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A Hopf algebra is called pointed if all its simple left (or right) comodules are one-dimensional. See for example this question for a discussion.

Now every Hopf algebra $H$ admits a one-dimensional comodule, namely the trivial comodule for a field $\mathbb{K}$ where $$ \delta: \mathbb{K} \to \mathbb{K} \otimes H, ~~~~ k \mapsto k \otimes 1_H. $$ Is there a name for a Hopf algebra that admits no one-dimensional comodule other than the trivial comodule?

Jake Wetlock
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3 Answers3

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Q: Is there a name for a Hopf algebra that admits no one-dimensional comodule other than the trivial comodule?

A: Not in the literature, but if you would like to coin a specific name for such a Hopf algebra, at the opposite extreme of pointed, you could call it blunt.

A class of blunt Hopf algebras is formed by the Hopf algebras with free fusion semirings, see remark 1.3 in arXiv:1212.4763. Section 1.4 contains several examples of compact quantum groups associated with these Hopf algebras.

Carlo Beenakker
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I am not really sure if this what the OP is looking for but i guess that a closely relevant notion here is that of connected Hopf algebras (i.e HAs which are connected as coalgebras). These are Hopf algebras which have no simple subcoalgebras other than $k\cdot 1_H$.

Let me attempt to explain that: The definition of pointed HAs can be stated equivalently as: HAs for which all simple subcoalgebras are 1-dim. This is due to the following well-known fact:

There is a 1-1 correspondence between the set of isomorphism classes of simple right $H$-comodules and the set of simple subcoalgebras of $H$.

(See: Indecomposable coalgebras, simple comodules, and pointed Hopf algebras, S. Montgomery, Proc. Amer. Math. Soc. 123 (1995), 2343-2351)

So, if you have a unique one-dimensional comodule (the trivial one) then under the above correspondence the unique subcoalgebra is $k\cdot 1_H$. But this is the definition of the connected HAs (Connected coalgebras essentially are: pointed, irreducible coalgebras. At the level of HAs, connected HAs are the same thing as irreducible HAs. Universal enveloping algebras and most of their deformations are examples of connected HAs).

Edit: At a second read, and taking into account the way the OP is stated, maybe the notion of connected is more restrictive than desired: in the sense that for connected HAs, the trivial one-dim comodule is the unique simple comodule; while the OP asks for it to be the unique 1-dim comodule (leaving thus open the possibility that higher dim simple comodules may exist). But if this is the intention of the OP then i do not think -modulo my knowledge of course- that such a characterization exists.

Konstantinos Kanakoglou
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  • You are right that connectedness is too restricted. Let $C$ denote the complex numbers, and consider the Hopf algebra $C[A_5]^$, the $C$-linear dual of the group algebra of the alternating group $A_5$. One-dimensional $C[A_5]^$-comodules are given by one-dimensional complex representations of $A_5$, and from the character table (or from simplicity of $A_5$) you know that there is only one, up to isomorphism. So $C[A_5]^$ is of the kind that the original question asks for. But $C[A_5]^$ is not a connected Hopf algebra, since any grading on it must put the entire Hopf algebra in degree zero. –  May 29 '21 at 04:34
  • @A.S. thanks for mentioning an example. yes this is what i meant: there are indeed other HAs (non-connected) which have the desired property. But in any case the connected ones still form a concrete class of HAs which has the required property (though in a more special form). – Konstantinos Kanakoglou May 29 '21 at 12:38
  • @A.S., motivated by your comment i was thinking .. do you know if there is a name for such algebras? I mean algebras satisfying the dual property to that described in the OP; that is algebras admitting no non-trivial 1-dim modules (but possibly admitting higher dim simple modules)? – Konstantinos Kanakoglou Jun 02 '21 at 20:07
  • I do not know a name for them, but I spent a while thinking through Dmitry Vaintrob's answer to this question, and while I did not attempt to write out a careful proof, I came away thinking that Dmitry's answer appears correct, at least for finite-dim'l Hopf algebras (and for finite-type graded Hopf algebras). Dualizing Dmitry's answer, I would expect that "algebra with nilpotent abelianization" is the name you're asking for. –  Jun 02 '21 at 20:37
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Unless I'm mistaken you can say "coalgebra with conilpotent coabelianization" (the Hopf algebra structure doesn't seem relevant here). Namely, first note that any one-dimensional representation of a coalgebra facors through the coabelianization $H^{ab}$ (the maximal commutative sub-coalgebra) and a commutative coalgebra with a unique one-dimensional corepresentation is called conilpotent. This is equivalent to asking that the dual commutative ring $R = (H^{ab})^\vee$ viewed as a topological ring has a closed topologically nilpotent ideal $I$ with quotient $\mathbb{K}$.

Dmitry Vaintrob
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