Representation of a Lie algebra

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In mathematics, a representation of a Lie algebra is a way of writing a Lie algebra as a set of matrices (or endomorphisms of a vector space) in such a way that the Lie bracket is given by the commutator.

The notion is closely related to that of a representation of a Lie group. Roughly speaking, the representations of Lie algebras are the differentiated form of representations of Lie groups, while the representations of the universal cover of a Lie group are the integrated form of the representations of its Lie algebra.

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[edit] Formal definition

A representation of a Lie algebra \mathfrak g is a Lie algebra homomorphism

\rho\colon \mathfrak g \to \mathfrak{gl}(V)

from \mathfrak g to the Lie algebra of endomorphisms on a vector space V (with the commutator as the Lie bracket). Explicitly, this means that

\rho_{[x,y]} = [\rho_x,\rho_y] = \rho_x\rho_y - \rho_y\rho_x\,

for all x,y in \mathfrak g. The vector space V, together with the representation ρ, is called a \mathfrak g-module. (Many authors abuse terminology and refer to V itself as the representation).

One can equivalently define a \mathfrak g-module as a vector space V together with a bilinear map \mathfrak g \times V\to V such that

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

for all x,y in \mathfrak g and v in V. This is related to the previous definition by setting x\cdot v = \rho_x(v).

[edit] Infinitesimal Lie group representations

If \phi:G\to H is a homomorphism of Lie groups, and \mathfrak g and \mathfrak h are the Lie algebras of G and H respectively, then the induced map \phi_*: \mathfrak g \to \mathfrak h on tangent spaces is a Lie algebra homomorphism. In particular, a representation of Lie groups

\phi: G\to \mathrm{GL}(V)\,

determines a Lie algebra homomorphism

\phi_*: \mathfrak g \to \mathfrak{gl}(V)

from \mathfrak g to the Lie algebra of the general linear group GL(V), i.e. the endomorphism algebra of V.

A partial converse to this statement says that every representation of a finite-dimensional (real or complex) Lie algebra lifts to a unique representation of the associated simply connected Lie group, so that representations of simply-connected Lie groups are in one-to-one correspondence with representations of their Lie algebras.

[edit] Properties

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Representations of a Lie algebra are in one-to-one correspondence with representations of the associated universal enveloping algebra. This follows from the universal property of that construction.

If the Lie algebra is semisimple, then all reducible representations are decomposable. Otherwise, that's not true in general.

If we have two representations, with V1 and V2 as their underlying vector spaces and ·[·]1 and ·[·]2 as the representations, then the product of both representations would have V_1\otimes V_2 as the underlying vector space and

x[v_1\otimes v_2]=x[v_1]\otimes v_2+v_1\otimes x[v_2] .

If L is a real Lie algebra and \rho:L\times V\rightarrow V is a complex representation of it, we can construct another representation of L called its dual representation as follows.

Let V∗ be the dual vector space of V. In other words, V∗ is the set of all linear maps from V to C with addition defined over it in the usual linear way, but scalar multiplication defined over it such that (z\omega)[X]=\bar{z}\omega[X] for any z in C, ω in V∗ and X in V. This is usually rewritten as a contraction with a sesquilinear form 〈·,·〉. i.e. 〈ω,X〉 is defined to be ω[X].

We define \bar{\rho} as follows:

〈\bar{\rho}(A)[ω],X〉 + 〈ω,ρA[X]〉 = 0,

for any A in L, ω in V∗ and X in V. This defines \bar{\rho} uniquely.

[edit] Algebra representation of a Lie superalgebra

If we have a Lie superalgebra L, then, a (not necessarily associative) Z2 graded algebra A is an algebra representation of L if as a Z2graded vector space, A is a vector space rep of L and in addition, the elements of L acts as derivations/antiderivations.

More specifically, if H is a pure element of L and x and y are a pure elements of A,

H[ab] = (H[a])b + (−1)aHa(H[b])

Also, if A is unital, then

H[1] = 0

Now, for the case of a representation of a Lie algebra, we simply drop all the gradings and the (−1) to the some power factors.

Given a vector space which happens to be an associative algebra and a Lie algebra at the same time, and in addition, as an associative algebra, it is a rep of itself as a Lie algebra? We then have a Poisson algebra. And what about the corresponding case for an associative superalgebra? We have a Poisson superalgebra.

A Lie (super)algebra is an algebra and it has an adjoint representation of itself. Now what does the (anti)derivation rule say? It is the superJacobi identity.

[edit] See also

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