Lie algebra representation

Where the story starts: vector fields

If I had to point to the object from which the whole theory grows, I would point to the set $X (M)$ of all vector fields on a manifold $M$ . It carries a bracket,

[X, Y] = X Y - Y X,

the commutator of two derivations, and this bracket is bilinear, alternating and satisfies the Jacobi identity. In other words, $X (M)$ is the first and most natural example of a Lie algebra — for me, the original one. Vector fields are not just an example among others: they are the concrete thing the abstract definition was distilled from (see about Lie algebras and involutive sets of vector fields).

The abstraction of the notion of Lie algebra has advantages. But abstraction has also a price: once the elements of $g$ are merely symbols obeying identities, we lose the feeling of what they do. A vector field pushes points along its flow; an abstract $x \in g$ does nothing at all.

The theory of representations is the answer to a single question:

Given an abstract Lie algebra $g$ , can I make its elements act on something again, as vector fields act on a manifold?

First answer: let them act on a vector space (the module picture)

The simplest thing to let them act on is a vector space $V$ . We ask for a bilinear action

\cdot : g \times V \to V,

so that each $x \in g$ moves a vector $v \in V$ to a new vector $x \cdot v$ . But we cannot ask for any action. The bracket in $g$ was extracted from a commutator of movements, so the action must remember that origin. The compatibility condition has to be

[x, y] \cdot v = x \cdot (y \cdot v) - y \cdot (x \cdot v) .

Read the right-hand side as an instruction: "do $x$ , then do $y$ , then subtract the same in reverse order." The axiom says that doing $[x, y]$ is exactly that difference. When it holds, we say $V$ is a $g$ -module (module over a Lie algebra).

Second answer: linear operators (the representation picture)

The rule $x \mapsto (v \mapsto x \cdot v)$ takes an element of $g$ and produces a linear operator on $V$ , i.e. an element of $gl (V)$ , the Lie algebra of operators under the commutator bracket. Writing $f (x)$ for that operator, we have built a map

f : g \to gl (V), f (x) (v) = x \cdot v .

The dictionary between the two pictures is precisely this equation. And the module axiom, translated, becomes

f ([x, y]) = f (x) f (y) - f (y) f (x) = [f (x), f (y)],

which is exactly the statement that $f$ is a Lie algebra homomorphism. This is the usual definition: a representation of $g$ on $V$ is a homomorphism $f : g \to gl (V)$ . Differentiating a group representation $p : G \to GL (V)$ yields one of these, and conversely, when $G$ is simply connected and $g$ is its Lie algebra, every such $f$ can be "exponentiated" back to a representation $p$ of $G$ with $d p = f$ .

Third answer: the vector-field picture, which is where we began

Now we close the circle. What is $gl (V)$ really? Its elements are linear operators, and a linear operator on $V$ is nothing but a linear vector field on $V$ , seen as a manifold $M = V$ : to $A \in gl (V)$ associate

X_{A} : v \mapsto - A v,

that is, $X_{A} = - \sum_{i} (\sum_{j} A_{i j} v_{j}) \partial_{v_{i}}$ in coordinates (compare the example in Lie algebra action). The minus sign is a convention, chosen so that the correspondence respects commutators: $[X_{A}, X_{B}] = X_{[A, B]}$ . So a representation $f : g \to gl (V)$ is the same thing as a map

A : g \to X (V), A (x) = X_{f (x)},

a Lie algebra action of $g$ on the manifold $V$ by vector fields with linear coefficients. The abstract algebra has regained its geometric meaning: the elements of $g$ push points of $V$ around again. And the bracket has come back home, because $f ([x, y])$ is now the Lie bracket of the vector fields $f (x)$ and $f (y)$ .

The cleanest instance of this is the adjoint representation

ad : g \to gl (g),

with ${ad}_{x} (y) = [x, y]$ . When $g$ is concretely realized as a set of vector fields, ${ad}_{x} (y) = [x, y]$ is literally the Lie bracket of vector fields — no translation needed.

Why the grounding matters

It would be a mistake to read the vector-field picture as a mere analogy. Ado's theorem says that every finite-dimensional Lie algebra has a faithful finite-dimensional representation. Composing that with the identification above, every finite-dimensional Lie algebra is (isomorphic to) a Lie algebra of linear vector fields; subalgebra of the general algebra vector fields $X (V)$ . So the abstract definition of a Lie algebra is not more fundamental than the vector-field picture.