The Exponential Map

Notice from last time that the matrix exponential maps from the Lie algebra to the Lie group. It took an arbitrary matrix (i.e. a tangent vector at the identity) and mapped it to a non-singular matrix (i.e. a matrix in GL_n(\mathbb{R})). Now we want a map exp: \frak{g}\to G that does the same basic idea. Let’s unravel what that idea was.

It was sort of hidden, but we said that the one-parameter subgroup of GL_n(\mathbb{R}) generated by A, is F(t)=e^{tA}. So it took a line through the origin: t\mapsto tA, and sent it to a one-parameter subgroup F(t).

Let’s sort of work backwards. We have a one-parameter subgroup generated by X, say F(t) from the last post (the integral curve starting at the identity). From our matrix example, we also saw that if we flowed along this for 1 time unit, F(1)=e^X. Thus we’ll define exp(X)=F(1), where F is the one-parameter subgroup generated by X. Thus exp will be mapping from the Lie algebra to the Lie group.

Now we need to check the line through the origin property. i.e. Is F(s)=exp(sX) the one-parameter subgroup generated by X? Well, it is really a simple matter of rescaling. Let \overline{F}(t)=F(st) by rescaling. Then \overline{F}(t) is the integral curve of sX starting at e, so exp(sX)=\overline{F}(1)=F(s).

Now there are lots of properties we should check. I’ll go through a few this time and the rest tomorrow before moving on. First off, we hope that exp is a smooth map. In other words, in terms of the flow, we need \theta_X^e(1) to depend smoothly on X.

Define a vector field on G\times\frak{g} by Y_{(g, X)}=(X_g, 0). Note we made the natural identification T_{(g, X)}(G\times \frak{g})\cong T_gG\oplus T_X\frak{g}. Let X_1, \ldots, X_k be a basis for \frak{g} and (x^i) the coordinates for \frak{g}, x^iX_i. Also, let (w^i) be smooth coordinates for G.

Let f\in C^{\infty}(G\times \frak{g}). Then in coordinates Y(f(w^i, x^i))=\sum_j x^jX_jf(w^i, x^i). Each X_j is a derivative in the w^j direction, so the expression depends smoothly on (w^i, x^i). Hence Y is a smooth vector field.

Now the flow of Y, say \Theta, is \Theta_t(g, X)=(\theta_X(t, g), X). Now the fact that \Theta is a flow of a smooth map shows that it is smooth. Let \pi_G:G\times \frak{g}\to G be projection. Then exp(X)=\pi_G(\Theta_1(e, X)) and hence is a composition of smooth maps so is itself smooth.

That was sort of exhausting, so I’ll just do one quick other property before quitting: exp((s+t)X)=exp(sX)exp(tX). This is just because t\mapsto exp(tX) is a one-parameter subgroup and hence homomorphism. The group structure on \mathbb{R} is additive and the Lie group operation is multiplicative: exp((s+t)X)=F(s+t)=F(s)F(t)=exp(sX)exp(tX).

Alright, one more quick one, since that one shouldn’t count. This one is a big one, in that it corresponds with a very characteristic property of the complex valued exponential. If we identify T_0\frak{g} and T_eG with \frak{g}, then d\exp_0:T_0\frak{g}\to T_eG is the identity map!

Let \gamma(t)=tX. Then \gamma'(0)=X. Thus dexp_0(X)=dexp_0(\gamma'(0))=(exp\circ\gamma)'(0)=\frac{d}{dt}\big|_{t=0}exp(tX)=X (the equality comes from the fact that exp is the flow).

I guess this ran a bit long. Oh well, time is running out.


One thought on “The Exponential Map

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s