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1. Context

Obtaining (co)monoids from dual objects
Let $(C, \otimes, I, a, l,r)$ be a monoidal category. To simplify notation (and work with string diagrams) we assume that $C$ is strict. Let $V \in C$ be a right dualizable object, i.e. there exists an object $V^* \in C$ and morphisms $b_V: I \rightarrow V \otimes V^*$, $d_V: V^* \otimes V \rightarrow I$ that satisfy the zigzag-identities. It seems, this data alone induces the structure of a monoid object $(V \otimes V^*, \mu, \eta)$ where $\mu = (r_V \otimes id_{V^*})\circ (id_V \otimes d_V \otimes id_{V^*})$ and $\eta =b_V$. This can be verified by using the zigzag identities. Analogously, it seems we have the structure of a comonoid object $(V^* \otimes V, \Delta, \epsilon)$ where $\Delta:(id_{V^*} \otimes b_V \otimes id_V)\circ (r^{-1}_{V^*} \otimes id_V)$ and $\epsilon=d_V$.

Two motivating examples

  • The category of endofunctors $End(C)$ of any small category $C$. It becomes a monoidal category in the following way: Composition of functors is the monoidal product. The monoidal unit is given by the identity functor on $C$. As the composition of functors is associative this category is strict. A right dual to an object $F \in End(C)$ is a right adjoint functor to that functor $F$. (Co)monads are (co)monoid objects in the category of endofunctors. Hence, above construction shows how one can obtain a (co)monad from a pair of adjoint functors (i.e. by suitably composing the pair of adjoint functors, and defining the respective natural transformations as described above.)

  • Consider the monoidal category of finite dimensional vector spaces (over a field) with tensor product of vector spaces as the monoidal product. This category is rigid. (The dual vector space is precisely the right/left dual object. Evaluation and coevaluation are the morphisms $d$ and $b$ respectively.) Let $V$ be an object in that category. We then have the identification $End(V) \cong V \otimes V^*$. The above construction hence endows $End(V)$ with the structure of a unital, associative algebra.

2. Questions

  • This algebra structure is the same as the the algebra structure on $End(V)$ given by the composition of maps (multiplication) and $\eta (1_{\mathbb k})=id_V$ (unit). Correct?
  • By the above construction we can turn $V \otimes V^* \cong End(V)$ into a coalgebra. Is the induced coproduct $\Delta:End(V) \rightarrow End(V) \otimes End(V)$ simply the diagonal map $\Delta(f)=f \otimes f$? What is the counit specified on a basis of $End(V)$?
  • What are other (enlightening or interesting) examples of the above construction (obtaining (co)monoids from dual objects) in other monoidal categories from the ones mentioned?
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    $\begingroup$ The counit $\operatorname{End}(V)\to k$ should be the trace. $\endgroup$
    – jgon
    Aug 4, 2020 at 0:55
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    $\begingroup$ The diagonal map isn't linear. $\endgroup$ Aug 4, 2020 at 1:30

1 Answer 1

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  1. Yes, the algebra structure on $\text{End}(V)$ is the familiar one. I don't know a super clean way to see this off the top of my head but you can just work it out by picking a basis.

  2. No, the diagonal map isn't linear. The counit $\text{End}(V) \to k$ is the trace. The comultiplication $\Delta : \text{End}(V) \to \text{End}(V) \otimes \text{End}(V)$ is given by inserting the unit into the middle, so explicitly in a basis $e_i$ of $V$ and a dual basis $e_i^{\ast}$ of $V^{\ast}$ this means

$$\Delta \left( e_i \otimes e_j^{\ast} \right) = e_i \otimes \left( \sum_{k=1}^n e_k^{\ast} \otimes e_k \right) \otimes e_j^{\ast}.$$

  1. (cont) Honestly I don't know a super clean way of thinking about this other than as the dual of the algebra structure on $\text{End}(V^{\ast})$ (or $\text{End}(V)$ itself, I suppose). I suppose you can think of it as a "path coalgebra" structure, where if $e_i \otimes e_j^{\ast}$ denotes an edge between two vertices $i$ and $j$ in the complete multigraph on $n$ vertices (so including edges between each vertex and itself, and edges are directed) then the comultiplication sends it to a sum over all paths of length $2$ between $i$ and $j$, and repeated comultiplication is a sum over paths of longer length.

  2. I'm actually not aware of substantially different examples than these. Note that the monad / comonad construction is actually more general, when generalized to 2-categories; see this blog post for more. The string diagrams look almost exactly the same.

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  • $\begingroup$ Thank you, that was helpful. The space $End(V)$ doesn't become a bialgebra in this way, i.e. $\Delta$ is not an algebra morphism - correct? Follow up: Can $End(V)$ with above coalgebra structure be made into a bialgebra, or vice versa $End(V)$ with above algebra structure into a bialgebra? $\endgroup$ Aug 6, 2020 at 10:42
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    $\begingroup$ @M.C.: correct. Your questions are equivalent and the answer to both of them is no (except in dimension $1$) - thinking of it as an algebra, no suitable candidate for the counit exists because of simplicity. $\endgroup$ Aug 6, 2020 at 20:30
  • $\begingroup$ Sorry to reopen ... do you mean the following by your last comment? $End(V)$ is a simple algebra (since $dim_{k} (V)=n \in \mathbb N$, $End(V)$ is isomorphic to $M_{n \times n} (k)$ as an algebra. It is easy to see that $M_{n \times n} (k)$ is a simple algebra – even holds for $k$ a divison ring.) If the algebra $End(V)$ could become a bialgebra, its counit $\epsilon$ would be an algebra morphism, hence $ker \epsilon$ a two-sided ideal with dimension either $n$ or zero (due to simplicity). The zero map doesn‘t obey counitality. So by rank-nullity, unless $n=1$, there exists no suitable counit. $\endgroup$ Sep 17, 2020 at 17:27
  • $\begingroup$ @M.C.: yes, that's what I mean, except I don't understand how you're computing the dimension of the kernel. The kernel of the counit must have codimension $1$ so dimension $n^2 - 1$, and in any case it doesn't matter because simplicity forces it to be zero unless $n = 1$. $\endgroup$ Sep 17, 2020 at 17:28
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    $\begingroup$ …right, thanks a lot. It was a typo – it was supposed to read "either $n^2$ or zero (due to simplicity.)" $\endgroup$ Sep 17, 2020 at 17:36

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