# Why is enveloping algebra called enveloping algebra?

What does the enveloping algebra of $$\mathfrak{g}$$ have to do with envelopes? If $$\mathfrak{g}$$ is a Lie algebra, we take tensor algebra on $$\mathfrak{g}$$ and make quotient through ideal of T, so we put elements of $$\mathfrak{g}$$ into some equivalence classes determined by the ideal. How is this "enveloping" the $$\mathfrak{g}$$? What is the intuition?

(Maybe there is no interesting interpretation of this and I am just being too curious, but had to try it).

Remark 1: I am refering to the enveloping algebra of $$\mathfrak{g}$$, which is sometimes called universal enveloping algebra. It is the same.

Remark 2: I am using the definition from Dixmier: • The purpose is to get an associative algebra $U(L)$ with the same representation theory as the Lie algebra $L$, and which "envelopes" the Lie algebra $L$ for the canonical mapping $L\rightarrow U(L)$. Sep 8, 2021 at 14:08
• The idea is to embed a Lie algebra into a larger (associative) algebra, so the name is self-explanatory. Sep 8, 2021 at 14:09
• The verb 'envelop' means 'cover' or 'wrap'. Sep 8, 2021 at 14:48
• Thank you all! Why we want the Lie algebra to be associative though? Sep 8, 2021 at 14:58
• @TerezaTizkova We don't "want a Lie algebra to be associative." Very few Lie algebras are associative. We are very comfortable with associative algebras, so they are a useful environment to embed Lie algebras into. Sep 8, 2021 at 18:30

1. The very first thing that is crucial to have in mind is that every associative algebra (where we have a vector space addition $$+$$, and a multiplication $$\cdot$$) can be turned into (or let's better say, can also be viewed as) a Lie algebra. Namely, we can use the same vector space structure, and "make" a Lie bracket by defining

$$[a,b] := a\color{red}{\cdot} b-b \color{red}{\cdot} a$$

i.e. use the commutator from the associative multiplication $$\cdot$$ to define a Lie bracket $$[ , ]$$. It's a standard exercise that this really is a Lie bracket i.e. satisfies the axioms of bilinearity, anticommutativity and the Jacobi identity.

So this means: We can view every associative algebra also as a Lie algebra. And it gives tons of standard examples of Lie algebras. For example, the associative algebra $$M_n(K)$$ of $$n \times n$$-matrices over a field $$K$$, when viewed this way, becomes a Lie algebra, which is usually called $$\mathfrak{gl}_n(K)$$.

2. Once we understood 1, the question arises if every Lie algebra is such an "associative algebra in disguise"? (And note this is actually highly related to your recent question https://math.stackexchange.com/q/4241787/96384 .) No it is not. For a slick argument, if in our definition of associative algebra we demand that it has a multiplicative unit $$1 \neq 0$$, then by definition this element $$1$$ commutes with all elements of the algebra, so we would have $$[1,x]=0$$ for all $$x$$ in our algebra; i.e. the Lie algebra has a non-trivial center. But there are many Lie algebras with trivial center, e.g. $$\mathfrak{sl}_n(\mathbb C)$$, or the Lie algebra $$\mathbb R^3$$ with bracket given by the cross product.

3. But wait: Is not my example $$\mathfrak{sl}_n(\mathbb C)$$ given by some matrices, and the Lie bracket is the matrix commutator? Yes, but look closely: E.g. $$\mathfrak{sl}_n$$ consists only of the $$n \times n$$ matrices with trace $$0$$. Not only does this not include a multiplicative unit $$1$$, it is not even closed with respect to the usual matrix multiplication at all.

For example, both $$a=\pmatrix{1&0\\0&-1}$$ and $$b=\pmatrix{-1&0\\0&1}$$ are in $$\mathfrak{sl}_2(\mathbb C)$$ -- but $$a \cdot b = b \cdot a = \pmatrix{1&0\\0&1}$$ is not.

On the other hand, $$[a,b] = ab-ba = \pmatrix{0&0\\0&0}$$ is contained in $$\mathfrak{sl}_2(\mathbb C)$$. But for this computation we had to "secretly leave" the traceless matrices, and compute $$a\cdot b$$ and $$b \cdot a$$ in the "bigger" full matrix algebra $$M_2(\mathbb C)$$.

In other words, the subspace $$\mathfrak{sl}_n(\mathbb C) \subset M_n(C)$$ is not an associative subalgebra (it is not closed under multiplication); but it is a Lie subalgebra. That means, though, that although we cannot view $$\mathfrak{sl}_n$$ itself as an "associative algebra in disguise", we can place it inside an associative algebra -- an "enveloping algebra" --, and compute the Lie bracket in $$\mathfrak{sl}_n(\mathbb C)$$ as the commutator of the associative multiplication in that enveloping algebra.

[Why is that helpful? Because historically, we already knew a lot about associative algebras before Lie algebras were even invented. So it made sense, to understand Lie algebras, to "envelop" them with some associative algebras, and then use what we know about those, to infer something about the Lie algebra. Also, one really does not want to, but if nothing else works, one can always do a matrix computation ...]

4. We just saw that one obvious "enveloping associative algebra" for $$\mathfrak{sl}_n(\mathbb C)$$ is the full matrix algebra $$M_n(\mathbb C)$$, whose dimension is just one bigger than that of the Lie algebra it "wraps up".

What about the other example, $$\mathbb R^3$$ with the cross product? Well, as a Lie algebra this is isomorphic to $$\mathfrak{su}_2 = \{ \pmatrix{ai & b+ci\\-b+ci&-ai}: a,b,c \in \mathbb R\}$$, where the Lie bracket is given by the matrix commutator. Again, check that the usual matrix multiplication of two elements in there "leaves" the algebra itself, but at least it stays inside the $$2\times2$$ complex matrices. So in this case, we have found $$M_2(\mathbb C)$$ as an enveloping associative algebra of our Lie algebra. Note that while the dimension of our real Lie algebra was $$3$$, this enveloping algebra has real dimension $$8$$, so it is significantly bigger!

Now the next thing to note is that any given Lie algebra can have many different such "enveloping associative algebras", and they can become much bigger than the Lie algebra we started with. For example, I claim that $$\mathfrak{sl}_2(\mathbb C)$$ is isomorphic to

$$\{\pmatrix{x&0&0&0&y&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\z&0&0&0&-x&0\\0&0&0&0&0&0}: x,y,z \in \mathbb C\}$$

which quite visibly can be put inside a big enveloping algebra $$M_6(\mathbb C)$$. For this one you could say, well but I would find a "tighter fit" of an enveloping associative algebra in there, namely a certain associative subalgebra in that matrix algebra (for starters, discard the last row and column), of dimension less than $$36$$. That is true. But I can do better: I claim that $$\mathfrak{sl}_2(\mathbb C)$$ is also isomorphic to

$$\{ \pmatrix{3x&\sqrt 3 y&0&0\\ -\sqrt 3z&x&2y&0\\ 0&-2z&-x&\sqrt 3 y\\ 0&0&-\sqrt 3z&-3x} : x,y,z \in \mathbb C \}$$

and the tighest associative envelope of this that you find in there is indeed the full, $$16$$-dimensional matrix algebra $$M_4(\mathbb C)$$! This comes from here and has to do with irreducible representations of $$\mathfrak{sl}_2(\mathbb C)$$.

Indeed, from this viewpoint, the entire point of representation theory of a (say, simple) Lie algebra $$\mathfrak g$$ is: What are all possible ways to view $$\mathfrak g$$ as being "enveloped" by some matrix algebra $$M_n(\mathbb C)$$?

And in this sense, the construction of the universal enveloping algebra $$U(\mathfrak g)$$ -- which, as the name suggests, is the "biggest associative envelope" of a given Lie algebra $$\mathfrak g$$, insofar as all the most interesting "associative envelopes" of $$\mathfrak g$$ can be viewed as quotients of $$U(\mathfrak g)$$ -- contains "all" the representation theory of $$\mathfrak g$$. Of course, not in an obvious way; but in a way that allows one to view things from a different perspective, and make use of classical tools of associative rings and modules. For example, did you know that $$\mathfrak{su}_2$$ can also be viewed as the space of pure quaternions, i.e. the quaternions are an associative envelope of $$\mathfrak{su}_2$$ -- and hence, a (tiny) quotient of the (huge) universal envelope $$U(\mathfrak{su}_2)$$?

(For a few other applications cf. Applications and usefulness of universal enveloping algebra .)

• Wonderful answer. I really enjoyed reading this. Sep 8, 2021 at 22:46
• Thank you. I for my part am happy that my unconscious somehow managed to insert several subliminal cheers for the ABBA reunion into this answer. Sep 9, 2021 at 0:48

In addition to @Torsten Scheneberg's great explanation, it is possible to give an explanation in terms of "goals" or "intentions".

First, the so-called "tensor algebra" of a vector space $$V$$ is more aptly called "the universal associative algebra" attached to $$V$$. The fact that it can be constructed in terms of tensor powers is a useful fact, but is incidental to its characterization. Namely, the universal associative algebra $$AV$$ attached to a vector space $$V$$ (over some fixed base field $$k$$) is the (provably unique-up-to-unique-isomorphism) associative algebra $$AV$$ so that every vector space map $$V\to B$$ to an associative algebra $$B$$ factors (uniquely) through $$AV$$. The existence of such a thing can be proven via the direct sum of tensor powers, yes.

Then the universal enveloping algebra of a Lie algebra $$\mathfrak g$$ is a quotient of $$A\mathfrak g$$, that respects the Lie algebra operations in $$\mathfrak g$$ and those in $$A\mathfrak g$$ that any associative algebra has... Much more can be said.

So it's not just "a definition", but is an idea meant to fulfill a need, an intention.