,
and $L^2(S^1, W)$ the space of maps from $S^1$ to $W$ with the
$U(1)$-action by translation as defined earlier. 
{\bf \red Then $W$ can be realized
as a sub-representation of $L^2(S^1, W)$}.

\proof
For any $x\in W$ consider 
$\alpha_x\in L^2(S^1, W)$ taking
$t\in U(1)=\R/\Z$ to $\rho(t)(x)$.
Clearly, {\bf \purple $x\mapsto \alpha_x$
defines a homomorphism of representations.}
\endproof

\theorem
Let $W$ be a Hermitian representation of $U(1)$.
{\bf \red Then $W$ admits a weight decomposition
$W = \widehat{\bigoplus_{i\in \Z} W_i}$.}

\proof
We realize $W$ as a subrepresentation in
 $L^2(S^1, W)$, and use the Fourier series to
obtain the weight decomposition of  $L^2(S^1, W)$.
\endproof

\newpage

{\bf \blue Weight decomposition for $T^n$-action}

\exercise
Consider the $n$-dimensional torus $T^n$ as a Lie group,
$T^n = U(1)^n$. {\bf \purple Prove that any finite-dimensional 
Hermitian representation of $T^n$ is a direct sum
of 1-dimensional representations,} with action of $T^n$ given by 
$\rho(t_1, ..., t_n)(x)= \exp(2\pi\1\sum_{i=1}^np_it_i) x$,
for some $p_1, ..., p_n\in \Z^n$, called {\bf \blue the
weights} of the 1-dimensional representation.

\definition
Let $V$ be a Hermitian space (possibly infinitely-dimensional)
equipped with an action of $T^n$,
and $V_\alpha\subset V$ weight $\alpha$ representations, $\alpha\in \Z^n$.
The direct sum $\bigoplus_{\alpha\in \Z^n} V_\alpha$ is called {\bf \blue
the weight decomposition} for $V$ if it is dense in $V$.

\theorem
Let $W$ be a Hermitian vector space. Then
{\bf \red the Fourier series provide the weight decomposition
on $L^2(T^n, W)$.}
\endproof

\theorem
Let $W$ be a Hermitian representation of $T^n$.
{\bf \red Then $W$ admits a weight decomposition
$V = \widehat{\bigoplus_{\alpha\in \Z^n} W_\alpha}$.}

\proof
We realize $W$ as a subrepresentation in
 $L^2(T^n, W)$, and use the Fourier series to
obtain the weight decomposition of  $L^2(T^n, W)$.
\endproof


\newpage

{\bf \blue Weight decomposition for $T^n$-action on differential forms}

\remark
Let $M$ be a manifold with the $T^n$-action,
and \[ \Lambda^*(M)= \hat\bigoplus_{\alpha \in \Z^n} \Lambda^*(M)_{p_1, ..., p_k}\]
be the weight decomposition on the differential forms.
Then the {\bf \red de Rham differential preserves each term
$\Lambda^*(M)_{p_1, ..., p_k}$.} Indeed, {\bf \purple $d$ commutes
with the action of the Lie algebra of $T^n$,
and $\Lambda^*(M)_{p_1, ..., p_k}$ are its
eigenspaces.}


\remark
The weight decomposition $\alpha= \sum \alpha_{p_1, ...,  p_k}$ 
converges, generally speaking, only in $L^2$, however,
if action of $T^n$ is smooth, it converges uniformly
in $t\in T^n$ because {\bf \purple the Fourier serie of a smooth
function converge uniformly}.

\remark
Let $\alpha= \sum \alpha_{p_1, ..., p_k}$ be the weight decomposition.
{\bf \purple The forms $\alpha_{p_1, ..., p_k}$ 
are obtained by averaging
\[ 
e^{2\pi\1\sum_{i=1}^np_it_i}\alpha
= \Av_{T^n} e^{2\pi\1\sum_{i=1}^n-p_it_i}\alpha
\]
hence they are smooth.}


\newpage

{\bf \blue De Rham cohomology and $T^n$-action}


\theorem
Let $M$ be a smooth manifold, and $T^n$ a torus
acting on $M$ by diffeomorphisms. Denote
by $\Lambda^*(M)^{T^n}$ the complex of 
$T^n$-invariant differential forms.
{\bf \red Then the natural embedding
$\Lambda^*(M)^{T^n}\hookrightarrow \Lambda^*(M)$
induces an isomorphism on de Rham cohomology.}

\pstep
Let $\alpha\in \Lambda^*(M)$ be a form
and $\alpha= \sum \alpha_{p_1, ..., p_n}$ its weight
decomposition, with $\alpha_{p_1, ..., p_n}\in \Lambda^*_{p_1, ..., p_n}(M)$ 
a form of weight $p_1, ... , p_n$.
 Since $T^n$-action
commutes with de Rham differential, 
these forms are closed when $\alpha$ is closed.

{\bf \green Step 2:}
Let $r_1, ..., r_n$ be the standard 
generators of the Lie algebra of $T^n$
rescaled in such a way that 
$\Lie_{r_k}(\exp(2\pi\1\sum_{i=1}^np_it_i))=\1 p_k$,
and $i_{r_k}:\; \Lambda^i(M)\arrow \Lambda^{i-1}(M)$
the convolution operator.
Since $\Lie_{r_k}= \{d, i_{r_k}\}$, we have
$p_k\alpha_{p_1, ..., p_n}= d (i_{r_k}\alpha_{p_1, ..., p_n})$
whenever $\alpha_{p_1, ..., p_n}$ is closed.
Therefore, {\bf \purple all terms in the weight 
decomposition $\alpha= \sum \alpha_{p_1, ..., p_n}$ 
are exact except $\alpha_{0,0, ...,0}$.}

{\bf \green Step 3:}
In the direct sum decomposition of the de Rham complex
\[
\Lambda^*(M)= \Lambda^*(M)^{T^n}\oplus 
\hat\bigoplus_{p_1,  ..., p_k\neq (0,0,..., 0)}\Lambda^*_{p_1, ..., p_k}(M)
\]
the second component has trivial cohomology, because
$\Lie_{r_k}$ is invertible on 
$\bigoplus_{p_k\neq 0}\Lambda^*_{p_1, ..., p_n}(M)$
{\bf \purple (deduce it from
$p_k\alpha_{p_1, ..., p_k}= d (i_{r_k}\alpha_{p_1, ..., p_k})$),} and\\
$\Lie_{r_k}(\text{closed form})$ is exact.
\endproof


\end{document}





