Introductory Riemannian Geometry 2: Curvature

Topic
Curvature. Part of the series outlined at Introductory Riemannian Geometry

Seminar
 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&gt;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-&gt;M. 17:01:47 kommodore: go ahead 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)-&gt;(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 &lt;R(@_i,@_j)@_k,@_l&gt;. 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]-&gt;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&gt;0 (depending on t), the tangent vectors {v=gamma'(s):|s-t|&lt;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 &lt;(d exp_p)_{tu}(v),gamma_{(p,u)}'(t)&gt; =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|&lt;epsilon(p), if we choose |v|&lt;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&gt;0, providing t is                   sufficiently small. 17:23:03 kommodore: (actually for all nonzero c, but we won't need c&lt;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 &lt;(d exp_p)_{tu}(v),gamma_{(p,u)}'(t)&gt; =0, whenever exp_p(tu) is                   defined. //// 17:26:02 kommodore: Let's decipher this a little. exp_p is a map T_pM -&gt; M, so its derivative at tu is a map T_pM=T_{tu}T_pM -&gt; 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)|&gt;=|r'(t)|, and integrating gives L(gamma([0,t]))&gt;=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 &quot;full&quot; version of the exponential map Exp: TM --&gt; M\times M; v|-&gt;(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 &quot;the Gaussian curvature at p to                   the surface exp(small neighbourhood of 0 in sigma)&quot;. 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-&gt;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)=&lt;R(u,v)v,u&gt; 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 &quot;sub-sub-representation&quot; 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 &quot;traceless Ricci&quot; 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 &quot;conformal invariant&quot; (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                   &quot;conformally flat&quot;, 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&gt;=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: &lt;R(X,Y)Y,X&gt;=&lt;-[[X,Y],Y],X&gt;/4=&lt;[Y,[X,Y,X&gt;/4                                               =Y&lt;[X,Y],X&gt;/4-&lt;[X,Y],[Y,X]&gt;/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: &lt;(D/@t)R(J,gamma')gamma',W&gt;                                       = (d/dt)&lt;R(J,gamma')gamma',W&gt;-&lt;R(J,gamma')gamma',W'&gt; 17:52:18 kommodore:                    = -(d/dt)&lt;R(W,gamma')gamma',J&gt; 17:52:41 kommodore:                    = -&lt;(D/dt)R(W,gamma')gamma',J&gt;-&lt;R(W,gamma')gamma',J'&gt; 17:52:51 kommodore:                    = -&lt;R(J',gamma')gamma',W&gt; 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: &lt;J,J&gt;(0)=0 17:53:45 kommodore: &lt;J,J&gt;'(0)=2&lt;J,J'&gt;(0)=0 17:53:45 kommodore: &lt;J,J&gt;(0)=2&lt;J',J'&gt;(0)+2&lt;J,J&gt;(0)=|v|^2 17:53:45 kommodore: &lt;J,J&gt;(0)=2&lt;J,J&gt;(0)+6&lt;J',J&gt;(0)=0 17:53:45 kommodore: &lt;J,J&gt;^{iv}(0)=2&lt;J,J'&gt;(0)+8&lt;J',J&gt;(0)+6&lt;J,J''&gt;(0)=8&lt;R(v,u)u,v&gt;                                 =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                    length(circle radius r)/(2*pi)=r-Kr^3/6+higher order. 17:54:47 kommodore: Since sectional curvature controls how quickly geodesics &quot;converges&quot; 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)-&gt;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(&lt;J_i(t),J_j(t)&gt;: 1&lt;=i,j&lt;=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&gt;=0 18:05:45 kommodore: (actually &gt;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 &lt;J,J&gt; 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