In this note we try to understand the relations between two different notions of exponential map, one from Riemannian geometry, and the other coming from Lie group theory.
where is the unique one-parameter subgroup (i.e. ) such that .
On the other hand, if is equipped with a Riemannian metric , then at any point , there is a Riemannian exponential map defined by
where is the unique geodesic such that . [Technically the domain of should be an open subset of instead, and may not be well-defined on the whole , but these are not important here.] It is interesting to know if these two concepts are related, and why are they called exponential map at all?
- For , this is the simplest example of a compact Lie group. Its tangent space , the Lie group exponential map is actually given by
where RHS is defined by the ordinary exponential map . Thus we see that the notation “exponential map” is reasonable in this case. Actually, by and for matrix , this example can be generalized for matrix Lie group. i.e. (after taking care of the domain), the matrix exponential map is the Lie group exponential map for matrix Lie groups, e.g. etc.
- Actually there is a Lie theory viewpoint for the ordinary exponential map for real numbers. Indeed, consider as a (non-compact) Lie group with ordinary multiplication. Its tangent space at is just , and both the left translation by and its differential map are just the multiplication by itself. The Lie group exponential map must satisfy: and for ,
So we must have
But this is nothing but the definition of (well at least it can be taken as a definition). So we have .
Now, if is a Lie group with a bi-invariant metric , i.e.
for all and , where are the left and right translations respectively. Then there is a Riemannian exponential map associated with this metric. We would like to know if they are equal in the following sense:
It is also interesting to know when a Lie group has a bi-invariant metric. It is quite obvious that any Lie group has a left-invariant (and a right-invariant) metric by left(or right)-translating an inner product at . I only know recently that any compact Lie group has a bi-invariant metric which is unique up to scaling.
Perhaps Ken can supply a proof of this , using modular function. Following Ken’s argument, we can prove the following two results.
where is the Riemannian exponential map.
For fixed , the differential map of at is then . We will denote this map as (adjoint action). Note that , i.e.
By an Ad-invariant inner product on we mean an inner product such that for all and . We have the following
Proof: For any bi-invariant metric on , clearly its restriction on is an Ad-invariant inner product: .
Conversely, for any Ad-invariant inner product on , we left translate it to a Riemannian metric on . Clearly this metric is left-invariant by definition. We check that this is also right-invariant. For , let and , then
where we have used (1) on line 3.
We are now ready to prove Theorem 1.
Proof: [Proof of Theorem 1]
Let be the right-invariant Haar measure on . By lemma 3, it is equivalent to find an Ad-invariant inner product on . Let be any inner product on , then the desired inner product is defined by
To show that this metric is Ad-invariant, we note that is a homomorphism, in other words,
This can be seen by taking the differential of the relation where is the conjugate action.
So we have
To prove Theorem 2 we need the following
is an isometry. Also . (The differential of the map at the point . )
For , we claim that at ,
To see this, if , then . Differentiating gives
As , the claim is proved. Thus we have
by the bi-invariance of the metric. The second assertion is trivial.
for all small enough and . By extending beyond any interval by , we see that can indeed by extended (smoothly as a geodesic) to the whole . And by a standard argument (of chopping into “small pieces”), from (2), we indeed have
Thus is a one-parameter subgroup in . As and , by definition of ,