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# Basics on Hermitian Symmetric Spaces

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Diploma-Thesis
Basics on Hermitian Symmetric Spaces
Tobias Strubel
ETH ZÄurich,Wintersemester 2006/2007
2
Contents
0 Introduction 3
1 Basics 5
1.1 Basics in Riemannian Geometry.........................5
1.2 Riemannian Symmetric Spaces..........................13
1.3 Involutive Lie Algebras...............................19
1.4 Hermitian Symmetric Spaces...........................26
2 Some Topics on Symmetric Spaces 31
2.1 Complexi¯cation and Cartan Decomposition...................31
2.2 Totally Geodesic Submanifolds..........................32
2.3 Rank.........................................34
2.4 Bounded Symmetric Domains...........................35
2.5 Embedding.....................................36
2.5.1 The Borel Embedding...........................36
2.5.2 The Harish-Chandra Embedding.....................41
2.7 Symmetric Cones and Jordan Algebras......................50
2.7.1 Cones....................................50
2.7.2 Jordan Algebras..............................52
2.7.3 Correspondence between Cones and Jordan Algebras..........52
A Appendix 54
A.1 Riemannian geometry...............................54
A.2 Roots........................................56
A.3 Algebraic groups..................................57
A.4 Details........................................59
3
0 Introduction
Consider a space on which we have a notion of length and curvature.When should one call it
symmetric?Heuristically a straight line is more symmetric than the graph of x
5
¡x
4
¡9x
3
¡
3x
2
¡2x+7,a circle is more symmetric than an ellipse and a ball is more symmetric than an
egg.Why?Assume we have a ball and an egg,both with blue surface without pattern.If you
close your eyes and I rotate the ball,you can't reconstruct the rotation.But if I rotate the
egg,you can reconstruct at least the rotations along two axes.The reason is the curvature.
The curvature of the ball is the same in each point (i.e.constant),the curvature of the egg
is not.Should we call a space symmetric if its curvature is constant?No,because there are
to less.Every manifold of constant curvature can be obtained by one of the following:from
a n-dimensional sphere if the curvature is positive,from a n-dimensional euclidian space if
the curvature is zero and from a n-dimensional hyperbolic space if the curvature is negative.
So we go a step back and consider spaces on which a\derivative"of the curvature vanishes.
This gives an interesting class of spaces,called locally symmetric spaces.They are de¯ned as
Riemannian manifold,where the covariant derivative of the curvature tensor vanishes.We'll
show that this happens if and only if there exists for every point x a local isometry which
¯xes x and acts by multiplication by ¡1 on the tangent spaces at x.A Riemannian manifold
is called symmetric if this local symmetry extends to a global isometry.Surprisingly its group
of automorphisms acts transitively on the symmetric space and it is homogeneous.The study
and classi¯cation can be reduced to the study of Lie algebras equipped with an involutive
automorphism.We treat this relation explicitly in the ¯rst three sections.

Elie Cartan used
this fact in the early 20th century to classify them.In the 60th Koecher studied them via a
relation between symmetric cones and Jordan algebras.We will sketch this relation in Section
2.7.2.
This text should be an introduction to symmetric spaces readable for a third year student.
We presuppose only the knowledge of an introduction to di®erential geometry and basic
knowledge in the theory of Lie groups and Lie algebras.
The ¯rst chapter gives an introduction to the notion and the basic theory of symmetric
spaces.The ¯rst section is a short introduction to Riemannian geometry.We present there in
short the theory needed later in the text.In the second section we de¯ne symmetric spaces as
Riemannian manifold whose curvature tensor is invariant under parallel transport.Further
we give some examples and deduce from the de¯nition that a symmetric space has the form
G=K for a Lie group G (the automorphism group) and a compact subgroup K.The Lie
algebra g of G decomposes in a natural way into k +p.In the third section we discuss the
decomposition of Lie algebras,which can be uses to decompose symmetric spaces.The fourth
and last section treats Hermitian symmetric spaces.They are symmetric spaces which have
a complex structure.There can be characterized as the spaces G=K where the center of K is
non-trivial.
In the second chapter we treat some topics on symmetric spaces.The ¯rst section explains
the notion of a Cartan decomposition of a Lie algebra,which is generalization of decomposition
of g = k+p.The second and the third section introduce the rank.The next two chapters show
that Hermitian symmetric spaces of the non-compact type are exactly the bounded domains
in C
n
.In the sixth section gives an application of Hermitian symmetric spaces as the moduli
space of variations of Hodge structures on a vector space V.The last chapter explains a
one-to-one correspondence between algebraic objects (Jordan algebras) and geometric ones
(symmetric cones).
4
A very good introduction to Riemannian geometry is Lees Book [Le97].THE standard
book for symmetric spaces is Helgason [He78].It is complete and can be used as an intro-
duction to Riemannian geometry too,but it is technical and its structure is not very good.
A short and precise treatment of symmetric spaces can be found in [Bo98].Further good
introductions are the book of Wolf [Wo67] and the text of Kor¶anyi [Ko00].The latter is a
very nice introduction for the short reading.The texts [Ma06] and [Pa06] should be read
together.The ¯rst gives an introduction from the di®erential geometers point of view,the
latter from the algebraists point of view.Delignes course notes [De73] are very interesting,
since his way to symmetric spaces is di®erent from the usual ways.But the text is not so
easy to read,since the proof are discontinuous and he omitted big parts without telling what
he omits.
ZÄurich,March 2007 Tobias Strubel
.tstrubel@math.ethz.ch
1 Basics
1.1 Basics in Riemannian Geometry
In this section we will introduce notions and some general facts from Riemannian geometry.
We will need them for the de¯nition and the study of symmetric spaces.A very well readable
and detailed introduction is [Le97].
All manifolds and vector ¯elds are assumed to be C
1
.
De¯nition 1.1.1.
Let M be a manifold and X(M) the vector space of vector ¯elds on M.
An a±ne connection r assigns to each X 2 X(M) a linear mapping r
X
of X(M) into itself,
satisfying the following two conditions:
(i)
r
fX+gY
= fr
X
+gr
Y
;
(ii)
r
X
(fY ) = fr
X
Y +(Xf)Y.
for f;g 2 C
1
(M) and X;Y 2 X(M).The second condition is sometimes called Leibniz rule.
The operator r
X
is called covariant derivative along X.
Let r be a a±ne connection on M and'a di®eomorphism of M.Put
r
0
X
Y:= d'
¡1
r
d'X
d'Y:
One can easily check that this de¯nes a connection.The a±ne connection ris called invariant
under',if r
0
= r.In this case'is called an a±ne transformation of M.
A tensor of type (r;s) over a vector space V is an element of (V
¤
)

r
V

s
.A tensor of type
(r;s) over a manifold is a section in the bundle (TM
¤
)

r
(TM)

s
.By abuse of notation
we call them tensor,too.The tensors T of type (2;1) and R of type (3;1) are de¯ned by
T(X;Y ):= r
X
Y ¡r
Y
X ¡[X;Y ];R(X;Y ):= r
X
r
Y
¡r
Y
r
X
¡r
[X;Y ]
:
T is called torsion,R is the curvature.
Let (U;x) be a local chart on M.We write @
i
for
@
@x
i
and with this notation we can write
r
@
i
@
j
=
X
k
¡
k
ij
@
k
;
with ¡
k
ij
2 C
1
(U).The ¡
k
ij
are called Christo®el symbols.
Remark 1.1.2.
The vector (r
X
Y )
p
depends only on the values of X and Y on a neighbor-
hood of p.We show this for Y,for X the proof works similarly.Let Y and
~
Y be vector
¯elds which coincide on a neighborhood U of p.We want to prove that r
X
Y
p
= r
x
~
Y
p
.By
linearity this is true if and only if Y ¡
~
Y = 0 on U implies r
X
(Y ¡
~
Y )
p
= 0.Now let Y be
a vector ¯eld which vanishes on a neighborhood of p.We show that r
X
Y
p
= 0.To do this,
choose a bump function'with support in U and with'(p) = 1.Note that'Y ´ 0.With
the linearity we have r
X
('Y ) = 0.And by De¯nition 1.1.1 (ii) we get
0 = r
X
('Y ) = (Y ¢')X +'r
X
Y:
The ¯rst term of the right hand side is zero,since Y vanishes on the support of'.Therefore
(r
X
Y )
p
= 0.
6
Now we use this result to show,that (r
X
Y )
p
depends only on the values of Y in a
neighborhood of p and on X(p).We choose local coordinates and we can write X =
P
X
i
@
i
.
Again by linearity we can assume that the X
i
(p) = 0.Since we proved that (r
X
Y )
p
depends
only on the values of X and Y in a neighborhood of p we get with De¯nition 1.1.1 (ii)
(r
X
Y )
p
= (r
P
X
i
@
i
Y )
p
=
X
X
i
(p)(r
@
i
Y )
p
= 0:
De¯nition 1.1.3.
Let ° be a curve in M and X a vector ¯eld on M with X(°(t)) = _°(t).
Such a vector ¯eld exists always,because on can construct X with the help of local charts in
a neighborhood of ° and a partition of unity.A vector ¯eld Y is called parallel (along °) if
(r
X
Y )
°(t)
= 0 for all t.
A geodesic is a curve for which every vector ¯eld X with X(°(t)) = _°(t) is parallel along °,
i.e.
r
_°(t)
_°(t) = 0 8t:
Let ° be a curve.Avector ¯eld along ° is a smooth map Y:I ½ R!TM with Y (t) 2 T
°(t)
M
for all t.
Remark 1.1.4.
The de¯nition of parallelity makes sense,since by Remark 1.1.2 the vector
(r
X
Y )
p
depends only on Y and the value of X in p.Therefore the de¯nition is independent
of the extension of X on M.
Let ° be a curve and X and Y a vector ¯elds,where _°(t)
i
= X
i
(°(t)).In local coordinates
X and Y can be written as
P
X
i
@
i
respectively
P
Y
i
@
i
and we have
r
X
Y =
X
k
0
@
X
i
X
i
(@
i
Y
k
)@
k
+
X
i;j
X
i
Y
j
¡
k
ij
@
k
1
A
:
Since on ° the _°
i
(t) = X
i
(°(t)),for the vector ¯eld Y being parallel is equivalent to
dY
k
dt
+
X
i;j
¡
k
i;j
dx
i
dt
Y
j
= 0:(1)
for all k.
Proposition 1.1.5.
Let p 2 M and X 6= 0 in T
p
M.There exists a unique maximal geodesic
° in M such that
°(0) = 0 and _°(0) = X:
We will denote this geodesic by °
X
.
One can ¯nd straight from the de¯nitions of ¡
k
ij
and the geodesic a system of di®erential
equation for a geodesic (the geodesic equation):
d
2
x
k
dt
2
+
X
i;j
¡
k
ij
dx
i
dt
dx
j
dt
= 0:
By the existence and uniqueness theorem for linear di®erential equations it has a unique
solution.See [He78,Ch.I.6] for details.
7
Remark 1.1.6.
Let X 2 T
p
M and ° be a geodesic with °(0) = p.Then there exists a vector
¯eld along ° with X(°(0)),which is parallel along °.Indeed let (x
1
;:::;x
m
) be a local chart
with (x
1
(p);:::;x
m
(p)) = 0.By the existence and uniqueness theoremfor ordinary di®erential
equations and Equation (1) we know that there exists,for every vector X = (X
1
;:::;X
m
) in
T
p
M,functions
Y
i
(t) ='
i
(t;X
1
;:::;X
m
)
with Y
i
(0) ='
i
(0;X
1
;:::;X
m
)) = X
i
.The vector ¯eld (Y
1
;:::;Y
m
) is therefore parallel
with respect to °.The mapping Y (0) 7!Y (t) is linear,since'
i
(0;X
1
;:::;X
m
) is linear
in X
1
;:::;X
m
and'
i
is unique.From the uniqueness also follows that Y (0) 7!Y (t) is in
isomorphism from T
°(0)
M to T
°(t)
M.We call this map parallel transport (see [He78,Ch.
I.5]).
Let X be a vector ¯eld and denote by ¿
s;p
the parallel transport from T
p
M to T
°(s)
M along
the geodesic °
X(p)
.Let A be a tensor of type (l;k).Since ¿
s;p
is an isomorphism between
T
p
M and T
°(s)
M we can transport the tensor A
p
on T
p
M to T
°(0)
M via
(¿
s;p
A
p
)(a
1
;:::;a
l
;b
1
;:::;b
k
):= A
p
(¿
¡1
s;p
a
1
;:::;¿
¡1
s;p
a
l
;¿
¤
s;p
b
1
;:::;¿
¤
s;p
b
k
);
for a
1
;:::;a
l
2 T
°(s)
M and b
1
;:::;b
k
2 T
¤
°(s)
M.De¯ne
(r
X
A)
p
:= lim
s!0
1
s
((¿
¡1
s;p
A
°(s)
) ¡A
p
):
This is the covariant derivative for tensor ¯elds.A tensor is called invariant under parallel
transport,if ¿
s
A = A for all s.This is the case if and only if r
X
A = 0 for all vector ¯elds X.
De¯nition 1.1.7.
Let p be a point in M.For every X 2 T
p
M there exists a geodesic °
X
.
This de¯nes a function from T
p
M to M
exp
p
:X 7!°
X
(1)
if °
X
(1) is de¯ned.It is called the exponential map.Its di®erential exists and it can be seen
as a map T
p
M!T
p
M and a direct calculation shows that it is the identity.By the local
inversion theorem exp
p
is a local di®eomorphism between non-empty open sets U ½ T
p
M and
V ½ M.Its inverse is noted log
p
.
A neighborhood N
0
of 0 2 T
p
M is called normal if exp
p
is a di®eomorphism of N
0
onto an
open neighborhood N
p
of p and for all X 2 N
0
and 0 · t · 1 then tX 2 N
0
.N
p
is also called
normal.
For a more precise treatment of the exponential map see [He78,Ch.I.6] or [Le97,Ch.5].
Let X be a vector ¯eld.De¯ne the 1-forms
!
i
(X
j
):= ±
i
j
!
i
j
=
X
k
¡
i
kj
!
k
:
The following theorem shows that these two forms are determined by the torsion and cur-
vature tensor.Since they de¯ne the connection r,the torsion and curvature tensor determine
the connection.
8
Theorem 1.1.8.
Let T be the torsion tensor and R the curvature tensor.Then
d!
i
= ¡
X
p
!
i
p
^!
p
+
1
2
X
j;k
T
i
jk
!
j
^!
k
d!
i
l
= ¡
X
p
!
i
p
^!
p
l
+
1
2
X
j;k
R
i
ljk
!
j
^!
k
:
These equations are called the Cartan structure equations.
Let Y
1
;:::;Y
m
be a basis for the tangent space T
p
M and let N
0
be a normal neighborhood of
the origin in T
p
M and N
p
:= expN
0
.Let Y
¤
1
;:::;Y
¤
m
be the vector ¯elds on N
p
Y
1
;:::;Y
m
n
!M be given by
1
;:::;a
m
)!exp(ta
1
Y
1
+:::+ta
m
Y
m
):
¤
!
i
¤
!
j
i
are given by
¤
!
i
= a
i
dt + ¹!
i

¤
!
i
l
= ¹!
i
l
;
where ¹!
i
and ¹!
i
j
are 1-forms in da
1
;:::;da
m
.They are given by
@¹!
i
@t
= da
i
+
X
k
a
k
¹!
i
k
+
X
j;k
T
i
jk
a
j
¹!
k
:¹!
i
t=0
(t:aj;da
k
) = 0:(2)
@¹!
i
l
@t
=
X
j;k
R
i
ljk
a
j
¹!
k
¹!
i
l
(t;a
j
;da
k
)
t=0
= 0 (3)
For a proof see [He78,I.8].
Theorem1.1.9.
Let M and M
0
be di®erentiable manifolds with a±ne connections r and r
0
.
Assume that the torsion tensors T and T
0
and the curvature tensors R and R
0
are invariant
under parallel transport,i.e.:
rT = 0 = r
0
T
0
;rR = 0 = r
0
R
0
:
Let x 2 M and x
0
2 M
0
and A:T
x
M!T
x
0
M
0
be an isomorphism of the tangent spaces
at x and x
0
sending T and R onto T
0
and R
0
.Then there exists a local isomorphism
1
a:
(M;r;x)!(M
0
;r
0
;x
0
) whose di®erential in x is A.
Proof.
De¯ne
a(y):= exp
x
0
(Alog
x
y):
It seems to be a good candidate for our local isomorphism.It is in fact a local di®eomorphism.
Now we have to check,that it is an a±ne transformation.Before we show that,we take a
closer look on what it means to be an a±ne transformation.On M respectively on M
0
we
have
r
X
i
X
j
=
X
¡
k
ij
X
k
r
0
X
0
i
X
0
j
=
X
¡
0k
ij
X
0
k
;
1
An isomorphism is a di®eomorphism which is an a±ne transformation in both directions.
9
if X
i
and X
0
i
are local basis'for vector ¯elds in M respectively M
0
.A map a:M!M
0
is an
a±ne transformation if and only if
r
0
X
Y = dar
da
¡1
X
da
¡1
Y:
If X
0
i
= daX
i
and a is a±ne,we have
X
k
¡
k
ij
X
k
= r
X
i
X
j
= r
da
¡1
X
0
i
da
¡1
X
0
j
= da
¡1
³
X
¡
0k
ij
X
0
k
´
= da
¡1
³
r
0
X
0
i
X
j
´
=
X

0k
ij
± a)X
k
:
hence ¡
0k
ij
±a = ¡
k
ij
is a su±cient and necessary condition for a to be an a±ne transformation.
Lets check this for our a.
Choose normal neighborhoods N
0
and N
0
0
of the origins in T
x
M and T
x
0
M
0
and a basis
Y
1
;:::;Y
m
of T
x
M.Put N
x
:= exp(N
0
) and N
0
x
0
:= expN
0
0
.Then AY
1
;:::;AY
n
is a basis
of T
x
0
M
0
and the coe±cients of T and T
0
with respect to these basis'are constant,since all
derivatives vanishes by rT = 0.Furthermore they are the same for T and T
0
since A maps
T to T
0
.This also holds for the coe±cients of R and R
0
.Put
1
;:::;a
m
) = exp
x
(ta
1
Y
1
+¢ ¢ ¢ ta
m
Y
m
) (4)
0
(t;a
1
;:::;a
m
) = exp
x
0
(ta
1
Y
1
+¢ ¢ ¢ ta
m
Y
m
):(5)
0
= a ± ©.The Cartan structure equations say
¤
!
i
= a
i
dt + ¹!
i
0
)
¤
!
0
i
= a
i
dt + ¹!
0i
¤
!
i
j
= ¹!
i
j
0
)
¤
!
0
i
j
= ¹!
0i
j
By equations (2) and (3) in Theorem1.1.8 ¹!
i
= ¹!
0i
and ¹!
0i
j
= ¹!
i
j
,since both sides are solutions
of the same di®erential equation because the coe±cients of T and T
0
respectively R and R
0
are the same.With (4),(5) and ©

¤
± a
¤
we get
¤
!
i
= a
i
dt + ¹!
i
= a
i
dt + ¹!
0i
0
)
¤
!
0
i
¤
± a
¤
(!
0
i
)
and with a similar calculation
¤
!
i
j
¤
± a
¤
!
0
i
j
:
Putting t = 1 we obtain
!
i
= a
¤
(!
0
i
);!
i
j
= a
¤
(!
0
i
j
):
Finally we have
X
k
¡
i
kj
!
k
=!
i
j
= a
¤
(!
0
i
j
) =
X
k

0
i
kj
± a)a
¤
(!
0
k
) =
X
k

0
i
kj
± a)!
k
;
so ¡
i
kj
= ¡
0i
kj
± a.
The theorem above is Lemma IV.1.2 in [He78].
10
Proposition 1.1.10.
Under the hypotheses of Theorem 1.1.9 assume that M and M
0
are
simply connected and geodesically complete.Then a can be continued to a unique global
isomorphism.
Proof.
Let a:= exp
0
p
(Alog
p
) and x be a point in M.We are searching for a point y in M
0
,
such that y = a(x).Since M is geodesically complete,we can join p and x by a geodesic °
(see Theorem A.1.1),such that p = °(0) and x = °(1).A transports the initial vector in
T
p
M to a vector in T
p
0
M which generates a geodesic ~° in M
0
.Put a(x):= ~°(1).Since M and
M
0
are simply connected,this is wellde¯ned and unique.See [He78,Thm.IV.5.6] or [Wo67,
Ch.1.8] for details.
Corollary 1.1.11.
Let (M;r) be such that rT = 0 = rR and M simply connected and
geodesically complete.Let °(t) be a geodesic (i.e.r
_°(t)
_°(t) = 0 for every t).There exists a
unique one-parameter group of automorphisms ¿(°;u) of M that induces translation °(t) 7!
°(t +u) on ° and displaces the tangent bundle on ° by parallel transport.
Proof.
Fix t 2 R and denote by#the parallel transport from T
°(t)
M to T
°(t+u)
.Since#is
an isomorphism we have by Theorem 1.1.9 and Proposition 1.1.10 a unique automorphism
¿(°;u) which maps °(t) to °(t+u) and d¿(°;u)
°(t)
=#.Being an automorphism¿(°;u) maps
geodesics on geodesics.Since it acts as parallel transport fromT
°(t)
M to T
°(t)
M if maps _°(t)
to _°(t +u),hence ° on itself.Using the geodesic equation one can show that ¿(°;u) maps
°(s) to °(s +u) for all s.By the uniqueness of the solution of the Equation (1) ¿(°;u) acts
by parallel transport on the tangent bundle of °.
Remark 1.1.12.
Conversely if such a one-parameter group of automorphisms exists,M is
geodesically complete.If a geodesic ° exists on [¡1;1],then we have °(n +a) = ¿(°;n)°(a),
if n 2 Z and a 2 [0;1).
Corollary 1.1.13.
Let M a manifold with an a±ne connection r.The following conditions
are equivalent:
(i)
T = 0 = rR;
(ii)
for all x 2 M there exist s
x
,a local automorphism of (M;r) inducing t 7!¡t on T
x
M.
Proof.(i))ii)
Use Theorem 1.1.9 with A = ¡1.
(ii))i)
Fix x 2 M.s:= s
x
is a local automorphism of (M;r),i.e.we have
(r
X
Y )
y
= (ds
¡1
(r
dsX
dsY ))
s(y)
for all vector¯elds X and Y and y in a neighborhood of x.Using this fact and
the de¯nition of R and T we see that r
Z
R(X;Y ) = ds
¡1
r
ds
¡1
Z
R(dsX;dsY )ds and
T(X;Y ) = ds
¡1
T(dsX;dsY ).ds
x
= ¡1,hence
T(X;Y )
x
= ¡T(X;Y )
x
r
Z
R(X;Y )
x
= ¡r
Z
R(X;Y )
x
;
since r is invariant under s.But x was arbitrary,so rR ´ 0 ´ T.
11
De¯nition 1.1.14.
Let M be a manifold and g a (2;0) tensor ¯eld on M such that g
p
is
positive de¯nite for all p 2 M,i.e.g
p
is a scalar product on T
p
M.Such a g is a Riemannian
metric.A connection r on a manifold equipped with a Riemannian metric is called Levi-
Civita-Connection
2
if
(i)
r
X
Y ¡r
Y
X = [X;Y ] (torsion free)
(ii)
r
X
g ´ 0 for all X (invariant under parallel transport).
By a Riemannian manifold we mean a triple (M;g;r) where M is a connected manifold,g
a riemannian metric and r a Levi-Civita-Connection.Usually we simply write M.
Remark 1.1.15.
The condition (ii) for the Levi-Civita Connection is equivalent to X ¢
g(Y;Z) = g(r
X
Y;Z) +g(Y;r
X
Z):
The Levi-Civita Connection exists and it is unique.From the de¯nition one can get the
formula
2g(X;r
Z
Y ) = Z¢g(X;Y )+g(Z;[X;Y ])+Y ¢g(X;Y )+g(Y;[X;Y ])¡X¢g(Y;Z)¡g(X;[Y;Z]):
This formula and the fact that g is non-degenerate can be used to prove uniqueness and
existence (see [Le97,Ch.5]).
The tensor g is called metric,because one can use it to measure the length of a path ° by
L(°):=
Z
°
q
g
°(t)
( _°(t);_°(t))dt:
For p;q 2 M we de¯ne
d(p;q):= inffL(°)j ° a path joining p and qg:
This de¯nes a distance function on M.
De¯nition 1.1.16.
Let M be a Riemannian manifold and p 2 M a point.Let S be a
two-dimensional subspace of T
p
M.The sectional curvature is de¯ned by
K(S) = ¡
g
p
(R
p
(Y;Z)Y;Z)
jY ^Zj
2
;
(c.f.[He78,Thm.I.12.2]) where Y and Z are linearly independent vectors in S and jY ^ Zj
denotes the area of the parallelogram spanned by Y and Z.
For details see [He78,I.12]).
For the proof of the next proposition we need a technical lemma,it is Lemma I.12.4 from
[He78].
Lemma 1.1.17.
Let E be a vector space over F (¯eld of characteristic 0).Suppose B:
E £E £E £E!F is quadrilinear and satis¯es the identities
(i)
B(X;Y;Z;T) = ¡B(Y;X;Z;T);
(ii)
B(X;Y;Z;T) = ¡B(X;Y;T;Z);
2
Sometimes it is called Riemannian connection.
12
(iii)
B(X;Y;Z;T) +B(Y;Z;X;T) +B(Z;X;Y;T) = 0;
(iv)
B(X;Y;X;Y ) = 0;
then B ´ 0.
Proof.
The proof is a pure calculation without tricks.
Proposition 1.1.18.
Let M be a Riemannian manifold.The following conditions are equiv-
alent:
(i)
rR = 0;
(ii)
the sectional curvature is invariant under parallel transport;
(iii)
for all x 2 M,y 7!exp
x
(¡log
x
(y)) is a local isometry.
Proof.
(i) ) (ii) Let p 2 M and S a two dimensional subvectorspace of T
p
M.The sectional
curvature K in p is de¯ned as
K(S) = ¡
g
p
(R
p
(Y;Z)Y;Z)
jY ^Zj
2
:
The metric g and the curvature R are invariant under parallel transport (since rR ´ 0) and
the same holds for the length of the two vectors and the angle between them,since length
and angle are measured with the help of the invariant g.Hence K is invariant under parallel
transport.
ii) ) iii) Denote by ¿ the parallel transport from p to q along a geodesic.By the invariance
of K we have
g
p
(R
p
(X;Y )X;Y ) = g
q
(R
q
(¿X;¿Y )¿X;¿Y ) (6)
and by the invariance of g
g
p
(R
p
(X;Y )X;Y ) = g
q
(¿R
p
(X;Y )X;¿Y ) (7)
holds for every X;Y 2 T
p
M.Let
B(X;Y;Z;T):= g
q
(R
q
(¿X;¿Y )¿Z;¿T) ¡g
q
(¿R
p
(X;Y )Z;¿T)
If we showthat B ´ 0,then ¿R
p
= R
q
,hence rR = 0,since g is non-degenerate.To showB ´
0 we use Lemma 1.1.17.Lets check the assumptions.(i) is true,since R is skew-symmetric.
Checking (ii) needs a straightforward calculation which uses the de¯nition of R and the fact
that g is invariant under parallel transport and hence Z¢ g(X;Y ) = g(r
Z
X;Y ) +g(X;r
Z
Y ).
(iii) is exactly the Bianchi-identity.(iv) is clear since the di®erence of the equations (6) and
(7) is zero.By Corollary 1.1.13'(y):= exp
q
(¡log
q
(y)) is a local automorphism of (M;r).
It remains to show,that it is an isometry.Multiplication with ¡1(= d'
q
) is an isometry on
T
q
M.Then
g
p
(X;Y ) = g
q
(¿X;¿Y ) = g
'(q)
(d'
q
¿X;d'
q
¿Y ):
This last equals g
'(p)
(d'
p
X;d'
q
Y ) because'is an automorphism of (U;r),i.e.the parallel
transport ¿ commutes with d'.
(iii) ) (i) follows from Corollary 1.1.13.
13
1.2 Riemannian Symmetric Spaces
De¯nition 1.2.1.
If a Riemannian manifold M satis¯es the equivalent conditions of Propo-
sition 1.1.18,it is called locally symmetric.We call the map y 7!exp
x
(¡log
x
y) the geodesic
symmetry in x.If M is connected and for all x 2 M there exists an isometric involution s
x
with x as an isolated ¯xed point,M is called (Riemannian) symmetric.
Remark 1.2.2.
Since x is an isolated ¯xed point,ds
x
j
x
acts as multiplication with ¡1 on
the tangent space in x.
A symmetric space is complete.Indeed note ¯rst that for every point x on a geodesic ° the
symmetry s
x
maps the geodesic on itself because s
x
is an isometry.Therefore every geodesic
is either closed or can be continued to in¯nity.By the Hopf-Rinow-Theorem (Theorem A.1.1)
two arbitrary points can be joined by a geodesic arc and M is complete.Conversely if M is
locally symmetric,complete and simply connected it is symmetric,because by Hopf-Rinow-
Theorem the completeness implies that M is geodesically closed and therefore by Proposition
1.1.10 such a local symmetry extends to a global one.
Let M be a complete locally symmetric space.Its universal cover is locally symmetric too.
Since it is complete an simply connected,it is globally symmetric.Therefore complete,locally
connected spaces are quotients of global ones.
Example 1.2.3.
The euclidian spaces E
n
,the spheres S
n
and the hyperbolic spaces H
n
seen
as Riemannian manifolds are symmetric spaces.There curvatures are constant,hence they are
locally symmetric by Proposition 1.1.18.They are simply connected and complete,therefore
symmetric.For n = 2 it is easy to see how the geodesic symmetry acts on the space.Since it
acts as multiplication with ¡1 on the tangent space,s
p
is the rotation by ¼ around p in E
2
.
In S
2
the symmetry is rotation by ¼ around the axis through p and ¡p.If one looks at the
upper half plane model for H
2
,it is the inversion at the halfcircle through p perpendicular to
the x-axis with radius =p,composed with inversion at the geodesic line parallel to the y-axis
through p.
Example 1.2.4.
Let P:= P(n;R),the set of the symmetric positive de¯nite matrices in
SL(n;R).It is clearly an open subset of the vector space of the symmetric matrices.Its
dimension is n(n +1)=2.First we equip it with a Riemannian metric.On T
p
P (which is the
vector space of symmetric matrices) we de¯ne a scalar product by
< X;Y >
p
:= tr(p
¡1
Xp
¡1
Y ):
This is a Riemannian metric.
SL(n;R) acts on P via
g ¢ p:= gpg
>
;
where g 2 SL(n;R) and p 2 P.The di®erential is X 7!gXg
>
.The metric given above is
invariant under this action since
< gXg
>
;gY g
>
>
g¢p
= tr((gpg
>
)
¡1
gXg
>
(gpg
>
)
¡1
gY g
>
) = tr(g
>
p
¡1
Xp
¡1
Y (g
>
)
¡1
)
= tr(p
¡1
Xp
¡1
Y ) =< X;Y >
p
:
This explains why we de¯ned the metric in this way.
Now we want for each point p 2 P a symmetry s
p
.Put s
p
(q):= p(q
>
)
¡1
p.It is clearly an
involution which ¯xes p.We need to show that s
p
is an automorphism.It is a composition
14
of the maps ¾:q 7!q
¡1
and q 7!p ¢ q.We showed above,that the latter is an isometry.To
show that ¾ is an isometry we calculate its derivative.Let X 2 T
p
P be a tangent vector.It
is the initial vector of a curve p +tX.Let p
¡1
+t
¹
X be it inverse curve.Then
1 = (p +tX)(p
¡1
+t
¹
X) = 1 +t(Xp
¡1
+p
¹
X) +O(t
2
);
hence d¾
p
(X) =
¹
X = ¡p
¡1
Xp
¡1
.Therefore
<
¹
X;
¹
Y >
p
¡1
= tr(p
¹
Xp
¹
Y ) = tr(Xp
¡1
Y p
¡1
) = tr(p
¡1
Xp
¡1
Y ) =< X;Y >
p
;
where we have the general fact tr(AB) = tr(BA).The di®erential of the map q 7!p ¢ q is the
map itself and the composition of the two di®erentials is multiplication with ¡1.Hence P is
a symmetric space.
Proposition 1.2.5.
Let M be a Riemannian symmetric space ° be a geodesic.Denote by s
x
the symmetry in x 2 M.We write s
t
for s
°(t)
.Then:
i)
s
t
(°(t +u)) = °(t ¡u);
ii)
¿(°;u) = s
u=2
± s
0
and
s
t
± ¿(°;u) ± s
¡1
t
= ¿(°;¡u):
Proof.i)
The geodesics °(t+u) and °(t¡u) are generated by _°(t) respectively by ¡_°(t) =
ds
t
_°(t).By uniqueness of this geodesics we have s
t
(°(t +u)) = °(t ¡u).
ii)
With i):s
u
(°(t)) = s
u
(°(u+(t¡u))) = °(u¡(t¡u)) = °(2u¡t).Hence s
u=2
±s
0
(°(t)) =
s
u=2
(°(¡t)) = °(u+t) = ¿(°;u)(t).And s
t
±¿(°;u);±s
t
(°(r)) = s
t
±¿(°;u)(°(2t¡r)) =
s
t
(°(2t ¡r +u) = °(r ¡u) = ¿(°;¡u)°(r).
Let M be a symmetric Riemannian space and G its group of isometries.We endow G with
the compact-open topology It is generated by sets W(C;U):= ff 2 Gjf(C) ½ Ug,where
C ½ G is compact and U ½ G is open.It is the coarsest topology such that the evaluation
map is constant.It turns G into a topological group which acts continuously on M.It is a
¯nite dimensional Lie group (see Theorem A.1.3).For details on this topology see [Ha02,p.
529f] and [He78,Ch.IV.2].
It acts transitively on M.Indeed take two points p and q and join them by a geodesic °
so that p = °(0) and q = °(1).The symmetry in °(1=2) maps one to the other,since it is an
isometry.Further ¿(°;1) maps p to q.Since ¿ is a one parameter subgroup of G,(write ¿
u
for ¿(°;u)) for u;v 2 R ¿
v+u
= ¿
v
± ¿
u
.
Now we want to show that the identity component of G,denoted by G
0
,acts transitively
on M.To do that let V =
T
l
i=1
W(C
i
;U
i
) be an open subset of the identity component which
contains the identity.Since e 2 V we have C
i
½ U
i
.The metric d(x;¿
u
(x)) measures how far
a point x is moved by ¿
u
.It depends continuously of x,because d and ¿
u
are continuous in
x.It has for a ¯xed u a maximum on C
i
,since C
i
is compact.This maximum is continuous
as a function of u.We call this function m and we have m(0) = 0.Furthermore we can cover
C
i
by a ¯nite number of open balls which are contained in U
i
and such that the open balls
15
with half radius still cover C
i
.Denote by r the minimal radius of these balls.If one chooses
u
i
such that m(u
i
) < r=2,one has ¿
u
i
(C
i
) ½ U
i
.For all 0 < u < u
i
for all i the map ¿
u
is
contained in V.Since ¿ is a one-parameter group,we have ¿
u+v
= ¿
u
± ¿
v
.Therefore G
0
acts
transitively on M.
Let K be the stabilizer of a point x 2 M.It is compact (Theorem A.1.2).Hence
M = G=K (see Theorem A.1.4).On G there exists a natural involution ¾:g 7!s
x
gs
x
.For
k 2 K we have d(s
x
ks
x
) = (¡1)dk(¡1) = dk and since K injects into O(T
x
M) (Lemma
A.1.5) we have s
x
ks
x
= k and all elements of K are left ¯xed by ¾.Therefore d¾ is the
identity on k:= Lie(K).Let p ½ Lie(G) be the set of in¯nitesimal generators of the one
parameter subgroups ¿(°;u),where ° is a geodesic through x.The involution ¾ acts on p by
multiplication with ¡1 because of Proposition 1.2.5 (ii).The vector space p can be identi¯ed
with T
x
M,because for two geodesics ° and ~° with °(0) = ~°(0) and _°(0) = s
_
~°(0) with s > 0
we have °(t) = ~°(st) and therefore ¿(¾;u) = ¿(~°;u=s).Hence ¿(°;u) is uniquely determined
by a vector in T
x
M,its direction gives ° and its length u and vice versa.By dimension reasons
we have g = k ©p where p is the eigenspace of ¾ to the eigenvalue ¡1 and k is the eigenspace
to the eigenvalue 1.If not otherwise stated,in the following G denotes the isometry group of
the Riemannian symmetric space M and K the stabilizer of a point p 2 M.The Lie algebra
g of G admits always the decomposition k ©p with respect to the involution ¾.
This shows
Theorem1.2.6.
Let M be a symmetric space and G its group of isometries.Then M = G=K
where K is the stabilizer of a point x 2 M.
Example 1.2.7.
Lets look how our examples are as homogeneous spaces.The isometry
group of R
n
is R
n
nO(n) and we have O(n +1) for S
n
.For H
n
we have O(n;1)
+
.It is the
subgroup of O(n;1) of index 2,which preserves the upper sheet H
n
.The group of isometries
of P(n;R) is SL(n;R).The point-stabilizer in the ¯rst three cases is O(n),in the last case it
is SO(n).Therefore we have:
R
n
= R
n
nO(n)=O(n)
S
n
= O(n +1)=O(n)
H
n
= O(n;1)
+
=O(n)
P(n;R) = SL(n;R)=SO(n)
For details for the calculation of the isometry groups see [Br99,Ch.I].
Remark 1.2.8.
Now we want to see how we can build a symmetric space from a given Lie
group G and a compact subgroup K.We assume that
(i)
M = G=K is connected and
(ii)
there exists an involution ¾ of G such that Lie(G)

e
= Lie(K) and K
¾
= K.
We want to put a structure of a Riemannian symmetric space on M.First we choose Q a G-
invariant Riemannian structure.Such a structure exists always.We can de¯ne a K-invariant
inner product on T
e
M via
< a;b >=
Z
K
(dk ¢ a;dk ¢ b)d¹(k);
16
where (¢;¢) is an arbitrary positive de¯nite bilinear form on T
e
M and d¹(k) is the Haar-
measure
3
on K.The integral exists since K is compact and acts continuously on M.Now
h
gK
(X;Y ):=< dg
¡1
X;dg
¡1
Y >
de¯nes an invariant riemannian structure on M.It is well-de¯ned since < ¢;¢ >is K-invariant,
hence it does not depend on the representant of gK.
Now we are searching for a suitable point-symmetry for our M.Our ¾ is an involution,
therefore g decomposes into the direct sum k +p,where k is the eigenspace to the eigenvalue
1 (and the Lie algebra of K) and p is the eigenspace to the eigenvalue ¡1.Therefore it
has eK as an isolated ¯xed point and it is a symmetry in eK.Since M is homogeneous
this gives a symmetry for every gK 2 M,putting s
pK
(gK):= p¾(p
¡1
q)K.It is clearly an
involution and it is an automorphism of M,since ¾ and multiplication with an element of G
are automorphisms.
In Proposition 1.3.10 we will see that if G is semi-simple,connected and acts faithfully on
M,it is the identity component of the group of isometries of M.
Example 1.2.9.
Let G = SO(2;n) (with n > 3) the group of matrices in GL(2+n;R) which
leave the bilinear form of type (2;n) invariant,which is given by
(xjy):= x
1
y
1
+x
2
y
2
¡x
3
y
3
¡:::¡x
n+2
y
n+2
for x = (x
1
;:::;x
n+2
) and y = (y
1
;:::;y
n+2
).This form is de¯ned with respect to a basis
(e
1
;:::;e
2+n
).We can use this basis to de¯ne the standard scalar product and therefore a
norm on R
2
and R
n
.We denote the scalar product by h¢;¢i and the norm by k ¢ k.
The Lie algebra of G consists of matrices of the form
µ
X
1
X
2
X
>
2
X
3

where X
1
2 M(2;2;R) skew-symmetric,X
2
2 M
(
2;n;R) arbitrary and X
3
2 M
(
n;n;R)
skew-symmetric.One sees directly,that the subalgebra k consisting of matrices
µ
X
1
0
0 X
3

is compact.The corresponding Lie group K is S(O(2) £ O(n)).Let ¾:G!G the map
which maps g to (g
>
)
¡1
.It is clearly an involution.The ¯xed points of ¾ are exactely the
matrices in K.Its di®erential acts as X 7!¡X
>
on the Lie algebra g of G.The ¯xed points
are exactly the matrices in k.Therefore G=K is a symmetric space.
Amodel of this symmetric space is the space G of the two dimensional subspaces of R
2
n
on which (¢j¢) is positive de¯nite.The group G acts on it linearly.We show now that the
action is transitive.We have clearly V
0
:= R
2
©f0g 2 G.If we can show,that every point in G
can be mapped to V
0
,we are done.Let V 2 G.By V
?
we denote the orthogonal complement
of V with respect to (¢j¢).Since this bilinear form is non-degenerated on V we can decompose
R
2
n
?
.If we show,that our bilinear form is negative de¯nite on V
?
,we can
¯nd a orthogonal basis fw
1
;:::;w
n
g of V
>
and a orthogonal basis fv
1
;v
2
g of V.The matrix
(v
1
;v
2
;w
1
;:::;w
n
) is by orthogonality in SO(2;n) and it maps V
0
to V.Hence the action is
3
The Haar-measure exists since K is compact.
17
transitive.It remains to show that,(¢j¢) is negative de¯nite on V
>
.Lets do this:let fv
1
;v
2
g
be a orthogonal basis of V.Lets decompose them as v
1
= »
1
+ ´
1
and v
2
= »
2
+ ´
2
with
»
i
2 R
2
and ´
i
2 R
n
and we have k»
i
k > k´
i
k.A vector » +´ 2 V
>
with » 2 R
2
and ´ 2 R
n
satis¯es (u
1
+u
2
j®v
1
+¯v
2
) = 0,hence h»;®»
1
+¯´
2
i = h´;®´
1
+¯´
2
i.Since f»
1

2
g form a
basis of R
2
we have
k»k = sup
k®»
1
+¯»
2
k
jh»;®»
1
+¯»
2
ij = sup
k®»
1
+¯»
2
k
jh´;®´
1
+¯´
2
j
· sup
k®»
1
+¯»
2
k
k´k ¢ k®´
1
+¯´
2
k < sup
k®»
1
+¯»
2
k
k´k ¢ k®»
1
+¯»
2
k = k´k:
We used in the ¯rst step a characterization of the norm,the third step is the Cauchy-Schwarz
inequality and the fourth step uses ®v
1
+¯v
2
2 V.It is important that we have there\<"
and not\·"!
The space G can be realized as a subset of M(2;n;R).Let v = v
1
+ v
2
2 V with
v
1
2 R
2
and v
2
2 R
n
.The projection on R
2
is injective since v
1
= 0 if and only if v
2
= 0.
By dimension reasons it is also surjective and we can choose a basis f»
1

2
g of V with
»
1
= f1;0;Z
11
;:::;Z
1n
g and »
2
= f0;1;Z
21
;:::;Z
2n
g.By de¯nition of G we have
P
j
Z
2
ij
< 1
for j = 1;2.Conversely every 2 £n matrix de¯nes an element of G via
V:= f(v;Zv)jv 2 R
2
g:
Therefore we have a one-to-one correspondence between G and the set of matrices M:= fZ =
(Z
1
;Z
2
) 2 M
2;n
(R)jZ
1
¢ Z
1
< 1;Z
2
¢ Z
2
< 1g.
Example 1.2.10.
Let G:= SL(n;R),K:= SO(n;R) and ¾:G!G with ¾:g 7!(g
>
)
¡1
.
M:= G=K is connected,since G and K are.¾ is an involution and its ¯x-point set is by
de¯nition exactly K.If the ¯xed point set of d¾
e
is Lie(K) we know,that M is a symmetric
space.Let X 2 Lie(G).It generates a path e +tX in G.The path ¾(e +tX) = (e +tX
>
)
¡1
can be written in the form e +td¾
e
X and is satis¯es therefore
e = (e +tX
>
)(e +td¾
e
X) = e +t(X
>
+d¾
e
X) +O(t
2
):
This means d¾
e
X = ¡X
>
.The ¯xed point set of this involution is the vector space of skew-
symmetric matrices,which is equal to the Lie algebra of SO(n;R).Therefore M = G=K is a
symmetric space.
In Example 1.2.4 we constructed a symmetric space P as the set of symmetric positive
de¯nite matrices in SL(n;R).The identity matrix I is clearly in P.The involution on the
isometry group of P given by ¾:g!s
I
gs
I
can be calculated explicitly.Let q 2 P:
s
I
(g ¢ (s
I
(q))) = s
I
(g ¢ (q
>
)
¡1
) = s
I
(gq
>
)
¡1
g
>
) = (g
¡1
)
>
qg
¡1
= (g
>
)
¡1
¢ q:
Hence ¾ maps g to (g
>
)
¡1
.This is the same action as above.Hence G=K = P.
Proposition 1.2.11.
Let R denote the curvature tensor of the space Riemannian symmetric
space G=K with G and K as usual and let Q be the Riemannian structure on G=K.Then at
the point o 2 G=K
R
o
(X;Y )Z = ¡[[X;Y ];Z];for X;Y;Z 2 p:
18
This proposition is Theorem IV.4.2 in [He78].In the following section we will use the
fact that a symmetry s
x
in a symmetric space M gives in a natural way an involution on
its group of automorphisms.We will use its di®erential,which is an involutive Lie algebra
automorphism,to study the structure of symmetric spaces.
Proposition 1.2.12.
Let G and K be as above.The following conditions are equivalent
(i)
G acts faithfully on M = G=K;
(ii)
K acts faithfully on p = Lie(G)=Lie(K)=tangent space to M at the origin.
Proof.
(i)) (ii).K < G acts faithfully on M and K is the stabilizer of the point eK 2 M.
If K acts not faithfully on p there exists k
1
;k
2
2 K with k
1
6= k
2
and dk
1
= dk
2
in T
eK
M.
But this is by Lemma A.1.5 a contradiction.
(ii) ) (i).Take h 2 G with h(gK) = gK for all g 2 G.This holds in particular for g = e
and have h 2 K.From h = id on M follows that dh = id on p.K acts faithfully on p and
therefore h = e.Hence G acts transitively on M.
If the two conditions hold,¾ is the identity on K,because it acts as identity on Lie(K).
Now we need some notions and theory of Lie algebras.
De¯nition 1.2.13.
Let g be a Lie algebra.By ad we denote the adjoint representation of g
in itself by
with X;Y 2 g.
The Killing form of g is de¯ned as
We denote the Killing form of a Lie algebra always by B or by B
g
if there is danger of
confusion.
Proposition 1.2.14.
The Killing form has the following properties:
(i)
It is a symmetric bilinear form,i.e.B(X;Y ) = B(Y;X) for all X;Y 2 g.
(ii)
It is in¯nitesimally invariant under ad,i.e.for all X;Y;Z 2 g we have
This is equivalent to B([X;Y ];Z) = B(X;[Y;Z]).
(iii)
Let i ½ g be an ideal.Then the Killing form on i equals the Killing form on g restricted
to i £i.
Proof.(i)
The bilinearity comes from the bilinearity of tr(XY ) and the linearity of ad.
The symmetry is a consequence of the general fact that for matrices A and B we have
tr(AB) =
P
A
ij
B
ji
=
P
B
ij
A
ji
= tr(BA).
19
(ii)
With the Jacobi-Identity we get:
ad([Z;Y ])(A) =[A;[Y;Z]] = ¡[Y;[Z;A]] ¡[Z;[A;Y ]]
Therefore
(iii)
Since i is a subvectorspace of g,one can complete a basis of i to a basis of g.Since it is
an ideal in g (i.e.ad(x)g ½ i) for x 2 i the matrix of ad(x) in g has the form
µ
A ¤
0

where A is the matrix of ad(x) in i.
De¯nition 1.2.15.
g
j
e
where
c
g
:h!ghg
¡1
is the conjugation by g.
We have for k 2 K and a geodesic through e
k¿(°;u)k
¡1
= ¿(k°;u):
Since ° is generated by a vector X = _°(0) 2 p and K acts faithfully on p via k 7!dk,we see
that k° is generated by dkX.Hence the adjoint representation of K on p is faithful.
Further we have that ad(X) is a skew-symmetric matrix (Appendix Lemma A.1.6),if
X 2 k.Therefore it is diagonalisable with eigenvalues in iR (Appendix Lemma A.1.7) and
2
) = ¡
P
¸
2
j
· 0,where ¸
j
2 R and i¸
j
are the eigenvalues of
2
) = 0 then the only eigenvalue of ad(X) is zero and therefore ad(X) = 0.
But then
and since the action of K on p is faithful,X is zero.This shows that Bj
k£k
is negative de¯nite.
De¯nition 1.2.16.
A Lie algebra is semi-simple if the Killing form is non-degenerate.
1.3 Involutive Lie Algebras
We have seen,that one can introduce on the group of automorphims G of a symmetric space
M an involution in a natural way.We ¯xed a point x 2 M and set ¾:g 7!s
x
gs
x
.This is
clearly an involution,since s
x
is.This involution gives an involution on g (the Lie algebra
of G) by taking the di®erential at the identity d¾j
e
.We have also seen,that we can write g
as k ©p where k and p are the eigenspaces to the eigenvalues 1 respectively ¡1 of d¾
e
.We
will always write p and k for these eigenspaces.By abuse of notation we write ¾ instead of

e
.First we de¯ne two properties and show that they are ful¯lled for Lie algebras that come
from symmetric spaces.
20
De¯nition 1.3.1.
An involutive Lie algebra (g;¾) is a Lie algebra together with an involutive
Lie algebra automorphism
4
¾.
Let (g;¾) be an involutive Lie algebra.By Remark A.1.8 ¾ is diagonalisable with eigen-
values 1 and ¡1.Therefore we can decompose g into k ©p where k and p are the eigenspaces
to the eigenvalues 1 respectively ¡1 of ¾.We have
[k;k] ½ k;[k;p] ½ p;[p;p] ½ k:(8)
because k and p are de¯ned as eigenspaces of ¾.k is a subalgebra of g.Since ¾ is an Lie
algebra automorphism it commutes with the bracket and we have ad(¾X) = ¾ad(X)¾
¡1
.As
a general fact we know that the trace is invariant under conjugation.Therefore we have for
k 2 k and p 2 p
¡1
¡1
=¡B(k;p);
hence B(p;k) = 0.Lets start our discussion of (g;¾).
Our main reference here is [Bo98].
De¯nition 1.3.2.
An involutive Lie algebra (g;¾) is said to be reduced,if k contains no
non-zero ideal of g.
Let i ½ k be an ideal of g.Then by de¯nition of an ideal we have [p;i] ½ i ½ k and by from
(8) we deduce [p;i] ½ p.Since p\k = f0g the ideal i is contained in the centralizer of p in k.
Proposition 1.3.3.
An involutive Lie algebra (g;¾) is reduced if and only if the representation
of k in p given by k 7!ad
g
(k)j
p
is faithful.
Proof.
The kernel of this representation is an ideal contained in k and it is by de¯nition of the
representation equal to the centralizer of p in k,denoted by z(p)
k
.To show that the kernel is
in fact an ideal,we need to show,that for r in the kernel and g = k +p 2 k ©p we have that
[g;r] is in the kernel,i.e.[[g;r];~p] = 0 for each ~p 2 p.We use the Jacobi-Identity:
[[k +p;r];~p] = [[k;r];~p] = [k;[r;~p]] +[r;[~p;k]]:
The right hand side is zero because ~p and [~p;k] are in p.
If (g;¾) is reduced the kernel of the adjoint representation is zero,since it is an ideal in k and
vice versa.
Fromthis proposition we see that an involutive Lie algebra which comes froma symmetric
space is reduced,since by Proposition 1.2.12 the representation k 7!ad
g
(k)j
p
is faithful.
Since z(p)
k
is the greatest ideal of g contained in k,it is invariant under ¾ and ¾ induces an
involution on g=z(p)
k
,which becomes a reduced involutive Lie algebra,denoted the reduced
involutive Lie algebra associated to (g;¾).
De¯nition 1.3.4.
An orthogonal involutive Lie algebra is an involutive Lie algebra (g;¾) on
which there exists a positive non-degenerate quadratic form which is invariant under ¾ and
in¯nitesimally invariant
5
g
k.
4
A vector space isomorphism with ¾([X;Y ]) = [¾(X);¾(Y )] for all X;Y 2 g.
5
21
Proposition 1.3.5.
An involutive Lie algebra is orthogonal if and only if k is compact and
p
leaves a positive non-degenerate quadratic form in¯nitesimally invariant.
Proof.
If (g;¾) is orthogonal,there exists Q a positive non-degenerate quadratic form invari-
ant under ¾.By de¯nition of k and p as eigenspaces of ¾ we get Q(k;p) = 0 and Q = Q
1
+Q
2
where Q
1
= Qj
k£k
and Q
2
= Qj
p£p
form Q
1
invariant,k is compact.Q
2
Conversely if k is compact,it leaves a positive non-degenerate quadratic form Q
1
in¯nitesi-
mally invariant.Denote by Q
2
the positive non-degenerate form on p invariant under ad(k).
Q:= Q
1
+Q
2
is by construction invariant under ¾ and in¯nitesimally invariant under ad(k),
hence (g;¾) is orthogonal.
Let (g;¾) be the involutive Lie algebra that comes froma symmetric space.We know from
our discussion above,that k is compact,hence ad(k) contains only skew-symmetric matrices.
Therefore we can construct on p a positive non-degenerate quadratic form Q invariant under
g
k by ¯xing a basis of p and take Q as the usual quadratic form (x;y):= x
>
¢ y.We have
shown,that ad(k) is skew-symmetric for each k 2 k,therefore this bilinear formhas all desired
properties and (g;¾) is orthogonal.
De¯nition 1.3.6.
A Lie algebra is said to be °at if its Lie bracket is constantly zero.
Theorem 1.3.7.
Let (g;¾) be a reduced orthogonal involutive Lie algebra.Then (g;¾) is the
direct product of a reduced orthogonal involutive °at Lie algebra and of reduced semi-simple
irreducible orthogonal involutive Lie algebras (g
i

i
),(i = 1;:::;a).This decomposition is
unique up to the order of the factors and z(p) = p
0
.
Proof.
Let Q be a positive non-degenerate quadratic form on p,in¯nitesimally invariant
g
k.Q exists since (g;¾) is orthogonal.Let A be the linear map of p into itself,
de¯ned by Q(Ax;y) = B(x;y) for every x;y 2 p.If Q and B are the matrices of the
corresponding bilinear forms A = Q
¡1
B,where Qis invertible since it is non-degenerate.Since
Q and B are both symmetric forms,we have Q(Ax;y) = B(x;y) = B(y;x) = Q(Ay;x) =
Q(x;Ay).Therefore A is symmetric,hence diagonalisable.We have a decomposition of p
into eigenspaces
p =
r
M
i=0
q
i
;
such that
Aj
q
i
= c
i
¢ id (c
i
2 R;c
0
= 0;c
i
6= c
j
for j 6= i):
q
0
is the kernel of the Killing form,i.e.it contains every x 2 g with B(x;y) = 0 for all y 2 q.
It is contained in p since B is non-degenerate on k.The q
i
are subvectorspaces of p.For
x 2 q
i
and y 2 q
j
,we have B(x;y) = c
i
Q(x;y) = c
j
Q(x;y) = 0 if i 6= j,hence
B(q
i
;q
j
) = Q(q
i
;q
j
) = 0;(i 6= j);B(q
0
;p) = 0
and
Bj
q
i
= c
i
Qj
q
i
:
Since B and Q are non-degenerate and in¯nitesimally invariant under ad
g
k,A commutes
g
k.Therefore ad(k) leaves the eigenspaces of A ¯xed and hence we have
22
[k;q
i
] ½ q
i
.Let U ½ p be a subvectorspace invariant under ad(k) and U
?
its orthogonal
complement in p with respect to Q.Then we have for all u 2 U;v 2 U
?
;k 2 k:
Hence ad(k)v is contained in U
?
,because it is orthogonal to u.Therefore we can decompose
q
i
in subvectorspaces invariant minimal under adk and there exists a direct sumdecomposition
of
L
r
i=1
q
i
in
L
a
i=1
p
i
,where each p
i
is invariant minimal under adk.Each p
i
is contained in
a q
j
and p
i
and p
l
are orthogonal with respect to Q if l 6= i.
Furthermore we have [p
i
;p
j
] = 0 for i 6= j,since for k 2 k:
B(k;[p
i
;p
j
]) = B([k;p
i
];p
j
) ½ B(p
i
;p
j
) = 0;
since p
i
and p
j
are orthogonal with respect to Q.B is non-degenerate on k,therefore [p
i
;p
j
] =
0.The same argumentation shows that [q
i
;q
j
] = 0,if i 6= j.We will need this fact later.
Let now
g
i
:= [p
i
;p
i
] +p
i
;(i = 1;:::;a):
[p
i
;p
i
] is contained in k because ¾ acts as the identity on it,thus the sumis direct.Furthermore
we have
[g
i
;g
j
] = 0;(i 6= j);[k;g
i
] ½ g
i
(1 · i;j · a)
and
[p
0
;g
i
] = 0;[p;g
i
] = [p
i
;g
i
] ½ g
i
:
g
i
is an ideal of g which follows from the Jacobi identity and the equations above.Therefore
the Killing form on g
i
is the Killing form on g restricted to g
i
£ g
i
.It is non-degenerate
on the p
i
,since it is a multiple of Q and it is non-degenerate on [p
i
;p
i
] ½ k.Hence it is
non-degenerate on g
i
and therefore g
i
is semi-simple.We have,again by the above formulae
and the Jacobi-identity for i 6= j
B(g
i
;g
j
) = B(p
i
;p
j
) +B(p
i
;[p
j
;p
j
]) +B(p
j
;[p
i
;p
i
]) +B([p
i
;p
i
];[p
j
;p
j
]) = 0:
Therefore the g
i
are linearly independent.Let m be their sum.It is a semi-simple ideal of g
invariant under ¾,therefore g = g
0
£m,with g
0
= z(g).Denoting by ¾
i
the restriction of ¾
to g
i
we have:
(g;¾) = (g
0

0
) £(g
1

1
) £:::(g
a

a
);
with (g
i

i
) semi-simple,irreducible for i ¸ 1 and (g
i

i
) reduced for all i's.(g
0

0
) is
°at.
This proof shows that for i 6= 0,Bj
p
i
is either negative de¯nite or positive de¯nite,since
it is a multiple of the positive de¯nite bilinear formQ.Therefore either g
i
is compact (if Bj
p
i
is negative de¯nite) or g
i
is non-compact and [p
i
;p
i
] is a maximal compact subalgebra.In
the ¯rst case we say that it is of compact type,in the second it is of non-compact type.This
property is directly related to the sectional curvature.
Corollary 1.3.8.
Let (g;¾) an orthogonal involutive Lie algebra,G a Lie group with Lie
algebra g and K a subgroup corresponding to k.Assume that K is connected and closed.Let
Q be an arbitrary K-invariant Riemannian structure on G=K.Then
i)
If G=K is of Euclidean type,then G=K has sectional curvature everywhere = 0.
23
ii)
If G=K is of compact type,then G=K has sectional curvature everywhere ¸ 0.
iii)
If G=K is of non-compact type,then G=K has sectional curvature everywhere · 0.
Proof.
The tangent space at o can be identi¯ed with p.Let S be two-dimensional subspace of
T
p
M and X and Y two orthonormal vectors in S.Proposition 1.2.11 tells us that R(X;Y )X =
¡[[X;Y ];X] and by the de¯nition of the sectional curvature (De¯nition 1.1.16,we get
K(S) = ¡Q
0
(R(X;Y )X;Y ) = Q
0
([[X;Y ];X];Y )
since the latter is zero in the Euclidean case,we have proved i).
The other two cases are not di±cult.We assume that g is semi-simple and use the results
of Theorem 1.3.7.We decomposed p into subspaces q
i
as eigenspaces to the eigenvalues c
i
of
the matrix A de¯ned by Q
0
(AX;Y ) = B(X;Y ) and for X
i
;Y
i
2 q
i
we have
B(X
i
;Y
i
) = c
i
Q
0
(X
i
;Y
i
):
And [q
i
;q
j
] = 0 holds if i 6= j.With X =
P
X
i
and Y =
P
Y
i
(X
i
;Y
i
2 q
i
we get
[X;Y ] =
P
[X
i
;Y
i
] and [[X
i
;Y
i
];X] = [[X
i
;Y
i
];X
i
].Hence
K(S) = Q
0
([[X;Y ];X];Y ) =
X
i
Q
0
([[X
i
;Y
i
];X
i
];Y
i
) =
X
1
c
i
B([X
i
;Y
i
];[X
i
;Y
i
]):(9)
If G=K is of compact type,by de¯nition the c
i
are all negative and the sectional curvature is
negative.If G=K is of non-compact type,the c
i
are all positive and the sectional curvature
is positive.
Proposition 1.3.9.
For an involutive semi-simple reduced Lie algebra (g;¾)
holds.
Proof.
First note,that the Killing-form B on g is non-degenerate,since g is semi-simple.
Therefore we can decompose g in a direct sum of p + [p;p] and c:= (p + [p;p])
>
.Since
[p;p] ½ k,the ¯rst sum is also direct.c,being orthogonal to p,is contained in k.If we show,
that c is zero,we are done,because B is non-degenerate.To do that we will show that it is
an ideal of g contained in k and must therefore be zero.
Fix c 2 c and let p;~p 2 p and g
0
= p
0
+k
0
B(g
0
;[p;c]) = B([g
0
;p];c) = B([p
0
;p]
|
{z
}
2[p;p]
;c) +B([k
0
;p]
|
{z
}
p
;c) = 0;
hence [p;c] = 0,since g
0
was arbitrary and B is non-degenerate.With the Jacobi-identity we
get immediately that [[p;~p];c] = 0.We are not far from our result,since
B([g
0
;c];p) = B(g;[c;p]) = 0
B([g
0
;c];[p;~p]) = B(g;[c;[p;~p]]) = 0:
This shows that c is an ideal of g contained in k which must be zero,since g is reduced.
Therefore
holds.
24
This property characterizes the isometry group of a symmetric space.In Remark 1.2.8 we
announced the following proposition:
Proposition 1.3.10.
Let G be a connected semi-simple Lie group,K a compact subgroup
and ¾ an involution on G.Assume that
i)
M = G=K is connected,
ii)
there exists an involution ¾ of G such that Lie(G)

e
= Lie(K) and K
¾
= K and
iii)
G acts faithfully on M.
Then G is the identity component of the group of isometries on M.
Proof.
Let G
0
denote the group of isometries of M.M can be written in the form G
0
=K
0
where K
0
is a compact subgroup of G
0
.G acts by de¯nition of M isometrically on M and
since it is connected,it can be identi¯ed with a subgroup of the identity component of G
0
.
g
0
,the Lie algebra of G
0
,is reduced since,by Proposition 1.2.12 K
0
and therefore by Lemma
A.1.5 k act faithfully on p'T
p
M.Furthermore g
0
is semi-simple,since G
0
has no abelian
factor.Hence g
0
The Lie algebra g of G is a subalgebra of g
0
and it decomposes therefore into eigenspaces ~p©
~
k
with respect to ¾.Since
~
p'T
p
M'p and g is by assumption reduced and semi-simple,we
have
g = p ©[p;p] = g
0
and therefore G is the identity component of G
0
.
Remark 1.3.11.
Helgason classi¯es in [He78] irreducible orthogonal involutive Lie algebras,
where a Lie algebra is irreducible,if ad
g
(k) acts irreducibly and faithful on p.By assumption
in our case here,(g;¾) is reduced,hence ad
g
(k) acts faithful on p.Let i ½ k
i
be an ideal of g
i
,
i.e.[i;g
i
] ½ i.Then i is an ideal of g =
L
a
i=0
g
i
since [g
i
;g
j
] = 0 if (i 6= j).We have assumed
(g;¾) to be reduced,therefore i is trivial and hence g
i
is reduced.By construction of p
i
the
g
k is irreducible,hence the (g
i

i
) are irreducible in the sense of Helgason.
By Theorem 1.3.7 we can assume that (g;¾) is either °at or a semi-simple reduced or-
thogonal involutive Lie algebra.If (g;¾) is °at,it corresponds to an euclidean space,since
we have a vector space with vanishing Lie bracket.
A compact semi-simple Lie algebra g (ignoring the involution) decomposes in a product of
simple ideals
g =
n
M
i=1
h
i
:
¾(h
i
) is simple,hence ¾(h
i
) = h
j
.We have two cases:
i)
i = j.In this case (h
i
;¾j
h
i
) is a simple reduced orthogonal involutive Lie algebra.
h
i
= k
i
i
where k
i
and p
i
are the eigenspaces with respect to the eigenvalues 1 and
¡1.
ii)
i 6= j.Then h
j
and h
i
are isomorphic via ¾ and (h
i
i
);¾j
h
i
i
)
) is a reduced
orthogonal involutive Lie algebra.¾ acts on it via (x;y) 7!(y;x).
25
They are both invariant under ¾.This gives us a new decomposition of g
g =
M
i
l
i
;
where l
i
is an ideal in g,invariant under ¾.It can therefore be decomposed into eigenspaces
l
i
= k
i
+ p
i
.The l
i
are ideals,hence [l
i
;l
j
] ½ l
i
for all i and j.Therefore [l
i
;l
j
] ½ l
i
\l
j
.
The last is trivial if i 6= j,therefore [k
i
;p
j
] = [k
i
;k
j
] = 0 if i 6= j.Since k acts by assumption
i
)(p
j
) = 0 for i 6= j we have p = p
i
for some i.We can assume i = 1
without loss of generality.For i 6= 1 we have [k
j
;p
1
] = 0,hence k
j
is an ideal of g in k.But
ad(k) acts faithfully on p,hence k = k
1
.This shows that g = l
1
.
An irreducible Lie algebra g of compact type has the form
I.
(g;¾) is a compact simple Lie algebra and ¾ is any involution on it.
II.
g is the direct sum l
1
2
and ¾ acts on it via (x;y) 7!(y;x).
If (g;¾) is a non-compact irreducible orthogonal involutive Lie algebra,we can construct a
dual compact Lie algebra by putting
~
C
:
~
g is also irreducible,since taking this dual does not change the action of k on p respectively
ip.
Let
~
g be in case I.Assume g to be not simple.Then there exists a decomposition in non-zero
ideals n
1
and n
2
.They can be decomposed into k
1
+p
1
and k
2
+p
2
,hence (k
1
+ip
1
)+(k
2
+ip
2
)
is a decomposition of
~
g into non-zero ideals.Since this is impossible,g is simple.A similar
argumentation shows that the complexi¯cation g
C
of g is simple,too.
If
~
g is in case II.,its dual admits a complex structure (see [He78,Thm.V.2.4]) and it is
simple.Therefore we have the following two possibilities for the non-compact case.
III.
g is a simple,noncompact Lie algebra over R,the complexi¯cation g
C
is a simple Lie
algebra over R and ¾ is an involutive automorphism of g such that the ¯xed points form
a compactly imbedded subalgebra.
IV.
g is a simple Lie algebra over C.Here ¾ is the conjugation of g with respect to a
maximal compactly imbedded subalgebra.
Theorem 1.3.12.
Let M be a simply connected symmetric space of compact or non-compact
type.Then M is product
M = M
1
£:::£M
n
;
where the factors M
i
are irreducible.
Proof.
We proofed that the Lie algebra g of its group of isometries decomposes into semi-
simple factors (Theorem 1.3.7)
g = g
1
£:::£g
n
with g
i
= p
i
i
.Hence G = G
1
£:::£ G
n
where g
i
is the Lie algebra of G
i
.Further
K = K
1
£:::£K
n
and k
i
is the Lie algebra of K
i
.Therefore M = M
1
£:::£M
n
.
26
1.4 Hermitian Symmetric Spaces
De¯nition 1.4.1.
A Hermitian manifold is a di®erentiable Riemannian manifold with the
Riemannian connection r whose tangent bundle is endowed with a Hermitian structure,i.e.
a (1;1)-tensor J with J
2
= ¡1,which leaves the metric invariant (g(JX;JY ) = g(X;Y )).
Like in the study of riemannian symmetric spaces we start with a local property.
Proposition 1.4.2.
Let M be a Hermitian manifold.The following conditions are equivalent:
i)
rR = 0 = rJ;
ii)
for all x 2 M,s
x
:y 7!exp
x
(¡log
x
y) is a local automorphism at x of M,i.e.s
x
leaves
r and J invariant.
Proof.
i))ii).From Corollary 1.1.13 we know that s
x
is a local di®eomorphism which leaves
r invariant,because rR = 0.J is invariant under parallel transport,since rJ = 0.Since
ds
x
acts by multiplication with ¡1 on T
x
M,it commutes with J
x
and hence J and s
x
are
compatible.
ii))i).If ii) is satis¯ed every canonical tensor of odd degree is zero,because it is invariant by
s
x
.Therefore rJ and rR are both zero.(Same argumentation as in Corollary 1.1.13.)
De¯nition 1.4.3.
A Hermitian manifold M is locally symmetric if the equivalent conditions
of Proposition 1.4.2 are satis¯ed.It is called Hermitian symmetric if it is connected and for
all x 2 M there exits an involutive automorphism s
x
of M with x as isolated ¯xed point.
Everything we said about Riemannian symmetric spaces applies to Hermitian symmetric
spaces,too.
Corollary 1.4.4.
A locally symmetric Hermitian manifold is KÄahler.If g is the metric on
M,then
!(X;Y ):= g(X;JY )
is a KÄahler form,i.e.!is a symplectic,closed 2-form.
The following fact is a central fact in our study of hermitian symmetric spaces.
Corollary 1.4.5.
Let M be a simply connected symmetric Hermitian space and let G be
the identity component of its group of automorphisms.Let x be a point of M and let K its
stabilizer.There exists a homomorphism u
x
:U
1
!K such that u
x
(z) induces multiplication
6
by z on T
x
M.
Proof.
We use Theorem 1.1.9 and Proposition 1.1.10 to construct u
x
.For a given z 2 U
1
,
we have in the notation of Theorem 1.1.9 M = M
0
,x = x
0
,r = r
0
and A is multiplication
with z.By Proposition 1.4.2 and the de¯nition of the Riemannian connection we have rT =
0 = rR.If we show that R is invariant under the automorphism v 7!zv on T
p
M (i.e.
¹zR(zX;zY )zZ = R(X;Y )Z),then we are done.Note that
g(R(X;Y )Z;T) = g(R(Z;T)X;Y ) (10)
6
U
1
:= fz 2 C:jzj = 1g and for v 2 T
x
M and z = a +ib 2 U
1
we de¯ne z ¢ v by av +bJv.
27
holds.A proof of this fact can be found
7
in [He78,Lem.I.12.5].
By assumption J is invariant under parallel transport.Hence it commutes with the con-
travariant derivative r
X
for all X.Therefore
R(X;Y )(JZ) = r
X
r
Y
(JZ) ¡r
Y
r
X
(JZ) ¡r
[X;Y ]
(JZ) = JR(X;Y )Z:
Now we can show what we wanted:
g(¹zR(zX;zY )zZ;T) = g(R(zX;zY )¹zzZ;T) = g(R(Z;T)zX;zY ) = g(¹zR(Z;T)zX;Y )
= g(R(Z;T)¹zzX;Y ) = g(R(X;Y )Z;T):
Remember the maps ¿(°;u) introduced in Corollary 1.1.11.They are composites of sym-
metries and are therefore automorphisms of M.As before,we have
M = G=K and Lie(G) = k ©p
where G is the identity component of the Lie group of automorphisms of M and K the
stabilizer of a x 2 M.¾ denotes the automorphism of G induced by s
x
.
Before we continue,we have a look at an example.
Example 1.4.6.
Let V be a n-dimensional complex vector space endowed with a non-
degenerate Hermitian formh of signature (p;q),i.e.p is the maximal dimension of a subspace
L ½ V such that hj
L
is positive de¯nite.We can choose a basis (e
1
;:::;e
p
;e
p+1
;:::;e
p+q
) of
V such that h is of the form
h(z;w) =
p
X
i=1
z
i
¹w
i
¡
q
X
j=1
z
j+p
¹w
j+p
:
Denote W
+
the space spanned by e
1
;:::;e
p
.Its complement with respect to h is spanned
by e
p+1
;:::;e
p+q
.Denote by X the space of p-dimensional subspaces W ½ V with hj
W
is
positive de¯nite.X is an open subset of the Grassmannian Gr
p
(V ),hence a manifold.Fix
W 2 X.Since hj
W
is non-degenerated,we can decompose V into the direct sum of W and
W
?
,the orthogonal complement with respect to h.An element x of V can be decomposed
into w +w
?
.Denote s
W
the map which maps w +w
?
to w ¡w
?
.It is the re°ection with
respect to W.For two orthogonal vectors x;y 2 V (i.e.h(x;y) = 0) we have
h(x +y;x +y) =h(x;x) +h(y;y) +h(x;y) +h(y;x) = h(x;x) +h(y;y) ¡h(y;x) ¡h(x;y)
=h(x ¡y;x ¡y):
Therefore the re°ection s
W
lies in SU(p;q),the subgroup of SL(n;C) which contains all
matrices which leaves h invariant and s
W
induces a map from X into X,also denoted by s
W
.
It is involutive with W as an isolated ¯xed point.This makes X into a symmetric space.
Now we want to see how X can be realized as a homogeneous space G=K.Let G:=
SU(p;q) and ¯x W
+
2 X as a base point.G acts transitively on X,since one can choose
7
Or one can proof it oneself by using the de¯nition of R and the invariance of g under paralleltransportation
in the form X ¢ g(Y;Z) = g(r
X
Y;Z) +g(Y;r
X
Z) to show that g(R(X;Y )Z;T) = ¡g(R(X;Y )T;Z) and use
this,the Bianchi identity and Lemma 1.1.17 to prove Equation 10.
28
a basis of W and one of W
?
and write down a matrix for h with respect to this basis.
This matrix is hermitian,hence diagonalizable and by scaling it can be brought to the form
diag(1;:::;1;¡1;:::;¡1) with p-times 1 and q-times ¡1.Therefore W can be mapped onto
W
+
by an element of G and G acts transitively on X.
Let g be in G.Write g in the form
µ
A B
C D

with A 2 M(p;p;C),B 2 M(p;q;C),C 2
M(q;q;C) and D 2 M(q;p;C).As a direct calculation shows the following conditions must
hold:
A
¤
A¡C
¤
C = id
p
B
¤
B ¡D
¤
D = ¡id
q
B
¤
A¡D
¤
C = 0
det(g) = 1:
Such a matrix g maps W
+
into itself if and only if C = 0.Fromthe third equation follows that
in this case B = 0 because by the ¯rst equation A is invertible.The ¯rst two equations tell
us that A and D are hermitian,since they leave a positive-de¯nite Hermitian form invariant.
Therefore the stabilizer of W is equal to S(U(p) £U(q)) the matrices of U(p) £U(q) with
det(u
p
) det(u
q
) = 1 for u
p
2 U(p) and u
q
2 U(q).We can write X as the quotient G=K =
SU(p;q)=S(U(p) £U(q)).A discussion of SU(1;1) and the associated symmetric space,the
hyperbolic plane,can be found in Example 2.5.3 and Proposition A.4.7.
Remark 1.4.7.
The complex structure on M induces on p(= T
x
M) a complex structure J,
which is invariant under K,i.e.J commutes with dkj
e
for k 2 K.Conversely,if M = G=K
is Riemannian and symmetric,every complex K-invariant structure on p endows M with a
hermitian symmetric structure,since the structure can transported along the geodesics to
every point and this is well-de¯ned since the hermitian structure on p is K-invariant.
Remark 1.4.8.
If J
e
is a k-invariant complex structure on p,i.e.J([k;p]) = [k;Jp] then the
p
i
from theorem 1.3.7 are complex subspaces,because they are disjoint representations of k
and hence [k;Jp
i
] ½ p
i
.By Theorem 1.3.7 the Lie algebra of the group of automorphisms
decomposes into a direct sum of a °at Lie algebra and semi-simple involutive Lie algebras g
i
which can be decomposed as k
i
+p
i
.There exists subgroups K
i
corresponding to k
i
.Hence
M decomposes into Hermitian factors
M = M
0
£
Y
M
i
;
where M
0
is °at.
Remark 1.4.9.
Let G be the identity component of the group of isometries of M.Assume
that it is semi-simple.Then it has the same Lie algebra [p;p]©p as the analogous group for the
universal covering
~
M of M (Proposition 1.3.9).Therefore we have by Corollary 1.4.5 a map
u
x
:U
1
!
~
K where
~
K is the stabilizer of a point x 2
~
M.Composing u
x
with the canonical
projection
~
M!M gives a map from U
1
to K the stabilizer of a point in M.Since u
x
acts as
multiplication with
z
on
~
p
and since the canonical projection is a local di®eomorphism,the
composition acts as multiplication with z on p.
Proposition 1.4.10.i)
The subgroup K is the centralizer of u
x
(U
1
) in G and K is con-
nected.
29
ii)
The center of G is trivial.
iii)
The Hermitian symmetric space M is simply connected.
iv)
If (g;¾) is indecomposable,then u
x
(U
1
) is the center of K.
Proof.i)
Let K
0
be the centralizer of u
x
(U
1
).First we show K
0
½ K using Lie(K
0
) ½
Lie(K).Remark that the Lie algebra of K is the ¯xed point set of the action of d¾
e
in g.As ¾ is multiplication with ¡1 on p,we have ¾ = Adu
x
(¡1).By de¯nition
x
(¡1)(= ¾) is the di®erential of g 7!u
x
(¡1)gu
x
(¡1)
¡1
,which is the identity on
K
0
.Hence d¾
e
leaves Lie(K
0
) ¯xed and therefore Lie(K
0
) ½ Lie(K).K
0
is the centralizer
of a torus,thus it is connected (see Appendix,Proposition A.4.1).Therefore K
0
is a
subgroup of K.
Conversely,note that the representation of k on p and the complex structure J on
p commute and and that the representation of k on p is faithful.Therefore we have
K ½ K
0
,hence K = K
0
and K is connected.
ii)
We have seen above that k¿(°;u)k
¡1
= ¿(k°;u) for all k 2 K,where ° is a geodesic
starting in x.By i) the center Z of G must be contained in K.If k is in Z it must
commute with ¿(°;u) and u 2 R and hence ¿(°;u) = ¿(k°;u) for all k 2 Z.But since
K acts faithful on p this is only possible for k = e,i.e.Z is trivial.
iii)
Let ¼:
~
G!G be the universal covering of G.The Lie algebras of
~
G and G are both
~
G exists an involution ~¾ with d~¾ = d¾.Denote by
~
K the ¯xed
point group of
~
G,it is connected.We have ¼(
~
K) = K,since the Lie algebras of
~
K
and K are equal and they are both connected.From Remark 1.2.8 we know that
~
G=
~
K
is a symmetric space.We can apply i) and
~
K contains the center of
~
G.Therefore
¼
¡1
(K) =
~
K.Hence
~
G=
~
K = G=K = M;
is simply connected,since
~
G is and
~
K is connected.
iv)
Since g is indecomposable,the representation of K in p is irreducible.The commutator
of K in End(p) the same as the commutator of K in End
C
(p),since the complex structure
on J commutes with the K-action.By Schur's lemma the center of K consists only of
multiples of the identity matrix.Since K is compact,the center is,as a closed subgroup,
compact to.Hence the center is u
x
(U
1
).
Proposition 1.4.11.
Let M be a symmetric Riemannian space and let G be the identity
component of its group of automorphisms,x 2 M,K its stabilizer in G and ¾ the involution
on G generated by conjugation with s
x
.Assume that G is semi-simple,K is connected and
(g;¾) is indecomposable.The following conditions are equivalent:
i)
the representation p of k is not absolutely simple (i.e.p
C
is not simple);
ii)
the center of k is non-zero;
iii)
M admits the structure of a symmetric Hermitian space (compatible with its Riemannian
structure).
30
Proof.
i)) iii).By assumption p
C
decomposes into a direct sum of two non-trivial complex
subrepresentations p
¡
and p
+
.We can embed p into p
C
= p © ip and project p
C
onto p
§
.
Composing these maps we get two maps
p!p
C
!p
§
:
Note that p 2 p ½ p
C
can be written uniquely as p
+
+ p
¡
with p
¡
2 p
¡
and p
+
2 p
+
.
Since p\p
¡
= f0g = p\p
+
this maps are injective.Therefore p
+
and p
¡
must have the
same dimension,hence the maps are isomorphisms.Therefore we can transport the complex
structure of p
+
to p.Since p can be identi¯ed with the tangent space of M in some point this
tangent space can be endowed with a complex structure too.By Remark 1.4.7 this induces a
complex structure on M.
iii) ) i).Let J denote the complex structure on p.Since J
2
is symmetric it can be diago-
nalized over C with eigenvalues i and ¡i.The eigenspace to i is an ad(k) invariant subspace
of the complexi¯cation of p,therefore the representation is not absolutely simple.
iii) )ii) follows from Proposition 1.4.10.
ii) ) iii) By Schur's Lemma (Lemma A.3.11) the center of K acts on p
C
as multiplication
with complex diagonal matrices.Since K is compact,it contains only multiples of the identity
matrix with U
1
.Since the representation is faithful,we get a complex structure on p and
therefore one on M.
Example 1.4.12.
This shows that P(n;R) is a Hermitian symmetric space if and only if
n = 2.We saw that P(n;R) = SL(n;R)=SO(n;R) and the stabilizer of a point equals
SO(n;R),which has non-trivial center if and only if n = 2 (see [Hi91,p.328]).
Example 1.4.13.
Remember Example 1.2.9.We saw there that the symmetric space
SO(2;n)=S(O(2) £O(n))
can be realized as the space G of two dimensional subspaces of R
2
n
on which the bilinear
form of signature (2;n) is positive de¯nite.Since we are dealing with two-dimensional real
subspaces of R
n+2
it seems possible,that we can put a complex structure on G.And this
is in fact true:since the center of S(O(2) £ O(n)) contains at least O(2),it is non-trivial
and by Proposition 1.4.11,there exists a complex structure on G.There exists a holomorphic
di®eomorphism of
D:= fz 2 C
n
j1 ¡2hz;zi +jzz
>
j > 0;jzj < 1g
to M.
Remark 1.4.14.
Let M = G=K be a compact simple symmetric hermitian space.For x 2 M
let u
x
:U
1
!G be as before.In the representation Adu
x
on Lie(G)
C
it acts only through
the characters 1 (on k
C
) and z and ¹z (on p
C
).
Conversely,if G is a compact adjoint simple group and u:U
1
!G acts like above
on Lie(G)
C
.Its centralizer K is connected since it is the centralizer of a torus.It has for
Lie algebra the subspace ¯xed by the involution Adu(¡1) of Lie(G) and G=K is Hermitian
symmetric.
The classi¯cation problem of irreducible Hermitian symmetric spaces is therefore reduced
to the classi¯cation of compact adjoint simple groups with a morphismu:U
1
!G such that
its representation on Lie(G)
C
acts only through the characters 1 (on k
C
) and z and ¹z (on p
C
).
31
2 Some Topics on Symmetric Spaces
2.1 Complexi¯cation and Cartan Decomposition
Let V be a ¯nite dimensional vector space over R.A complex structure of V is a R-linear
endomorphismJ with J
2
= ¡1.This endomorphismis something like the multiplication with
i in a complex vector space.A vector space V with a complex structure can be turned into
a vector space
~
V over C by putting
(a +ib)X:= aX +bJX;
for a;b 2 R and X 2 V.The dimension of
~
V over C is
1
2
dim
R
V.
A Lie algebra v over R is said to have a complex structure,if it has a complex structure
J as a vector space and
[X;JY ] = J[X;Y ] (11)
holds for X;Y 2 v.Remark that the Lie bracket on a complex Lie algebra is C-linear,hence
it commutes with i.If (11) would not hold,it would not be possible to turn it into a complex
Lie algebra like we did it with the vector space V.Bur if (11) hold,it is possible as one can
easily check.
Now let W be a ¯nite dimensional vector space over R (without further structure).If one
wants to extend W to a complex vector space W
C
of the same dimension
8
,i.e.make a scalar
extension,one has several possibilities (which all yield the same result).The most obvious
way may be to take a basis fe
1
;:::;e
n
g of W and de¯ne
W
C
:=
n
X
i=1
Ce
i
:
This works but one has to check that W
C
does not depend on the chosen basis.Another
possibility uses the observation that for a;b 2 R we have i(a+ib) = ¡b+ia,i.e.multiplication
with i maps the vector (a;b) to (¡b;a).De¯ning J:W £ W!W £ W by (X;Y ) 7!
(¡Y;X) for X;Y 2 W gives us a complex structure on W £W which we also denote by W
C
.
Furthermore one can write W
C
:= W +iW as a formal sum and de¯ne multiplication with i
in the obvious way.Last but not least one can use some theory and put W
C
:= W
R
C.All
four possibilities are isomorphic.The vector space W
C
is called complexi¯cation of W.
The complexi¯cation works for Lie algebra too.Let g be a Lie algebra and g
C
its com-
plexi¯cation (as a vector space).It consists of all symbols X + iY with X;Y 2 g.For
X +iY;Z +iT 2 g
C
we de¯ne
[X +iY;Z +iT]:= [X;Z] ¡[Y;T] +i([Y;Z] +[Z;T]):
This makes g
C
to a complex Lie algebra.
Let g be a Lie algebra over C.A real form of g is a subalgebra g
0
of g considered as a
real Lie algebra,s.t.
g = g
0
0
as a R Lie algebra.
In this case each Z 2 g can be written as X + iY with unique X;Y 2 g
0
.The map
¾:X +iY 7!X ¡iY is called conjugation with respect to g
0
.
An important theorem is the following:
8
dim
R
W = dim
C
W
C
32
Theorem 2.1.1.
Every semi-simple Lie algebra over C has a real form which is compact.
See [He78,Thm.III.6.3] for a proof.It uses a root decomposition to write down explicitly
the compact real form.
De¯nition 2.1.2.
Let g
0
be a semi-simple Lie algebra over R.Denote by g its complexi¯ca-
tion an by ¾ the conjugation in g with respect to g
0
.A decomposition of g
0
g
0
= k
0
+p
0
where k
0
is a subalgebra of g

format: plain


text:
# Careful observation

Chemistry Level 2

The degree of unsaturation in $$\ce{C3H3Cl3}$$ is $$x$$,

and the degree of unsaturation in $$\ce{C3H4O}$$ is $$y$$,

and the degree of unsaturation in $$\ce{C4H5N}$$ is $$z$$.

Find $$x+y+z$$.

TAKE UP THE CHALLENGES!!

##### This is a part of my set Aniket's Chemistry Challenges (BEST OF JEE - ADVANCED)
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text:
# point set topology: closed points dense

Let $X$ be an irreducible finite dimensional Jacobson scheme (i.e. the closed points lie dense and the underlying topological space is sober). If one chooses for every closed point $x \in |X|$ an open neighbourhood $U_x$, is the union $\bigcup_{x\in|X|}U_x = X$ necessarily the whole space?

This is e.g. true if $\dim{X} = 1$ since in this case the only non-closed point is the generic point, which lies in every non-empty open subset.

I am already interested in the case $\dim{X} = 2$ if this simplifies things. For this, one only has to show that for $x$ of codimension $1$ (i.e. $\dim\overline{\{x\}} = 1$), there is a closed point $y$ such that $x \in U_y$. Isn't this automatic by the same argument as with the generic point and $\dim{X} = 1$?

-
To translate this for "normal topologist": $X$ is just a space, $|X|$ is the set of $x$ such that $\{x\}$ is closed, and we just assume that $|X|$ is dense in $X$? No hidden assumptions in the words "scheme" etc.? –  Henno Brandsma May 3 '14 at 11:49
"Scheme" implies that the topological space is sober, but I think this is the only assumption. –  user5262 May 3 '14 at 14:20
what does sober translate to topologically? Just that irreducible closed sets are closures of singletons? –  Henno Brandsma May 3 '14 at 14:27


text:
# Count substrings with each character occurring at most k times in C++

C++Server Side ProgrammingProgramming

We are given a string str. The goal is to count the number of substrings in str that have each character occurring utmost k times. For example if input is “abc” and k=1, the substrings will be “a”, “b”, “c”, “ab”, “bc”, “abc”.

Let us understand with examples.

Input − str=”abaefgf”

Output − Count of substrings with same first and last characters are &mmius; 9

Explanation − Substrings will be

“a”, “b”, “a”, “e” ,”f”, “g”, “f”, “aba”, “fgf”. Total 9.

Input − str=”abcdef”

Output − Count of substrings with same first and last characters are: 6

Explanation − Substrings will be -

“a” , “b” , “c”, “d”, “e” ,”f”. Total 6

## The approach used in the below program is as follows

There can be multiple approaches to solve the given problem i.e. naive approach and efficient approach. So let’s first look at the naive approach. We will pass substrings of every length to a function check(). If that substring starts and ends with the same character then increment the count.

• Take string str. Calculate the length as str.size().

• Function check(string str) takes substring str and checks if it’s first and last character are the same. ( str[0]==str[length-1] ). If true, return 1.

• Function check_Start_End(string str, int length) takes str and its length as input and returns the count of substrings of str which start and end with the same character.

• Take the initial count as 0.

• Traverse str using two for loops. From i=0 to i<length and inner j=1 to j=length-1.

• We will get all substrings using substr(i,j) of all lengths. Pass each substring to check(). If it returns 1, increment count.

• At the end of both for loops count will have a number of substrings of str with the same start and end character.

• Return count as the desired result.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int count_k(string str, int len, int k){
int count = 0;
int arr[26];
for (int i = 0; i < len; i++){
memset(arr, 0, sizeof(arr));
for (int j = i; j < len; j++){
arr[str[j] - 'a']++;
if (arr[str[j] - 'a'] <= k)
{ count++; }
else
{ break; }
}
}
return count;
}
int main(){
string str = "bbddehj";
int k = 1;
int length = str.length();
cout<<"Count of substrings with each character occurring at most k times are: "<<count_k(str,
length, k);
return 0;
}

## Output

If we run the above code it will generate the following output -

Count of substrings with each character occurring at most k times are: 14
Published on 01-Dec-2020 12:12:12


text:
# Definition:Calculus of Variations/Historical Note

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## Historical Note on Calculus of Variations

Some sources suggest that, in a sense, the earliest problem in the calculus of variations arose in one of the legends of the founding of Carthage; the city was granted as much land as could be enclosed by a given length.

The calculus of variations emerged as a branch of mathematics as a result of investigations into the cycloid in the $18$th century.

The first systematic investigation of the topic was given by Joseph Louis Lagrange in his earliest and most important works, together with Leonhard Paul Euler, who coined the term in $1766$.

Karl Weierstrass ushered in a new era of precise reasoning with his lectures in $1879$ on the subject. One of his students, Oskar Bolza, took on the subject and developed the Chicago school of the calculus of variations.


text:
# NEET Practice Questions

• Subject:

...

• Chapter:

...

Four wires of equal length and of resistances 10 ohms each are connected in the form of a square. The equivalent resistance between two opposite corners of the square is

(1) 10 ohm

(2) 40 ohm

(3) 20 ohm

(4) 10/4 ohm


text:
# Must all types of clocks agree on the time that has passed since the spaceship started accelerating?

The above would be true if the spaceship were not accelerating, because SR says that the physical laws are all same in inertial frames of references. If there were a gap in the time of the 2 clocks, a person on the spaceship could measure that gap and calculate his velocity, hence contradicting SR. So, the time on all clocks must be same, irrespective of mechanism, in an inertial frame. But does the same hold true for non-inertial frames?

Edit : The clocks are at the same height.

• Could you please expend a little bit on what you mean by an atomic clock, vs a light clock? It is not immediately clear what you mean. – Mikael Fremling Oct 8 '16 at 11:28
• @MikaelFremling my focus isn't on the type of clock, but on the reason why all clocks must be affected the same way. Apologies for the misunderstanding – ryannmahajan Oct 8 '16 at 16:04

This depends on exactly what you are asking. Clocks at different heights in an accelerating spaceship run at different rates. This is discussed in the question Which clock is the fastest inside an accelerating body?.

All clocks are affected by the time difference in the same way, so it doesn't matter whether the clock is atomic, a light clock, a mechanical clock or whatever, all clocks at the same height in the accelerating spaceship will run at the same rate. However a light clock is traditionally quite large i.e. the light travels some large distance then reflects back, and we measure the time by the duration of the light beam's round trip. If the light beam travels up the accelerating spaceship then it will pass through regions of measurably different time dilation and the time it measures will differ from the time measured by a more compact clock.

So if the light clock is small, or if it's arranged so the light beams stay at the same height as they travel, then the light clock will measure the same time as all the other clocks. If the light clock is large and the light beams travel vertically then there will be a difference.

• Why must all clocks be affected in the same way? – ryannmahajan Oct 8 '16 at 16:00
• @ryannmahajan: time dilation is a geometrical property of spacetime and not dependent on any details of how individual clocks work. For more on this see What is time dilation really?. – John Rennie Oct 8 '16 at 16:27
• You're absolutely right. Look at this way : We generally prove gravitational time dilation by showing that a clock is faster at a height or that photons lose energy and their frequency is lowered and so on. The point is, we can't possibly show it for all the mechanisms (can't show that thought processes and cancer development will be faster). We can, by argument, prove it when acceleration is not involved, as I have in the question's description. But is there such an argument for when acceleration is involved? – ryannmahajan Oct 10 '16 at 10:40
• @ryannmahajan: time dilation is not due to photons losing energy. It arises from the geometrical properties of spacetime. This is what I discuss in the article I linked in my comment above. This approach treats both accelerated and non-accelerated motion. – John Rennie Oct 10 '16 at 15:42
• Off topic. Just like to say thanks for the help. I'm unsubbing from physics. It's a waste of time for experimentalists and the moderation is starting to annoy me. More in meta, but it really doesn't matter. regards Dirk – user56903 Oct 12 '16 at 7:44

The answer depends on the geodesic of a photon in the accelerated frame. The accelerated frame is the Rindler wedge which is illustrated below. The metric in that frame is $$ds^2~=~dT^2~-~dX^2~-~dy^2~+~dz^2$$ with a transformation to the $t,~x$ coordinates of the flat Minkowski spacetime as $$T~=~x~sinh(at),~X~=~x~cosh(at),$$ for $a$ the acceleration. The line element for the path of an accelerated observer in Minkowski spacetime is then $$ds^2~=~a^2x^2dt^2~-~dx^2~-~dy^2~-~dz^2.$$ Let us look at geodesics in the Rindler wedge, which we can find by a variation on the above metric. This leads to $$\frac{d^2t}{ds^2}~+~2x^{-1}\frac{dx}{ds}\frac{dt}{ds}~=~0$$ $$\frac{d^2x}{ds^2}~+~x\left(\frac{dt}{ds}\right)^2~=~0,$$ and we have $d^2y/ds^2~=~d^2z/ds^2~=~0$ From these last two we have $y~=~y_0$ $z~=~z_0$ as constants. The first geodesic equation gives $dt/ds~=~kx^{-2}$, for $k$ a constant. The differential equation for the $x$ coordinate is then $$\frac{dx}{ds}~=~\sqrt{s~+~kx^2~-~y_0^2~-~z_0^2}$$ A solution for the $x$ coordinate position is $$x~=~\sqrt{s^2~+~2s\sqrt{k^2~-~x_0^2(y_0^2~+~z_0^2)}~-~s^2(y_0^2~+~z_0^2)},$$ which is a semicicular arc. This shows that geodesics and null geodesics are semicicular arcs that emerge and enter perpendicular to the Rindler horizon.

With some further geometry one can look at this arc and find its radius. If one is on an accelerated frame a photon clock that measures time should have the photon bounce between mirrors spaced a distance much smaller than the radius of this arc. In that way one can satisfy the criterion for a proper local frame. As the distance between the mirrors increases and approaches scale of these perpendicular arc the photon clock time deviates from the proper time. We may think of these semi-circular arcs as virtual photon loops, where the accelerated observer witnesses the vacuum as the production and destruction of photons across the Rindler horizon. The radius of these loops will be on the order to the distance $d~=~c^2/a$ the horizon is from the observer. For the mirrors in the photon clock a distance $d_{cl}~<<~d$ this means the observer is measuring time in intervals at the UV cutoff in photons produced in Unruh radiation.


text:
# Article

Full entry | PDF   (0.1 MB)
Keywords:
Čech-Stone extension; coloring; Tychonov plank
Summary:
We complement the literature by proving that for a fixed-point free map $f: X \to X$ the statements (1) $f$ admits a finite functionally closed cover $\Cal A$ with $f[A] \cap A =\emptyset$ for all $A \in \Cal A$ (i.e., a coloring) and (2) $\beta f$ is fixed-point free are equivalent. When functionally closed is weakened to closed, we show that normality is sufficient to prove equivalence, and give an example to show it cannot be omitted. We also show that a theorem due to van Mill is sharp: for every $n \geq 2$ we construct a strongly zero-dimensional Tychonov space $X$ and a fixed-point free map $f: X \to X$ such that $f$ admits a closed coloring, but no coloring has cardinality less than $n$.
References:
[1] Aarts J.M., Fokkink R.J., Vermeer J.: Variations on a theorem of Lusternik and Schnirelmann. Topology 35 (1996), 4 1051-1056. MR 1404923
[2] Engelking R.: General Topology. revised and completed ed., Sigma series in pure mathematics, vol. 6, Heldermann Verlag, 1989. MR 1039321 | Zbl 0684.54001
[3] Krawczyk A., Steprāns J.: Continuous colourings of closed graphs. Topology Appl. 51 (1993), 13-26. MR 1229497
[4] van Douwen E.K.: $\beta X$ and fixed-point free maps. Topology Appl. 51 (1993), 191-195. MR 1229715 | Zbl 0792.54037
[5] van Hartskamp M.A., Vermeer J.: On colorings of maps. Topology Appl. 73 (1996), 2 181-190. MR 1416759 | Zbl 0867.55003
[6] van Mill J.: Easier proofs of coloring theorems. Topology Appl. 97 (1999), 155-163. MR 1676678 | Zbl 0947.54019

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