# For which category (if any) are Lie algebras the algebras of a monad?

I was reading about monads recently, and it came to me that the purpose of the category of algebras of a monad seems to be to switch to a "representation" which is easier for computations. Soon after I realized that I heard similar things were true for Lie algebras (i.e. that they simplify calculations for Lie groups), as well as for the matrix representations of groups.

I know that (free) monoids are the algebra for the monad on $Set$ which given a set $X$ returns the set $X^*$ of words in elements of $X$. Since Lie algebras are monoids, I figured that they might also be the algebra of some monad. In other words, I would like to know if it is possible to make the above analogy rigorous.

Do Lie algebras (over $\mathbb{R}$ or $\mathbb{C}$) form the category of algebras for any monad $T$?

What I tried to do to find an answer (these efforts were not successful)

Ideally I would probably want to find some adjunction that generates an appropriate monad, but since I am new at this, I decided to Google instead. Here is the most relevant result I found, from page 133, edited by Paul Gregory Goerss and Stewart Priddy, of the collection of papers:

Specifically the result I am looking at, Proposition 1.2.16, says the following:

We assume that the ground ring $\mathbb{K}$ is a field of characteristic $p >0$. Recall that an algebra over the monad $\mathcal{S(L)}$ is a Lie algebra $\mathcal{G}$ together with a Lie bracket $[-,-]:\mathcal{G}\otimes\mathcal{G} \to \mathcal{G}$ such that $[x,y]=-[y,x]$, for all $x,y \in \mathcal{G}$. An algebra over the monad $\Lambda(\mathcal{L})$ is a Lie algebra $\mathcal{G}$ together with a Lie bracket $[-,-]: \mathcal{G} \otimes \mathcal{G} \to \mathcal{G}$ such that $[x,x]=0$. An algebra over the monad $\Gamma(\mathcal{L})$ is a restricted Lie algebra $\mathcal{G}$.

Since I don't really understand the article, I am not quite sure what all of the notation means, except that $S$ is a functor which takes operads to monads (whatever operads are). Also this might not be a result I am interested in, since it seems to only hold for fields of characteristic $p>0$, i.e. not for Lie algebras over the real or complex numbers.

Most of the other Google search results I found for "Lie algebra monad" also did not seem to answer my question. I thought this question might help: Do adjoint functors really define monads?, but I'm not sure how to interpret it -- was the result that Lie algebras cannot be/are not the algebra of any monad?

• The article concerns an issue that occurs in positive characteristic, which is that there are a few different possible notions of Lie algebra. Each of these is described by a slightly different monad. The notion I have in mind is the second notion in your quote out of the three. The third one only makes sense in positive characteristic, and the first two agree in characteristic $0$. – Qiaochu Yuan Jul 25 '16 at 4:46
• It's not just free monoids that are algebras for the monoid monad! – Kevin Arlin Jul 26 '16 at 16:01
• @QiaochuYuan I'm curious what the possible definitions should be in positive characteristic? I naively thought that the only issue would be in characteristic 2, where the alternating property of the bracket implies antisymmetry, but not vice versa. Or are there other issues as well? – ಠ_ಠ Nov 30 '17 at 3:57
• @ಠ_ಠ: yes, there are other issues. The following special thing happens in characteristic $p$: if $D : A \to A$ is a derivation on an algebra, then so is $D^p$. This is extra structure that appears on natural Lie algebras (e.g. an algebra under commutator bracket, or the Lie algebra on an algebraic group) so it should be taken into account: see en.wikipedia.org/wiki/Restricted_Lie_algebra (this is the third definition above). Without this extra structure there is no reason to expect, for example, a version of the PBW theorem to hold. – Qiaochu Yuan Nov 30 '17 at 4:11

## 1 Answer

Yes, Lie algebras (over any commutative ring) are the algebras of a monad over $\text{Set}$. This is true more generally for any kind of algebraic object defined by some operations satisfying some equational identities. The monad is built out of the free object functor. In other words, the forgetful functor

$$\text{LieAlg} \to \text{Set}$$

has a left adjoint, sending a set to the free Lie algebra on this set, and this adjunction is monadic.

There are various gadgets that can be used to describe various kinds of algebraic objects which, in particular, can be turned into monads; operads are an example (and Lie algebras are also the algebras of an operad), and Lawvere theories are another (and Lie algebras are also the models of a Lawvere theory).

• OK that makes sense -- I didn't even realize there was such a thing as free Lie algebras -- something to look into. Thank you so much for your help! – Chill2Macht Jul 25 '16 at 2:46