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.

Let be a Lie group, then there is an exponential map defined by

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?

Example 1

- 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 groupexponential 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.

Theorem 1[Ken’s theorem, 2011] If is a compact Lie group, then it has a bi-invariant Riemannian metric.

Theorem 2[Ken’s theorem, 2011] If is Lie group with a bi-invariant metric (for example is compact), thenwhere is the Riemannian exponential map.

To prove Theorem 1 we first introduce some notations. Firstly, for , we can define a conjugate action of by the composition , i.e.

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

Lemma 3

There is a one-one correspondence between Ad-invariant inner products on and bi-invariant Riemannian metrics on .

*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

Lemma 4

For a Lie group with a bi-invariant metric, the map defined by

is an isometry. Also . (The differential of the map at the point . )

*Proof:*

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.

*Proof:* [Proof of Theorem 2]

Let , , be a geodesic starting at with initial vector . By the Lemma 4, as and is an isometry, is a geodesic. Thus we have, by the uniqueness of geodesic,

For small , if we define , then is a geodesic (for small ) with and , by uniqueness of short geodesic (joining two nearby points), we must have

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 ,

How does the (bi-, left- or right-) invariance of the metric relate to the Haar measure of the Lie group? Can the metric in some sense be given in terms of integration against the Haar measure?

A (bi-, left- or right) invariant metric induces a (bi-, left- or right) Haar measure of the Lie group. One knows that (left- or right-) Haar measure of a locally compact group is unique up to positive scalars, indeed, when it is a Lie group, such Haar measures always come from volume forms on the underlying differential manifolds. In particular, a (bi-, left- or right) invariant metric will give a Riemannian volume form (namely the Hodge dual of the 0-form, constant function 1).

Can we do something to utilize the fact that when G is compact, for any finite-dimensional representation , V admits a G-invariant inner product given by

to the case of the adjoint representation of G on its Lie algebra?

What you are doing is to show that there is a G-invariant metric on its Lie algebra and then you can use translation to move the metric everywhere. Indeed, proveided that it is a connected Lie group, bi-invariant metric exists iff it is a direct product of a compact Lie group and a vector space.

Great! I have updated the note accordingly.

Your equation defining Ad-invariance has its RHS missing~ [Corrected. Thanks! -KKK]

KKK, perhaps left-invariant metric is induced by right translations, as long as left and right actions commute. Isn’t it so?

Why are you calling these result’s “Ken’s theorem” and claiming they were proven in 2011? These have been known for a long time.