The kernel of a Hopf algebra map $\phi:H_1 \to H_2$ is in general not a Hopf sub-algebra of $H_1$. Is there some replacement or alteration of the notion of a kernel in the Hopf algebra setting. Same question for cokernels. So can we construct an abelian category from the category of Hopf algebras (over a fixed field $k$).


2 Answers 2


$\DeclareMathOperator\Hker{Hker}\DeclareMathOperator\Hcoker{Hcoker}\DeclareMathOperator\Im{Im}\DeclareMathOperator\coIm{coIm}\DeclareMathOperator\Id{Id}$The category $\mathcal{H}$ of finite dimensional, commutative, cocommutative hopf algebras is an abelian category.
The set $\mathcal{H}(F,G)$ of all hopf algebra maps $F\to G$ is an abelian group with sum given by the convolution product. (The convolution inverse for $f\in F$ is: $f\!\circ\! S_F=S_G\!\circ\! f$ and the neutral element is $\eta_G\!\circ\!\varepsilon_F$).
We have also product of maps, given by composition (and composition distributes over convolution).

This is a classic result. I have been told it was first shown by Grothendieck. I know you can find details in Sweedler's book, ch. XVI, p. 314 (basically the whole chapter, p.303–315 presents a detailed proof of the above).

Remark 1: You are right to mention that in the general case the definitions of the kernels and the cokernels seem to pose a problem when one tries to figure out which hopf algebras would be good to make an abelian category from.
For $F$, $G$, f.d., commut, cocommut, hopf algebras and $\omega:F\to G$ a hopf algebra map, then in $\mathcal{H}$, kernels are defined through: $$\Hker\omega=\{f\in F|(I\otimes\omega)\Delta (f)=f\otimes 1\}$$ which -due to cocommutativity- is the same with $\{f\in F|(\omega\otimes I)\Delta (f)=1\otimes f\}$,
and cokernels through: $$\Hcoker\omega=G\big/\big(\omega (F^+)G\big)$$ where $F^+$ is the augmentation ideal and $\omega (F^+)G$ denotes the right ideal (which by commutativity is a two-sided ideal), of $G$ generated by $\omega(F^+)$.
Under the above defs, it can be shown that $\Hker\omega$ is a sub-hopf algebra of $F$, $\Hcoker\omega$ is a quotient hopf algebra of $G$ (i.e. $\omega (F^+)G$ is a hopf ideal) and any hopf algebra map $\omega\in\mathcal{H}(F,G)$ "factorizes" as: $$ \Hker\omega \overset{i_{\omega}}{\hookrightarrow} A \overset{\pi_{i_{\omega}}}{\twoheadrightarrow} \coIm\omega \cong \Im\omega \overset{i_{\pi_{\omega}}}{\hookrightarrow} B \overset{\pi_{\omega}}{\twoheadrightarrow} \Hcoker\omega $$ where:

  • $\Hker\omega \overset{i_{\omega}}{\hookrightarrow} A$ is the kernel of $A\overset{\omega}{\to}B$,
  • $B\overset{\pi_{\omega}}{\twoheadrightarrow} \Hcoker\omega$ is the cokernel of $A\overset{\omega}{\to}B$,
  • $A\overset{\pi_{i_{\omega}}}{\twoheadrightarrow} \coIm\omega$ is the cokernel of $\Hker\omega \overset{i_{\omega}}{\hookrightarrow} A$ and
  • $\Im\omega\overset{i_{\pi_{\omega}}}{\hookrightarrow} B$ is the kernel of $B \overset{\pi_{\omega}}{\twoheadrightarrow} \Hcoker\omega$.

Remark 2: There are even more detailed things to be said about the category $\mathcal{H}$ of finite dimensional, commutative, cocommutative hopf algebras over a field: We can construct a functor $\mathcal{G} : \mathcal{H} \Rrightarrow \mathcal{Ab}_{\text{fin}}$ from $\mathcal{H}$ to the category $\mathcal{Ab}_{\text{fin}}$ of finite, abelian groups (assigning to each hopf algebra of $\mathcal{H}$ the group of its grouplikes) and another functor $\mathcal{F} : \mathcal{Ab}_{\text{fin}} \Rrightarrow \mathcal{H}$ (assigning to each fin, abelian group its group hopf algebra). It is not difficult to show that these functors satisfy $$ \begin{array}{cccc} \mathcal{G} \mathcal{F} = \Id_{\mathcal{Ab}_{\text{fin}}} & & & \mathcal{F} \mathcal{G} \cong \Id_{\mathcal{H}}\\ \end{array} $$ constituting thus an equivalence of the categories $\mathcal{H}$, $\mathcal{Ab}_{\text{fin}}$.

  • $\begingroup$ Is remark 2 definitely correct? If these categories are equivalent then they have the same isomorphism classes of objects, but there is only one finite abelian group of order $p$ and typically more than one isomorphism class of fin.dim. com. cocom. hopf algebras of order $p$ (for example take the group algebra and its dual over characteristic $p$)? Forgive me if I have missed something basic. $\endgroup$
    – James
    Commented Oct 10, 2020 at 22:11
  • $\begingroup$ @James, although i am not sure i have fully understood what you mean with "hopf algebras of order $p$... " (maybe you mean hopf algebra of dim $p$?), i think you have a point in the following sense: remark 2 is meanignful for hopf algebras over an algebraically closed field of char $0$. Not over arbitrary fields. One can further relax these conditions (to include finite fields), but i think they guarantee that things will work ok (under such circumstances, the group algebra and its dual are isomorphic). $\endgroup$ Commented Oct 15, 2020 at 22:37

This is a delicate matter, and there are in principle many possible answers depending on exactly what you wish to apply things to.

Frequently, one of two variations of the Hopf subalgebra mentioned by Konstantinos is used: Given $\pi\colon H\to K$ a morphism of Hopf algebras, we have the left coinvariants and right coinvariants (of $\pi$) defined respectively by: $$ {}^{\text{co}\,\pi}H = \{ h\in H \ | \ (\pi\otimes\operatorname{id})\Delta(h) = 1\otimes h\},\\ H^{\text{co}\,\pi} = \{ h \in H \ | \ (\operatorname{id}\otimes\pi)(\Delta(h))= h\otimes 1\}.$$ These need not be the same subobjects of $H$, however.

One could then define a short exact sequence of Hopf algebras (over $k$) $$ k\longrightarrow K \overset{i}{\longrightarrow} H \overset{\pi}{\longrightarrow} L\longrightarrow k$$ to be a sequence of morphisms of Hopf algebras such that

  1. $i$ is injective and $\pi$ is surjective;
  2. $\ker(\pi) = H\, i(K)^+$;
  3. $i(K) = {}^{\text{co}\,\pi}H$.

Note that the more classical notion of kernel is still involved: it will generally not be enough to conduct any of the usual arguments you'd like to do with a "short exact sequence" to assume only the first and third (or first and second) conditions. The first and third are enough when $H$ is faithfully coflat, and the first and second are enough when $H$ is faithfully flat.

This is a good definition for short exact sequences and results that normally rely on them, such as looking for Jordan-Holder analogues. As such, coinvariants-as-kernels will be prominent when working in a category of Hopf algebras. But as I said it's not the only answer, and in other categorical contexts (generalizing/lifting to functors between representation categories, namely) it is the wrong answer. The categorical kernels and cokernels of $\pi\colon H\to K$ are $$ \text{Hker}(\pi) = \{ h\in H \ | \ h_{(1)}\otimes \pi(h_{(2)})\otimes h_{(3)} = h_{(1)}\otimes 1 \otimes h_{(2)}\},\\ \text{Hcoker}(\pi) = K/(K \pi(H^+) K).$$ And this kernel (and cokernel) is not necessarily the same thing as the left or right coinvariants (and associated cokernel). By applying counits to the left/right of the defining relation, we see that $$\text{Hker}(\pi)\subseteq {}^{\text{co}\,\pi}H\cap H^{\text{co}\,\pi},$$ at least. Though equality (of all three subobjects) can happen, and assuring it does is usually the key to making sure attempts at defining exact sequences of tensor categories works out well.

Sonia Natale recently posted a review of these notions, and the related issues with tensor categories, that should be helpful.

  • $\begingroup$ Two questions: First, in the commutative, cocommutative case, the left and right coinvariants are the same: ${}^{\text{co}\,\pi}H = H^{\text{co}\,\pi}$. Are they also equal to your second definition of the kernel: $\text{Hker}(\pi)$? Are there other interesting situations in which the second definition is identified with the one posted in my answer? $\endgroup$ Commented Apr 16, 2020 at 17:29
  • $\begingroup$ And second, you claim that "categorically speaking (the first def) is the wrong answer". I guess you refer to the non-(co)commutative case. what is the advantage of the second definition for $\text{Hker}(\pi)$ in the general case? Does it enable one, to enlarge somehow the class of hopf algebras which make an abelian category? $\endgroup$ Commented Apr 16, 2020 at 17:35
  • $\begingroup$ @KonstantinosKanakoglou Such Hopf algebras would be faithfully flat and coflat, so everything would coincide nicely. Proposition 4.2 and subsequent remark in the linked paper should answer the remaining matters. $\endgroup$ Commented Apr 16, 2020 at 17:37
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    $\begingroup$ If I am correct $S({}^{co\pi}H)=H^{co\pi}$ and I wonder whether $\mathrm{Hker}(\pi)$ can be characterized thorugh some $S$-closedness condition. After all if you really want to be Hopf the antipode should play a role here... $\endgroup$ Commented Jun 19, 2020 at 8:22
  • $\begingroup$ @NicolaCiccoli Well, when one moves from Hopf algebras to representations thereof to more general categorical settings like fusion categories, the antipode sort of disappears as a core feature. What it normally gives us is the existence of duals in the representation category, which as a categorical property is called "rigid", but it doesn't do much else for the category. And in the categorical setting the rigid property is typically assumed, rather than derived, and you don't necessarily need functors to "respect" underlying antipodes. $\endgroup$ Commented Jun 19, 2020 at 8:55

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