The Killing form; factorising non-Abelian Lie groups

It could be fun to try and define a "dot product" on a Lie algebra.

You know, you might've already realised that the cross product is a Lie bracket of sorts -- you know, given its antisymmetry and the whole $a^\mu b^\nu - a^\nu b^\mu$ representation of the wedge product and all that. It's a short exercise to verify that the Lie algebra $\mathfrak{so}(3)$ of $SO(3)$ is the algebra of skew-symmetric matrices, and with the Lie bracket $XY-YX$ is isomorphic to $\mathbb{R}^3$ with the cross product.

Well, the dot product on $\mathbb{R}^3$ has an interesting connection to $SO(3)$ -- it is precisely the form that is invariant under the action of $SO(3)$. Well, but that's $SO(3)$ acting on $\mathbb{R}^3$ -- what is that action in the notation of $\mathfrak{so}(3)$? As it turns out (and you can work this out), it is precisely the adjoint map $\mathrm{Ad}_gX:=gXg^{-1}$ which corresponds to this "rotating $X$ by $g$". It's not really that unexpected, if you ask me -- conjugation is always the natural way to transform matrices in linear algebra when vectors are multiplied on the left.

So the "dot product" is an $\mathrm{Ad}$-invariant bilinear form. In fact, adding a symmetricity requirement allows us to just bother with norms (as a symmetric inner product can be determined from the norm, through the cosine rule). Conjugation basically allows you to determine the "contours" of this norm or inner product. The question is: can we determine the bilinear form -- up to scaling -- just from "$\mathrm{Ad}$-invariant symmetric bilinear form" alone?

This is equivalent to asking "is the orbit of some non-zero $X$ under conjugation by $G$ equal to $\mathfrak{g}$?" (so that the norm of that $X$ would suffice to determine all norms -- do you see why?) Well, this is equivalent to asking "is $X$ contained in some non-trivial ideal?" (prove that these are equivalent!), and this is equivalent to asking "does $\mathfrak{g}$ have any non-trivial ideals?" (do you see why?)

A Lie algebra without nontrivial ideals is called a simple Lie algebra. Our demonstration above shows that a simple Lie algebra has a unique $\mathrm{Ad}$-invariant symmetric bilinear form, determined by the value of $\langle X, X\rangle$ for some non-zero $X$.

Even before we actually derive what this form must look like, we can derive one important consequence of automorphism invariance: $\langle X, [X, Y]\rangle = 0$ (prove it!), i.e. the tangent to an automorphism curve is perpendicular to the position vector at every point. The understanding of the group as acting as a "rotation group" on its Lie algebra in the adjoint representation really makes sense!

Someone tell me if they know how one may "derive" the trace-form formula from this characterisation rather than pulling it out of the blue and then proving it is the unique $\mathrm{Ad}$-invariant symmetric bilinear form. Here's something I started to write:

Here's an idea for the base length (i.e. to define the scaling): $X$ has length 1 iff the length of $[X,Y]$ equals the length of $Y$ for all $Y$ perpendicular to $X$ -- equivalently: $\forall V\in\mathfrak{g}, |[X,[X,V]]|=|[X,V]|$. We need to check that this condition is well-defined, i.e. that:

1. Given an $X$, $|[X,[X,U]]|=|[X,U]|$ for some $U$ not a multiple of $X$ implies that $|[X,[X,V]]|=|[X,V]|$ for all $V$.
2. $X$ satisfying $|[X,[X,V]]|=|[X,V]|$ implies that all conjugates $gXg^{-1}$ of it satisfy it too. This is trivial from considering $V=gV'g^{-1}$ (since the identity is true for all $V$).

Is the first one even true outside $\mathfrak{so}(3)$ -- for all simple Lie algebras?

One may come up with the idea of defining a form $\langle X, Y\rangle = \mathrm{tr}[X,[Y,\cdot]]$ (example of some weak motivation -- the vector triple product $x\times(x\times v)$ has as eigenvectors the vectors $v$ perpendicular to $x$ and the eigenvalues depend on the length of $x$) and check that this is indeed an $\mathrm{Ad}$-invariant symmetric bilinear form, and is thus unique up to scaling for simple Lie algebras. This form is called the Killing form.

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Factorisation of Lie groups

We have seen the classification of connected Abelian Lie groups: they are products of circles and lines. We wonder if such a classification is possible for more general Lie groups.

The natural way to "factorise" groups by taking quotients over normal subgroups -- we wonder if this means that all Lie groups can be written as direct products of simple Lie groups (groups that don't have a nontrivial connected normal subgroup -- can you see why "connected" matters?). Well, not really -- the quotients need not be subgroups at all, after all. Instead, the "factorisation" takes the form of what is known as a group extension. A group for which it is a direct product is called a reductive Lie group -- and its Lie algebra is the direct sum of simple Lie algebras, or a reductive Lie algebra.

It is more conventional in the literature to define a simple Lie algebra excluding the one-dimensional/abelian case. In this definition, direct sums of simple Lie algebras are semisimple Lie algebras, and reductive Lie algebras are direct sums of semisimple and abelian Lie algebras.

TBC: Cartan's criterion, solvability, nilpotency

Lie group topology

I'll assume you have a basic understanding of general topology -- if not, consult the topology articles here. Most of the abstract stuff and "weird" cases are not really important, because it is easy to see that Lie groups are manifolds.

We need to be careful while studying the topology of Lie groups, because we already have an intuitive picture of a Lie group, and we need to be careful to prove all the things we just "believe" to be true.

The main point of the topology of a Lie group is that the group elements define the "flows" on the manifold. What this means is that left-multiplication is a homeomorphism, and it's not absurd to say that inversion is a homeomorphism, because it represents a "reflection" of the manifold. That these conditions make sense is confirmed by looking at the proofs of the following "obvious" facts.

(1) In a connected group, a neighbourhood of the identity generates the entire group, i.e. $H\le G\land H\in N(1)\implies H=G$ for connected $G$.

Let's think about why this is true. Why does $H$ need to be a neighbourhood -- why must it contain an open set containing the identity? Suppose instead we just knew it contained a set $Q$ that looked like this:

Well, $H$ still contains the orange point, but we cannot say it contains the purple point, because it's perfectly happy not containing it -- it's not like we have some vertical element in the Lie group that if you multiplied to some point in $Q$, you'd get the purple point. But instead if $Q$ was an open neighbourhood of the identity:

Then the purple point has to be in $H$, because $Q$ contains flows in "all directions" on the group. To actually prove that every point will be contained in $H$ -- well, we know that the point is (will eventually be) that $H$ is the connected component of $G$ (and since $G$ is connected, $H=G$) -- let's just show that $H$ is both open and closed, i.e. nothing in $H$ touches its exterior, and nothing in its exterior touches $H$. Here's the proof:

• Nothing in $H$ touches anything -- Suppose $\exists x\in H, x\in\mathrm{cl}(H')$. Then $xQ$ contains a point in $H'$.
• Nothing outside $H$ touches it -- Suppose $\exists x\in H', x\in\mathrm{cl}(H)$. Then $xQ$ contains a point in $H$, so $x$ must be in $H$.

We're really just formalising the notion of "translating $Q$ to its edges to extend $H$ further and further". The key fact we've used here is, of course, that left-multiplication is a homeomorphism, so $xQ$ is still an open set.

(2) The connected component of the identity is a subgroup.

The idea is that taking two elements $g,h$ of the connected component, their product should remain in the connected component. Once again, this follows from the continuity of left-multiplication -- considering the action of left-multiplication by $g$ on the connected component, its continuity implies that the image must remain connected.

(3) If a subgroup contains a neighbourhood of the identity, it contains the connected component of the identity.

Corollary to (1) and (2).

(4) The connected component of the identity is a normal subgroup.

Conjugation is a continuous map.

(5) Open subgroups are closed.

Corollary to (3). Alternate proof: the complement is the union of some cosets, which are open sets too. A weaker theorem can be made of closed sets -- closed subgroups with finite index are open.

What this means: any open subgroup is a union of connected components.

(6) Intuition for compact subgroups

How can a Lie group possibly "close in on itself"? Surely we keep "extending" an open neighbourhood $W$ of the identity by observing that $xW$ must be in the subgroup? The idea is that these translations of $W$ form an open cover of the group, if it has a finite subcover, then it makes sense for the group to close in on itself. By playing around with different open neighbourhoods $W$ and taking some suitable unions, one can see that this is equivalent to the condition that every open cover has a finite subcover, i.e. the group is compact.

(7) A compact, connected Abelian Lie group is a torus.

This is a generalisation of "a finite Abelian group is the direct product of cyclic groups".

The idea behind the proof is that in the Abelian case, the exponential map is a homomorphism from the Lie algebra to the Lie group, but the Lie algebra cannot detect compactness in the Lie group -- the kernel of the exponential map can. We know from our study of the exponential map that it has a discrete kernel, and in the Abelian case is surjective -- thus the Lie group is homeomorphic to $\mathbb{R}^n/\mathbb{Z}^n$, which is an $n$-torus.

(8) A connected Abelian Lie group is a cylinder (direct product of a torus and an affine space)

Analogous to above, except $\mathbb{R}^m/\mathbb{Z}^n$ where $m\ge n$.

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: $df([Y,X])=[df(Y),df(X)]$.

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The adjoint map

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.

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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:

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

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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).