# Smooth homomorphisms

## Homomorphisms

### Recap

Recall that a homomorphism is a map $F\colon G\to H$ between two groups such that $F(g_{1}g_{2})=F(g_{1})F(g_{2})$ for all $g_{1},g_{2}\in G$ and $F(1)=1$ . In other words, $F$ is a map which intertwines the group structures on $G$ and on $H$ . Homomorphisms play a key role in group theory, and we will focus on the case when $G$ and $H$ are matrix groups.

Example:

The determinant gives us a homomorphism $\det\colon GL(n,\mathbf{R})\to GL(1,\mathbf{R})$ , by sending a matrix $M$ to the 1-by-1 matrix $(\det M)$ . This is a homomorphism because $\det(M_{1}M_{2})=\det(M_{1})\det(M_{2})$ and $\det(I)=1$ .

### Smoothness

In this course we are trying to use tools from calculus to study groups and homomorphisms, so we would like to be able to differentiate $F$ . You know how to differentiate a function. You know how to differentiate a map from $\mathbf{R}^{n}$ to $\mathbf{R}^{m}$ by taking partial derivatives of components. But how can we differentiate a map between two matrix groups?

Definition:

A homomorphism $F\colon G\to H$ of matrix group is smooth if it is smooth (infinitely-differentiable) when written in local exponential charts.

We have exponential maps $\exp\colon\mathfrak{g}\to G$ and $\exp\colon\mathfrak{h}\to H$ which (as was proved in this optional video) are locally invertible. When we view a map $F\colon G\to H$ in local exponential charts, it means that we get a (locally-defined) map $f\colon\mathfrak{g}\to\mathfrak{h}$ which makes this diagram commute:

In other words, $f=\log\circ F\circ\exp$ wherever that makes sense, i.e. on some neighbourhood of the zero-matrix in $\mathfrak{g}$ . This is a map between open subsets of vector spaces $\mathfrak{g}$ and $\mathfrak{h}$ so it is infinitely-differentiable if all possible iterated partial derivatives of all components exist.

Remark:

Note that $F(\exp X)=\exp(f(X))$ for all $X$ in the neighbourhood where $f$ is defined. You can see this by taking $\exp$ on both sides of $f=\log\circ F\circ\exp$ .

### Main result

Theorem:

If $F\colon G\to H$ is a smooth homomorphism of matrix groups then:

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

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

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

Remark:

We'll prove this theorem in a later video, but let us remark that this third result is not unexpected: we obtain $F_{*}$ by "taking $F$ inside the exponential", and the Baker-Campbell-Hausdorff formula told us that the group product $\exp(X)\exp(Y)$ is determined by $X$ , $Y$ and iterated brackets between them, so if $F$ preserves group multiplication, it's not surprising that $F_{*}$ preserves the bracket.

### Example

Example:

Since $\det$ is a homomorphism, this theorem tells us that there is a linear map $\det_{*}\colon\mathfrak{gl}(n,\mathbf{R})\to\mathfrak{gl}(1,\mathbf{R})$ such that $\det(\exp X)=e^{\det_{*}(X)}$ for all matrices $X$ . It turns out that $\det_{*}(X)=\mathrm{Tr}(X)=X_{11}+X_{22}+\cdots+X_{nn}$ . So $\det(\exp X)=e^{\mathrm{Tr}(X)}.$

Why is it true that $\det_{*}=\mathrm{Tr}$ ? By the theorem, $\det_{*}$ is linear, so it suffices to check $\det_{*}=\mathrm{Tr}$ on a basis for $\mathfrak{gl}(n,\mathbf{R})$ .

Let's use the basis of $n^{2}$ matrices whose entries are all zero except for one which is 1: $\begin{pmatrix}1&0&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{pmatrix},\quad\begin{pmatrix}0&1&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{pmatrix},\quad\cdots\quad,\begin{pmatrix}0&0&\cdots&1\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{pmatrix},\quad\cdots\quad,\begin{pmatrix}0&0&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1\end{pmatrix}$ First suppose that we pick a basis element with a 1 on the diagonal, like $X=\begin{pmatrix}1&0&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{pmatrix}$ . Then $\det\exp X=\det\begin{pmatrix}1+1+1/2+1/3!+\cdots&0&\cdots&0\\ 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1\end{pmatrix}=e=e^{1}=e^{\mathrm{Tr}(X)}$ as required.

## Pre-class exercise

Exercise:

If $X$ has a 1 on the off-diagonal, I'll leave it as an exercise to check that $\det(\exp(X))=e^{\mathrm{Tr}(X)}$ .