Is it true that any monomorphism of commutative Hopf algebras over a field is injective? Moreover, is it true that any epimorphism of commutative Hopf algebras over a field is surjective?
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4 Answers
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It seems that Ben is nevertheless right that the answer to the first question is "NO". Let $G=SL(2,\Bbb C)$, and $B$ be the subgroup of lower triangular matrices. Then the inclusion $B\to G$ is an epi, since every algebraic representation of $B$ that extends to $G$ does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation $V$ of the Lie algebra $sl(2)$, the operator $e$ is determined by $f$ and $h$. Indeed, the kernel $K$ of $e$ is spanned by vectors $v$ satisfying $hv=mv$ and $f^{m+1}v=0$ for some integer $m\ge 0$, and since $V=\Bbb C[f]K$, the operator $e$ on $V$ is uniquely determined.
So one might guess that a morphism of complex affine algebraic groups $\phi: H\to G$ is an epi if and only if $G/\phi(H)$ is connected and proper (but I did not check this).
answered Feb 5, 2010 at 5:47
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$\begingroup$ I'm not sure if this argument is 100% correct, but it seems to work under some mild technical assumptions. Either way I am convinced now that the answer is "no". As I have understood the article of Chirvasitu mentioned below, the reason for this should be the lack of faithful coflatness. $\endgroup$ Commented Feb 6, 2010 at 21:50
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$\begingroup$ Can you please explain what may be a problem with this argument and which technical assumptions do you mean? $\endgroup$ Commented Feb 6, 2010 at 23:20
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$\begingroup$ Your argument basically depends on the fact, that an irreducible lie representation of sl(2) on a finite dimensional vector space $V$ is completely determined by the endomorphisms representing e and h: Take $v$ any eigenvector of $h$ with $ev = 0$, then we can construct canonically a basis of $V$, where the action of $f$ is a priori known. However splitting a given representation into irreducible ones, or even determining whether it is irreducible seems to depend on the whole action of sl(2). $\endgroup$ Commented Feb 7, 2010 at 13:37
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$\begingroup$ I don't think I used that the representation of $sl(2)$ is irreducible. I claimed that if V is any representation of $sl(2)$ and we are given the operators $h$ and $f$ on this representation, then we can uniquely reconstruct the action of $e$. Namely, 1) if $v$ is a nonzero vector with $hv=mv$ and $f^{m+1}v=0$, then $ev=0$ (this is true for any finite dimensional $sl(2)$-module); 2) if $u=f^nv$, where $v$ is as in (1), then $ef^nv$ is computed in the usual way using $[e,f]=h$. Do you disagree with this? $\endgroup$ Commented Feb 7, 2010 at 17:26
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$\begingroup$ I'd like to add that even if only $f$ is given, you can determine the decomposition of $V$ into irreducible $sl(2)$-modules. Namely, the irreducible summands correspond to Jordan blocks of $f$. In particular, $V$ is irreducible iff $f$ is a regular nilpotent matrix, i.e. $f^{dim(V)-1}$ is nonzero. $\endgroup$ Commented Feb 7, 2010 at 17:34
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No. The category of commutative Hopf algebras over a field is opposite to the category of affine group schemes, so your question is "is every epi of affine $k$-group schemes surjective?" But there are maps of finite groups that are epis, but not surjective (this happens when the image has normal closure the whole group), for example the inclusion of a transposition into $S_3$.
I suspect every epi in commutative Hopf algebras is surjective. Certainly one can't construct an example from finite groups. I might be forgetting some funniness about group schemes.
answered Jan 31, 2010 at 23:22
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1$\begingroup$ The last claim in your frst paragraph is happening in the category of finite groups? For the whole category, one has [C. E. Linderholm. A Group Epimorphism is Surjective. The American Mathematical Monthly, Vol. 77, No. 2 (Feb., 1970), pp. 176-177.] $\endgroup$ Commented Feb 1, 2010 at 1:42
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$\begingroup$ Ah. In fact, the paper also proves that an epi in the category of finite groups is surjective too. I had completely forgotten about that... $\endgroup$ Commented Feb 1, 2010 at 2:00
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I haven't read through yet but the following paper by Chirvasitu seems to answer your second question and discusses closely related problems. See discussion in page 7 after Proposition 2.5.
http://arxiv.org/abs/0907.2881
Namely an epimorphism of Hopf algebras over a field $$f : H \longrightarrow K$$ is necessarily surjective if $K$ is commutative.
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Yes, the category of commutative Hopf algebras over k is antiequivalent to the category of (pro-)affine group schemes over k, but this equivalence does not respect the obvious functors to (Set). I really meant injective as maps between algebras, not between maximal spectrum, prime spectrum or rational points.
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$\begingroup$ The example of Ben cannot be correct, if I understand him correctly: The category of finite groups is embedded as a full subcategory of the category of affine group schemes over k via the constant group functor. Any epimorphism in the surrounding category remains of course an epimorphism in the smaller one. If the inclusion of a transposition into $S_3$ were an epimorphism of algebraic groups, it be so as abstract groups. But as was already mentioned, every epimorphism of finite groups is surjective (in fact this statement remains true without 'finite' and is an exercise in Saunders MacLane). $\endgroup$ Commented Feb 3, 2010 at 11:45
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$\begingroup$ Nicolas, yes you are right. I misunderstood Ben's answer and the comment following it. I have deleted my incorrect "contribution" to the conversation. So the answer to your question, combining Ben's first sentence, with Mariano's answer is, in fact, yes? $\endgroup$ Commented Feb 3, 2010 at 12:04
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$\begingroup$ By the way, Ben won't see your further questions unless you comment to his answer. $\endgroup$ Commented Feb 3, 2010 at 12:05
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$\begingroup$ For some reason I cannot comment on his answer, I can only add comments to my own (?). $\endgroup$ Commented Feb 3, 2010 at 14:09
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$\begingroup$ Nicolás, you need X reputation points for some X > 1 to comment on other people's answers. There are two accounts (mathoverflow.net/users/3679/nicolas-schmidt and mathoverflow.net/users/3693/nicolas-schmidt) under your name. I'll ask on meta for them to be joined. $\endgroup$ Commented Feb 5, 2010 at 5:54