Optional: Local exponential charts

Local coordinates for GL(n,R)

The exponential map exp : 𝔤 𝔩 ( n , 𝐑 ) G L ( n , 𝐑 ) is not invertible, but we have seen that there are neighbourhoods U 𝔤 𝔩 ( n , 𝐑 ) of the zero-matrix and V G L ( n , 𝐑 ) of the identity matrix such that exp ( U ) = V and exp | U : U V is invertible, with inverse log : V U .

We can think of this as providing for us coordinates near I G L ( n , 𝐑 ) , namely exp ( a 11 a 1 n a n 1 a n n ) gives a parametrisation of V , so we can think of the matrix entries a i j as coordinates on V : anything in V can be written in this form for a unique collection of numbers a i j .

Local coordinates for matrix groups

We could like the same to work for any matrix group G G L ( n , 𝐑 ) . Namely, we would like to show:


If G is a matrix group and 𝔤 is its Lie algebra then there are neighbourhoods U 𝔤 of the zero matrix and V G of the identity such that exp ( U ) = V and exp | U : U V is invertible.


Let U = U 𝔤 and V = V G . First, we note that exp does indeed go from 𝔤 to G by definition of 𝔤 , and exp ( U ) = V , so exp ( U ) V .

So what is left to prove? The map exp : U V is invertible, hence injective, so its restriction exp | U must also be injective. But is exp | U surjective? In other words, is it clear that log ( V ) U ?

This figure shows what you might imagine going wrong (we will later show that this doesn't happen, at least if you shrink U and V ). It shows a cartoon of a subgroup G G L ( n , 𝐑 ) which "wraps back on itself" and gets very close to the identity but never quite gets there. You can now imagine that when you intersect with a very small V (in red), exp ( U ) (in blue) could end up missing this appendage of G which wraps back towards the identity (in green), because to get to this appendage you have to exponentiate something very large. In other words, exp ( U ) is the blue bit and V is everything which is blue or green. In the end, we will show this doesn't happen, so this is a cartoon picture of something which doesn't happen. You therefore shouldn't be too annoyed if the picture doesn't make sense.

A fictitious subgroup G which wraps back on itself and comes close to the identity. This shows how it might be that exp(U') is smaller than V'.

If surjectivity of exp | U fails then there's an element g V = G V such that g exp ( U ) . We could try to fix this by shrinking U and V , but let's suppose that doesn't help us. This will mean there is a sequence g i V such that g i I and g i exp ( U ) for all i (you should imagine a sequence of matrices on the green appendage, tending to the origin in the picture).

Let's assume that there is such a sequence and aim to derive a contradiction.

  1. Recall that 𝔤 𝔤 𝔩 ( n , 𝐑 ) is a subspace. Pick a vector space complement W for 𝔤 , that is 𝔤 W = { 0 } and 𝔤 + W = 𝔤 𝔩 ( n , 𝐑 ) . In the figure, 𝔤 is supposed to be the horizontal axis (tangent to G at the identity) and W is supposed to be the vertical axis.

    I claim that the map F : 𝔤 W G L ( n , 𝐑 ) defined by F ( v , w ) = exp ( v ) exp ( w ) is locally invertible like exp , i.e. there exists a neighbourhood N 1 of 0 𝔤 W and a neighbourhood N 2 of I G L ( n , 𝐑 ) such that F ( N 1 ) = N 2 and F : N 1 N 2 is invertible. This is proved using the inverse function theorem, just like for exp (compute the derivative of F at the zero map and show this derivative is invertible). I leave it as an exercise to fill in the details.

    Therefore, if g i is sufficiently close to the identity (which it is for large i ), then g i = exp ( v i ) exp ( w i ) for some sequence v i 𝔤 and w i W .

  2. Our sequence g i is not in exp ( U ) , so w i 0 for all i . In particular, we can divide w i by its matrix norm to get a matrix w i / w i W with norm 1. Since the set of matrices with norm 1 is closed and bounded (compact), the sequence w i / w i W converges to some matrix w W with norm 1 (in particular, w 0 ).

  3. We are going to prove that w 𝔤 ; this will give us a contradiction, as w W and W is a complement for 𝔤 . For this, we need to show that exp ( t w ) G for all t 𝐑 .

    Fix t . Consider t / w i and take its integer and fractional parts t / w i = n i + ϵ i , n i , ϵ [ 0 , 1 ) . Since g i I as i , we know that w i 0 , so w i 0 , hence t / w i (as t is fixed) and hence n i .

  4. We want to show that exp ( t w ) G . First note that exp ( w i ) G for all i because exp ( w i ) = exp ( - v i ) g i and exp ( - v i ) and g i are both in G . Because G is a group, ( exp ( w i ) ) n i G and ( exp ( w i ) ) n i = exp ( n i w i ) because w i commutes with itself. The sequence exp ( n i w i ) = exp ( t w i / w i - ϵ i w i ) converges to exp ( t w ) because w i / w i w and ϵ i w i 0 .

    Because G is topologically closed, this limit lives in G , so exp ( t w ) G . This argument works for every t , so we're done.

The outcome of all this is that, via the exponential map, local coordinates on 𝔤 near the zero matrix give us local coordinates on G near the identity. We call this an exponential coordinate chart.