## FANDOM

39 Pages

This seminar by kommodore took place on 10th August 2008 16:00 UTC, in #mathematics channel of irc.freenode.net.

The timestamps displayed in the IRC text are UTC+1.
Any explanation or errors regarding this seminar should be recorded in the discussion page.

## Topic Edit

Curvature. Part of the series outlined at Introductory Riemannian Geometry

## Seminar Edit

17:00:01 ChanServ changed the topic of #mathematics to: SEMINAR IN PROGRESS.  If you want to
ask a question, say ! and wait to be called
17:00:46 kommodore: For this seminar, our M is assumed to be connected, of dimension n>1,
ask a question, say ! and wait to be called unless otherwise specified.
17:01:05 kommodore: I won't be introducing extrinsic curvature this time.
17:01:10 ness: !
17:01:16 kommodore: Recall from last time: we have defined the _curvature form_ of a
connexion on a vector bundle E->M.
17:02:13 ness: just for clarity, M is a differentiable manifold, i.e. a topologic manifold
with a differentiable structure, right?
17:02:45 kommodore: M is a differential manifold with Riemannian metric g
17:03:32 kommodore: If M is Riemannian, and E=TM its tangent bundle, we saw there is a unique
torsion-free connexion compatible with the metric g.  So we may ask: is
there anything special about the curvature form of the Levi-Civita
connexion, which is a tensor on M (the Riemann curvature tensor)?
17:03:52 kommodore: Indeed there is.  It plays a really fundamental part in almost everything
we do in Riemannian geometry.
17:04:02 kommodore: However, this definition of the curvature tensor is sometimes a bit
awkward to work with.  So we will seek some alternative definitions.
17:04:30 kommodore: Recall the Levi-Civita connexion defines a covariant derivative
D_X:(sections of E)->(sections of E) for every vector field X.
Therefore, a natural question to ask is: do these covariant derivatives
commutes?
17:04:51 kommodore: Unfortunately, the result of (D_X D_Y Z - D_Y D_X Z)(p) does not only
depend on the values of Z^{i}, @_{j}Z^{i}, @_{jk}Z^{i}, X^{i}, Y^{i} at p,
but also (@_{j}X^{i})(p), (@_{j}Y^{i})(p).  So we could not hope for,
even in the simplest case of (R^n, usual metric), to have
D_X D_Y Z = D_Y D_X Z.
17:05:16 kommodore: Fortunately, this dependecy on the first order derivatives of X and Y
are easy to get rid of.
17:05:28 kommodore: Lemma: D^2_{X,Y}Z=D_X D_Y Z - D_{D_X Y} Z is tensorial in X,Y, for *any*
connexion D on M.
17:05:49 kommodore: Proof: Tensoriality in X is clear by definition of connexion.  So
suffices to check Omega^0(M)-linearity on Y.
17:05:59 kommodore: Using Leibniz rule and R-linearity, we get
17:06:15 kommodore: D_X D_{fY} Z - D_{D_X fY} Z = D_X(fD_Y(Z)) - D_{fD_X(Y)+X(f)Y} Z
17:06:25 kommodore: = fD_X(D_Y(Z))+X(f)D_Y(Z)-fD_{D_X(Y)}Z+X(f)D_Y(Z)
17:06:33 kommodore: = f[D_X D_Y Z - D_{D_X Y}Z].  ////
17:06:52 kommodore: Consequently, D^2_{X,Y}Z-D^2_{Y,X}Z is tensorial in X,Y, i.e.
D_X D_Y Z - D_Y D_X Z - D_{[X,Y]}Z is tensorial in X,Y.  In fact,
17:07:19 kommodore: Theorem: D^2_{X,Y} Z - D^2_{Y,X} Z is tensorial in X,Y *and* Z, if D is
torsion-free.
17:07:29 kommodore: Proof: This is just a matter of calculation.
17:07:39 kommodore: D^2_{X,Y}fZ - f D^2_{X,Y}Z
= D_X D_Y(fZ)-D_{D_X(Y)}(fZ) - f D_X D_Y(Z) + fD_{D_X(Y)}Z
17:07:50 kommodore: =D_X (f D_Y Z + Y(f)Z) - f D_{D_X(Y)}Z - (D_X(Y))(f) Z - f D_X D_Y(Z)
+ f D_{D_X(Y)}Z
17:08:00 kommodore: =X(f)D_Y(Z) + X(Y(f))Z + Y(f) D_X Z - (D_X(Y))(f)Z
17:08:08 kommodore: Swapping X,Y
17:08:19 kommodore: D^2_{Y,X}fZ - f D^2_{Y,X}Z = Y(f)D_X(Z) + Y(X(f))Z + X(f) D_Y Z
- (D_Y(X))(f)Z
17:08:30 kommodore: So subtracting,
17:08:39 kommodore: [D^2_{X,Y}-D^2_{Y,X}](fZ)-f[D^2_{X,Y}-D^2_{Y,X}](Z)
= [X,Y](f)Z - (D_X(Y)-D_Y(X))(f)Z
17:08:54 kommodore: But D is torsion-free, so D_X(Y)-D_Y(X)=[X,Y].  So
D^2_{X,Y}Z-D^2_{Y,X}Z is tensorial in Z too.  ////
17:09:21 kommodore: Finally, note that this is exactly what is going on when we take the
curvature form
17:09:47 kommodore: the dA part is antisymmetric
17:10:18 kommodore: and the A\wedge A part is what is added to make it invariant under
change of coordinates
17:10:34 kommodore: So we can define the curvature tensor as
17:10:40 kommodore: R(X,Y)Z=D^2_{X,Y}Z-D^2_{Y,X}Z.
17:12:30 kommodore: Before presenting another way to look at this curvature tensor, we have
some symmetry of R that is worth mentioning:  For brevity, write
R_{ijkl} for <R(@_i,@_j)@_k,@_l>.  Then we have
17:12:47 kommodore: (1) R_{ijkl}=-R_{jikl}.
17:12:54 kommodore: (2) R_{ijkl}=R_{klij}.
17:13:08 kommodore: The first two implies R can be thought of as a symmetric linear map on
Lambda^2T^*_p.
17:14:28 kommodore: (3) R_{ijkl}+R_{jkil}+R_{kijl}=0.
17:14:36 kommodore: (4) D_{h}R_{ijkl}+D_{i}R_{jhkl}+D_{j}R_{hikl}=0.
17:14:52 kommodore: These are the first and second Bianchi identity, although the first is
actually due to Ricci.  (3) essentially boils down to the Jacobi
identity, while (4) is proved last time.
17:15:15 kommodore: From the first three, we conclude that the number of linearly
independent components of R_{ijkl}:
17:15:33 kommodore: i,j,k,l distinct: 24 permutations, comes in 3 groups of 8 (within each
group are +/- each other by (1) and (2)).  Then (3) gives a linear
relation between the three groups, so only 2 linearly independent choices.
17:16:33 kommodore: 1 pair in i,j,k,l:  WLOG consider 1,1,2,3.  Then only one choice:
R_{1213}, the rest are forced.
17:17:05 kommodore: 2 pairs: Again, WLOG consider 1,1,2,3.  Only one choice: R_{1212}, the
rest are forced
17:17:15 kommodore: So the number of linearly independent components are
2*(n choose 4)+n*(n-1 choose 2)+(n choose 2)=n^2(n^2-1)/12.
This is, of course, assuming we have no other additional constraints.
17:18:32 kommodore: Yet another way of looking is by taking sectional curvature, which is
essentially how Riemann did it back in 1854.  For this, I'll need to
introduce geodesics and the exponential map.
17:18:49 kommodore: Definition: the length of a C^1 curve gamma:[0,1]->M is
17:18:56 kommodore: \int_0^1 |gamma'(t)| dt
17:19:03 kommodore: where |gamma'(t)|^2=g_{gamma(t)}(gamma'(t),gamma'(t)).
17:19:19 kommodore: Definition: A C^1 curve gamma is a geodesic if D_{gamma'}gamma'=0.
17:19:36 kommodore: Note 1: for this to make sense, we need to extend, for each t and some
epsilon>0 (depending on t), the tangent vectors
{v=gamma'(s):|s-t|<epsilon} to a vector field on some neighbourhood of
gamma(t).  But it is easy to see the result doesn't depend on how we
make the extension.
17:20:04 kommodore: Note 2: Note that this implies |gamma'| is constant, because
17:20:05 kommodore: (d/dt)g(gamma'(t),gamma'(t))=(gamma')g(gamma'(t),gamma'(t))
=2g(D_{gamma'(t)}gamma'(t),gamma'(t))=0
17:20:31 kommodore: Note 3: All geodesics are smooth, because the equation
D_{gamma'}gamma'=0 is elliptic.
17:21:02 kommodore: Gauss' Lemma: let u,v be perpendicular in T_pM.
Then <(d exp_p)_{tu}(v),gamma_{(p,u)}'(t)> =0, whenever exp_p(tu) is
defined. ////
17:21:25 kommodore: oops, need to introduce exp_p first...
17:21:41 kommodore: For each point p in M and v in T_pM, we define the geodesic
gamma_{(p,v)}(t) to be the solution to D_{gamma'}gamma'=0 with initial
condition gamma(0)=p, gamma'(0)=v.
17:22:03 kommodore: By the usual existence of solutions to ODEs, we have gamma_{(p,v)}(t)
exists for |t|<epsilon(p), if we choose |v|<C(epsilon(p),p).
17:22:26 kommodore: Also, by uniqueness of solutions to ODEs, we have
gamma_{(p,v)}(ct)=gamma_{(p,cv)}(t) for all c>0, providing t is
sufficiently small.
17:23:03 kommodore: (actually for all nonzero c, but we won't need c<0 case)
17:23:19 kommodore: Thus, we make the definition:
17:23:57 kommodore: The Riemannian exponential map exp_p (terminology inspired by Lie
theory) is defined by exp_p(v)=gamma_{(p,v)}(1).  This map is defined at
least for v in a neighbourhood of 0 in T_pM, for all p.  Moreover, it is
easy to check (d exp_p)(0) is the identity map on T_pM (when we make the
usual identification of T_0 T_pM with T_pM).  If @_1,...,@_n is an
orthonormal basis of T_pM, the coordinates x^1,...,x^n is called
the (geodesic) normal
17:25:02 kommodore: Now we can make sense of Gauss's lemma, and justify why we put the
condition D_{gamma'}gamma'=0
17:25:24 kommodore: For reference again, here is Gauss's lemma
17:25:27 kommodore: Gauss' Lemma: let u,v be perpendicular in T_pM.
Then <(d exp_p)_{tu}(v),gamma_{(p,u)}'(t)> =0, whenever exp_p(tu) is
defined. ////
17:26:02 kommodore: Let's decipher this a little.  exp_p is a map T_pM -> M, so its
derivative at tu is a map T_pM=T_{tu}T_pM -> T_{exp_p(tu)}M.  Also,
gamma_{(p,u)}'(t) is the velocity vector of gamma_{(p,u)} at exp_p(tu),
so the two are both elements of T_{exp_p(tu)}M, so we can take their
inner product using g_{exp_p(tu)}.
17:27:06 kommodore: So Gauss' Lemma says rays and tangent vectors to geodesic spheres are
orthogonal
17:27:36 kommodore: (here rays take the usual Euclidean meaning, not the Riemannian meanning)
17:28:01 kommodore: Using Gauss's Lemma, we can reproduce the familiar argument:
If gamma(t)=r(t)xi(t) parametrises a curve, with r(0)=0, r(t) real,
xi(t) taking value in S^{n-1} in T_pM, then the radial and spherical
component of gamma'(t) are perpendicular.
17:28:36 kommodore: So |gamma'(t)|>=|r'(t)|, and integrating gives L(gamma([0,t]))>=r(t),
so geodesics (which are straight lines in normal coordinates) is
locally length-minimising, which is what we expect.
17:29:14 kommodore: Remark: by using the "full" version of the exponential map
Exp: TM --> M\times M; v|->(pi(v),exp_{pi(v)}(v)), we can show that in
a sufficiently small neighbourhood of any point p, any
two points p_1,p_2 can be joined by a minimising geodesic.
Such neighbourhoods are called _convex_ neighbourhood.
17:29:48 kommodore: Now, we recall we know how to define Gaussian curvature at a point of a
surface.  So we make the following definition:
17:30:07 kommodore: Definition: The _sectional curvature_ K(sigma) of a 2-plane sigma
in T_pM, p in M, is defined to be "the Gaussian curvature at p to
the surface exp(small neighbourhood of 0 in sigma)".
17:30:39 kommodore: Moreover, we define K(u,v)=K(span{u,v}) |u\wedge v|^2 for all u,v
tangent vectors in the same tangent space T_pM.  Then K:T_pMxT_pM->R
is symmetric bilinear (recall from linear algebra that the inner product
on V induces an inner product on all tensor powers, hence on exterior
powers, of V).
17:31:02 kommodore: K is related to our previous definition of the curvature tensor R by
K(u,v)=<R(u,v)v,u>
17:31:21 kommodore: Claim: for each symmetric bilinear K, there is exactly one R satisfying
the symmetry (1)--(3) (antisymmetric in first two, can swap 2 blocks of
2s, first Bianchi).
17:31:55 kommodore: Proof: polarise one of the variables is easy.  Then polarise the other
will leave you with some unpleasant terms, but they can be rid of using
the first Bianchi identity.  The full result is
17:32:15 kommodore: 6R(u,v,w,z)=[K(u+z,w+v)-K(u+z,v)-K(u+z,w)-K(u,v+w)-K(z,v+w)-K(v+z,u)
+K(u,w)+K(v,z)]
-[K(u+w,v+z)-K(u+w,v)-K(u+w,z)-K(u,v+z)-K(w,v+z)-K(u+w,v)
+K(v,w)+K(u,z)]
17:32:26 kommodore: ////
17:32:44 kommodore: Recall R has n^2(n^2-1)/12 components.  This is sometimes too many to
carry around (and too strict a condition).  So we may want to ask: are
there other simpler version that we can try?
17:33:14 kommodore: There is a natural action of O(n) on each tangent space, so we can
regard R as a representation of O(n).  So the first question is: is this
representation reducible, and what are its irreducible components?
17:33:43 kommodore: The Ricci curvature is defined by averaging:
17:33:51 kommodore: Ric(u,v)=sum_i R(u,e_i,e_i,v)/(n-1),
17:34:07 kommodore: where {e_i} is an orthonormal basis of T_pM.
17:34:16 kommodore: (Some books don't have the factor n-1.)
17:34:53 kommodore: This is a symmetric bilinear form on T_pM, so we also have
Ric_p(u)=Ric(u,u) for any tangent vector u in T_pM
17:35:11 kommodore: The scalar curvature is defined by a further averaging:
17:35:24 kommodore: scal=sum_{j} Ric(e_j,e_j)/n=sum_{i,j}R(e_j,e_i,e_i,e_j)/[n(n-1)].
(again, some book don't have the factor n)
17:35:48 kommodore: we can write down a decomposition of R explicitly, as
17:35:55 kommodore: R=scal+Ric_0+W
17:36:10 kommodore: where:
17:36:26 kommodore: Ric:=scal+Ric_0 is the Ricci curvature as a sub-representation
17:36:41 kommodore: scal is the "sub-sub-representation"
17:36:52 kommodore: and W is the Weyl curvature tensor.
17:37:05 kommodore: W is what is left after we split off these averages
17:37:21 kommodore: In dimension=4, W has a further decomposition into W^+ and W^-,
the self-dual and anti-self-dual part of W.
17:37:37 kommodore: We call Ric_0=Ric-scal*g the "traceless Ricci" curvature tensor.
17:38:01 kommodore: Gauss and Riemann identify R as the obstruction to writing
17:38:07 kommodore: g=(dx^1)^2+...+(dx^n)^2
17:38:17 kommodore: in some coordinates x^1,...,x^n, i.e. R=0 is a necessary and sufficient
condition for the existence of such x^1,...,x^n.  Such a space is
called flat.
17:39:11 kommodore: W has the nice feature of being "conformal invariant" (or equivariant if
you lower/raise some indices)
17:39:25 kommodore: We say two metrics g,g' on M are conformal if g'=e^(2f)g for some smooth
function f on M.  The conformal invariance of W is then expressed by
W^i_{jkl}=(W')^i_{jkl}
17:39:55 kommodore: Since the Weyl curvature is conformal invariant, it should not surprise
you that W=0 is a necessary and sufficient condition that the space is
"conformally flat", i.e. there exists smooth f, and coordinates
x^1,...,x^n, such that
17:40:08 kommodore: e^(2f)g=(dx^1)^2+...+(dx^n)^2.
17:40:43 kommodore: The scalar curvature carries very little information.  For example, if
dimension>=3, and f is a smooth function that changes sign, then there
exists a metric g on M with scal(g)=f.  So most local-to-global theorems
assume at least some condition on Ricci.
17:41:20 kommodore: So now let's have some example of computing sectional curvature...
17:41:31 kommodore: Example 0: R^n with the usual metric is flat.  Indeed, covariant
derivative is the usual derivative, which commutes.
17:41:51 kommodore: Example 1: spheres.
17:41:59 kommodore: The sphere S^n in R^{n+1} with the usual metric has constant sectional
curvature +1.  Indeed, given p,v with p perpendicular to v, |p|=|v|=1,
the geodesic is gamma_{(p,v)}(t)=cos(t)p+sin(t)v, so exp_p(2-planes) are
standard 2-spheres, which has curvature +1.
17:42:25 kommodore: Example 2: Poincar\'e disc.
17:42:33 kommodore: The open unit ball B^n in R^n equipped with the metric
17:42:34 kommodore: ds^2=[(dx^1)^2+...+(dx^n)^2]/(1-|x|^2)^2
17:42:34 kommodore: has constant sectional curvature -1.  Again, this reduces to the usual
calculation in 2-dimensional hyperbolic disc.
17:42:50 kommodore: Example 3: compact Lie groups with bi-invariant metrics.
17:43:14 kommodore: Let G be a compact Lie group with its bi-invariant metric.  Recall we
have seen D_X(Y)=[X,Y]/2 for left-invariant vector fields X,Y.  So, for
left-invariant vector fields X,Y,Z, we have
17:43:32 kommodore: R(X,Y)Z =D_X D_Y Z - D_Y D_X Z - D_{[X,Y]}Z
17:43:38 kommodore: =[X,[Y,Z]]/4-[Y,[X,Z]]/4-[[X,Y],Z]/2
17:43:49 kommodore: =([X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]])/4-[[X,Y],Z]/4
17:43:57 kommodore: =-[[X,Y],Z]/4
17:44:07 kommodore: and by tensoriality, this holds for all X,Y,Z.
17:44:19 kommodore: The sectional curvature is given by
17:44:27 kommodore: <R(X,Y)Y,X>=<-[[X,Y],Y],X>/4=<[Y,[X,Y]],X>/4
=Y<[X,Y],X>/4-<[X,Y],[Y,X]>/4
17:44:54 kommodore: the first term vanish because of Lie bracket of left-invariant vector
fields is left-invariant, and the metric is bi-invariant.  So
17:45:03 kommodore: K(X,Y)=|[X,Y]/2|^2
17:45:08 kommodore: is nonnegative.
17:45:54 kommodore: Now we want to give some concrete feeling of what these curvature
quantities really control
17:46:10 kommodore: What sectional curvature measures is how quickly two geodesics diverges.
17:46:25 kommodore: The easiest way is to do this by Jacobi fields
17:46:46 kommodore: Consider the parametrised surface f(s,t)=exp_p(tV(s)), where V(s) is a
curve in T_pM such that V(0)=u, V'(0)=v.  We would like to study the
vector field J(t)=(@f/@s)(0,t), which is called the Jacobi field.
17:47:08 kommodore: here, @f/@s is the vector field f_*(@/@s)
17:47:48 kommodore: Since s=0 corresponds to a geodesic exp_p(tu)
17:47:56 kommodore: 0=(D/@t)(@f/@t)
17:48:07 kommodore: The D reminds us that we are taking covariant derivatives, and @f/@t is
the vector field f_*(@/@t).
17:48:35 kommodore: Taking covariant derivative in direction @/@s, we have
17:48:45 kommodore: 0=(D/@s)(D/@t)(@f/@t)
17:48:59 kommodore: Because R measures the failure for covariant derivatives to commute,
17:49:15 kommodore: 0=(D/@t)(D/@s)(@f/@t)+R(@f/@s, @f/@t)(@f/@t)
17:49:28 kommodore:  =(D/@t)(D/@t)(@f/@s)+R(@f/@s, @f/@t)(@f/@t)
17:50:01 kommodore: The last line is because (D/@t)(@f/@s)=(D/@s)(@f/@t) from the symmetry
of the (usual) second derivative and D is torsion-free.
17:50:23 kommodore: Now putting J(t)=(@f/@s)(0,t), gamma'(t)=(@f/@t)(0,t), we get
17:50:35 kommodore: 0=J''(t)+R(J,gamma')gamma' --- Jacobi equation.
17:50:59 kommodore: Proposition: If J is the Jacobi field J(t)=(d exp_p)_{tu}(tv), then
|J(t)|^2= t^2v^2-t^4/3 K(u,v) + higher order.
17:51:17 kommodore: Proof: This is just a matter of differentiating enough times.
17:51:27 kommodore: The J in question satisfies J(0)=0, J'(0)=v, so
J''(0)=-R(J(0),gamma'(0))gamma'(0)=0.
17:51:47 kommodore: The last derivative we need is J'''(0).  For that, we need to compute
(D/@t)R(J,gamma')gamma' at t=0:
17:52:06 kommodore: <(D/@t)R(J,gamma')gamma',W>
= (d/dt)<R(J,gamma')gamma',W>-<R(J,gamma')gamma',W'>
17:52:18 kommodore:                    = -(d/dt)<R(W,gamma')gamma',J>
17:52:41 kommodore:                    = -<(D/dt)R(W,gamma')gamma',J>-<R(W,gamma')gamma',J'>
17:52:51 kommodore:                    = -<R(J',gamma')gamma',W>
17:53:03 kommodore: So J'''(0)=R(J'(0),gamma'(0))gamma'(0)=R(v,u)u.
17:53:16 kommodore: Thus:
17:53:44 kommodore: <J,J>(0)=0
17:53:45 kommodore: <J,J>'(0)=2<J,J'>(0)=0
17:53:45 kommodore: <J,J>''(0)=2<J',J'>(0)+2<J,J''>(0)=|v|^2
17:53:45 kommodore: <J,J>'''(0)=2<J,J'''>(0)+6<J',J''>(0)=0
17:53:45 kommodore: <J,J>^{iv}(0)=2<J,J''''>(0)+8<J',J'''>(0)+6<J'',J''>(0)=8<R(v,u)u,v>
=8K(u,v).
17:53:47 kommodore: ////
17:54:08 kommodore: Integrating, this gives us back the familiar characterisation of
Gaussian curvature K as the Taylor coeffient
17:54:47 kommodore: Since sectional curvature controls how quickly geodesics "converges" on
each other, it shouldn't be surprising that the Ricci curvature controls
how quickly the volume density sqrt(g_{ij}) dx^1...dx^n collapses.
17:55:06 kommodore: Proposition:
17:55:20 kommodore: The volume V(p,r) of ball B(p,r), where exp_p:B(0,r)->B(p,r) is a
diffeomorphism, is given by
17:55:31 kommodore: V(p,r) = \int_0^r \int_{u\in S^{n-1}} theta(t,u) du dt
17:55:58 kommodore: where theta is the volume density
17:56:06 kommodore: theta(t,u)^2=det(<J_i(t),J_j(t)>: 1<=i,j<=n)
17:56:16 kommodore: J_i(t) is the Jacobi field J(0)=p, J'(0)=e_i, {e_i} is an orthonormal
basis of T_pM.
17:56:43 kommodore: Proof: Expand. ////
17:57:13 kommodore: Using the previous expansion of J(t), we have
17:57:22 kommodore: theta(t,u)=t^{n-1}-(n-1)Ric_p(u)t^{n+1}/6+higher order.
17:57:37 kommodore: So after the S^{n-1}-integration, we see the scalar curvature controls
the behaviour of V(p,r).
17:57:53 kommodore: In fact, we can take the expansion of theta(t,u) as the definition of
the Ricci curvature, and the expansion of V(p,r) as the definition of
the scalar curvature.
17:59:07 kommodore: Perhaps I should end this by stating the Hadamard-Cartan theorem, which
shows what global property curvature can influence
17:59:25 kommodore: Hadamard-Cartan theorem: if M is complete with non-positive sectional
curvature for all 2-planes sigma, for all point p, then the universal
cover of M is diffeomorphic to R^n.
18:00:26 kommodore: here, completeness can be thought of as completeness under the induced
distance function, or as geodesic completeness (all geodesics can be
infinitely extended).  Either way is equivalent by the Hopf-Rinow
theorem
18:01:21 kommodore: I'd probably need another 5 minutes to prove H-C, but I think there are
enough stuff about curvature for today...
18:01:44 kommodore: Any questions?
18:02:48 _llll_: just a general vague one: if the H-C thing doesnt hold, can you nevertheless
know something about the universal cover form the curvature?
18:03:37 kommodore: well, you can't necessarily.  For example, if we replace it with Ricci,
the result is false
18:04:22 kommodore: But Ricci pinched below by a positive constant does imply the universal
cover is compact (corollary of Bonnet-Myers)
18:05:34 kommodore: Of course, there are manifolds where H-C does not hold, but the
universal cover is R^n, e.g. the paraboloid z=x^2+y^2 has K>=0
18:05:45 kommodore: (actually >0)
18:07:56 kommodore: Maybe I should just outline what is needed to prove H-C.  It is
surprisingly little.
18:08:45 kommodore: A point q is conjugate to p along geodesic gamma if p,q both lie on
gamma, and there is a nontrivial Jacobi field vanishing at both p and q
18:09:27 kommodore: the exponential map exp_p is a local diffeomorphism at all points where
it is not referring to a conjugate point
18:10:48 kommodore: so the H-C thing just boils down to nonvanishing of Jacobi fields, which
is done by computing second derivative of <J,J>
18:11:57 kommodore: so, if a point p is a pole (meaning no conjugate points), then we can
deduce M has universal cover = R^n and exp_p is a covering map
18:13:21 kommodore: The example of paraboloid z=x^2+y^2 is such an example, with the origin
being a pole

18:29:22 ChanServ changed the topic of #mathematics to: NEXT SEMINAR: Introduction to
Number Theory by pyninja on Sunday 17 August 16:00UTC | Transcript of
last seminar: http://www.freenode-math.com/index.php/Introductory_Riemannian_Geometry_2:_Curvature
| Other seminars (past and future):
http://www.freenode-math.com/index.php/Seminars