Kaplansky's second conjecture (on Hopf algebras) deals with "admissible" coalgebras: He calls a coalgebra admissible, if there is an algebra structure making it a Hopf algebra. The conjecture states that:

a coalgebra $C$ is admissible if and only if any finite subset of $C$ lies in a finite dimensional admissible subcoalgebra.

As far as I know there are counterexamples refuting the conjecture. (the first one was stated by Larson if I remember correctly).

Inspired by the above, let us lay the following definition of the notion of admissible algebra:

Definition: An algebra $A$ will be called admissible if there is a coalgebra structure on $A$ and a suitable linear map $S:A\to A$, such that all these data (the algebra, the coalgebra and the linear map $S$) constitute a Hopf algebra with antipode $S$.

Now, my question is whether there is some criterion (necessary, sufficient or both) regarding to when an algebra is admissible (or non-admissible) in the sense of the above definition. In other words:

  • Given an algebra $A$, under which conditions can a coalgebra structure be found on $A$ such that it will be turned into a Hopf algebra?
  • When would such a coalgebra structure be unique?
  • What could be examples of a non-admissible algebras, in the sense of the above definition?
  • $\begingroup$ I don't guess what you mean by an admissible algebra. $\endgroup$ – YCor Feb 18 '17 at 1:39
  • $\begingroup$ @YCor, the definition is layed in the question: I propose calling an algebra admissible, if there is (at least) one coalgebra structure -on the same set- and a suitable antipode , such that they altogether form a hopf algebra. $\endgroup$ – Konstantinos Kanakoglou Feb 20 '17 at 0:52
  • $\begingroup$ I've edited the post, hoping to become more clear. $\endgroup$ – Konstantinos Kanakoglou Feb 20 '17 at 1:56
  • $\begingroup$ There is a related question (and answer) at: mathoverflow.net/q/238270/85967 $\endgroup$ – Konstantinos Kanakoglou Jul 1 '17 at 18:42

In regards to your first question: I don't believe there is a way of telling in general whether or not an arbitrary algebra $A$ admits a Hopf structure. There are, however, in certain settings, some conditions an algebra $A$ must satisfy in order to be a Hopf algebra. \

For example, if $A$ is a Hopf algebra over an algebraically closed field of characteristic zero, its abelianisation $A_{ab}=A/\langle [A, A] \rangle$ is a commutative Hopf algebra (with structure inherited from $A$ - this part is true for any Hopf algebra over any field), and, by the Nullstellensatz, is the coordinate ring of a smooth algebraic group. In particular, $A_{ab}$ is a commutative domain. Thus, it must be that any algebra $B$ with an abelianisation $B_{ab}$ which is NOT a domain cannot be an admissible algebra in the above sense. An example of such an algebra would be $k[x,y]/\langle [x,y] = y^2 \rangle$. This example also answers your third question.

In regards to your second question about uniqueness: in general an algebra can admit several distinct Hopf structures. Take, for example, $H = k[x,y,z]$. Its Hopf structures are in 1-1 correspondence with 3-d unipotent groups, of which there are two: the abelian group and the 3-d Heisenberg group. These correspond to a Hopf structure generated by primitive elements in the former case, and a Hopf structure generated by two primitives ($x$ and $y$ say) and a non-primitive, $z$ say, such that $\Delta(z) = 1\otimes z + x\otimes y + z\otimes 1$.

You can read a bit more about the examples I've given here:


  • $\begingroup$ thank you for the feedback and the ideas contributed. I didn't yet find time to further study your paper but it seems interesting and relevant. thank you very much for citing it! $\endgroup$ – Konstantinos Kanakoglou Mar 21 '17 at 1:57
  • $\begingroup$ In your second paragraph you mention that "algebras whose abelianazation is not a domain are non-admissible". Do such algebras -like $k[x,y]/\langle [x,y] = y^2 \rangle$ for example- fall into the class of semisimple, non-separable algebras mentioned in my answer? $\endgroup$ – Konstantinos Kanakoglou Mar 21 '17 at 2:05
  • $\begingroup$ About your proposition in Konstantinos Kanakoglou's answer (Prop. 2.7 in arxiv.org/pdf/1601.06687v1.pdf): To my knowledge, the cohomology ring of a connected topological group is a (connected graded) Hopf algebra. The cohomology ring of the 3-torus $T=S^1\times S^1 \times S^1$ is the exterior algebra $H^\ast(T,\mathbb{C})=\bigwedge(x,y,z)$ with degree one generators. In particular, $xy=-yx$, contradicting the proposition. Am I missing something ? $\endgroup$ – tj_ Aug 28 '18 at 21:58
  • 1
    $\begingroup$ Matter cleared: In the realm of algebraic topology the product in $A \otimes A$ is defined to be graded, i.e. $(a_1 \otimes b_1)(a_2\otimes b_2)=(-1)^{|b_1||a_2|}a_1a_2\otimes b_1b_2$. For example, if $A=H^\ast(S^1,k)=k[x]/(x^2)$ then multiplication on $S^1$ induces the coproduct $\Delta: A \to A \otimes A, x \mapsto x\otimes 1 + 1 \otimes x$. The graded product then makes $\Delta(x)\cdot \Delta(x)=0$. However, if (as in your proposition) the product in $A \otimes A$ is taken to be ungraded, then $\Delta(x)\cdot \Delta(x)=2\cdot x \otimes x$ which is non-zero if $k$ is not of char. 2. ... $\endgroup$ – tj_ Aug 29 '18 at 16:31
  • $\begingroup$ Hence $(A, \Delta) $ is no Hopf algebra in this case if the characteristic of $k$ is not 2. $\endgroup$ – tj_ Aug 29 '18 at 16:34

A couple of examples of non-admissible algebras:

$\bullet$ It is known (see here, p.222) that a semisimple Hopf algebra is also separable (as an algebra). Consequently, a semisimple but non-separable algebra has to be non-admissible.
(An example of such an algebra could be the following: Consider a field $K$ and $L$ an inseparable extension. Then $L$ is a semisimple $K$-algebra but the algebra $L\otimes L$ is no longer semisimple).

$\bullet$ On the other hand, another example (outside the semisimple algebras) might be the case of Weyl algebras and their deformations. As far as I know, there is no known Hopf structure for the family of Weyl algebras. Furthermore, in an old paper of T.D. Palev, it is argued that it is impossible to even define a comultiplication on Weyl algebras. See: Is it possible to extend the deformed Weyl algebra $W_q(n)$ to a Hopf algebra? (the discussion of p.3-5 refers to the undeformed case).

$\bullet$ I have recently found an interesting result leading to some other classes of examples of non-admissible algebras:

Proposition: Let $A$ be a connected graded algebra and let $x,y$ be non-zero elements in $A$ such that $xy=qyx$ for some $q\in k\backslash \{1\}$. Then there is no Hopf algebra structure on $A$.

Thus, $A$ is non-admissible in the sense of the OP's definition. This is proposition 2.7 from Connected (graded) hopf algebras. One of the authors appears to be the P.Gilmartin (who has provided the other answer above). So I am adding this reference and the following examples for the shake of completeness. As a consequence of the above proposition we get the following families of examples of non-admissible algebras (i am copying Corollary 2.8 of the formerly mentioned paper):

  • Sklyanin algebras of any dimension
  • skew-polynomial rings $k_{p_{i,j}}[x_1,x_2,...,x_n]$ (except for the case $p_{i,j}=1$ for all $i,j$)
  • Quantum matrix algebras $\mathcal{O}_q(M_{n\times n})$ for $q\neq 1$
  • non-commutative Koszul Artin-Schelter regular algebras of dimension $\leq 4$

$\bullet$ Using the above proposition (for $q=-1$), we can conclude that the fermionic algebra or the algebra of anticommutation relations, generated as an algebra by the fermion creation-annihilation operators $f_i^+, f_i^-$ for $i=1,2,...$ modulo the following relations $$ \{f_i^-,f_j^+\}=\delta_{ij}, \ \ \{f_i^-,f_j^-\}=0, \ \ \{f_i^+,f_j^+\}=0 $$ (where $\{x,y\}=xy+yx$) for all $i,j=1,2,...$ is another example of non-admissible algebra i.e. of an algebra which does not admit a Hopf algebra structure. This might be of particular interest to mathematical physicists since this algebra describes the spin-$\frac{1}{2}$ particles.
(This last example is due to me and is not contained in the paper cited above. So if there is any disagreement ... put the blame on me!)


Note that a Hopf algebra is augmented. Let $A$ be an augmented algebra over a commutative ring $k$. A necessary condition for $A$ having a Hopf algebra structure is that the cohomology ring $$H^\ast(A,k) := Ext_A^\ast(k,k)$$ is graded commutative under Yoneda composition.

In fact, the coproduct of a Hopf algebra (or, more generally, of a bialgebra) induces a graded comutative cup product on cohomology and one can show that Yoneda composition and cup product agree on $Ext_A^\ast(k,k)$.

When I have time, I'll look for an example there this criterion applies.


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