Download FUCHSIAN GROUPS CLASS 7. Tangent bundles and topological

FUCHSIAN GROUPS CLASS
MAT 210A - FALL 2005 - DUCHIN
7. Tangent bundles and topological groups
Notational note: I am going to use kθ for the matrix/FLT
cos θ sin θ
− sin θ cos θ
.
7.1. Vector bundles and the tangent bundle.
Definition. A vector bundle is a structure given by the following data: topological spaces B and E called the base space and total space, a continuous surjection
π : E → B, and a vector space V , such that every fiber is a copy of the vector space
and they fit together well; that is:
• for all p ∈ B, π −1 (p) ∼
= V (isomorphic as vector spaces), and
• for all p ∈ B, there is a neighborhood U of p such that π −1 (U ) ∼
=U ×V.
Clearly E = B × V always works, where π is projection onto the first factor.
Thus, any bundle which is isomorphic to B × V is called a trivial bundle, or just
trivializable if the isomorphism is not specified.
Examples: The cylinder is a trivial line bundle over the circle. The Mobius strip,
given as E = I × R/(0, x) ∼ (1, −x), is a line bundle over S 1 , where π is again
projection to the first factor (because π(E) is the interval I with 0 and 1 glued,
which is a circle).
Definition. A section of a bundle is a map s : B → E such that π ◦ s = id.
So the cylinder has many non-vanishing sections (for example the constant sections s(α) = k 6= 0) but the Mobius strip has none. (Proof: say without loss of
generality s(0) > 0. Then the non-vanishing implies that s(x) > 0 for all x by
continuity, so s(1) > 0, but (1, s(1)) ∼ (0, −s(1)), which contradicts s(0) > 0.
Fact: the bundle E is trivial if and only if there exist n linearly independent
sections, where n = dim V .
Proof: ⇒ is clear: if E ∼
= B × V , take a basis e1 , . . . , en of V and the constant
sections si (p) = ei .
⇐ is not hard: Given si which are linearly independent at every point, it’s
immediate that {si (p)} are a basis for the vector space π −1 (p). A trivialization is
an
P isomorphism φ : E → B × V , so for each (p, v) ∈ E, write v in the basis (say v =
ai si (p)) and let φ(p, v) = (p, a1 , a2 , . . . , an ). This is clearly an isomorphism. Now, for a manifold M , its tangent bundle is just T M = {(p, v) : v ∈ Tp M }.
That is, the vectors at a point are the tangent vectors: these form a vector space.
(Local triviality follows from the existence of charts, which are local maps to Rn ;
the tangent vectors can be defined as tangents of curves in the charts, and the
tangent space to Rn is canonically identified with itself.) In particular, if M is an
n-dimensional manifold, dim Tp M = n, and in fact T M is a (2n)-manifold. (If the
atlas for M was (Uα , ϕα ), then the atlas for T M can be simply (Uα × V, ϕα × id).
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MAT 210A - FALL 2005 - DUCHIN
Definition. A section of T M is called a vector field on M : it is a choice of a
tangent vector at every basepoint.
Example: The cylinder is (isomorphic to) the tangent bundle of the circle S 1 ,
and the counter-clockwise pointing unit vector is an example of a vector field. That
is, s(θ) = (− sin θ, cos θ).
Definition. A manifold is called parallelizable if its tangent bundle is trivial.
Nice theorem: the sphere S n is parallelizable exactly when n = 1, 3, 7.
Another nice theorem: the sphere S n admits a nonvanishing vector field if and
only if n is odd.
For the second theorem, we can see the “if” direction readily: consider the sphere
S n as the set of unit vectors in Rn+1 . Then consider the vector field given by
s(x) = s(x1 , x2 , x3 , x4 , . . . , xn , xn+1 ) = (−x2 , x1 , −x4 , x3 , . . . , −xn+1 , xn ),
rigged so that s(x) is perpendicular to x via the usual dot product. Of course,
the entries only arrange in pairs when n + 1 is even, that is when n is odd. (In
particular, note this is exactly the section of T S 1 given above.)
The dimensions 1,3, and 7 come up in the first theorem because S 1 is the unit
sphere in C, S 3 is the unit sphere in the space of quaternions, and S 7 is the unit
sphere in the space of octonions– these (plus R itself, whose unit sphere has no
tangent bundle to speak of!) are the only examples of a division algebra structure on Rn . (A division algebra is a not-necessarily-associative algebraic structure
that has a bilinear composition law, an identity, and no zero divisors. Ordinary
multiplication on R and complex multiplication regarded as an operation on R2
are examples.) The idea of the trivialization is to take advantage of the algebraic
structure to produce vectors perpendicular to a given unit vector. For example,
in the quaternion case, every point has a representation p = a1 + a2 i + a3 j + a4 k
in the basis 1, i, j, k. But then p, pi, pj, and pk are linearly independent, so the
projections onto the tangent plane of pi, pj, and pk must be a basis for Tp (S 3 ).
This gives three nonvanishing sections, as required.
Back to the hyperbolic plane: it’s clear that T H is a trivial bundle, because the
constant sections i and 1 give a trivialization.
We will often work with the unit tangent bundle of H. We could either think
of this as unit tangent vectors in the hyperbolic metric, or as unit tangent vectors
in the trivalization (so with the usual Euclidean metric). Katok uses SH for the
former:
SH = {(z, ζ) : z ∈ H, kζkz = 1},
where the norm of ζ is |ζ|/ Im z because of how the Riemannian metric was defined.
That is, the unit vector at a point z are the ones whose Euclidean length is Im z.
The action of P SL2 R on H extends to an action on SH in the expected way:
the usual fractional linear transformation action on points of H and the derivative
action on tangent vectors. That is,
az + b
1
0
a
b
T.(z, ζ) = c d .(z, ζ) = (T z, T (z)ζ) =
,
ζ .
cz + d (cz + d)2
(Note that we can define a metric on SH by the length element (ds2 +dθ2 )1/2 and
a volume by dA dθ, and then check that the P SL2 R action on SH is by isometries.)
It is now easy to verify that, by associating an isometry T with where it takes a
fixed basepoint in the tangent bundle– say (i, ζ0 ), being i with the upward-pointing
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unit vector, we get a bijection of P SL2 R with SH. (To check this, just show that it
is injective and surjective; the key observation for both of those parts is that elliptic
element kθ rotates the tangent vector by 2θ, and the other two types of elements
do not change the tangent vector.)
7.2. P SL2 (R) as a topological group.
Definition. A topological group G is a topological space and also a group such
that those structures are compatible: composition (G × G → G) and the taking of
inverses (G → G) are continuous maps. A Lie group has the compatible structure
of a group and a differential manifold: in this case, the group laws have to be
differentiable maps.
Examples: All Lie groups are topological groups. (R, +) is a Lie group because
addition and x 7→ −x are clearly differentiable. A whole boatload of classical matrix
groups are Lie groups. Matrix groups are sometimes called linear groups because
of their action by linear transformations. One way to present the classical groups is
in terms of preserving certain bilinear forms. Here is a brief rundown: the standard
bilinear form on Rn is the usual dot product x · y = x1 y1 + x2 y2 + . . . + xn yn .
This can also be written hx, yi = xt In y, namely matrix multiplication where In is
the n × n identity matrix. If the bilinear form hx, yi is given instead by (x1 y1 +
. . . + xp yp ) − (xp+1 yp+1 + . . . xp+q yp+q ), we say it has signature (p, q). In this case,
we have hx, yi = xt Ip,q y, where Ip,q is the square diagonal matrix with its first p
diagonal entries equal to 1 and the following q entries equal to −1.
• GLn (R) is the group of nonsingular n × n matrices, as usual, and SLn (R)
is the subgroup with determinant 1—appending an S to the beginning of a
Lie group means, in general, restricting to the determinant one subgroup.
Appending a P to the beginning (as with P SL2 (R)) denotes the quotient
by the center of the group.
• the orthogonal group is
O(n) = {M ∈ GLn (R) : M t IM = I}.
This is the subgroup of GLn containing matrices which preserve the standard bilinear form:
hM x, M yi = (M x)t I(M y) = xt (M t IM )y = hx, yi.
Likewise, O(p, q) = {M ∈ GLp+q (R) : M t Ip,q M = Ip,q }.
• SO(n) is the subgroup of O(n) consisting of determinant one matrices: preserving the dot product means preserving norm, and positive determinant
means orientation-preserving, so this is exactly the collection of rotations
about the origin in Rn . The groups SO(n, 1) are called the Lorentz groups
and come up a great deal in physics, including relativity theory;
• the unitary group is the analogous group over the complex numbers:
U (n) = {M ∈ GLn (C) : M ∗ IM = I};
• if Jn,n is the 2n × 2n matrix −I0 n I0n then the symplectic group is the
matrices preserving its bilinear form:
Sp(n, C) = {M ∈ SL2n (C) : M t Jn,n M = Jn,n };
Note that the Lie group SO(2) is exactly the subgroup of SL2 (R) that we have
been calling K.
Remark. There are several Lie group “coincidences” in low dimensions: for instance,
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MAT 210A - FALL 2005 - DUCHIN
• SU (2) is homeomorphic to the sphere S 3 ;
• SO(3) is homeomorphic to real projective 3-space (so it is double-covered
by SU (2));
• SO(2, 1) ∼
= SL2 (R); and
• SO(3, 1) ∼
= SL2 (C).
We can topologize P SL2 R as the quotient of a subspace of R4 (map ac db to
(a, b, c, d) and use the subspace topology from R4 to get a topology on SL2 R, then
quotient by {±I}). In fact, there are any number of reasonable ways to topologize
the group, and they all turn out to be equivalent.
Claim. As topological groups, we have the isomorphisms K ∼
= S1, A ∼
= R, and
∼
N = R.
0
Proof. Let φ(kθ ) = θ, φ( k0 1/k
) = ln k, and φ( 10 n1 ) = n. These are group
homeomorphisms because in each case φ(ST ) = φ(S)φ(T ) and each of these is
obviously a bi-continuous map.
Please note, I’m going to use the notation K, A, N for the subgroups of the
specified forms no matter whether the setting is SL2 R or P SL2 R. None of the
claims is affected by the ambiguity.
Claim (Iwasawa Decomposition). P SL2 R = N AK as topological groups.
Proof. Consider the following map SH → N × A × K. Build up φ(z, ζ) by choosing
(1) the unique elliptic kθ such that d(kθ )i (ζ0 ) = ζ (the rotation such that the
derivative sends the upward pointing tangent vector to ζ); (2) the unique parabolic
T = 10 n1 in the upper half-plane model sending i to a point on the same Euclidean
line as 0 and z; (3) the unique hyperbolic T 0 that sends this translated point to z
by rescaling along that line. Then, by construction, T 0 T kθ (i, ζ0 ) = (z, ζ). Now via
the earlier bijection of SH with P SL2 R, we have an isomorphism as needed– it is
easy to see that it is a homeomorphism.
7.3. Hyperbolic n-space and the symmetric space structure. Recall the
definitions of orbits and stabilizers for group actions: if G acts on X, then the
G-orbit of x ∈ X is G.x = {gx : g ∈ G} and the stabilizer in G of x is StabG (x) =
{g ∈ G : gx = x}.
Important and obvious note: G.x is in bijective correspondence with G/ StabG (x).
That is because the orbit {gx}, as a set, is clearly {g} modulo the equivalence relation g ∼ h ⇐⇒ gx = hx. But then g ∼ id exactly when g ∈ Stab(x), so
G.x ∼
= G/ Stabg (x).
In particular, in the case of the P SL2 (R) action on H, the orbit of i is all of H
and the stabilizer of i is the subgroup K, so
H∼
= P SL2 (R)/K = SL2 (R)/K = SL2 (R)/SO(2).
(The first equality follows from observing that ±I ∈ K.) In this case, it is clearly a
homeomorphism. Also note that since SL2 (R) ∼
= N AK, this identification can be
rewritten H ∼
= N A—since a tranformation in N A is of the form z 7→ k 2 z + n, this
is especially obvious: any point in H can be written as k 2 i + n for some real k and
n.
Note that in quite the same way, the sphere S n can be realized as
SO(n + 1)/SO(n). Just think of the rotations in Rn+1 . The orbit of the north pole
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is the unit sphere S n , and the stabilizer of the north pole is the rotations that fix
its axis; these are rotations of the copy of Rn which is orthogonal to that axis.
The rest of this section is strictly intended for flavor; I’m just going to survey
some facts.
Hyperbolic n-space (or, fully, real hyperbolic n-space, since there are complex
and quaternionic versions as well) can be given an upper
√ half-space model via
dx2 +...dx2
n
1
Hn = {x ∈ Rn : xn > 0} with the length element ds =
. It will turn
xn
2
out that the geodesics, as in H , are orthocircles. Also, via a conformal map, the
upper half-space can be mapped to an open unit ball such that the geodesics are,
again, orthocircles.
Fact: The group of orientation-preserving isometries of Hn is isomorphic to the
Lie group SO+ (n, 1), which is the component of SO(n, 1) containing the identity
matrix. It acts doubly transitively on the boundary: any geodesic in Hn can be
taken to any other geodesic by some isometry.
Fact: The stabilizer of any point in Hn is conjugate to SO(n) (and should be
thought of as rotation around the origin in the unit ball model where the point is
mapped to the origin).
It follows that Hn ∼
= SO+ (n, 1)/SO(n). From the low-dimensional coincidences
listed above for Lie groups, we see that H2 ∼
= SL2 (R)/SO(2) and H3 ∼
= SL3 (C)/SU (2).
Definition. A homogeneous space is any space of the form G/H for G a Lie
group and H a closed subgroup. A symmetric space is any space of the form
G/K for G a Lie group and K a maximal compact subgroup.
Alternatively, a homogeneous space is one in which there is a transitive action
by some Lie group: the space is the orbit of any one basepoint, by transitivity, so
it gets its structure G/H as the group modulo the stabilizer of the basepoint. You
should think about this as any point being equivalent to any other: if you stand at
a point and survey the geometric landscape, you will not be able to tell where you
are standing.
A symmetric space is alternatively defined as one which has a certain symmetry
condition: every point has a neighborhood where there is an involutive (order two)
isometry with p as an isolated fixed point. That this is equivalent to the other
definition is a nontrivial theorem.
It is clear that the hyperbolic plane has such a symmetry at every point: pick a
geodesic going through the point and choose the isometry that reverses its two endpoints. It follows that the isometry reverses the orientation on all of the geodesics
through the point; this clearly has order two. In fact, this works for Hn and spheres
S n as well, so we should expect them to have the structure of symmetric spaces,
and indeed they do. (See above.)
Since compact groups are closed, all symmetric spaces are clearly homogeneous
spaces: in general, this means we should be able to build a transitive group action
out of having symmetries at every point. For geodesic spaces, this isn’t hard: to
get a map from p to q, choose the geodesic between them and do a geodesic-reversal
based at the midpoint.