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In Tom Leinster's book on operads, he gives Ab(V), the category of abelian groups in a symmetric monoidal category V, as an example of a multicategory that doesn't arise from a monoidal category, since Ab(V) will not generally have a tensor product. Example 2.1.5 on page 37.

I can't quite understand this. Why do you need colimits in V to define tensor product in Ab(V)? What is the definition exactly of a tensor product? I know it as the universal object representing bilinear morphisms in some multicategory, and from that definition it's not clear how colimits in some category help it exist.

Actually I have a more basic question about this setup: Can you define group objects in an arbitrary symmetric monoidal category? I thought you could only define monoid objects, and you needed a Cartesian monoidal category to define a group object. Cartesian monoidal categories have the diagonal morphism which you need to define inverses, whereas general monoidal categories do not, right?

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Concerning group objects in arbitrary symmetric monoidal categories: there is a sense in which this can be done, and what you get is the notion of Hopf algebra! The idea is to regard the codiagonal as additional algebra data, namely a comultiplication. In the cartesian case, the comultiplication is unique and given by the codiagonal. Check out… for more information. – Tobias Fritz Nov 27 '12 at 21:35
up vote 6 down vote accepted

It is easy to define the tensor product as being the object that represents the bilinear maps functor, but to prove that tensor products exist requires something extra. If you have free abelian groups, then it is enough to have coequalisers of abelian groups to construct the tensor product, but even the construction of coequalisers of abelian groups can be quite non-trivial.

I'm more familiar with the case where the base category $\mathcal{V}$ is cartesian closed, so I'll briefly mention the extra structures we need to get tensor products in $\textbf{Ab}(\mathcal{V})$:

  1. If $\mathcal{V}$ has countable colimits and is a well-powered regular category, then the forgetful functor $\textbf{Ab}(\mathcal{V}) \to \mathcal{V}$ is strictly monadic. (In particular, it has a left adjoint.) A proof can be extracted by chasing the references in theorem 2.4.21 of my notes. Moreover these hypotheses suffice to show $\textbf{Ab}(\mathcal{V})$ has coequalisers for all parallel pairs: combine proposition 2.4.7 (ii) and theorem 2.4.21 (ii).

  2. Once we know that $\textbf{Ab}(\mathcal{V})$ has coequalisers of reflexive pairs and that $\textbf{Ab}(\mathcal{V}) \to \mathcal{V}$ is monadic, we can construct tensor products the usual way using generators and relations.

As for your second question: you are quite right in saying that we need some kind of diagonal map $\delta_X : X \to X \otimes X$ in order to even define inverses. But that's OK: we just build that into the definition of a group object in a symmetric monoidal category. More accurately, what we are doing is expressing the axioms for a Hopf algebra in diagrammatic form: so a group object in a general symmetric monoidal category will consist of

  • an object $X$, and
  • morphisms $\eta : I \to X$, $\mu : X \otimes X \to X$, $\epsilon : X \to I$, $\delta : X \to X \otimes X$, and $\sigma : X \to X$

satisfying the following axioms:

  • $(X, \eta, \mu)$ is a monoid.
  • $(X, \epsilon, \delta)$ is a comonoid.
  • $\sigma$ gives left inverses: $\mu \circ (\sigma \otimes \textrm{id}_S) \circ \delta = \eta \circ \epsilon$
  • $\sigma$ gives right inverses: $\mu \circ (\textrm{id}_S \otimes \sigma) \circ \delta = \eta \circ \epsilon$

This is a generalisation of the usual definition of group object in a cartesian monoidal category because every object in a cartesian monoidal category is a comonoid in a unique way. (This is a fun exercise.) The coordinate ring of any affine group scheme (over $\operatorname{Spec} A$) will also give examples in the category of $A$-modules.

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Re point 2. This in fact works for any commutative monad whose algebra category has coequalisers of reflexive pairs. This is due to Anders Kock in the 1970s, see and – Chris Heunen Nov 27 '12 at 23:07
Indeed – that's theorem 3.4.11. in my notes. Actually, if we only want representability of the bihomomorphism functor then we don't even need commutativity – see theorem 3.3.17. As far as I can tell we need some condition on the preservation of (regular) epimorphisms in order to prove the theorem. – Zhen Lin Nov 27 '12 at 23:15
The explanation for the Hopf algebra axioms that the comonoid and antipode structures are precisely the thing you need to make group objects work is lovely and surprising. Thank you! As for the construction of the tensor product out of reflexive coequalisers, that pretty much went over my head. But the answer to my question is clear enough: tensor objects need not exist, but some extra axioms, including the existence of the right kinds of colimits, will guarantee them. Thank you also for the link to your notes on algebraic theories. They look very useful to me. – ziggurism Nov 28 '12 at 17:59
When you say "once we know that $\textbf{Ab}(\mathcal{V})$ has coequalisers ... we can construct tensor products in the usual way", I'm imagining you mean something like: set $A\otimes B=\text{Coker}\{I\to F(A\times B)\}$, where $F$ is the free abelian group object functor (which exists because $\textbf{Ab}(\mathcal{V})\to \mathcal{V}$ is monadic?) and I is some subobject of $F(A\times B)$ generated by elements like $(a+a',b)−(a,b)−(a',b)$, and here "element" means "morphism $1\to A$". And we have 1 and $\times$ since $\mathcal{V}$ is Cartesian. Is that right? – ziggurism Nov 28 '12 at 18:26
@ziggurism Something like that, but not quite. A morphism $1 \to A$ is called a "global element" (by analogy with what happens in the topos of sheaves) but we need to make sure the tensor product does the right thing to all "elements". A more precise description would say something like, we have a diagram $I \rightrightarrows U F (A \times B)$ in $\mathcal{V}$ tabulating all the equations we want to be true in $A \otimes B$, and then transposing across the adjunction $F \dashv U$, we build $A \otimes B$ as the coequaliser of $F I \rightrightarrows F (A \times B)$ in $\textbf{Ab}(\mathcal{V})$. – Zhen Lin Nov 28 '12 at 19:55

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