Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^∗$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^∗$ has a natural Hopf algebra structure.

Question: Is the Hopf dual of the Hopf dual of $A$ isomorphic to $A$. It is not obvious for me that it is. If not, then do we know in which cases it is true?

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    $\begingroup$ @SamHopkins: That would make the question trivial. $\endgroup$ Aug 17, 2018 at 15:51
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    $\begingroup$ Chapter 9 in Susan Montgomery's book is entirely dedicated to exploring properties of the Hopf dual, including several isomorphisms and density results. I'm not sure right now if it contains any results that are immediately applicable to the question at hand, but if you haven't already done so it may be worth looking through. $\endgroup$ Aug 18, 2018 at 4:08
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    $\begingroup$ @darijgrinberg For $A=k[x]$ you can use Example 9.1.7 in Susan's book. The Hopf dual of $A$ is the linearly recursive functions on $k[x]$, and she provides another characterization and isomorphisms. $\endgroup$ Aug 18, 2018 at 4:25
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    $\begingroup$ @darijgrinberg for $k=\mathbb{C}$ and letting $G$ be the group $(\mathbb{C},+)$, I think the Hopf dual of $k[x]$ is $k[x^*]\otimes k[G]$ where $x^*$ is defined by $x^*(x^n)=\delta_{1,n}$. $\endgroup$
    – Adrien
    Aug 18, 2018 at 9:00
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    $\begingroup$ @Adrien: Thank you! This is a beautiful counterexample. I've worked it into my answer now. $\endgroup$ Aug 18, 2018 at 20:44

3 Answers 3


I am going to give three counterexamples to your first question. (The third counterexample is courtesy of @Adrien, who did most of the job.) While none of them leads to a full answer of your second question, at least they strongly restrict the possibilities.

1. The first counterexample: binate groups

I will denote the Hopf dual of a Hopf algebra (or coalgebra) $A$ by $A^o$.

The Hopf dual of the Hopf dual of a Hopf algebra $A$ is not, in general, isomorphic to $A$. Better yet:

Proposition 1. Let $k$ be any field. There exists a Hopf algebra $H$ such that $H$ is infinite-dimensional but $\left(H^o\right)^o$ is $1$-dimensional.

Proof. I hope the following is true -- I am using results from the literature I have never checked myself.

The paper A. J. Berrick, The acyclic group dichotomy, arXiv:1006.4009v1, Journal of Algebra, Volume 326, Issue 1, 15 January 2011, pp. 47--58 includes a survey of known results about binate groups. In particular, it says (Theorem 2.7 (a)) that every group embeds in a universal binate group. Thus, there exists at least one infinite binate group. Fix such a group, and denote it by $G$. Also, Theorem 2.2 (b) in this paper says that binate groups have no nontrivial finite-dimensional representations over any field. Thus, $G$ has no such representations.

Now, let $H$ be the group ring $k\left[G\right]$ regarded as a Hopf algebra. Let $f \in H^o$. By the definition of $H^o$, this means that $f$ is a $k$-linear map $H \to k$ that vanishes on some two-sided ideal $I$ of $H$ of finite codimension. Consider this $I$. The finite-dimensional quotient space $H/I$ is an $H$-module, thus a representation of $G$, and therefore must be the trivial representation of $G$ (since $G$ has no nontrivial finite-dimensional representations). Thus, each $g \in G$ acts as the identity on $H/I$. In other words, each $g \in G$ and $h \in H$ satisfy $gh \equiv h \mod I$. Applying this to $h = 1$, we conclude that each $g \in G$ satisfies $g \equiv 1 \mod I$. In other words, each $g \in G$ satisfies $g - 1 \in I$. Therefore, the counit $\varepsilon$ of $H$ satisfies $\operatorname{Ker}\varepsilon \subseteq I$ (since the vector space $\operatorname{Ker}\varepsilon$ is spanned by the $g-1$ for $g \in G$). Hence, the map $f$ vanishes on $\operatorname{Ker}\varepsilon$ (since it vanishes on $I$), and therefore factors through the projection map $H \to H/\operatorname{Ker}\varepsilon$. But factoring through this projection map is tantamount to factoring through $\varepsilon : H \to k$ (since $H/\operatorname{Ker}\varepsilon \cong k$). Thus, $f$ factors through $\varepsilon : H \to k$. Therefore, $f$ is a multiple of $\varepsilon$.

Now, forget that we fixed $f$. Thus, we have shown that every $f \in H^o$ is a multiple of $\varepsilon$. Hence, the Hopf dual $H^o$ of $H$ is spanned by $\varepsilon$ (indeed, it is easy to see that $\varepsilon$ indeed lies in $H^o$). Therefore, $H^o$ is $1$-dimensional, and isomorphic to the trivial Hopf algebra $k$. Thus, its dual $\left(H^o\right)^o$ is isomorphic to the Hopf algebra $k^o \cong k$, hence also $1$-dimensional. But the group $G$ is infinite, and thus its group ring $H$ is infinite-dimensional. This proves Proposition 1. $\blacksquare$

The $H$ constructed in this proof is a fairly wild object by the criteria of Hopf algebra theory or even combinatorics. In particular, $G$ is not finitely generated (again, see the above-cited paper), whence the algebra $H$ is not finitely generated either. A first step in improving the above proposition would be to see if requiring $H$ to be finitely generated helps. Finitely generated groups can still be fairly perverse -- e.g., the Higman group has no faithful finite-dimensional representation, so one would expect $H^o$ to "forget" some part of $H$, but this is no longer completely automatic.

2. The second counterexample: the Higman group

There is another way of proving Proposition 1 in the case when $k = \mathbb{C}$. I will actually show the following stronger fact in this case:

Proposition 2. Let $k$ be a subfield of $\mathbb{C}$. Then, there exists a Hopf algebra $H$ such that $H$ is infinite-dimensional but $\left(H^o\right)^o$ is $1$-dimensional, and furthermore, $H$ is finitely generated as an algebra.

Proof of Proposition 2. Let $G$ be the Higman group; this is the group with four generators $a,b,c,d$ and four relations \begin{equation} ab = ba^2, \quad bc = cb^2, \quad cd = dc^2, \quad da = ad^2 . \end{equation} This is the same group as what the Wikipedia article calls "Higman group", except that our generators $a,b,c,d$ correspond to $a,d,c,b$ in that article. Theorem 1 in Terry Tao's post Finite subsets of groups with no finite models shows that this group $G$ is infinite. But Remark 2 in the same post shows that this group $G$ has no non-trivial finite-dimensional representations. (Note that the proof Tao gives relies on asymptotics of powers of the matrices representing $a, b, c, d$; this is why I required $k$ to be a subfield of $\mathbb{C}$. But I wouldn't be surprised if the argument can be tweaked to work over any field of characteristic $0$.) From this point on, we can use the very same argument that we made in the proof of Proposition 1 to show that $H$ is infinite-dimensional but $\left(H^o\right)^o$ is finite-dimensional. Finally, the group $G$ is generated by four generators $a, b, c, d$; thus, its group algebra $H = k\left[G\right]$ is generated by eight generators $a, b, c, d, a^{-1}, b^{-1}, c^{-1}, d^{-1}$. Hence, $H$ is finitely generated. This proves Proposition 2. $\blacksquare$

3. The third counterexample: $\mathbb{C}\left[x\right]$

The final counterexample is mostly due to @Adrien. It is, in a sense, the most striking since it shows that $\left(H^o\right)^o \not\cong H$ can happen even if $H$ is a univariate polynomial ring -- more or less the simplest case that isn't finite-dimensional!

Proposition 3. Let $k = \mathbb{C}$. Let $H$ be the polynomial ring $k\left[x\right]$ with its usual Hopf algebra structure (in which $x$ is primitive). Then, $H$ has countable dimension (as $k$-vector space) whereas $\left(H^o\right)^o$ has uncountable dimension.

Proof of Proposition 3. Let $G$ be the additive group $\left(\mathbb{C}, +\right)$ written multiplicatively. For each $\lambda \in \mathbb{C}$, let $\left[\lambda\right]$ be the corresponding element of $G$; thus, $\left[0\right]$ is the identity element of $G$, and $\left[\alpha+\beta\right] = \left[\alpha\right]\left[\beta\right]$ holds for any $\alpha, \beta \in \mathbb{C}$.

For each nonnegative integer $n$, let $f^{\left(n\right)}$ denote the $k$-linear map $H \to k$ sending each monomial $x^m$ to $\delta_{n, m}$ (Kronecker delta). It is well-known that $f^{\left(n\right)} \in H^o$ (since $f^{\left(n\right)}$ annihilates the finite-codimensional ideal $\left(x^{n+1}\right)$ of $H$). Actually, $\left(f^{\left(n\right)}\right)_{n \geq 0}$ is a basis of the graded dual of the graded Hopf algebra $H^{\operatorname{gr} *}$ of $H$. Thus, this family $\left(f^{\left(n\right)}\right)_{n \geq 0}$ spans a Hopf subalgebra of $H^o$, which we denote by $H_P$. Note that this Hopf subalgebra is itself isomorphic to $k\left[x\right]$ (using the isomorphism that sends each $f^{\left(n\right)}$ to $x^n / n!$).

The Hopf dual $H^o$, however, is larger than this. Namely, Example 9.1.7 in Susan Montgomery, Hopf Algebras and Their Actions on Rings shows that $H^o \cong H_P \otimes k\left[G\right]$. More precisely, for each $\left[\lambda\right] \in G$, we can define a map $\phi_{\lambda} : H \to k$ which sends each polynomial $p \in H$ to $p\left(\lambda\right)$. This $\phi_{\lambda}$ is a $k$-algebra homomorphism (it is just the evaluation homomorphism at $\lambda$), and thus is a grouplike element of $H^o$. Moreover, each grouplike element of $H^o$ has the form $\phi_\lambda$ for some $\lambda \in G$ (because a grouplike element of $H^o$ is the same as a $k$-algebra homomorphism $H \to k$, but all $k$-algebra homomorphisms from the polynomial ring $H = k\left[x\right]$ are evaluation homomorphisms). The $k$-linear map \begin{equation} \phi : k\left[G\right] \to H^o, \qquad \left[\lambda\right] \mapsto \phi_{\lambda} \end{equation} is a $k$-algebra homomorphism (this boils down to the identity $\phi_\alpha * \phi_\beta = \phi_{\alpha+\beta}$, which in turn boils down to the binomial formula). Moreover, the elements $\phi_\lambda$ for $\lambda \in G$ are distinct grouplike elements of $H^o$, and thus are linearly independent (due to the known fact that any set of distinct grouplike elements of a coalgebra is linearly independent). Hence, the map $\phi$ is injective. Now, the $k$-algebra homomorphism \begin{equation} A : H_P \otimes k\left[G\right] \to H^o, \qquad f \otimes g \mapsto f \phi\left(g\right) \end{equation} turns out to be an isomorphism of Hopf algebras. This is the full version of the $H^o \cong H_P \otimes k\left[G\right]$ statement I mentioned above. (Montgomery writes it as $H^o \cong H_P \otimes \mathbf{k} G\left(H^o\right)$; here, $G\left(H^o\right)$ denotes the span of the grouplike elements of $H^o$, which is isomorphic to our $k\left[G\right]$ because the grouplike elements of $H^o$ are all of the form $\phi_\lambda$.)

Let $\varepsilon$ denote the counit of the Hopf algebra $H_P$; it sends each $f^{\left(n\right)}$ to $\delta_{n,0}$. Recall again that $k = \mathbb{C}$, so we can use transcendental tools like the exponential map. Now, for each $\lambda \in \mathbb{C}$, we define a $k$-linear map \begin{equation} \psi_\lambda : H_P \otimes k\left[G\right] \to k, \qquad \alpha \otimes \left[\beta\right] \mapsto \varepsilon\left(\alpha\right) \exp\left(\lambda \beta\right) . \end{equation} It is easy to see that this $\psi_\lambda$ is well-defined and a $k$-algebra homomorphism. In view of the Hopf algebra isomorphism $H^o \cong H_P \otimes k\left[G\right]$, we can thus consider $\psi_\lambda$ as a $k$-algebra homomorphism $H^o \to k$. In other words, we consider $\psi_\lambda$ as a grouplike element of the Hopf algebra $\left(H^o\right)^o$. These grouplike elements $\psi_\lambda$ for varying $\lambda \in \mathbb{C}$ are all distinct (indeed, $\psi_\lambda$ sends $\phi_\mu \in H^o$ to $\exp\left(\lambda \mu\right)$, and if you know the values of $\exp\left(\lambda \mu\right)$ for all $\mu \in \mathbb{C}$, then you can recover $\lambda$), and thus are linearly independent (due to the known fact that any set of distinct grouplike elements of a coalgebra must be linearly independent). Thus, we have found uncountably many linearly independent elements of $\left(H^o\right)^o$ (since there are uncountably many $\lambda \in \mathbb{C}$). Therefore, the vector space $\left(H^o\right)^o$ has uncountable dimension. But the vector space $H = k\left[x\right]$ has countable dimension. This proves Proposition 3. $\blacksquare$


I don't know a concrete answer to the second question, but I think this will rarely happen that those are the same (and in fact on the top of my head I can't think of an example where they are apart from the finite dimensional case).

So not a real answer but this is too long for a comment: it's useful to remember that a great deal of the asymmetry between algebras and coalgebras boils down to the fact that a coalgebra is characterized by its finite dimensional comodules, while an algebra is not characterized by its finite dimensional modules. This is relevant to this discussion because $A^\circ$ can also be characterized as the (unique up to iso) Hopf algebra whose category of f.d. comodules is equivalent as a monoidal category with fiber functor to f.d. modules over $A$. This somewhat explains why non-isomorphic Hopf algebra can have isomorphic Hopf duals.


Regarding your first question: the answer is generally no, the restricted dual of the restricted dual of $A$ is generally not isomorphic to $A$: $$ (A^{\circ})^\circ\ncong A $$ as has already been indicated by the counterexamples of darij grinberg's answer.

More counterexamples can be constructed when the restricted dual (or: Sweedler's dual) $A^\circ$ is trivial while $A$ is not:
Consider the case when the algebra $A$ has no finite dimensional representations, except the zero vector space $V=\{0\}$. Now, recall that

$f\in A^\circ$ $\Longleftrightarrow$ $\dim(A\rightharpoonup f)<\infty$

where the action of the $A\rightharpoonup f$ module is defined by $(a\cdot f)(b)=f(ba)$ for all $a,b\in A$ and $f\in A^\circ$ (this is a standard result, see for example Montgomery's book, Lemma 9.1.1, p. 149). Thus,

If $A$ has no finite dimensional representations apart from the zero vector space $V=\{0\}$, then $A^\circ=\{0\}$.

Regarding your second question, unfortunately i have no examples or conditions available (and i think it is a difficult task to find something similar in the literature).

  • $\begingroup$ But is the Weyl algebra Hopf? $\endgroup$ Aug 20, 2018 at 18:36
  • $\begingroup$ not as far as i know! you are right. i just mentioned this example as an easy case for which the restricted dual is trivial. $\endgroup$ Aug 20, 2018 at 18:47
  • $\begingroup$ Infinite field extensions aren't Hopf algebras (the counit must be an algebra homomorphism!). $\endgroup$ Aug 20, 2018 at 19:34
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    $\begingroup$ the Weyl algebra (in characteristic zero) can not be a Hopf algebra because it can not have a counit, because it has no finite dimensional representations. $\endgroup$ May 9, 2019 at 23:27
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    $\begingroup$ I'm saying that a counit is a finite dimensional representation. If an algebra has no finite dim reps at all, it can't be Hopf. $\endgroup$ Sep 9, 2021 at 12:11

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