# Linearity of homomorphisms in exponential charts

## Statement of theorem

In this video, we will prove the following theorem which was stated earlier:

Theorem:

If $F\colon G\to H$ is a smooth homomorphism of matrix groups and $f$ denotes the locally-defined map $\log\circ F\circ\exp\colon\mathfrak{g}\to\mathfrak{h}$ then:

1. $f$ is the restriction of a linear map $F_{*}\colon\mathfrak{g}\to\mathfrak{h}$ to a neighbourhood of $0\in\mathfrak{g}$ .

2. $F(\exp X)=\exp(F_{*}(X))$ for all $X\in\mathfrak{g}$ ,

3. $F_{*}[X,Y]=[F_{*}X,F_{*}Y]$ for all $X$ and $Y$ in $\mathfrak{g}$ .

## Proof of theorem

Proof:

Our first observation is that, for all $X\in\mathfrak{g}$ , $\exp(tX)$ is a 1-parameter subgroup. Since $F$ is a smooth homomorphism, and since a 1-parameter subgroup is just a smooth homomorphism from $\mathbf{R}$ , we deduce that $F(\exp(tX))$ is a 1-parameter subgroup in $H$ . By what we proved about 1-parameter subgroups, $F(\exp(tX))=\exp(tY)$ for some $Y\in\mathfrak{h}$ .

The figure above shows the groups $G$ and $H$ , the paths $\exp(tX)$ and $\exp(tY)$ , and (red) neighbourhoods of the identity in each on which the logarithm is well-defined. For small $t$ , $\exp(tX)$ and $\exp(tY)$ are both close to the identity, so live in the domain of the local logarithm.

We have $F(\exp(tX))=\exp(f(tX))$ for small $t$ and $F(\exp(tX))=\exp(tY)$ for all $t$ . Therefore, for small $t$ we can take logarithms and deduce that $f(tX)=tY$ .

Since $f$ is a smooth map, we can take the Taylor series $f(tX)=td_{0}f(X)+\mbox{higher order in }t$ , where $d_{0}f$ is the derivative of $f$ at $0$ . Because $f(tX)=tY$ , we see that $d_{0}f(X)=Y$ and all higher order terms vanish.

This means that $f(tX)=tY=td_{0}f(X)=d_{0}f(tX)$ for all $X$ and all sufficiently small $t$ . Therefore $f=d_{0}f$ on the domain of the exponential chart. We take $F_{*}=d_{0}f$ . This proves the first part of the theorem. Recall that the matrix for the linear map $d_{0}f$ is the matrix of partial derivatives of components of $f$ , $d_{0}f=\left(\frac{\partial f_{i}}{\partial x_{j}}\right).$

The second part follows from what we have already said: we showed that $F(\exp tX)=\exp(tY)$ (for all $t$ ), and $Y=d_{0}f(X)=F_{*}(X)$ . Setting $t=1$ we get $F(\exp X)=\exp(F_{*}(X))$ .

To prove the final part of the theorem, consider the expression $\exp(tX)\exp(tY)\exp(-tX)\exp(-tY)$ Use the Baker-Campbell-Hausdorff formula to multiply the first two and last two terms together and we get $\exp\left(t(X+Y)+\frac{1}{2}t^{2}[X,Y]+\cdots\right)\exp\left(-t(X+Y)+\frac{1}% {2}t^{2}[X,Y]+\cdots\right)$ Using the Baker-Campbell-Hausdorff formula to multiply these two together, after a little manipulation, we get $\exp(t^{2}[X,Y]+\mathcal{O}(t^{3})).$

If I apply $F$ to this whole expression, we get: $F(\exp(t^{2}[X,Y]+\mathcal{O}(t^{3})))=\exp(t^{2}F_{*}[X,Y]+\mathcal{O}(t^{3}))$ by definition of $F_{*}$ .

Because $F$ is a homomorphism, $F(\exp(tX)\exp(tY)\exp(-tX)\exp(-tY))=F(\exp(tX))F(\exp(tY))F(\exp(-tX))F(\exp% (-tY)).$ By definition of $F_{*}$ , this equals $\exp(tF_{*}X)\exp(tF_{*}Y)\exp(-tF_{*}X)\exp(-tF_{*}Y)$ and using the Baker-Campbell-Hausdorff formula in the same way we get $\exp(t^{2}[F_{*}X,F_{*}Y]+\mathcal{O(t^{3})})$

If we take $t$ small then we can take logarithms and we get $t^{2}[F_{*}X,F_{*}Y]=t^{2}F_{*}[X,Y]+\mathcal{O}(t^{3}).$ Differentiating twice and setting $t=0$ we get $F_{*}[X,Y]=[F_{*}X,F_{*}Y]$ as required.