The Exponential Map

September 1, 2009 by hilbertthm90 | 1 Comment

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.

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

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The Exponential Map

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