I'm reading Akhil Mathew's blog post on Formal Lie Theory in Characteristic Zero.

Let $H$ be cocommutative Hopf algebra over a field $k$. We can form $\mathfrak{g}$, the Lie algebra over $k$ consisting of the primitive elements of $H$. There is a canonical map $\phi : U(\mathfrak{g}) \rightarrow H$ from the universal enveloping algebra of $\mathfrak{g}$ to $H$. In fact, the functor $P : \textbf{Hopf} \rightarrow \textbf{Lie}$ sending a Hopf algebra over $k$ to its subspace of primitive elements (which forms a Lie algebra over $k$) is right adjoint to the universal enveloping algebra functor $U : \textbf{Lie} \rightarrow \textbf{Hopf}$, under certain finiteness conditions on both categories.

I am wondering,

(a) Under what conditions is $\phi$ (as above) surjective? Of course, if $H$ is generated by its primitive elements then this is true.

(b) When will $\text{im}(\phi)$ contain the group-like elements of $H$?

  • 1
    $\begingroup$ Regarding (a): If the field is algebraically closed and of zero characteristic, then, by the Cartier-Konstant-Milnor-Moore theorem (see: mathoverflow.net/q/255767/85967), it turns that $\phi$ is surjective if the set of grouplikes is trivial (and thus $H$ is generated by its primitives). Are you looking for some further conditions specifying this situation? $\endgroup$ – Konstantinos Kanakoglou Jul 26 '18 at 12:52
  • $\begingroup$ Ah, yes, that works! Post as an answer and I can give you some credit for your help. Unfortunately, this assumption is not true in my situation. (Let $H$ be a Hopf algebra with Lie algebra $\mathfrak{g}$). I am trying to find the map from $G$ to the group-like elements of $\hat{U(\mathfrak{g})}$. (We have a filtration on $U(\mathfrak{g})$ if we assume $\mathfrak{g}$ is finitely generated over a field of characteristic $0$- I want to assume this). $\endgroup$ – Kind Bubble Jul 26 '18 at 16:22
  • 1
    $\begingroup$ so you are interested in field of characteristic $0$ with $g$ assumed to be finitely generated ? Or are you interested in the case for which the set $G(H)$ of grouplikes is a non-trivial group ? $\endgroup$ – Konstantinos Kanakoglou Jul 26 '18 at 21:15
  • $\begingroup$ I would like to assume both. $\endgroup$ – Kind Bubble Jul 26 '18 at 21:26

Some thoughts, regarding question (a):

In the case of a pointed, cocommutative hopf algebra $H$ over a field $k$ of characteristic $0$, by the Cartier-Konstant-Milnor-Moore theorem (see: Classification of quasitriangular Hopf algebras) we have that $$ H\cong U\big(P(H)\big)\sharp kG(H) $$ where $P(H)$ is the lie algebra of the primitives, $U(.)$ its UEA and $G(H)$ the group of the group-likes. So, it turns, that a general condition for $\phi$ to be surjective, is the set of grouplikes to be trivial, in which case $H$ is generated by its primitives and $H\cong U\big(P(H)\big)$.
(this includes as a special case the cocommutative hopf algebras over algebraically closed fields, since these are pointed).

A similar result -again refering to a case with trivial grouplikes- is the situation of an irreducible, cocommutative hopf algebra $H$ over a field $k$ of characteristic zero: then we have the hopf algebra isomorphism $$ H\cong U\big(P(H\big)) $$ If the above examples are of interest for your purposes, maybe the article: Hopf algebras with one groupl-like element and these notes (especially sect. 9) will be useful.

Edit: Similar topics are discussed in most of the classical books on the subject (Sweedler's book, Abe's book, Montgomery's etc), but one of the most valuable references -imo- still remains the seminal paper of Milnor and Moore: On the structure of Hopf algebras (especially the discussion on p. 239).
(however it is important to keep in mind that in the above paper the authors define hopf algebras to be what is nowadays called "$\mathbb{Z}$-graded hopf algebras" and generally they use the terminology in a slightly different way than what is understood to be "mainstream" today).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.