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Groups can be defined abstractly as sets with a binary operation satisfying certain identities, or concretely as a collection of permutations of a set. Cayley's theorem ensures that these two definitions are equivalent: any abstract group acts as a collection of permutations of its underlying set, and this action is faithful.

Similarly, rings can be defined abstractly as sets with a pair of binary operations satisfying certain identities, or concretely as a collection of endomorphisms of an abelian group. There is a "Cayley's theorem" here as well: any abstract ring acts as a collection of endomorphisms of its underlying abelian group, and this action is faithful.

The situation for Lie algebras seems much less clear to me. The adjoint representation is not generally faithful, and Ado's theorem comes with qualifications and doesn't have the simplicity of the two theorems above. For me, the problem is that I don't have a good sense of what the concrete definition of a Lie algebra is supposed to be.

I suspect that a good concrete definition of a Lie algebra is as a space of derivations on some algebra closed under commutator. In that case, is it correct to say that a Lie algebra acts faithfully as derivations on its universal enveloping algebra? Is this a good analogue of Cayley's theorem?

(Motivation: in the books on Lie algebras I have read, the authors verify that Lie algebras which occur in nature satisfy alternativity and the Jacobi identity, but I have never seen any simple justification that these axioms are "enough" in the same way that Cayley's theorem tells you that the axioms for a group or a ring are "enough." There is just Ado's theorem which, again, comes with qualifications and is hard.)

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Two comments: a finite dimensional Lie algebra $\mathfrak g$ acts faithfully as derivations on $C^\infty(\exp \mathfrak g)$. As per the axioms being enough, the Poincare-Birkoff-Witt theorem needs the Jacobi identity. –  Eric O. Korman Aug 22 '10 at 15:29
    
Eric, I would prefer a theorem which holds in complete generality, with no assumption on dimension. This is why I don't like Ado's theorem. –  Qiaochu Yuan Aug 22 '10 at 15:40
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A lie algebra does not act faithfully through derivations on its enveloping algebra, as the center acts trivially. –  Mariano Suárez-Alvarez Aug 22 '10 at 16:55

4 Answers 4

up vote 12 down vote accepted

Here is a way of avoiding Ado's theorem, at the expense of using the Poincare-Birkhoff-Witt Theorem. The PBW theorem has no finite dimensionality or characteristic hypotheses, so you may like this better. Note, however, that I will realize finite dimensional Lie algebras as endomrophisms of infinite dimensional vector spaces.

Let's define a concrete Lie algebra to be a vector space $V$, and a vector subspace $\mathfrak{h}$ of $\mathrm{End}(V)$ closed under commutator.

Theorem: Every Lie algebra is isomorphic to a concrete Lie algebra.

Proof: Let $\mathfrak{g}$ be a Lie algebra and $U$ its universal enveloping algebra. The Lie algebra $\mathfrak{g}$ acts on $U$ by left multiplication, so this gives a map $\mathfrak{g} \to U$ taking bracket to commutator. We must prove this map is injective.

Choose a basis ${ v_i }$ for $\mathfrak{g}$. Suppose that left multiplication by $\sum a_i v_i$ is $0$. Then $\left( \sum a_i v_i \right) \cdot 1 = \sum a_i v_i$ would be zero in $U$. But, by the PBW theorem, the $v_i$ are linearly independent in $U$, a contradiction. QED

As far as I know, this special case of PBW is as hard as the whole theorem.

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Thanks, David. This looks like what I'm looking for, although I am still wondering about how to reconcile the difference between this definition of a concrete Lie algebra and the definition via derivations. –  Qiaochu Yuan Aug 23 '10 at 13:47
    
You get this result more directly: PBW states that if $\mathfrak{g}$ is free as a $k$-module, then $S\mathfrak{g} \cong \operatorname{gr}U\mathfrak{g}$ as graded algebras. Therefore $$\mathfrak{g} \cong S^1\mathfrak{g} \cong \operatorname{gr}_1 U\mathfrak{g} \cong \varepsilon(\mathfrak{g})$$ as modules, where $\varepsilon: \mathfrak{g} \to U \mathfrak{g}$ is the unit of $U \dashv L$. Thus $U$ is faithful, which then implies that $\varepsilon$ is a monomorphism by general nonsense. –  Alexei Averchenko Dec 8 '12 at 4:20
    
Edit: you can get this result more directly :) –  Alexei Averchenko Dec 8 '12 at 4:25
    
You're using PBW. The question was whether you could get this result without PBW. –  David Speyer Dec 8 '12 at 16:03

A finite-dimensional real Lie algebra always arises as the Lie algebra of a Lie group. Of course, the proof of this does use Ado's theorem, as well as some Lie group theory (a subalgebra of a Lie algebra of a Lie group $G$ is also a Lie algebra of a Lie group, but not necessarily the Lie algebra of a closed subgroup of $G$). The upshot is that over $\mathbb{R}$ the definition of (finite-dimensional) Lie algebra suffices to deal with all Lie groups.

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Again, I would prefer a theorem which holds in complete generality. Infinite-dimensional Lie algebras do not always occur as Lie algebras of "infinite-dimensional Lie groups." I would also prefer a theorem which does not depend on the base field; my point is that I do not know any reason why one would be interested in Lie algebras in complete generality. What if, for example, every finite-dimensional real Lie algebras satisfies some "hyper-Jacobi identity" and this is the reason they have such nice properties? I would like to know that this is not the case. (Or is it?) –  Qiaochu Yuan Aug 22 '10 at 15:50
    
(I'm not an expert, but) an abstract infinite-dimensional Lie algebra is indeed too broad notion, perhaps. There are a lot of theorems about finite-dimensional Lie algebras over $\mathbb C$ but do you know many theorems about all infinite-dimensional ones? People usually study some more concrete class — e.g. Kac–Moody algebras or smth. –  Grigory M Aug 22 '10 at 16:10

Due to the Peter-Weyl theorem every finite dimensional Lie group is isomorphic to a subgroup of the orthogonal group O(m) for some m. Therefore its Lie algebra is a subalgebra of the Lie algebra of the orthogonal group which is the same as the Lie algebra of the spin group Spin(m). Now, the Lie algebra of the spin group is the bivector subalgebra of the clifford algebra Cl(m). Thus every finite dimensional Lie algebra is isomorphic to a Lie subalgebra of the bivector subalgebra of a clifford algebra. This point of view seems analogous to the Cayley's theorem for the following reasons:

  1. There exists a projective representation of the symmetric group in the Clifford algebra: (a,b) --> e_a - e_b. Thus the Clifford algebra may be seen as a kind of quantization of the symmetric group.

  2. This construction generalizes at least to the case of affine Kac-Moody algebras which have similar realizations in the infinite dimensional Clifford algebra.

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Classically, the "concrete" definition of what a Lie algebra is is simply a subspace of $\mathrm{End}(V)$, for some vector space $V$, which is closed under commutators. It is a theorem, then, that such a thing is the same thing as a pair $(\text{vector space},\text{antisymmetric bracket})$ satisfying the Jacobi identity. This theorem is quite important when one develops the theory that way, because it means that one can describe abstractly Lie algebras through equations.

(This should be compared with the situation for Jordan algebras, as explained in Jacobson's amazing book on the subject: special Jordan algebras, those subspaces of $\mathrm{End}(V)$'s closed under the product $A\bullet B\dot=\tfrac12(AB+BA)$, are not characterizable through equations only; this is a consequence of the fact that the class of special Jordan algebras is not closed under quotients)

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Isn't this Ado's theorem, and doesn't it fail for infinite-dimensional Lie algebras? –  Qiaochu Yuan Aug 22 '10 at 16:56
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@Qiaochu, every Lie algebra acts (by left multiplication) on its enveloping algebra, and this action is faithful, so every Lie algebra is "concrete". (The point of Ado's theorem (once extended by Iwasawa and Harish-Chandra to avoid the hypothesis on the caracteristic) is that finite-dimensional Lie algebras have finite-dimensional faithful representations: that they have faithful representations is again trivial, because the enveloping algebra works; looking for faithful finite dimensional reps in the case of infinite-dimensional algebras is of course hopeless) –  Mariano Suárez-Alvarez Aug 23 '10 at 2:30
    
@Qiaochu, by the way: if you want a faithful action of a Lie algebra $\mathfrak g$ on an algebra $A$ through derivations, then let $A=T(\mathcal U(\mathfrak g))$ be the tensor algebra on the enveloping algebra $\mathcal U(\mathfrak g)$ of $\mathfrak g$, with $\mathfrak g$ acting with the unique action by derivations (given by one of the two universal properties of tensor algebras) which extends the multiplication action of $\mathfrak g$ on $\mathcal U(\mathfrak g)$. –  Mariano Suárez-Alvarez Aug 23 '10 at 2:37
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Of course, as David observes above (or below, depending on how you sort answers...) all this depends on the PBW theorem, and is the reason why all texts insist on the fact that the map $\mathfrak g\to\mathcal U(\mathfrak g)$ is injective. –  Mariano Suárez-Alvarez Aug 23 '10 at 12:25

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