This is an extension of this question about symmetric algebras in positive characteristic. The title is also a bit tongue-in-cheek, as I am sure that there are multiple "correct" answers.

Let $\mathfrak g$ be a Lie algebra over $k$. One can define the universal enveloping algebra $U\mathfrak g$ in terms of the adjunction: $$\text{Hom}_{\rm LieAlg}(\mathfrak g, A) = \text{Hom}_{\rm AsAlg}(U\mathfrak g, A)$$ for any associative algebra $A$. Then it's easy enough to check that $U\mathfrak g$ is the quotient of the free tensor algebra generated by $\mathfrak g$ by the ideal generated by elements of the form $xy - yx - [x,y]$. (At least, I'm sure of this when the characteristic is not $2$. I don't have a good grasp in characteristic $2$, though, because I've heard that the correct notion of "Lie algebra" is different.)

But there's another good algebra, which agrees with $U\mathfrak g$ in characteristic $0$. Namely, if $\mathfrak g$ is the Lie algebra of some algebraic group $G$, then I think that the algebra of left-invariant differential operators is some sort of "divided-power" version of $U\mathfrak g$.

So, am I correct that this notions diverge in positive characteristic? If so, does the divided-power algebra have a nice generators-and-relations description? More importantly, which rings are used for what?


The notions do indeed diverge in positive characteristic: there is the enveloping algebra, and then (in the case that $\mathfrak g$ is the Lie algebra of an algebraic group G) there is also the hyperalgebra of G, which is the divided-power version you mention. In characteristic 0 these two algebras coincide, but in positive characteristic they differ very much. In particular, the hyperalgebra is not finitely-generated in positive characteristic; see Takeuchi's paper "Generators and Relations for the Hyperalgebras of Reductive Groups" for the reductive case. There is also a good exploration of the hyperalgebra in Haboush's paper "Central Differential Operators of Split Semisimple Groups Over Fields of Positive Characteristic."

One can obtain the hyperalgebra as follows. I don't know in what generality the following construction holds, so let's say that $\mathfrak g$ is the Lie algebra of an algebraic group G defined over $\mathbb Z$. Then there is a $\mathbb Z$-form of the enveloping algebra of G (the Kostant $\mathbb Z$-form) formed by taking divided powers, and upon base change this algebra becomes the hyperalgebra. Alternatively, one can take an appropriate Hopf-algebra dual of the ring of functions on G (cf Jantzen's book "Representations of Algebraic Groups").

As for their uses, in positive characteristic the hyperalgebra of G captures the representation theory of G in the way that the enveloping algebra does in characteristic 0, i.e. the finite-dimensional representations of G are exactly the same as the finite-dimensional representations of the hyperalgebra. This fails completely for the enveloping algebra: instead, the enveloping algebra sees only the representations of the first Frobenius kernel of G.

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    $\begingroup$ Could you also remark on Theo's comment, namely that the definition of a Lie algebra is different in characteristic 2? I know that [x, x] = 0 in this case but is this due to the fact that one should specify what "[x, x]/2" is in characteristic 2? $\endgroup$ Nov 30 '09 at 16:15
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    $\begingroup$ One does have to take a different definition in characteristic 2 -- well, depending on one's usual definition of Lie algebra. You need to replace the axiom [x,y]=-[y,x] with the axiom [x,x]=0. Actually, though, regardless of characteristic, [x,x] = 0 for all x implies [x,y] = -[y,x] for all x and y, so in the definition of Lie algebra one could replace the axiom [x,y]=-[y,x] with the axiom [x,x]=0 and still get the right thing. That way you wouldn't have to give a separate definition in characteristic 2. $\endgroup$ Nov 30 '09 at 21:21
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    $\begingroup$ The reason I think this is a different definition is that in higher characteristics, all the axioms are (multi-)linear, whereas the axiom [x,x]=0 is quadratic, even though the [,] is a bilinear form, not a quadratic form. Put another way, the anti-symmetry axiom is portable to any symmetric monoidal category, whereas the [x,x]=0 axiom is not. $\endgroup$ Jan 4 '10 at 4:38
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    $\begingroup$ @Marc: Specifying in characteristic $2$ what $[x,x]/2$ is means exactly to specify a restricted Lie algebra structure so it is not directly related the definition of an ordinary Lie algebra. @Theo: I agree that a downside of setting $[x,x]=0$ is that we do not get multilinear axioms. On the other hand without it we can not embed the Lie algebra in an associative algebra, i.e., the map to the enveloping algebra vanishes on $[x,x]$. $\endgroup$ Apr 1 '10 at 15:59
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    $\begingroup$ The Kostant $\mathbb{Z}$-form depends on working with a Chevalley basis of a semisimple complex Lie algebra. On the other hand, for the Lie algebra of a linear algebraic group there is a naturally defined hyperalgebra in the spirit of Jantzen's book (some of this goes back many decades to people like Dieudonne). In prime characteristic such a Lie algebra is in a natural way a restricted Lie algebra, thus has a restricted enveloping algebra. But restricted Lie algebras don't have to arise from algebraic groups. So caution is needed. $\endgroup$ Apr 1 '10 at 16:41

This question popped up on the feed recently, and I wanted to add another, "brave new math"-style answer. Namely, the point of Lie algebras is that, in characteristic $0$, a Lie algebra (resp., an $L_\infty$-algebra) gives necessary and sufficient information to define a formal Lie group (resp., a cogroup object in CDGA). Global sections of this group form a commutative, co-associative Hopf algebra and the (topological/filtered/etc.) dual will be the universal enveloping algebra.

There is an analogous story to the $L_\infty$ story in characteristic $p$, proved by Akhil Mathew and Lukas Brantner, https://arxiv.org/abs/1904.07352. They prove that the structure of a derived connected formal group scheme (which determines a formal moduli problem) is determined by a so-called partition Lie algebra structure on the tangent space. This is no longer an "additive" structure (not determined by a multilinear operad unlike $L_\infty$), but it is "just" some nicely combinatorial algebraic structure on the tangent space. If you modify your definition of Lie algebra using this picture then to each (partition) Lie algebra you can associate a formal moduli problem, and to every formal moduli problem you can associate a universal enveloping algebra by taking the dual to a universal Hopf algebra associated to your formal moduli problem. From what I understand, the "$\pi_0$ operations" inherent in this structure are exactly Lie algebra operations and divided powers.


You forgot the third one: restricted enveloping algebra. Hence, in characteristic p we have 3 enveloping algebras with homomorphisms U->U_0->U_{dp} All 3 are Hopf algebras and can be used for different things. You need to formulate your question better. What do you want to use it for?

They are all different unless your Lie algebra is zero. If g is finite-dimensional, U is finitely generated but infinite-dimensional, U_0 is finite-dimensional and U_{dp} is not finitely-generated. If g is the Lie algebra of an algebraic group, you can describe U_{dp}(g) invariantly but you cannot really write generators and relations: U_{dp} is not finitely generated.

  • $\begingroup$ [ I had to split this up into two pieces because of size. ] Are you saying that one can construct an $U_{dp}$ from a restricted Lie algebra? At least if you want it canonically as a Hopf algebra this seems fishy. In the case of a restricted Lie algebra which over an algebraic closure is the Lie algebra of a torus (of dimension $n$) we should have the following situation: $\endgroup$ Apr 1 '10 at 17:39
  • $\begingroup$ [ cont'd ] The Lie algebra is classified by a representation of the Galois group of the base field in $(\mathbb Z/p)^n$. On the other hand $U_{dp}$ is classified by a representation of the Galois group of the base field in $(\mathbb Z_p)^n$. Many representations in $(\mathbb Z/p)^n$ do not lift to a representation in $(\mathbb Z/p)^n$. $\endgroup$ Apr 1 '10 at 17:40
  • $\begingroup$ Sorry, I am wrong, although I am not sure why all of your (\mathbb Z/p)^n look the same. Clearly two completely distinct algebraic groups may have the same restricted Lie algebras, so their U_{dp}-s will be different. I will edit in a sec. $\endgroup$
    – Bugs Bunny
    Apr 3 '10 at 10:33
  • $\begingroup$ Sorry, the last $(\mathbb Z/p)^n$ should be $(\mathbb Z_p)^n$. $\endgroup$ Apr 3 '10 at 11:53

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