10
$\begingroup$

If $k$ is an algebraically closed field of characteristic zero and $H$ is a cocommutative Hopf algebra, then $$ H \cong U(P(H)) \ltimes kG(H). $$ What happens if the field is not algebraically closed? Is the theorem still true or is there any counterexample? What about characteristic different from zero?

$\endgroup$
3
  • $\begingroup$ It would help to define what the symbol $G$ means here, and perhaps also to add a reference. Since you are asking at least two distinct questions, it's unclear what would count as an "answer". $\endgroup$ Jan 15, 2017 at 14:49
  • $\begingroup$ I guess the OP refers to the Cartier-Kostant-Milnor-Moore theorem in the same sense as in: mathoverflow.net/q/255767/85967 $\endgroup$ Jan 15, 2017 at 23:49
  • $\begingroup$ Details can also be found in Ch. $VIII$, $XIII$, sect. $13.1$ of Sweedler's book on Hopf algebras. $\endgroup$ Jan 16, 2017 at 17:54

2 Answers 2

10
$\begingroup$

When $k$ fails to be algebraically closed the theorem is false but the discrepancy can be understood in terms of Galois descent and so in principle understood in terms of Galois cohomology.

Suppose $H$ is a cocommutative Hopf algebra over $k$, not algebraically closed but characteristic $0$. Then the classification theorem applies to the base change $H \otimes_k \bar{k}$ of $H$ to the algebraic closure, which must therefore take the form of the semidirect product of a universal enveloping algebra and a group algebra. So $H$ is a $k$-form of such a thing, but need not be such a thing itself.

For example, suppose $H$ is finite-dimensional, so $H \otimes_k \bar{k}$ is a group algebra $\bar{k}[\Lambda]$ of a finite group $\Lambda$. The most obvious $k$-form of $\bar{k}[\Lambda]$ is, of course, $k[\Lambda]$, but other $k$-forms are possible. In this setting $H$ necessarily splits over a finite Galois extension $L$ in the sense that we must already have

$$H \otimes_k L \cong L[\Lambda].$$

Now, the Galois group $G = \text{Gal}(L/k)$ acts on the LHS by Hopf algebra automorphisms (over $k$), so also acts on the RHS by Hopf algebra automorphisms (over $k$), and hence acts on the group of grouplike elements $\Lambda$ on the RHS. If we already had $H \cong k[\Lambda]$ then this action would be trivial, but in fact every possible action of $G$ on $\Lambda$ occurs, and conjugacy classes of such actions classify the possible choices of $H$. Given such an action $\rho : H \to \text{Aut}(\Lambda)$ we can recover $H$ as

$$H \cong L[\Lambda]^G$$

where the action of $G$ on $L[\Lambda]$ extends the action of $G$ on $\Lambda$.

Very explicitly, let $k = \mathbb{R}, L = \mathbb{C}$, and $\Lambda = C_3$, where the Galois group $G = C_2$ acts via inversion. Then we have

$$H \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[C_3] \cong \mathbb{C}[x]/(x^3 - 1)$$

where $x$ is grouplike but the action of complex conjugation sends $x$ to $x^{-1} = x^2$. The fixed point subalgebra $H$ has a basis given by the identity $1$ and the "real and imaginary parts"

$$c = \frac{x + x^{-1}}{2}$$ $$s = \frac{x - x^{-1}}{2i}$$

(it helps to think of $x$ as secretly being $\omega = e^{\frac{2\pi i}{3}}$ here) of $x$, with comultiplication given by the sine and cosine angle addition formulas

$$\Delta(c) = c \otimes c - s \otimes s$$ $$\Delta(s) = c \otimes s + s \otimes c.$$

So neither $c$ nor $s$ is grouplike, and in fact $H$ has no nontrivial grouplike elements except the identity. In $H \otimes_{\mathbb{R}} \mathbb{C}$ the nontrivial grouplike elements are $c + is$ and $c - is$.

$\endgroup$
4
  • $\begingroup$ Cannot this (already quite neat) example be realized as (maybe some quotient of) the Hopf algebra holding the generic special orthogonal $2\times2$-matrix $\left[\begin{smallmatrix}c&s\\-s&c\end{smallmatrix}\right]$? It then is sort of dual of $k[SO(2)]$... $\endgroup$ Jan 18, 2017 at 9:29
  • 2
    $\begingroup$ @მამუკა ჯიბლაძე: you can repeat the above discussion verbatim but replacing $\Lambda = C_3$ with $\Lambda = \mathbb{Z}$. The result is that $\text{Spec } H$, thought of as an affine group scheme, is a real form of $\mathbb{G}_m$; I suppose it deserves the name $U(1)$? $\endgroup$ Jan 19, 2017 at 4:35
  • $\begingroup$ Thank you for your answer, it is very illuminating. Would it be also a ``counterexample'' of Milnor-Moore theorem for non algebraically closed fields? $\endgroup$
    – a213f
    Jan 19, 2017 at 12:36
  • $\begingroup$ Could you direct me to a reference on exploring this Galois descent for Hopf algebras? $\endgroup$
    – JamalS
    Mar 8, 2021 at 12:49
6
$\begingroup$

The case of irreducible, cocommutative Hopf algebras, over a field with $char(k)> 0$, is discussed in Sweedler's textbook on Hopf algebras, Ch.$XIII$, sect. $13.2$. (See prop. $13.2.2$, $13.2.3$).

For the case of cocommutative Hopf algebras, over an algebraically closed field $k$, with $char(k)\geq 3$ you can have a look at:

Modules of solvable infinitesimal groups and the structure of representation-finite cocommutative Hopf algebras, R.Farnsteiner, D.Voigt, Math. Proc. Cambridge Philos. Soc., v.127, p.441-459, 1999

and the references therein (among them, there is also an interesting older paper by the same authors discussing the case of cocommutative hopf algebras of finite representation type, over an algebraically closed field of $char(k)>0$).

Edit (July 2018): Since user's Qiaochu Yuan answer, provides a form of counter-example to the theorem, for non-algebraically closed fields, i think it would be of some added value to indicate cases in which the Cartier-Kostant-Milnor-Moore theorem remains valid for non-algebraically closed fields: this happens if we take as an assumption that we are dealing with pointed hopf algebras. So, the (somewhat more general) statement of the theorem goes like:

If $H$ is a pointed, cocommutative hopf algebra over a field $k$ of characteristic $0$, we have that $$ H\cong U\big(P(H)\big)\ltimes kG(H) $$ where $P(H)$ is the lie algebra of the primitives, $U(.)$ its UEA and $G(H)$ the group of the group-likes.

The algebraically closed field case (mentioned in the OP), may be considered a special case of the above, taking into account that cocommutative hopf algebras over algebraically closed fields, can be easily seen to be pointed.

$\endgroup$
0

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.