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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small lecture 7: Weitzenb\"ock formula}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, February 14, 2018 } 
\end{center}

\newpage

{\bf \blue REMINDER: de Rham algebra}

\definition
Let $\Lambda^* M$ denote the vector bundle
with the fiber $\Lambda^*T^*_xM$ at $x\in M$
($\Lambda^*T^*M$ is the Grassmann algebra of the cotangent
space $T^*_x M$). The sections of $\Lambda^i M$
are called {\bf \blue differential $i$-forms}.
 The algebraic operation ``wedge product'' defined
on differential forms is $C^\infty M$-linear; 
the space $\Lambda^* M$ of all differential forms
is called {\bf \blue the de Rham algebra}.

\remark $\Lambda^0 M = C^\infty M$.

\theorem
{\bf \red There exists a unique operator 
$C^\infty M\stackrel d \arrow \Lambda^1 M\stackrel d
\arrow \Lambda^2 M \stackrel d \arrow \Lambda^3 M \stackrel d \arrow ...$
satisfying the following properties}

1. On functions, $d$ is equal to the differential.\\
2. $d^2=0$ \\
3. $d(\eta \wedge \xi) = 
d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge
d(\xi)$,
where $\tilde \eta=0$ where $\eta\in \lambda^{2i}M$ is
{\bf \blue an
even form,} and $\eta\in \lambda^{2i+1}M$ is {\bf \blue odd.}


\definition
The operator $d$ is called {\bf \blue de Rham differential}.

\definition
A form $\eta$ is called {\bf \blue closed} if $d\eta=0$,
{\bf \blue exact} if $\eta in \im d$. The group $\frac{\ker d}{\im d}$
is called {\bf \blue de Rham cohomology} of $M$.

\newpage

{\bf \blue Supercommutator (reminder)}

\definition
A {\bf \blue supercommutator}  of pure operators
on a graded vector space is defined by a formula
$\{a,b\}= ab - (-1)^{\tilde a \tilde b}ba$.

\definition
A graded associative algebra is called {\bf \blue
graded commutative} (or ``supercommutative'')
if its supercommutator vanishes.

\example {\bf \purple The Grassmann algebra is supercommutative.}

\definition
{\bf \blue A graded Lie algebra} (Lie superalgebra)
is a graded vector space $\g^*$
equipped with a bilinear graded map 
$\{\cdot,\cdot\}:\; \g^*\times \g^* \arrow \g^*$
which is graded anticommutative:
$\{a,b\} = - (-1)^{\tilde a \tilde b}\{b,a\}$
and satisfies {\bf \blue the super Jacobi identity}
$\{c, \{a,b\}\} = \{\{c, a\},b\}+ (-1)^{\tilde a \tilde c}\{a,\{c, b\}\}$

\example
Consider the algebra $\End(A^*)$ of operators on a 
graded vector space, with supercommutator as above.
{\bf \purple Then $\End(A^*), \{\cdot,\cdot\}$ is a graded Lie algebra.}

{\bf \green Lemma 1:}
Let $d$ be an odd element of a Lie superalgebra, satisfying
$\{d,d\}=0$, and $L$ an even or odd element. {\bf \red Then $\{\{L, d\}, d\}=0$.}

{\bf \green Proof:} 
$0=\{L,\{d,d\}\}= \{\{L, d\}, d\}+
(-1)^{\tilde L}\{d,\{L, d\}\}=2\{\{L, d\}, d\}.$ \endproof

\newpage

{\bf \blue Hodge $*$ operator}

Let $V$ be a vector space. {\bf \blue A metric $g$ on $V$ induces
a natural metric on each of its tensor spaces:}
$g(x_1\otimes x_2 \otimes ... \otimes x_k, x_1'\otimes x_2' \otimes ... \otimes x_k') = g(
x_1, x'_1)g(x_2, x'_2) ... g(x_k, x'_k)$.

{\bf \purple This gives a natural positive definite scalar product
on differential forms over a Riemannian manifold $(M,g)$:}
$g(\alpha, \beta) := \int_M g(\alpha, \beta) \Vol_M$

Another non-degenerate 
form is provided by the {\bf \blue Poincare pairing:}\\
$\alpha, \beta \arrow \int_M \alpha \wedge \beta$.

\definition
Let $M$ be a Riemannian $n$-manifold.
Define {\bf \blue the Hodge $*$ operator}
$*:\; \Lambda^k M \arrow \Lambda^{n-k} M$ by the following relation:
{\bf \red $g(\alpha, \beta) = \int_M \alpha \wedge *\beta.$}

\remark {\bf \red The Hodge $*$ operator always exists.} It is defined
explicitly in an orthonormal basis $\xi_1, ..., \xi_n \in \Lambda^1 M$:
\[ * (\xi_{i_1}\wedge\xi_{i_2} \wedge ... \wedge\xi_{i_k}) =
(-1)^s \xi_{j_1}\wedge\xi_{j_2} \wedge ... \wedge\xi_{j_{n-k}},
\]
where $\xi_{j_1},\xi_{j_2}, ...,  \xi_{j_{n-k}}$ is a complementary
set of vectors to $\xi_{i_1}, \xi_{i_2}, ..., \xi_{i_k}$, and
$s$ the signature of a permutation $(i_1, ..., i_k, j_1, ..., j_{n-k})$.

\remark $*^2\restrict{\Lambda^k(M)}=(-1)^{k(n-k)}\Id_{\Lambda^k(M)}$

\newpage

{\bf \blue $d^*=(-1)^{nk}*d*$}

\claim
On a compact Riemannian $n$-manifold, one has
$d^*\restrict{\Lambda^k M} = (-1)^{nk} *d*$,
where $d^*$ denotes {\bf \blue the adjoint operator}, which
is defined by the equation $(d\alpha, \gamma) = (\alpha, d^*\gamma)$.

{\bf \green Proof:} Since 
\[ 
  0=\int_M d(\alpha\wedge \beta) = \int_M d(\alpha)\wedge \beta +
  (-1)^{\tilde \alpha} \alpha \wedge d(\beta),
\]
one has $(d\alpha, *\beta) = (-1)^{\tilde \alpha} (\alpha, *d\beta)$.
Setting $\gamma:= *\beta$, we obtain
\[
  (d\alpha, \gamma) = (-1)^{\tilde \alpha} (\alpha, *d(*)^{-1}\gamma)=
  (-1)^{\tilde \alpha} (-1)^{\tilde \alpha(\tilde n-\tilde\alpha)}
  (\alpha, *d*\gamma)= (-1)^{\tilde \alpha\tilde n}(\alpha, *d*\gamma).
\]
\endproof

\remark Since in all applications which we consider,
$n$ is even, {\bf \purple 
I would from now on ignore the sign $(-1)^{nk}$.}

\newpage

{\bf \blue Hodge theory}


\definition
The anticommutator $\Delta:=\{d, d^*\}= dd^* + d^* d$
is called {\bf \blue the Laplacian} of $M$. It is self-adjoint
and positive definite: $(\Delta x, x)= (dx, dx) + (d^*x, d^*x).$
Also, $\Delta$ commutes with $d$ and $d^*$ (Lemma 1).

\theorem {\bf \blue (The main theorem of Hodge theory)}\\
{\bf \red There is a basis in the Hilbert space $L^2(\Lambda^*(M))$
consisting of eigenvectors of $\Delta$.}

\theorem {\bf \blue (``Elliptic regularity for $\Delta$'')}
Let $\alpha\in L^2(\Lambda^k(M))$ be an eigenvector of $\Delta$.
{\bf \red Then $\alpha$ is a smooth $k$-form.}

{\bf \purple
These two theorems will be proven in the next lecture.}

\newpage

{\bf \blue De Rham cohomology (reminder)}

\definition 
The space 
$H^i(M):= \frac {\ker d\restrict{\Lambda^iM}}{d\left(\Lambda^{i-1}M\right)}$
is called {\bf \blue the de Rham cohomology of $M$}.

\definition
A form $\alpha$ is called {\bf\blue harmonic} if $\Delta(\alpha)=0$.

\remark Let $\alpha$ be a harmonic form. {\bf \red Then 
$(\Delta x, x)= (dx, dx) + (d^*x, d^*x),$} hence 
$\alpha \in \ker d \cap \ker d^*$

\remark {\bf \purple The projection ${\cal H}^i(M) \arrow H^i(M)$
from harmonic forms to cohomology is injective.} Indeed,
a form $\alpha$ lies in the kernel of such projection if
$\alpha=d\beta$, but then 
$(\alpha,\alpha)=(\alpha, d\beta) = (d^* \alpha, \beta) =0$.

\theorem
{\bf \red The natural map ${\cal H}^i(M) \arrow H^i(M)$ is an isomorphism}\\
(see the next page).

\remark {\bf \purple 
Poincare duality immediately follows from this theorem.}

\newpage

{\bf \blue Hodge theory and the cohomology (reminder)}

\theorem
{\bf \red The natural map ${\cal H}^i(M) \arrow H^i(M)$ is an isomorphism.}

{\bf \green Proof. Step 1:}
Since $d^2=0$ and $(d^*)^2=0$, one has 
$\{d, \Delta\}=0$.
This means that {\bf \red $\Delta$ commutes with the de Rham differential.}

{\bf \green Step 2:} Consider the eigenspace decomposition
$\Lambda^*(M) \tilde = \bigoplus_\alpha {\cal H}^*_\alpha(M)$,
where $\alpha$ runs through all eigenvalues of $\Delta$,
and ${\cal H}^*_\alpha(M)$ is the corresponding eigenspace.
{\bf \purple For each $\alpha$, de Rham differential defines a complex}
\[
{\cal H}^0_\alpha(M) \stackrel d \arrow 
{\cal H}^1_\alpha(M) \stackrel d \arrow 
{\cal H}^2_\alpha(M) \stackrel d \arrow ...
\]


{\bf \green Step 3:} On ${\cal H}^*_\alpha(M)$, one has
$dd^* + d^* d= \alpha$. When $\alpha \neq 0$, and $\eta$ closed,
this implies $dd^*(\eta) + d^* d(\eta)= dd^* \eta = \alpha\eta$,
hence $\eta= d\xi$, with $\xi:= \alpha^{-1}d^* \eta$.
This implies that {\bf \purple the complexes $({\cal H}^*_\alpha(M), d)$
don't contribute to cohomology.}

{\bf \green Step 4:} We have proven that
\[
H^*(\Lambda^* M, d) = \bigoplus_\alpha H^* ({\cal H}^*_\alpha(M),d)=
H^* ({\cal H}^*_0(M),d)={\cal H}^*(M).
\]
\endproof

\newpage

{\bf \blue The ring of symbols }

\theorem
Consider the filtration $\Diff^0(M)\subset
\Diff^1(M)\subset \Diff^2(M)\subset ...$ on the ring
of differential operators. Then
{\bf \red its associated graded ring is isomorphic to 
the ring $\bigoplus_i \Sym^i(TM)$.} \\
\proof Lecture 2. \endproof

\definition Let $D$ be a differential operator of order
$p$. Its class in $\Diff^p(M)/\Diff^{p-1}(M)$ is
called {\bf \blue symbol} of $D$. Symbol belongs
to $\Sym^p(TM)$. Similarly, for $D\in \Diff^p(F, G)$,
symbol is an element of 
$\Diff^p(F,G)/\Diff^{p-1}(F,G)=\Sym^p(TM)\otimes_{C^\infty M} \Hom(F, G)$.

\remark
$\symb(AB)=\symb(BA)$. Indeed, {\bf \purple the ring of symbols
\\ $\bigoplus_i \Diff^i(M)/\Diff^{i-1}(M)$ is commutative.}

\definition
Let $g\in \Sym^2(T^*M)$ be a Riemannian form. Using $g$
to identify $TM$ and $T^*M$, we can consider $g$
as an element in $\Sym^2(TM)$. 
This ``Riemannian bivector'' is denoted $g^{-1}$.

We are going to compute the symbol of the Laplacian operator and
the ``rough Laplacian'' $\nabla^*\nabla$.
Today we prove the following {\bf \blue ``Weitzenb\"ock formula'':}

\theorem
{\bf \red 
$\symb(\Delta) = \symb(\nabla^*\nabla)= g^{-1}\otimes \Id_{\Lambda^*(M)}$.}


\newpage

{\bf \blue Roland Weitzenb\"ock: 26 May 1885 - 24 July 1955}


\begin{center}
\epsfig{file=AmsterdamUniv1926_Korteweg_Weitzenbock_Sissingh.jpg,width=0.35\linewidth}\\[5mm]

{\green \scriptsize Left to right: Diederik Korteweg,
  Roland Weitzenb\"ock, \\
Remmelt Sissingh, 1926 in Amsterdam.}
\end{center}
{\em \scriptsize
...Weitzenb\"ock was elected member of the Royal Netherlands
Academy of Arts and Sciences (KNAW) in May 1924, but
suspended in May 1945 because of his attitude during the
war. Weitzenb\"ock had been a member of the National
Socialist Movement in the Netherlands.

In 1923 Weitzenb\"ock published a modern monograph on the
theory of invariants on manifolds that included tensor
calculus. In the Preface of this monograph one can read an
offensive acrostic. One finds that the first letter of the
first word in the first 21 sentences spell out:

    {\bf \red NIEDER MIT DEN FRANZOSEN}

He also published papers on torsion. In fact, in his paper
"Differential Invariants in Einstein's Theory of
Tele-parallelism" Weitzenb\"ock had given a supposedly
complete bibliography of papers on torsion without
mentioning \'Elie Cartan. 
}



\newpage

{\bf \blue Symbol of the connection}

\claim
Let $d:\; C^\infty M \arrow \Lambda^1 M$ be the differential.
{\bf \red Then its symbol $\symb(d)\in TM\otimes \Hom(C^\infty M, \Lambda^1 M)$
is identity:  
$\symb(d)=\Id_{\Lambda^1M}\in TM\otimes \Lambda^1 M=\End(\Lambda^1 M)$.}

\proof 
$d= \sum_i dx_i \frac{d}{dx_i}$, representing identity in
$\Lambda^1 M \otimes TM$.
\endproof

\remark
The same is true for the symbol of the connection 
$\nabla:\; B \arrow B\otimes \Lambda^1(TM)$:
\[ \symb(\nabla)=\Id_{\Lambda^1M}\otimes \Id_B
\]
Indeed, {\bf \purple in local coordinates the connection
is written as $\nabla=d + A$,} and $A$ is
a differential operator of order 0, hence it does
not contribute to $\symb$.

\exercise Let $D:\; B \arrow B\otimes \Lambda^1(TM)$
be a differential operator with $\symb(D)=\symb(\nabla)$.
{\bf \red Prove that it is a connection.}


\newpage

{\bf \blue Symbol of $d$ and $d^*$}

{\bf \green Claim 1:}
Let $A:\; F\arrow G$ be a linear operator,
and $D:\; G \arrow H$ a differential operator.
{\bf \purple Then $\symb(AD)=A(\symb(D))$.}
\endproof

{\bf \green Claim 2:}
Let $e:\; \Lambda^1(M)\otimes\Lambda^*(M)\arrow 
\Lambda^{*+1}(M)$ be the
multiplication operator, and $d:\; \Lambda^*(M)\arrow \Lambda^{*+1}(M)$
de Rham differential. 
{\bf \red Then $\symb(d)(\theta)= e(\theta)\in \End(\Lambda^*(M))$ for
any $\theta\in T^*M$.} Here we understand symbol as a map from $T^*(M)$
to $\End(\Lambda^*(M))$.

\proof In local coordinates, one has
$d= \sum_i e(dx_i) \frac{d}{dx_i}$.
\endproof

\definition
Let $i$ be the {\bf \blue ``interior multiplication'',}
\[
i:\; \Lambda^1(M)\otimes\Lambda^*(M)\arrow 
\Lambda^{*-1}(M)
\]
with $i(\theta):= (-1)^{nk} * e(\theta) *$. This is an operator
which takes a 1-form, uses Riemannian metric to produce
a vector field, and takes the convolution with this vector 
field.

\claim Let $d^*= (-1)^{nk} * d*$. Then
$\symb(d^*)(\theta)= i(\theta)\in \End(\Lambda^*(M))$.

\proof Follows from Claim 1, Claim 2.
\endproof

\newpage

{\bf \blue Symbol of the Laplacian}

\claim
Consider a Riemannian manifold $(M, g)$.
Let 
\[ e:\; \Lambda^1(M)\otimes\Lambda^*(M)\arrow 
\Lambda^{*+1}(M),\ \ \ \  i:\; \Lambda^1(M)\otimes\Lambda^*(M)\arrow 
\Lambda^{*-1}(M)
\] be the exterior and interior multiplication
operators defined above, and $x, y\in \Lambda^1 M$. 
{\bf \red Then the anticommutator 
$\{i_x, e_y\}$ is equal to a multiplication by
a function $\tilde g(x, y)$,} where $\tilde g=g^{-1}$
is the Riemannian form extended to $T^*M$ using the
natural isomorphism $T^*M=TM$.

\proof Let $x_1, ..., x_n$ be an orthonormal basis in $\Lambda^1(M)$.
Then $\{i_{x_1}, e_{x_1}\}$ takes a monomial $\alpha$ without $x_1$
to $i_{x_1} e_{x_1}\alpha= \alpha$ and takes a monomial
$x_1 \wedge \alpha$ to $e_{x_1} i_{x_1}(x_1 \wedge\alpha)= x_1 \wedge\alpha$.
Also, $i_{x_1}$ and $e_{x_2}$ anticommute on all monomials.
\endproof

\corollary
{\bf \red The symbol of $\Delta=\{d, d^*\}$, evaluated on $x\otimes y$, is equal
to $\{i_x, e_y\}= \tilde g(x, y)$.}

\proof Symbol is multiplicative:
$\symb(A)\symb(B) = \symb(AB)$. The symbol of $d$ is $e$, and
the symbol of $d^*$ is $i$. This gives
\[ \symb(\Delta)(x\otimes y) = 
   \{ \symb(d), \symb(d^*)\}(x\otimes y) = \{e_x, i_y\}= \tilde g(x, y).
\]
\endproof

\newpage

{\bf \blue Symbol of the rough Laplacian}

\claim
Consider a Riemannian manifold $(M, g)$, and let
$\nabla$ be a connection on a bundle $B$. {\bf \red Then
$\symb(\nabla^*\nabla)$, evaluated on 
$x\otimes y \in T^*(M)\otimes T^*(M)$, is equal to
$\tilde g(x,y)\Id_B$,} where $\tilde g=g^{-1}$
is the Riemannian form extended to $T^*M$ using the
natural isomorphism $T^*M=TM$.

\proof
The symbol of $\nabla$ takes $x\in T^*(M)$ to
$b \arrow b \otimes x$, and the symbol of
$\nabla^*:\; B \otimes \Lambda^1(M) \arrow B$ 
takes $x\in T^*(M)$ to an operator
$b\otimes y \arrow \tilde g(x, y) b$.
Using $\symb(A)\symb(B) = \symb(AB)$,
we obtain that $\symb(\nabla^* \nabla)$ evaluated
on $x\otimes y$ is equal to the multiplication by 
$\tilde g(x,y)$. \endproof


\end{document}