This is a sequel to the previous post on the exponential maps of Lie groups. In this note we record some results which are derived using Ken’s theorems. In particular we impose a topological restriction on Lie groups with bi-invariant metric.
Proposition 1 If is a Lie group with a bi-invariant metric, let be its Levi-Civita connection, then for any left-invariant vector field
where is the Lie bracket.
Proof: Note that at , because the integral curve of is exactly , and by Ken’s theorem, this is a geodesic, so . In general, for , as is an isometry, we have
For left-invariant ,
Together with the torsion-free condition , we can get the result.
For any vectors , there is a natural Lie bracket on defined as follows. Extend to be the unique left-invariant vector fields on , still denoted by . We can then define the vector field on . We then define the restriction of this vector field on to be (same notation as the vector field Lie bracket!). Thus we will always identify left-invariant vector fields with vectors in , which are by definition isomorphic as Lie algebra.
We first set up some notations. Recall that is defined as
Regarding as a map (indeed group homomorphism) , noting that the tangent space of at is [ is just a finite dimensional vector space, by fixing a basis, is just the space of invertible matrices, which is open in the space of all matrices, thus its tangent space is the space of all matrices, intrinsically(without basis) this is ], by taking the differential of at , we have a map
so that for any , there is a linear map
We briefly recall that the geometric meaning of is as follows:
where is the flow of .
Proof: Let where is the conjugate map by and . Then
So for a function ,
Consider the first term on RHS, and for fixed , is the integral curve of starting at , so
Thus by taking the derivative w.r.t. of the above,
For the second term on the RHS of (1), we switch the order of differentiation first and apply the same analysis to get . Therefore , we conclude that
We now prove Proposition 2.
Proof: [Proof of Proposition 2]
As is bi-invariant, is Ad-invariant. So for all ,
so if , by taking the derivative of above, and by Proposition 3
By the skew-symmetry of , rearranging, we can get the result.
We have the following important Jacobi identity.
Proof: (This of course can be verified directly by examining its action on a function . But this seems unnatural and is not easy to understand. We here show that this is just the familiar product rule as in calculus. ) Here we write instead of . We prove the following product law first:
For convenience let the one parameter family of vector field induced by the flow of be denoted by . It is easy to see that we have the invariance of Lie bracket:
Under local coordinates,
So using product rule, (with = derivative w.r.t. )
where . As is anti-symmetric, by rearranging the above equation,
Finally, we compute the curvature of with a bi-invariant metric. Recall that the Riemannian curvature tensor is defined as (by definition of , this is a tensor)
By an easy computation, this is equal to
Proof: Note that by invariance of Lie bracket, etc are also left-invariant. By Proposition 1,
where we have used the Jacobi identity (Proposition 4) on the forth line.
For with bi-invariant metric, for orthonormal , by the left-invariance of the metric, if we uniquely extend as left-invariant vector fields on , then and are still orthonormal at each point of , thus they span a two-plane in each tangent space . We define the sectional curvature of by
Theorem 6 [Ken’s theorem, 2011] If is a Lie group with a bi-invariant metric. Let be orthonormal and left-invariantly extend them. Then at each point
Proof: By Theorem 5,
By Proposition 2,
Corollary 7 [Ken’s theorem, 2011] If a Lie group has a bi-invariant metric (for example compact by Ken’s theorem), it must have non-negative sectional curvature.
Remark 1 There’re some topological restrictions on such . For example according to Soul theorem of Cheerger and Gromoll, since has non-negative curvature, it is diffeomorphic to the normal bundle of a compact totally geodesic (and thus also non-negatively curved by Gauss equation) submanifold. In particular must be homotopy equivalent to a compact manifold. [But does this submanifold have anything to do with the group structure I don’t know. ]