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

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:

*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:

- 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 $V$.
- $X$ 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 $V$).

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**.

**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

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