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Under what conditions do counital coalgebras have bases consisting entirely of grouplike elements? At least in the case of finite-dimensional coalgebras, or for bialgebras (or Hopf algebras in particular), is there a simple characterization?


  • A counital (coassociative) coalgebra is a vector space $V$ over a field $K$, together with operators $$\def\id{\mathrm{id}} \Delta : V \to V \otimes V \qquad\qquad \varepsilon : V \to K $$ such that the following equalities hold: $$\begin{gather*} (\Delta \otimes \id_V) \Delta \;=\; (\id_V \otimes \Delta) \Delta \;;\\[1ex] (\varepsilon \otimes \id_V) \Delta \;=\; \id_V \;=\; (\id_V \otimes \varepsilon) \Delta \;, \end{gather*}$$ which are the coassociative property of $\Delta$ (dual to the usual associative/distributive property of multiplication) and the counital property of $\varepsilon$ (dual to the property of being a multiplicative identity).

  • An element $\mathbf v \in V$ is grouplike if $\Delta(\mathbf v) = \mathbf v \otimes \mathbf v$, and $\mathbf v \ne \mathbf 0$. My question is about the conditions in which there exists a basis for $V$ consisting of such elements.


There are simple examples with and without a basis of grouplike elements. For instance, for an arbitrary field $K$ and $V$ a vector space generated by two basis vectors $ \def\r{\mathbf x} \def\i{\mathbf y} \r, \i$, if we choose $$ \begin{align*} \Delta(\r) &= \r \otimes \r &\quad \varepsilon(\r) &= 1 \\ \Delta(\i) &= \i \otimes \i & \varepsilon (\i) &= 1 \end {align*}$$ then $\{\r,\i\}$ itself is such a basis. In particular, we then have $$\Delta(a\r + b\i) = a(\r\otimes\r) + b(\i\otimes\i) \,,$$ which is a product if and only if either $a=0$ or $b=0$, so that $\{\r,\i\}$ is uniquely a basis of grouplike elements. On the other hand, a coalgebra need not have any grouplike elements at all: if we instead define $$ \begin{align*} \Delta(\r) &= \r \otimes \r - \i \otimes \i &\quad \varepsilon(\r) &= 1 \\ \Delta(\i) &= \r \otimes \i + \i \otimes \r & \varepsilon (\i) &= 0 \end {align*}$$ then $$\Delta(a\r + b\i) = a(\r \otimes \r) + b(\r \otimes \i) + b (\i \otimes \r) - a (\i \otimes \i)\,,$$ which is a product vector if and only if $a^2 = -b^2$, that is if $a = b = 0$ or $a = \pm bi$, where $i^2 = -1$. In particular, for fields such as $\mathbb R$ in which $x^2+1$ is irreducible, there are no non-trivial solutions.

Is there a characterisation of which coalgebras have such a basis? Again, if there is a simple characterization at least for the finite-dimensional case, or for bialgebras / Hopf algebras. (Of course, in the case of a bialgebra, at least the unit $\eta$ is a grouplike element.)

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What you call a preferred basis is usually called a «basis of group-like elements». – Mariano Suárez-Alvarez Dec 22 '12 at 19:21
@MarianoSuárez-Alvarez: thank-you, that's a very useful thing to know. I have revised my question accordingly. – Niel de Beaudrap Dec 23 '12 at 10:35
I'm not sure I really understand what would constitute an answer to this question. – Qiaochu Yuan Dec 23 '12 at 11:04
@QiaochuYuan: Is it because there's no criterion which suggests itself? The question is somewhat underdetermined, in that it admits uninformative answers; if the only answers that occur to you are trivial and uninformative, then perhaps the more meaningful answer is that there's no nice structure which captures this property. (Compare the situation between "What conditions characterize a field extension of $\mathbb R$ for which all polynomials factorize into linear factors?" versus the analogue for $\mathbb Q$; the latter lacks a good answer but the former has one in "$x^2+1$ factorizes".) – Niel de Beaudrap Dec 24 '12 at 0:07

A bialgebra has a basis of group-like elements iff it is a monoid algebra. A hopf algebra has a basis of group-like elements iff it is a group algebra. The point is that in a bialgebra the group-like elements form a submonoid and are automatically linearly independent. Thus you have a basis of group-likes iff the group-likes themselves form a basis in which case you have the monoid algebra of the set of group-likes. In a Hopf algebra the antipode restricts to an inverse on the monoid of group-likes and so you have a group algebra.

I am not sure about what happens in the coalgebra case.

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