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I want to prove that the cocommutative finite dimensional Hopf algebras over an algebraically closed field of characteristic zero are group algebras (for some finite group) and that the commutative f.d. Hopf algebras (over an algebraically closed field of characteristic zero) are dual to group Hopf algebras (for some finite group). More precisely, I want to prove the following proposition:

Let $k$ an algebraically closed field of characteristic zero and $Η$ a finite dimensional $k$-Hopf algebra. Then, we have the following Hopf algebra isomorphisms:

  • If $Η$ is commutative, then: $Η \cong (kG)^{*}$ (for some finite group $G$).
  • If $Η$ is cocommutative, then: $Η \cong kG$ (for some finite group $G$).

I understand that the above result can be extracted as a consequence of the the Cartier-Konstant-Milnor-Moore classification theorem, according to which, a cocommutative $k$-Hopf algebra is isomorphic to a smash product between the universal enveloping algebra of the Lie algebra of the primitives of $H$ and the group Hopf algebra of the grouplikes: $H\cong U(P(H))\sharp kG(H)$. ($k$ is of course considered alg. closed with zero characterictic).

However, I want to provide an independent proof. After working a bit on it, I have devised a proof (which I am posting below as an answer) which however makes use of a later result, the so-called Larson-Radford theorem. Could there be some different approach (involving or not the Larson-Radford theorem)?

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As requested...

Theorem 2.3.1 in Montgomery's book "Hopf algebras and their actions on rings", which is attributed to unpublished work of Harrision c. 1960 and independently to Cartier, reads as follows.

Suppose $H$ is a commutative semisimple Hopf algebra over a field $k$. Then there exists a separable field extension $E$ of $k$ and a (finite) group $G$ such that $H\otimes E\cong (E G)^*$.

The proof follows much as in the OP's argument. By Wedderburn, as an algebra $H$ is a direct product of separable field extensions of $k$. We can thus take $E$ to be any common separable extension, so that $H\otimes E\cong E^n$ as algebras, where $n=\text{dim}(H)$. Without loss of generality, for simplicity assume $k=E$. A basis of orthogonal idempotents for $H$ then dualizes to a basis of group-likes for the dual $H^*$; namely, the dual of such an element is in $\text{Alg}(H,k)$. These group-likes thus define a group $G$, and $H^*\cong k G$ as Hopf algebras, whence $H\cong (k G)^*$ as Hopf algebras. QED.

What one can then do is use the fact that every commutative or cocommutative Hopf algebra is involutive: $S^2=\text{id}$. So whenever the Larson-Radford theorem applies we may apply it to conclude that $H$ is both semisimple and cosemisimple, so the desired result follows from the above either immediately ($H$ is commutative) or by dualizing ($H$ is cocommutative).

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Let $k$ be a field. Recall:

Definition: 1. A $k$-algebra $A$ is called étale, if $A\otimes_k \bar{k} \cong \prod \bar{k}$ where $\bar{k}$ is an alg. closure of $k$.

  1. A group scheme $G$ over $k$ is called étale, if the associated Hopf algebra is étale.

A classical theorem of Cartier states:

Theorem (Cartier): Finite group schemes are étale in characteristic zero.

The essential part in your proof of the classification (where Larson-Radford was used) was to show $H \cong \prod k$ for $k$ alg and $H$ commutative. Since a commutative Hopf algebra defines a finite group scheme, this part also follows immediately from Cartier's theorem.

Hence each alternative proof of Cartier also yields a new proof of your classification. A proof that uses differentials (and no Larson-Radford) can be found for instance in Waterhouse: Introduction to affine group schemes, § 11.4.

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The Larson-Radford theorem states that:

Let $k$ a field of characteristic zero, and $Η$ a finite dimensional $k$-Hopf algebra, with antipode $S$. The following statements are equivalent:

  • $H$ is cosemisimple.

  • $H$ is semisimple.

  • $S^{2} = Id$

(As I understand it, what the above theorem essentially tells us, is that in finite dimension and over a field of characteristic zero, the notions of semisimplicity and cosemisimplicity are equivalent and they are direct consequences of either commutativity or cocommutativity).

Now I proceed as follows:

If $Η$ is either commutative or cocommutative then $S^{2} = Id$ and according to the above $Η$ will be semisimple and cosemisimple.

  • In case $Η$ is commutative and semisimple (and $k$ is alg. closed) then we will have the $k$-algebra isomorphism: $$ Η \cong k \times k \times \ldots \times k \equiv k^{(n)} $$ where $dim_{k}H = n$. Consequently, there will be exactly $n$ distinct $k$-algebra morphisms from $Η$ to $k$. On the other hand, $G(H^{*}) = \mathcal{A}lg(H, k) =\mathcal{A}lg(k^{(n)}, k)$ where $\mathcal{A}lg(H, k)$ are the $k$-algebra morphisms from $H$ to $k$. This implies that the dual $k$-Hopf algebra $H^{*}$ has exactly $n$ grouplikes. These are linearly independent, and since $dim_{k}H^{*} = dim_{k}H = n$ we end up in $Η^{*} \cong kG$ with $G = G(H^{*})$ thus: $$ Η \cong (kG(H^{*}))^{*} $$

  • If $Η$ is cocommutative then $Η^{*}$ is commutative (and semisimple). Using duality in finite dimension and applying the previous result, we get: $Η^{*} \cong (kG)^{*}$ with $G = G(H)$, thus: $$ Η \cong kG(H) $$ edit: an immediate consequence of the above is that if the f.d. hopf algebra $H$ is commutative and cocommutative then $G(H)$ is abelian hence $H$ is self-dual in the sense that: $H\cong H^*$.

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    $\begingroup$ This is basically just Theorem 2.3.1 in Mongtomery's book, which is attributed to Harrison and Cartier (independently), but using Larson-Radford to conclude that one of the assumptions can be relaxed. The result holds over an arbitrary field, up to a separable field extension. $\endgroup$ – zibadawa timmy Jan 10 '17 at 8:02
  • $\begingroup$ @zibadawa timmy: thank you for quoting this. I didn't knew that. $\endgroup$ – Konstantinos Kanakoglou Jan 11 '17 at 1:22
  • $\begingroup$ Actually, the way I have posed the question, I think your comment would deserve to be an answer. $\endgroup$ – Konstantinos Kanakoglou Jan 11 '17 at 3:06

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