### Lie group homomorphisms

Because a Lie group is fundamentally a group that is also a manifold, we'd like to define a Lie group homomorphism as one that is both a group homomorphism, and smooth. For this, though, we need to define what it means to differentiate a group homomorphism.

Recall that the general notion of a derivative is the idea of "how does the map work locally"? Letting a general function $f:G\to H$ map a curve $\gamma(t)$, it should be easy to see that $\gamma'(t)$ transforms as $(f\circ\gamma)'(t)$ (make sure that this makes sense -- think in terms of the chain rule, or write it out in limit form, or just in terms of the image of the curve).

Consequently this leads to the differential $df:dG\to dH$ (where $dG$ is the Lie Algebra of $G$) defined as $df(\gamma'(0))=(f\circ\gamma)'(0)$. Some short exercises:
• Confirm that this is equivalent to saying that $df(X)$ is the directional derivative of $f$ in the $X$ direction.
• Differentiate $f(xyx^{-1}y^{-1})$ with respect to $x$ in the $X$ direction at $x=1$ (hint: this is a direct application of the definition of the differential in reverse).
• Convince yourself that any derivative operator commutes with $df$, i.e. $D(df(X))=df(D(X))$.
It should be intuitively clear that if $f$ is a homomorphism, its local effect should be to act as a homomorphism of the Lie algebra as it should preserve all local structure. We can easily show that:
1. Since $df$ is a derivative of $f$, its value must be a linear map (like the Jacobian). This applies to the derivative as an operator on the tangent space of any manifold -- $f$ doesn't need to be a group homomorphism at all.
2. It preserves the Lie bracket. Take $f(xyx^{-1}y^{-1})=f(x)f(y)f(x)^{-1}f(y)^{-1}$ and differentiate it once with respect to $x$ in the $X$ direction at $x=1$, obtaining: $df(X-yXy^{-1})=df(X)-f(y)df(X)f(y)^{-1}$, simplify and differentiate it with respect to $y$ in the $Y$ direction at $y=1$ to get: $df([Y,X])=[df(Y),df(X)]$.

The Lie Bracket $[Y,X]$ is not the derivative of conjugation $gxg^{-1}$, so you don't have to worry -- the Lie Bracket is not a Lie algebra homomorphism (it doesn't preserve Lie Brackets), the derivative of conjugation at the identity is zero. That's unfortunate -- our explanation of the Jacobi identity ("a derivation acts through the Lie Bracket as a derivation on the space of derivations where multiplication is given by the Lie Bracket") really indicated that it has something to do with it.

The Lie Bracket is the derivative of conjugation $xgx^{-1}$. OK, so?

Here's the idea: $\mathrm{Ad}(x)(y)=xyx^{-1}$ defines a homomorphism $\mathrm{Ad}:G\to\mathrm{Aut}(G)$. Its differential $\mathrm{ad}:dG\to d\mathrm{Aut}(G)$ can be confirmed to be the Lie Bracket $\mathrm{ad}(X)(Y)=[X,Y]$. So preservation of the Lie Bracket means:

$$\mathrm{ad}([X,Y])=[\mathrm{ad}(X),\mathrm{ad}(Y)]$$
This is precisely the Jacobi identity! So the Lie bracket is a Lie algebra homomorphism, from a Lie algebra to the Lie algebra of half-filled Lie brackets.

There is indeed a relationship between this "homomorphism" understanding of the Jacobi identity and the "derivation" understanding. In general, given a curve $\phi:\mathbb{R}\to\mathrm{Aut}(G)$, differentiating $\phi(t)(gh)=\phi(t)(g)\phi(t)(h)$ at $t=0$ we see that its derivative $d\phi$ satisfies the product rule, i.e. is a derivation (in fact this is true even when $G$ is not a group -- often a Lie group arises this way, as the automorphism group of some object and these derivations then form its Lie algebra). This implies

$$d\mathrm{Aut}(G)\subseteq\mathrm{Der}(dG)$$
So $[X,\cdot]$ is a derivation, and the map from $X$ to $[X,\cdot]$ is a Lie algebra homomorphism $dG\to\mathrm{Der}(dG)$. This really does give us a much more general way to look at everything we talked about in the last article.

Wait -- shouldn't it be an equality? I thought all derivations were part of the Lie Algebra? Ah, but there the derivations on $M$ formed the Lie Algebra of $\mathrm{Aut}(M)$, i.e. $d\mathrm{Aut}(M)=\mathrm{Der}(M)$. So indeed $d\mathrm{Aut}(dG)=\mathrm{Der}(dG)$. This makes sense, indeed $\mathrm{Aut}(G)\subseteq \mathrm{Aut}(dG)$. It's interesting to think about when it is that the Lie algebra has "more" automorphisms than the Lie group.

One may wonder if all automorphisms of a group are a conjugation by something -- or equivalently, if all automorphisms of a Lie algebra are a derivation of some kind. We will later see a special classification of Lie group for which this is true -- in general, the conjugation automorphisms are called the innner automorphisms of the group and are denoted as $\mathrm{Inn}(G)$. The group of all endomorphisms (invertible linear transformations $dG\to dG$) of a Lie algebra, meanwhile are denoted as $\mathrm{End}(dG)$, and it's easy to see that this occurs iff the Lie algebra is Abelian.

Exercise: Show that the map $\mathrm{Ad}:G\to \mathrm{Aut}(G)$ is injective iff $G$ has a trivial center.

So if $G$ has trivial center and all its automorphisms are inner, it is isomorphic to $\mathrm{Aut}(G)$ and is called complete.

The determinant map

The determinant is a homomorphism $\det:GL_F(n)\to F$ from any matrix group. The first thing we'd like to do with this is find its differential $\det'$ (which will be an $F$-valued function on $M_F(n)$). By definition of the differential:

$$\det' A = \lim_{\varepsilon\to 0}\frac{\det (I+\varepsilon A)-1}{\varepsilon}$$
It's easy to prove by writing out the entries of the matrix as $\delta_{ij}+\lambda_{ij}\varepsilon$ and performing induction on the dimension of the matrix that this is equivalent to:

$$\det'A=\mathrm{tr} A$$

Lie algebra homomorphisms in detail: ideals

Well, Lie algebra homomorphisms are a specific category of vector space homomorphisms, aren't they? It's not enough that they preserve the linear structure, they must preserve the Lie bracket too. Well, let's study them in more detail -- like a crash course through linear algebra, but with Lie algebra instead.

What does the kernel of a Lie algebra homomorphism $A$ look like? Well, because the homomorphism preserves linear combinations, the kernel must be a linear subspace -- similarly because the homomorphism preserves the Lie bracket, we must have that $Av=0\implies \forall w\in\mathfrak{g}, A[v,w]=0$, i.e. the kernel must be closed under derivations from $\mathfrak{g}$: $[\mathfrak{g},\mathfrak{i}]\subseteq\mathfrak{i}$. Such a subalgebra is called an ideal.

Exercise: Show that the Lie algebra of a normal subgroup is an ideal (careful -- it's not as obvious as you might think -- but still pretty obvious).