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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hyperkahler manifolds, \\[15mm]
\small lecture 3: Spinors}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, September 21, 2019 } \\ [15mm]
\url{http://bogomolov-lab.ru/KURSY/HK-2019/}
\end{center}

\newpage

{\bf \blue Clifford algebras}

\definition
{\bf \blue The Clifford algebra} of a vector space $V$ with
a scalar product $q$ is an algebra generated by $V$ with 
a relation $xy+yx = -2q(x,y) 1$, that is, a quotient of
$T^{\otimes} V:= k \oplus V \oplus
V\otimes V \oplus ... \oplus T^{\otimes i} V$ by an ideal
generated by $xy+ yx= -2g(x,y)$ for all $x,y\in V$.

\newcommand{\Mat}{\operatorname{Mat}}

\example 
{\bf \purple If $g=0$, Clifford algebra is Grassmann algebra.}

\claim {\bf \red $\dim \Cl(V,g)= 2^{\dim V}$.}

\proof Consider $\Cl(V,g)$ as a filtered algebra with
$r$-th term of filtration given by $r$-th power of $V\subset \Cl(V)$.
{\bf \purple Its associated graded algebra is Grassmann algebra.}
\endproof

\remark 
{\bf \purple The Clifford algebra is $\Z/2\Z$ graded:}
$\Cl(V,g)= \Cl_\even(V,g)\oplus \Cl_\odd(V,g)$.

\definition
Let $A=A_\even\oplus A_\odd$ be a graded associative algebra.
Let $A^\bot$ be the same vector space 
with new multiplication 
$a \bullet a':= (-1)^{\tilde a \tilde a'} aa'$.

\exercise
{\bf \purple Prove that $\Cl(V,g)^\bot= \Cl(V,-g)$.}

\newpage

{\bf \blue Graded tensor product}

\definition
Let $A:=A_\even\oplus A_\odd$,
 $B:=B_\even\oplus B_\odd$ be graded associative algebras. Define
{\bf \blue the graded tensor product}
$A\tilde\otimes B$ as $A\otimes B$ with multiplication given by
$a\otimes b \cdot a' \otimes b' = (-1)^{\tilde b\tilde{a'}} aa' \otimes bb'$,
where $\tilde x$ denotes the parity of $x$.

\example
Graded tensor product of Grassmann algebras 
{\bf \purple gives the Grassmann algebra of a direct sum:}
\[ \Lambda^* V \tilde \otimes \Lambda^* W \cong \Lambda^*(V \oplus W)\]

\example {\bf \red The same is true for Clifford algebras:}
\[ \Cl(V,g) \tilde\otimes \Cl(V',g') =\Cl(V\oplus V',g+ g'). \]


\newpage

{\bf \blue Pseudoscalar}

{\bf \green LEMMA (*):}
Let $A:=A_\even\oplus A_\odd$,
 $B:=B_\even\oplus B_\odd$ be graded associative algebras.
Suppose that $B$ contains an even element {\bf \blue (pseudoscalar)}
$\epsilon$ with the following properties:
\[ \epsilon^2=1, \epsilon b  = (-1)^{\tilde b} b\epsilon.\]
{\bf \red Then $A\tilde\otimes B\cong A\otimes B$}
(the graded tensor product is isomorphic to the usual one).

\proof
Consider a subalgebra  $A' \subset A\tilde\otimes B$
generated by elements $a \tilde\otimes \epsilon^{\tilde a}$
and $B' = 1\otimes B\subset A\tilde\otimes B$.
Then

1. {\bf \purple $A'\cong A$ commutes with $B' \cong B$.}

2. {\bf \purple $A'\otimes B'= A\tilde\otimes B$ as a vector space.}
\endproof

{\bf \green REMARK (*):}
If in the definition of pseudoscalar we replace $\epsilon^2=1$
by $\epsilon^2=-1$, {\bf \purple Lemma (*) will give
 $A\tilde\otimes B\cong A^\bot \otimes B$.}

\newpage

{\bf \blue Unit pseudoscalar}

Let $(V,g)$ be an oriented real vector space with orthogonal basis
 $e_1, ..., e_n$ such that $g(e_i, e_i) =\pm 1$.
{\bf \blue Unit pseudoscalar} in $\Cl(V,g)$ is 
$\epsilon:= e_1e_2e_3 ... e_n$.

\exercise
{\bf \purple Prove $\epsilon e_i = (-1)^{n-1} e_i \epsilon$.}

\exercise
Prove that {\bf \purple $\epsilon^2= (-1)^{\frac{(n-1)(n-2)}2}(-1)^q$
if $g$ has signature $(p,q)$.}

\remark This gives
\[
 \epsilon^2 = (-1)^{n(n-1)/2}(-1)^q = (-1)^{(p-q)(p-q-1)/2} =
\begin{cases}+1 & p-q \equiv 0,1 \mod{4}\\ -1 & p-q \equiv 2,3 \mod{4}.\end{cases}
\]

\definition Denote {\bf \blue the Clifford algebra of a real vector
space of signature $(p, q)$ by $\Cl(p, q)$.}

\corollary
{\bf \purple $\Cl(p+m,q+m')\cong \Cl(p,q) \otimes \Cl(m,m')$
when $m+m'$ is even, and $m-m' \equiv 0 \mod{4}$.}

\proof
The pseudoscalar  $\epsilon$ in $\Cl(m,m')$ 
satisfies $\epsilon^2=1$. Applying Lemma (*), we obtain
$\Cl(p,q) \otimes \Cl(m,m')\cong \Cl(p,q) \tilde\otimes \Cl(m,m')$.
Then we apply the isomorphism
 $\Cl(V,g) \tilde\otimes \Cl(V',g') =\Cl(V\oplus V',g+ g')$.
\endproof


\newpage

{\bf \blue Bott periodicity over $\C$}

\corollary
Let $A[i]$ denote the tensor product 
$A \otimes \Mat(i)\cong \Mat(i, A)$.
{\bf \red Then $\Cl(p+1,q+1) \cong \Cl(p,q)[2]$.}

\proof Use the previous corollary and an isomorphism
$\Cl(1,1)=\Mat(2, \R)$ {\bf \purple (prove it).} \endproof


\theorem {\bf \blue (Bott periodicity over $\C$)}\\
Clifford algebra $\Cl(V, q)$ of a complex vector space
$V=\C^n$ with $q$ non-degenerate {\bf \red is isomorphic to
$\Mat\left(\C^{n/2}\right)$ ($n$ even) and 
$\Mat\left(\C^{\frac{n-1}{2}}\right)\oplus \Mat\left(\C^{\frac{n-1}{2}}\right)$
($n$ odd).}

\proof
Use the previous corollary and isomorphisms
$\Cl(\C)=\C \oplus \C$, $\Cl(0)=\C$.
\endproof

\newpage

{\bf \blue Bott periodicity over $\R$.}

\corollary
{\bf \red $\Cl(p+m,q+m') \cong \Cl(q,p) \otimes \Cl(m,m')$
if $m+m'$ is even, and $m-m' \equiv 2 \mod{4}$.}

\proof
In $\Cl(m,m')$ the pseudoscalar $\epsilon$ satisgies 
$\epsilon^2=-1$. Applying Remark (*), we obtain
$\Cl(p,q)^\bot  \otimes \Cl(m,m')\cong \Cl(p,q)\tilde\otimes \Cl(m,m')
\cong\Cl(p+m,q+m')$. Then we use an isomorphism
$\Cl(p,q)^\bot=\Cl(p,q)$. \endproof

\corollary
{\bf \red 
$\Cl(p+2,q)\cong \Cl(q,p)[2]$ and $\Cl(p,q+2)\cong \Cl(q,p)\otimes {\Bbb H}$.}

\proof
We use the previous corollary and the isomorphisms
$\Cl(2,0)=\Mat(2, \R)$, $\Cl(0,2)={\Bbb H}$.
\endproof

\corollary {\bf \blue (Bott Periodicity modulo 4):}\\
The previous corollary immediately gives
$\Cl(p+4,q)\cong \Cl(q,p+2)[2]=\Cl(p,q) \otimes \Mat(2, {\Bbb H})$ 
and $\Cl(p,q+4)\cong \Cl(q+2,p)\otimes {\Bbb H}= 
\Cl(p,q) \otimes \Mat(2, {\Bbb H})$.


\newpage

{\bf \blue Bott periodicity modulo 8.}

\exercise
{\bf \purple 
Prove the isomorphism  ${\Bbb H} \otimes_\R {\Bbb H} = \Mat(4, \R)$.}


\corollary {\bf \blue (Bott Periodicity modulo 8):}\\
This isomorphism and the previous corollary give
 $\Cl(p+8,q) = \Cl(p,q)[16]$,
$\Cl(p,q+8) = \Cl(p,q)[16]$.
\[
\rowcolors{1}{red!10}{blue!10} 
\begin{array}{||c||c|c|c|c|c|c|c|c|c|}
\hline\noalign{\smallskip}\hiderowcolors
& 1 & 2 &3 &4 &5 &6 &7 &8 \\ \showrowcolors \noalign{\smallskip} \hline\noalign{\smallskip
}
\Cl(i,0) &\R^2 & \R[2] & \C[2] & {\Bbb H}[2] & {\Bbb H}[2]\oplus {\Bbb H}[2]& 
{\Bbb H}[4] & \C[8] & \R[16]\\\noalign{\smallskip} \hline\noalign{\smallskip}
\Cl(0,i) &\C & {\Bbb H} & {\Bbb H}\oplus {\Bbb H} & {\Bbb H}[2] & 
\C[4]& \R[8] & \R[8]\oplus\R[8] &\R[16]\\ \hline
\end{array}
\]

\newpage

{\bf \blue Pseudoscalar on an odd-dimensional space}

\definition
For any odd-dimensional space $V$, the pseudoscalar
$\epsilon= e_1e_2 ... e_{2n+1}$ commutes with 
a multiplication by generators of $\Cl(V)$,
hence defines an automorphism of $\Cl(V)$.
If $V$ were a complex vector space, we 
can always chose the basis $e_1,e_2, ... ,e_{2n+1}$
in such a way that $\epsilon^2=1$.
{\bf \purple This gives the eigenvalue decomposition
$\Cl(V)= \Cl^+(V) \oplus \Cl^-(V)$.}

\claim
{\bf \red Each of the algebras $\Cl^+(V)$, $\Cl^-(V)$
is isomorphic to $\Mat(\C^r)$.}

\proof
Eigenvalues of $\epsilon$ acting
on $\Cl(V)$ are equal to $\pm1$ because $\epsilon^2=1$.
On the other hand, an automorphism of $V$
which exchanges $e_1$ and $e_2$ maps $\epsilon$
to $-\epsilon$, hence permutes the eigenspaces.
Therefore, {\bf \purple the subalgebras
$\Cl^+(V)$, $\Cl^-(V)$ are isomorphic.}
We obtain that {\bf \purple the decomposition
$\Cl(V)=\Mat(2^n, \C) \oplus \Mat(2^n, \C)$
coincides with the eigenspace decomposition
defined by $\epsilon$.}

\remark
The center of $\Cl(V)$ is isomorphic to $\C \oplus \C$.
{\bf \purple The orthogonal group $O(V)$ acts on $\Cl(V)$ and on $\Z$ by
automorphisms, and maps $\epsilon$ to $\pm \epsilon$.}
In particular, {\bf \purple $SO(V)$ acts on $\Cl^\pm(V)$
by automorphisms}.

\newpage

{\bf \blue Automorphisms of matrix algebra}

\exercise Let $V$ be a vector space over a field
of characteristic 0.
Prove that {\bf \red the automorphism group
$\Aut(\Mat(V))$ is isomorphic to 
$PGL(V)$} (the quotient of $GL(V)$ by its center).



\newpage

{\bf \blue Spinorial group $\Spin(2n)$}

\definition
Let  $V=\C^{2n}$ be a vector space over $\C$ with non-degenerate
scalar product. The group $SO(V)$ acts on $\Cl(V)$
be automorphisms, giving an action
\[ SO(V) \hookrightarrow \Aut(\Mat(2^n, \C))= PGL(2^n, \C)\]
as shown above.

\definition
{\bf \green (Elie Cartan, 1913)}\\
{\bf \blue Spinor representation}
of the Lie algebra $\goth{so}(V)$
is its representation on $\C^{2^n}$ induced 
by the isomorphism $\goth{pgl}(2^n)= \goth{sl}(2^n)$.

\definition
{\bf \blue Spinor group} $\Spin(2n+1)$ is a 
double cover of $SO(2n+1)$ obtained as a
Lie group of $\goth{so}(V)$ acting
on its spinorial representation.


\newpage

{\bf \blue Spinorial group $\Spin(2n)$}

\definition
Let $V= \C^{2n}$ be a vector space over $\C$ with non-degenerate
scalar product. The group $SO(V)$ acts on $\Cl(V)$
be automorphisms, and defines a homomorphism
\[ SO(V) \hookrightarrow \Aut(\Mat(2^n, \C))= PGL(2^n, \C).\]
\definition 
{\bf \green (Elie Cartan, 1913)}\\
{\bf \blue Spinor representation}
of the Lie algebra $\goth{so}(V)$
is its representation on $\C^{2^n}$ induced 
by the isomorphism $\goth{pgl}(2^n)= \goth{sl}(2^n)$.

\definition
{\bf \blue Spinor group} $\Spin(2n)$ is a 
double cover of $SO(2n)$ obtained as a
Lie group of $\goth{so}(V)$ acting
on its spinorial representation.

\exercise
In even- and odd-dimensional case, prove that
{\bf \red $\Spin(r)$ is, indeed, a double cover of $SO(r)$.}

\newpage

{\bf\blue Principal bundles}

\definition
Let $G$ be a Lie group.
{\bf\blue Principal $G$-bundle} over a manifold $M$
is a smooth fibration $P\mapsto M$ with a smooth $G$-action
which acts freely and transitively on fibers.

\example
{\bf\blue Frame bundle} on a smooth $n$-manifold $M$
is the bundle of all frames (basises) in $T_x M$, for 
all $x\in M$. 

\definition
Let $H\arrow G$ be a group homomorphism, and
$P$ a principal $H$-bundle. Then the quotient
$P_G:=P \times G/H$ (with $H$ acting on both components
in a natural way) is called {\bf \blue an associated 
principal bundle}, and $P$ is called {\bf \blue
reduction of the principal $G$-bundle
$P_G$ to the group $H$}.

\definition
Let $G$ be a Lie group, and $G\arrow GL(n,\R)$
a group homomorphism. {\bf \blue A $G$-structure
on a manifold $M$} is a reduction of the principal
frame bundle to $G$.

\definition
Let $G$ be a Lie group, $V$ its representation,
and $P$ a principal $G$-bundle on $M$.
The quotient $P \times V/G$ is a vector bundle
over $M$, called {\bf\blue the associated vector bundle}.

\newpage



{\bf \blue Spin-structures and spinor bundles}

\definition {\bf \blue A spin-structure} on an oriented $n$-manifold $M$
is a reduction of its structure group to $\Spin(n)$.
A manifold is called {\bf \blue spin} if it admits
a spin-structure.

\remark {\bf \purple This happens precisely when the second 
Stiefel-Whitney class $w_2(M)$ vanishes.}

\definition
{\bf \blue A bundle of spinors} 
on a spin-manifold $M$ is a vector bundle
associated to the principal $\Spin(n)$-bundle
and a spin representation.

\remark
The Levi-Civita connection {\bf \purple is naturally extended
from a connection on the bundle of orthogonal
frames to its double cover.} This defines
the Levi-Civita connection on the spinor bundle.

\newpage



{\bf \blue Spinor bundles and Dirac operator}


\definition
Consider the map $TM \otimes \Spin \arrow \Spin$
induced by the Clifford multiplication.
One defines {\bf \blue the Dirac operator} 
$D:\; \Spin \arrow \Spin$
as a composition of 
$\nabla:\; \Spin \arrow\Lambda^1 M\otimes\Spin = TM \otimes\Spin$
and the multiplication.

\definition
A {\bf \blue harmonic spinor} is a spinor $\psi$ such that
$D(\psi)=0$.

\theorem 
(Bochner's vanishing)  A harmonic spinor $\psi$
on a compact manifold with vanishing scalar curvature
$Sc= Tr(\Ric)$ {\bf \red satisfies $\nabla\psi=0$.}

{\bf \green Proof:} The {\bf \blue coarse Laplacian} $\nabla^* \nabla$
is expressed through the Dirac operator using the {\bf \blue 
Lichnerowitz formula}
$\nabla^* \nabla - D^2 = - \frac 1 4 Sc$. 
When these two operators are equal, {\bf \red any 
harmonic spinor $\psi$ lies in $\ker\nabla^* \nabla$, giving
$(\psi, \nabla^* \nabla\psi) = (\nabla\psi, \nabla\psi)=0$.}
\endproof

\newpage


{\bf \blue Bochner's vanishing on Kaehler manifolds}

\remark {\bf \purple A Kaehler manifold is spin if and only if
$c_1(M)$ is even,} or, equivalently, if there exists
a square root of a canonical bundle $K^{1/2}$.

\remark On a Kaehler manifold of complex 
dimension $n$, {\bf \red one has a natural
isomorphism between the spinor bundle 
and $\Lambda^{*,0}(M)\otimes K^{1/2}$} (for $n$ even)
and $\Lambda^{2*,0}(M)\otimes K^{1/2}$ (for $n$ odd).

\remark On a K\"ahler manifold, the Dirac operator corresponds
to $\6 + \6^*$.

\corollary {\bf \purple On a Ricci-flat K\"ahler manifold,
all $\alpha \in \ker (\6 + \6^*)\restrict{\Lambda^{*,0}(M)}$
ara parallel.}

\remark $\ker \6 + \6^*= \ker \{\6, \6^*\}$, where $\{\cdot, \cdot\}$
denotes the anticommutator. However, $\{\6, \6^*\}= \{\bar\6, \bar\6^*\}$
as K\"ahler identities imply. Therefore, {\bf \purple on a K\"ahler
manifold, harmonic spinors are holomorphic forms}.

\theorem {\bf \blue (Bochner's vanishing)}
Let $M$ be a Ricci-flat Kaehler manifold,
and $\Omega\in \Lambda^{p,0}(M)$ a holomorphic
differential form. {\bf \red Then $\nabla \Omega=0$.}


\end{document}

