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  \tiny M. Verbitsky }}
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf K\"ahler geometry \\[15mm]
\small lecture 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf University of Science and Technology China, Hefei
\\[2mm]  July 09, 2012
}
\end{center}

\newpage

{\bf \blue Some textbooks}

Arthur L. Besse, Einstein Manifolds, Springer-Verlag (Berlin, 1987).

P. Griffiths, J. Harris, Principles of Algebraic Geometry,
Wiley-Interscience, New York (1978).

D. Joyce, Compact manifolds with special holonomy, Oxford
Mathematical Monographs, Oxford University Press, 2000


Jean-Pierre Demailly,
Complex analytic and differential geometry, \\
\url {http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf}

Claire Voisin, Hodge theory and complex algebraic
geometry, Cambridge University Press, 2007

D. Huybrechts, Complex geometry: an introduction, Springer
Universitext, 2004







\newpage

{\bf \blue REMINDER: Complex structure on vector spaces}

\definition
Let $V$ be a vector space over $\R$, and $I:\; V \arrow V$
an automorphism which satisfies $I^2 = - \Id_V$. 
Such an automorphism is called {\bf\blue a complex structure
operator} on $V$.

\definition The vector space over $\C$ with the same
basis is called {\bf \blue a complexification} of $V$,
denoted $V \otimes_\R \C$.

\claim 
For an appropriate basis in $V \otimes_\R \C$, the
complex structure operatorcan be written as
{\scriptsize\[
I=\begin{pmatrix}
\1 &0&0 &\hdotsfor{1} &0&0&0\\
0&\1 &0 &\hdotsfor{1} &0&0&0\\
0&0&\1 &\hdotsfor{1} &0&0&0\\
\vdots&\vdots&\vdots&
\ddots
&\vdots&\vdots&\vdots\\
0&0&0 &\hdotsfor{1} &-\1&0&0\\
0&0&0 &\hdotsfor{1} &0&-\1&0\\
0&0&0 &\hdotsfor{1} &0&0&-\1
\end{pmatrix},
\]}
with the eigenspaces of equal dimension.



\newpage

{\bf \blue REMINDER: Hermitian structures}

\definition 
Let $(V,I)$ be a real vector space with a complex structure.
A scalar product is called {\bf \blue $I$-invariant},
if $g(Ix,Iy)=g(x,y)$. An $I$-invariant positive definite
scalar product on $(V,I)$ is called
{\bf \blue an Hermitian metric on $V$}, and $(V,I,g)$ -- 
an Hermitian space.


\remark
Let $I$ be a complex structure operator on a real 
vector space $V$, and $g$ -- a Hermitian metric. Then
{\bf \red
the bilinear form $\omega(x,y) := g(x, Iy)$ is skew-symmetric.}
Indeed, $\omega(x,y) = g(x, Iy) = g(Ix, I^2y) = -g(Ix, y) = -\omega(y, x)$.

\definition
A skew-symmetric form
 $\omega(x,y)$ is called {\bf \blue an Hermitian form on $(V,I)$}.

\remark 
In the triple $I, g, \omega$, each element can 
recovered from the other two.

\newpage

{\bf \blue REMINDER: tensor product}

\definition
Let $V, W$ be vector spaces over a field 
$k=\R$ or $\C$, and $V^*, W^*$ the dual spaces.
Denote by $\Bil(V\times W, k)$ the space of bilinear
maps from $V,W$ to $k$, and let $V \otimes W$ 
denote $\Bil(V^*\times W^*, k)$, where $V^*,W^*$ are
dual spaces to $V, W$. The space $V \otimes W$ is called
{\bf \blue the tensor product} of $V, W$. 

\definition Given $v, w\in V, W$, one has an element
$v\otimes w \in  \Bil(V^*\times W^*, k)$, mapping
a pair of functionals $\lambda\in V^*, \mu \in W^*$
to $\lambda(v)\mu(w)$. This vector is called {\bf \blue the tensor
product} of $v$ and $w$.

\claim
If $\{v_i, i=1,..., n\}$ is a basis in $V$,
$\{w_j, j = 1, ..., m\}$ a basis in $W$, {\bf \purple the vectors
$\{v_i \otimes w_j, i=1,..., n, j = 1, ..., m\}$
give a basis in $V \otimes W$.} In particuler,
{\bf \red $V \otimes W$ is $(\dim V \cdot \dim W)$-dimensional.}

\remark The tensor product is uniquely defined by the following
{\bf \blue universal property}. Each bilinear map
$B:\; (V, W) \arrow k$ can be extended uniquely
to the map $B^\otimes:\; V \otimes W \arrow k$, in such a way
that $B^\otimes(v \otimes w) = B(v,w)$.

\exercise {\bf \purple Prove this}.

\newpage

{\bf \blue REMINDER: tensor algebra}

\definition
Given several vector spaces $V_1, ..., V_n$, their {\bf \blue tensor product}
$V_1\otimes V_2\otimes ... \otimes V_n$ is defined as
$V_1\otimes (V_2 \otimes (V_2 \otimes ... V_n))...)$.

\claim 
The tensor product operation {\bf \red is commutative and associative.}
Moreover, {\bf \purple the space $V_1\otimes V_2\otimes ... \otimes V_n$
is isomorphic to the space $B(V_1^*, V_2^*, ..., V_n^*)$}
of polylinear maps from $V_1^*, V_2^*, ..., V_n^*$ to $k$.


\exercise {\bf \purple Prove this}.

\remark 
Given vectors in $v\in V_1\otimes V_2\otimes ... \otimes V_n$
and $w\in W_1\otimes W_2\otimes ... \otimes W_m$,
the tensor product $v \otimes w$ sits in 
$V_1\otimes V_2\otimes ... \otimes V_n\otimes
W_1\otimes W_2\otimes ... \otimes W_m$.

\claim This defines the structure of an algebra
on $T^\otimes V= k \oplus V \oplus V\otimes V \oplus ... \oplus V^{\otimes n}$,
where $V^{\otimes n}$ is a tensor product of $n$ copies of $V$.

\exercise {\bf \purple Prove this}.

\definition
The algebra $T^\otimes V$ is called {\bf \blue the tensor
  algebra}, or {\bf \blue free algebra} generated by $V$.


\newpage

{\bf \blue REMINDER: The Grassmann algebra}

\exercise Prove that {\bf \purple any algebra generated 
by $V$ can be obtained as a quotient of $T^\otimes V$ by
an ideal.}

\exercise
Let $v_1, ..., v_n\in V$ be a basis.
Prove that {\bf \purple the polynomial algebra $k[v_1, ..., v_n]$
is a quotient of $T^\otimes V$ by an ideal generated by 
$x\otimes y - y\otimes x$, for all $x,y \in V$.}

\definition
{\bf \blue A Grassmann algebra} 
$\Lambda^* V$ is a quotient of $T^\otimes V$ by an ideal generated by 
$x\otimes y + y\otimes x$, for all $x,y \in V$.
The multiplication in $\Lambda^* V$ is denoted by 
$x, y \arrow x\wedge y$, called {\bf \blue the
wedge product}.

{\bf \green Properties of Grassmann algebra:}

1. $\dim \Lambda^i V:= \binom{\dim V}{i}$, $\dim \Lambda^* V=2^{\dim V}$.

2. $\Lambda^*(V \oplus W)= \Lambda^*(V) \otimes \Lambda^*(W)$.

\newpage

{\bf \blue REMINDER: Vector fields}

\definition
Let $X$ be the vector field on a manifold $M$, and $f$ a
function. Denote by $\Lie_X f$ {\bf \blue the derivatiive} of $f$ along
$X$. 

\definition
{\bf \blue A derivation} on a commutative ring is a map
$R \stackrel d \arrow R$ satisfying {\bf \blue the Leibniz
  identity} $d(xy) = d(x) y + x d(y)$.

\theorem
Each derivation of the ring $C^\infty M$ of smooth 
functions on $M$ is given by a vector field $X$;
{\bf \red this correspondence is bijective.}

\remark This can be used as a definition of a vector field.

\exercise
Prove that {\bf \red a commutator of two derivations is again a
derivation.}

\remark Vector fields are the same as derivations of
$C^\infty M$.
This allows us to define {\bf \blue the commutator of two
vector fields} as the commutator of the corresponding derivations. 

\definition
Denote by $TM$ the bundle of vector fields, and by
$\Lambda^1 M$ or $T^*$ the dual bundle, called
{\bf \blue the bundle of 1-forms}.
For any $f\in C^\infty M$,
the operation $X \arrow \Lie_X f$ is linear as a function
of $X$, hence it defines a section of $T^*M$. We denote 
this section $df$, and call it {\bf \blue the 
differential} of $f$.


\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 Grassman 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}.

\exercise
{\bf \purple Prove it}.

\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 REMINDER: Complex manifolds}

{\bf\green DEFINITION:} Let $M$ be a smooth manifold. 
An {\bf \blue almost complex structure} is an operator
$I:\; TM \arrow TM$ which satisfies $I^2 = - \Id_{TM}$.

{\bf \purple The eigenvalues of this operator are $\pm \1$.}
The corresponding eigenvalue 
decomposition is denoted $TM=T^{0,1}M\oplus T^{1,0}(M)$.


{\bf\green DEFINITION:}
An almost complex structure is {\bf \blue integrable}
if $\forall X,Y \in T^{1,0}M$, one has $[X,Y]\in T^{1,0}M$.
In this case $I$ is called {\bf \blue a complex structure operator}.
A manifold with an integrable almost complex structure
is called {\bf \blue a complex manifold}. 

{\bf\green THEOREM:} (Newlander-Nirenberg)\\
{\bf \red This definition is equivalent to the usual one.}

\remark The commutator defines a $\C^\infty M$-linear map\\
$N:=\Lambda^2(T^{1,0})\arrow T^{0,1}M$, called {\bf \blue 
the Nijenhuis tensor} of $I$. {\bf \purple One can represent $N$ as a section 
of $\Lambda^{2,0}(M) \otimes T^{0,1}M$.}

\example {\bf \red Symmetric spaces.}

\example {\bf \red $\C P^n$.}


\newpage


{\bf \blue K\"ahler manifolds}

{\bf\green DEFINITION:} An Riemannian metric $g$ on
an almost complex manifiold $M$ is called 
{\bf \blue Hermitian} if $g(Ix, Iy)= g(x,y)$.
In this case, $g(x, Iy)= g(Ix, I^2y) = - g(y, Ix)$,
hence $\omega(x,y):= g(x, Iy)$ is skew-symmetric.

{\bf\green DEFINITION:} The differential 
form $\omega\in \Lambda^2(M)$ is called
{\bf \blue the Hermitian form} of $(M,I,g)$.

\remark It is $U(1)$-invariant, hence {\bf \purple of Hodge type (1,1)}.

{\bf\green DEFINITION:} A complex Hermitian manifold $(M,I,\omega)$
is called {\bf \blue K\"ahler} if $d\omega=0$. 
The cohomology class $[\omega]\in H^2(M)$ of a form $\omega$ 
is called {\bf \blue the K\"ahler class} of $M$, and
$\omega$ {\bf \blue the K\"ahler form}. 


{\bf \green Definition:} Let $M=\C P^n$ be a complex projective
space, and $g$ a $U(n+1)$-invariant Riemannian form. It is called
{\bf \blue Fubini-Study form on $\C P^n$}. The Fubini-Study
form is obtained by taking arbitrary Riemannian form
and averaging with $U(n+1)$ using the Haar measure on $U(n+1)$.

\exercise
Prove that {\bf \red the Fubini-Study form is unique} (up to a constant
multiplier).

\newpage

{\bf\blue Examples of K\"ahler manifolds.}

{\bf \green Remark:} For any $x\in \C P^n$, the stabilizer
$St(x)$ is isomorphic to $U(n)$. Fubini-Study form on
$T_x\C P^n= \C^n$ is $U(n)$-invariant, hence unique up to a constant.

{\bf \green Claim:} {\bf \red Fubini-Study form is K\"ahler.}
Indeed, $d\omega\restrict x$ is a $U(n)$-invariant 3-form
on $\C^n$, but such a form must vanish, because $-\Id\in U(n)$

\remark 
{\bf \purple The same argument works for all symmetric spaces.}

\definition
{\bf \blue An almost complex submanifold} $X \subset M$
of an almost complex manifold $(M,I)$ is a smooth submanifold
which satisfies $I(TX)\subset TX$.

\exercise
Let $X \subset M$ be an almost complex submanifold of $(M,I)$,
where $I$ is integrable. {\bf \purple Prove that $(X, I\restrict{TX})$
is a complex manifold.}

\definition
In this situation, $X$ is called {\bf \blue a complex submanifold} of $M$.

{\bf \green Corollary:} {\bf \red Every projective manifold 
(complex submanifold of $\C P^n$) is K\"ahler.}
Indeed, a restriction of a closed form is again closed.

\newpage

{\bf \blue Connections}

{\bf \green Notation:}
Let $M$ be a smooth manifold, $TM$ its tangent bundle,
$\Lambda^iM$ the bundle of differential $i$-forms,
$C^\infty M$ the smooth functions. {\bf \purple The space of sections 
of a bundle $B$ is denoted by $B$.}

\definition
A {\bf\blue connection} on a vector bundle $B$ is
a map $B \stackrel \nabla \arrow \Lambda^1 M \otimes B$ which
satisfies \[ \nabla(fb) = df \otimes b + f \nabla b\]
for all $b\in B$, $f\in C^\infty M$.

\remark A connection $\nabla$ on $B$ gives
a connection $B^* \stackrel {\nabla^*} \arrow \Lambda^1 M \otimes B^*$
on the dual bundle, by the formula
\[
d(\langle b, \beta\rangle) = \langle \nabla b, \beta\rangle+
\langle b, \nabla^*\beta\rangle
\]
These connections are usually denoted {\bf \red by the same letter $\nabla$.}

\remark
For any tensor bundle 
${\cal B}_1:=
B^*\otimes B^* \otimes ... \otimes B^* \otimes B\otimes B \otimes ... \otimes B$
{\bf \green a connection on $B$ defines a connection on ${\cal B}_1$}
using the Leibniz formula:
\[
\nabla(b_1 \otimes b_2) = \nabla(b_1) \otimes b_2 + b_1 \otimes \nabla(b_2).
\]

\newpage

{\bf \blue Torsion}

\definition
{\bf \blue The torsion} of a connection 
$\Lambda^1 \stackrel \nabla \arrow \Lambda^1 M \otimes \Lambda^1M$
is a map $\Alt \circ \nabla - d$,
where $\Alt:\;  \Lambda^1 M \otimes \Lambda^1M\arrow \Lambda^2 M$
is exterior multiplication. It is a map 
$T_\nabla:\; \Lambda^1M \arrow \Lambda^2 M$.

\exercise
{\bf \red Prove that torsion is a $C^\infty M$-linear.}

\remark
The dual operator $x, y \arrow \nabla_x Y - \nabla_y X-[X,Y]$
is also called {\bf \blue the torsion of $\nabla$}. It is a map
$\Lambda^2 TM \arrow TM$.

\exercise
{\bf \purple Prove that these two tensors are dual.}

\definition
Let $(M, g)$ be a Riemannian manifold. A connection $\nabla$ 
is called {\bf \blue orthogonal} if $\nabla(g) =0$.
It is called {\bf \blue Levi-Civita} if it is torsion-free.

\theorem (``the main theorem of differential geometry'')\\
{\bf \red For any Riemannian manifold, the
Levi-Civita connection exists,\\ and it is unique}.

\newpage

{\bf \blue Levi-Civita connection and K\"ahler geometry}

{\bf \green THEOREM:} Let $(M,I,g)$ be an almost complex Hermitian
manifold. {\bf \purple Then the following conditions are equivalent.}

(i) {\bf \red The complex structure $I$ is integrable, and 
the Hermitian form $\omega$ is closed.}

(ii) One has {\red $\nabla(I)=0$,} where $\nabla$ is the Levi-Civita connection.

\remark {\bf \purple The implication (ii) $\Rightarrow$ (i) is clear.}
Indeed, $[X,Y]=\nabla_X Y - \nabla_Y X$, hence it is
a $(1,0)$-vector field when $X, Y$ are of type (1,0), and 
then {\bf \purple $I$ is integrable}. Also, {\bf \purple $d\omega=0$, because
$\nabla$ is torsion-free,} and $d\omega= \Alt(\nabla\omega)$.

The implication (i) $\Rightarrow$ (ii) is proven by the same argument
as used to construct the Levi-Civita connection.

\newpage


{\bf \blue Holonomy group}

\definition (Cartan, 1923)
Let $(B,\nabla)$ be a vector bundle with connection over $M$.
For each loop $\gamma$ based in $x\in M$, let 
$V_{\gamma, \nabla}:\; B\restrict x \arrow B\restrict x$
be the corresponding parallel transport along the connection.
The {\bf \blue holonomy group} of $(B,\nabla)$
is a group generated by $V_{\gamma, \nabla}$,
for all loops $\gamma$. If one takes all contractible
loops instead, $V_{\gamma, \nabla}$ generates
{\bf \blue the local holonomy}, or {\bf \blue
the restricted holonomy} group.

\remark A bundle is {\bf \blue flat} (has vanishing curvature)
{\bf\purple if and only if its restricted holonomy vanishes.}

\remark If $\nabla(\phi)=0$ for some tensor 
$\phi\in B^{\otimes i}\otimes (B^*)^{\otimes j}$,
{\bf \red the holonomy group preserves $\phi$.}

\definition {\bf \blue Holonomy of a Riemannian manifold}
is holonomy of its Levi-Civita connection.

\example Holonomy of a Riemannian manifold lies in
$O(T_x M, g\restrict x)=O(n)$.

\example  Holonomy of a K\"ahler manifold lies in
$U(T_x M, g\restrict x, I \restrict x)=U(n)$.

\remark The holonomy group {\bf \red does not depend
on the choice of a point $x\in M$.}

\newpage

{\bf \blue Curvature of a connection}

Let $M$ be a   manifold, $B$ a bundle, $\Lambda^i M$ the
differential forms, and
$\nabla:\; B \arrow B \otimes \Lambda^1M$ 
a connection. We extend $\nabla$ to 
$B \otimes \Lambda^iM\stackrel\nabla \arrow B \otimes \Lambda^{i+1}M$ 
in a natural way, using the formula
\[
\nabla(b \otimes \eta) = \nabla(b)\wedge \eta + b \otimes d\eta,
\]
and define {\bf \blue the curvature $\Theta_\nabla$} of $\nabla$ as 
$\nabla\circ \nabla:\; B \arrow B\otimes \Lambda^2M$.

\claim {\bf \red This operator is $C^\infty M$-linear.}

\remark We shall consider $\Theta_\nabla$ as an element of 
$\Lambda^2M \otimes \End B$, that is, an $\End B$-valued 2-form.


\remark
Given vector fields $X, Y\in TM$, the curvature can be written
in terms of a connection as follows
\[
\Theta_\nabla(b)= \nabla_X\nabla_Yb - \nabla_Y\nabla_X B - \nabla_{[X,Y]}b.
\]

\claim Suppose that the structure group of $B$ is reduced
to its subgroup $G$, and let $\nabla$ be a connection
which preserves this reduction. This is the same as to say
that the connection form takes values in $\Lambda^1 \otimes \goth g(B)$.
{\bf \purple Then $\Theta_\nabla$ lies in $\Lambda^2M \otimes \goth g(B)$.}

\newpage



{\bf \blue The Lasso lemma}

\definition
A {\bf \blue lasso} is a loop of the following 
form:

{\begin{center}
\epsfig{file=Lasso.png,width=0.35\linewidth}
\end{center}}

The round part is called {\bf \blue a working part}
of a loop.

\remark {\bf \blue (``The Lasso Lemma'')} Let $\{U_i\}$ be a covering
of a manifold, and $\gamma$ a loop. Then {\bf \purple
any contractible loop  $\gamma$ is a product
of several lasso, with working part of each inside some $U_i$.}

\newpage


{\bf \blue The Ambrose-Singer theorem}

\definition
Let $(B, \nabla)$ be a bundle with connection,
$\Theta\in \Lambda^2(M)\otimes \End(B)$ its curvature,
and $a,b\in T_x M$ tangent vectors. An endomorphism
$\Theta(a,b)\in \End(B)\restrict x$ is called
{\bf \blue a curvature element}.

\theorem {\bf \blue (Ambrose-Singer)}
The restricted holonomy group of $B, \nabla$ at $z\in M$ 
is a Lie group, {\bf \red with its Lie algebra generated by all 
curvature elements $\Theta(a,b)\in \End(B)\restrict x$
transported to $z$ along all paths.}

\remark Its proof follows from the Lasso lemma.

\newpage

{\bf \blue Holonomy representation}

\definition
Let $(M,g)$ be a Riemannian manifold,
$G$ its holonomy group. A {\bf\blue holonomy representation} 
is the natural action of $G$ on $TM$.

\theorem (de Rham) 
Suppose that the holonomy representation is
not irreducible: $T_xM=V_1 \oplus V_2$.
Then $M$ locally splits as $M=M_1 \times M_2$,
with $V_1= TM_1$, $V_2=TM_2$.

{\bf\green Proof. Step 1:}
Using the parallel transform, we extend $V_1 \oplus V_2$
to a {\bf \purple splitting of vector bundles $TM = B_1 \oplus B_2$,
preserved by holonomy.}

{\bf\green Step 2:} The sub-bundles $B_1$, $B_2 \subset TM$
{\bf \purple are integrable:} $[B_1, B_1] \subset B_i$ 
(the Levi-Civita connection is torsion-free)

{\bf\green Step 3:} Taking the leaves of these
integrable distributions, {\bf \purple we obtain a local decomposition
$M=M_1 \times M_2$, with $V_1= TM_1$, $V_2=TM_2$. }


{\bf\green Step 4:} Since the splitting $TM = B_1 \oplus B_2$
is preserved by the connection, {\bf \purple the leaves $M_1, M_2$
are totally geodesic.} 

{\bf\green Step 5:} Therefore, {\bf \red locally $M$
splits (as a Riemannian manifold)}: \\
$M=M_1\times M_2$, where $M_1, M_2$ are any 
leaves of these foliations. \endproof


\newpage


{\bf \blue The de Rham splitting theorem}

\corollary 
Let $M$ be a Riemannian manifold, 
and $\Hol_0(M)\stackrel \rho \arrow \End(T_xM)$
a reduced holonomy representation. Suppose that $\rho$ is reducible:
$T_xM = V_1\oplus V_2 \oplus ...\oplus V_k$. {\bf \red Then $G=\Hol_0(M)$ also
splits: $G= G_1\times G_2 \times ...\times G_k$,}
with each $G_i$ acting trivially on all $V_j$ with $j\neq i$.

{\bf \green Proof:} Locally, this statement follows from
the local splitting of $M$ proven above. To obtain it globally
in $M$, use the Lasso Lemma. \endproof

\theorem (de Rham) A complete, simply connected  
Riemannian manifold with non-irreducible holonomy 
{\bf \red splits as a Riemannian product.}

\remark It is easy to find non-complete or non-simply
connected counterexamples to de Rham theorem.



\theorem (Simons, 1962)
Let $M$ be a manifold with irreducible holonomy.
{\bf \red Then either $M$ is locally symmetric, or $\Hol(M)$
acts transitively on the unit sphere in $T_xM$.}


\newpage

{\bf \blue Berger's theorem}

\theorem (Berger's theorem, 1955)
Let $G$ be an irreducible holonomy group of a
Riemannian manifold which is not locally symmetric. {\bf \red Then
$G$ belongs to the Berger's list:}

{
\begin{tabular}{|l|l|}
\hline
\multicolumn{2}{|c|}{\bf \color[rgb]{0,0,0.6}Berger's list}\\[1mm]
\hline
\it Holonomy  & \it Geometry\\[1mm]
\hline
$SO(n)$ acting on $\R^n$ & Riemannian manifolds\\[1mm]
\hline
$U(n)$ acting on $\R^{2n}$ & K\"ahler manifolds\\[1mm]
\hline
$SU(n)$ acting on $\R^{2n}$, $n>2$ & Calabi-Yau manifolds\\[1mm]
\hline
$Sp(n)$ acting on $\R^{4n}$ & hyperk\"ahler manifolds\\[1mm]
\hline
$Sp(n)\times Sp(1)/\{\pm 1\}$ & 
quaternionic-K\"ahler\\[1mm] acting on $\R^{4n}$, $n>1$ &  manifolds\\[1mm]
\hline
$G_2$ acting on $\R^7$ & $G_2$-manifolds \\[1mm]
\hline
$Spin(7)$ acting on $\R^8$ & $Spin(7)$-manifolds\\[1mm]
\hline
\end{tabular}
}


\remark There is one more group acting transitively on
a sphere: $Spin(9)$ acting on $S^{15}\subset \R^{16}$. In 1968, D. Alekseevsky
has shown  that {\bf \purple a manifold with holonomy $Spin(9)$ is automatically
locally symmetric.}

{\small \remark A similar list exists for non-orthogonal 
irreducible holonomy without torsion 
(Merkulov, Schwachh\"ofer, 1999).}



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