Download Lecture 9, October 17. The existence of a Riemannian metric on a C

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Cross product wikipedia , lookup

Fundamental theorem of algebra wikipedia , lookup

Dual space wikipedia , lookup

Euclidean space wikipedia , lookup

Oscillator representation wikipedia , lookup

Linear algebra wikipedia , lookup

Euclidean vector wikipedia , lookup

Vector space wikipedia , lookup

Laplace–Runge–Lenz vector wikipedia , lookup

Matrix calculus wikipedia , lookup

Basis (linear algebra) wikipedia , lookup

Bra–ket notation wikipedia , lookup

Cartesian tensor wikipedia , lookup

Covariance and contravariance of vectors wikipedia , lookup

Four-vector wikipedia , lookup

Transcript
Lecture 9, October 17.
The existence of a Riemannian metric on a C ∞ manifold.
This is given by Proposition 2.10 on p. 43 of do Carmo (we give his proof at the end of today’s
notes).
Theorem 1 If Mn is a C ∞ manifold whose topology is Hausdorff and has an atlas with a countable
number of charts, then there exists a C ∞ Riemannian metric g on M. In particular, each closed
manifold admits a Riemannian metric.
Lie groups.
Recall that a Lie group G is a set G with · : G × G → G such that (G, ·) is a group and with a
manifold structure on G such that the map φ : G × G → G defined by φ (x, y) + x · y −1 is C ∞ .
Let x ∈ G. Define the right and left multiplication maps
C∞
Rx : G → G
and Lx : G → G
by
Rx (y) = yx and Lx (y) = xy,
respectively. It is easy to see that Rx and Lx are diffeomorphisms with inverses Rx−1 and Lx−1 ,
respectively.
A C ∞ vector field X on G is left-invariant if
dLx (Xy ) = Xxy ∈ Txy G
for each x, y ∈ G. Right-invariant means
dRx (Xy ) = Xyx
for each x, y ∈ G.
Let e denote the identity element of G. Given a tangent vector Xe ∈ Te G, we define a corresponding
left-invariant vector field (LIVF) on G by
Xx = dLx (Xe )
(1)
for each x ∈ G.
Exercise 2
1. Show that X defined by (1) is left-invariant.
2. Show that if X and Y are LIVFs, then X = Y if and only if Xe = Ye .
3. Show that if X and Y are LIVFs, then [X, Y ] is a LIVF. So the set of LIVFs forms a Lie algebra.
So we may identify the set of left-invariant vector fields on G with Te G. In this way Te G is a Lie
algebra with Lie bracket defined by
[Xe , Ye ] + [X, Y ]e
for LIVFs X and Y.
Let h , ie be an inner product on Te G. Define a corresponding Riemannian metric h , i on G by
hu, vix = h(dLx−1 )x (u) , (dLx−1 )x (v)ie
1
(2)
for any x ∈ G and u, v ∈ Tx G. If a Riemannian metric g on G satisfies (2), then we say that g is a
left-invariant metric. Note that if u, v ∈ Ty G, then
D
E
= hu, viy .
(dLx )y (u) , (dLx )y (v)
xy
In other words, the left-invariant metric g = h , i satisfies
(dLx )∗ g = g.
(Recall that for a diffeomorphism ϕ : M → N , the pull back metric is defined by (ϕ∗ h) (X, Y ) =
h (dϕ (X) , dϕ (Y )) .) I.e., the diffeomorphisms Lx are isometries of g.
One similarly defines right-invariant metric on a Lie group.
Euclidean covariant derivative.
A vector field on Rn is equivalent
to a function Y : Rn → Rn , which we may write as an n-tuple of
functions Y = Y 1 , . . . , Y n , where each Y i : Rn → R. Given a vector Xp at a point p ∈ Rn , we may
define the Euclidean covariant derivative by
DXp Y = Xp Y 1 , . . . , Xp Y n ,
where Xp Y i is the directional derivative of Y i in the direction Xp . We have the linearity properties:
(1)
DXp +cWp Y = DXp Y + aDWp Y
for a vector field Y, vectors Xp , Wp ∈ Tp Rn = Rn and a ∈ R.
(2)
DXp (Y + bZ) = DXp Y + bDXp Z
for a vector fields Y and Z, a vector Xp ∈ Tp Rn = Rn and b ∈ R.
Of course, for functions f, g : Rn → R we have the product rule Xp (f g) = gXp f + f Xp g. So
(3)
DXp (f Y ) = Xp f Y 1 , . . . , Xp (f Y n )
= Xp f Y 1 + f Xp Y 1 , . . . , Xp f Y n + f Xp Y n
= Xp f Y + f DXp Y.
For vector fields X and Y and any function f : Rn → R,
(DX Y − DY X)p f = Xp Y 1 , . . . , Xp Y n f − Yp X 1 , . . . , Yp X n f
=
n
X
Xp Y i ei f − Yp X i ei f .
i=1
2
On the other hand,
[X, Y ]p f = Xp (Y f ) − Yp (Xf )
!
n
X
= Xp
Y i ei f − Yp
n
X
i=1
=
n
X
=
=
i=1
n
X
X i ei f
i=1
i
Xp Y ei f +
i=1
n
X
!
n
X
Ypi Xp (ei f )
−
n
X
i=1
i
Yp X ei f −
n
X
i=1
n
X
Xp Y i ei f − Yp X i ei f +
i=1
n
X
Ypi Xpj ej (ei f ) −
i,j=1
Xp Y i ei f − Yp X i ei f
Xpi Yp (ei f )
Xpi Ypj ej (ei f )
i,j=1
i=1
= (DX Y − DY X)p f.
Hence
(4)
[X, Y ] = DX Y − DY X.
Finally, given vector fields Y and Z on Rn , we have the inner product:
1
1
n
n
hY, Zi = Y Z + · · · + Y Z =
n
X
Y iZ i.
i=1
Thus, by the product rule,
(5)
Xp hY, Zi = Xp
n
X
!
Y iZ i
i=1
=
n
X
i
i
Z Xp Y +
i=1
n
X
Y i Xp Z i
i=1
= DXp Y, Z + Y, DXp Z .
Proof of Theorem 1. Fact. Under the above hypotheses, there exists a countable number of
coordinate charts {(Uα , xα )}α∈A such that {xα (Uα )}α∈A is a locally finite cover of M (i.e., for each
p ∈ M there exists a neighborhood V of p such that only a finite number of the xα (Uα ) intersect V)
and there exists a C ∞ partition of unity {ψα }α∈A subordinate to the cover {xα (Uα )}α∈A of M, i.e.,
1. ψα : M → [0, 1] is a C ∞ function with support in xα (Uα ) for each α ∈ A,
P
2.
α∈A ψα (p) = 1 (this is a finite sum) at each point p ∈ M.
Let {(Uα , xα )}α∈A and {ψα }α∈A be as above. For each α ∈ A, define the Riemannian metric g α
on xα (Uα ) by
D
E
−1
g α (u, v) = d x−1
(u)
,
d
x
(v)
α p
α p
n
R
).1
gα
for u, v ∈ Tp M, where p ∈ xα (Uα
We extend
to a symmetric bilinear form on each Tp M
for p ∈ M by setting it to be zero outside of xα (Uα ). Then ψα g α is a C ∞ symmetric bilinear
1
Note that g α
∂
, ∂
∂xiα ∂xjα
= δij .
3
form on each Tp M for p ∈ M. Note that ψα g α is positive definite wherever ψα 6= 0 (recall that
supp (ψα ) ⊂ xα (Uα )). Define
X
g+
ψα g α ,
α∈A
which at each point of M is a finite sum. g is a C ∞ symmetric bilinear form on each Tp M for
p ∈ M. Moreover, for any p ∈ M there exists β ∈ A such that ψβ (p) > 0 and p ∈ xβ (Uβ ). Then
for any u ∈ Tp M − {0} we have
X
g (u, u) =
ψα g α (u, u) ≥ ψβ g β (u, u) > 0.
α∈A
Of course, gp (0, 0) = 0 for any p ∈ M. Thus g is a C ∞ Riemannian metric on M.
4