The n-Category Café

I think that whenever He gives us an interesting algebraic structure, even if it arises ‘trivially’, there’s often going to be something useful you can do with it! At least it’s worth considering. Wouldn’t you agree?

For example, you say that the unique comonoid structure on a set is trivial. But together with a monoid on the same set, the structures interact to form a bimonoid. There’s a canonical monoidal faithful functor from Set to Hilb, and we can use this to transfer our bimonoid into Hilb. But bimonoids in Hilb have monoidal categories of modules – and so this tells us that our original monoid will have a monoidal category of representations.

Maybe you would say this is obvious anyway, but at least this gives a categorical way to see why it’s true.

Simon answered David’s question.

David says:

So on every vector space there’s a unique bialgebra structure?

Simon says:

It’s important to bear in mind that John is using $\oplus$ for his monoidal product on vector spaces, whereas the usual monoidal product is $\otimes$.

So: a bialgebra is a vector space with a map $V \otimes V \to V$ and a map $V \to V \otimes V$, fitting together nicely to make a bimonoid in the monoidal category $(Vect, \otimes)$. There are usually lots of ways to make a vector space into a bialgebra.

But I’m talking about a vector space with a map $V \oplus V \to V$ and a map $V \to V \oplus V$, fitting together nicely to make a bimonoid in the monoidal category $(Vect, \oplus)$. There is always exactly one way to make a vector space into this sort of bimonoid.

And I’m trying to get you guys to cough up the two really elegant (indeed, trivial) proofs of this last fact. Jamie proved almost everything except the part I was really interested in, namely this law:

Peter wrote:

I’ve always liked the fact that for $(\mathbf{C}, \otimes)$ symmetric monoidal, $\mathbf{ComMon}(\mathbf{C}, \otimes)$ is the universal way of making $\otimes$ into a coproduct.

Yes, that’s a nice and perhaps underappreciated fact. I think it’s even underappreciated that the tensor product of commutative rings is their coproduct. It’s not an obscure fact, but nevertheless I’ve known it to take people by surprise. Perhaps it’s something to do with the use of language: tensor products are thought of as more like products ($\times$) than sums ($+$).

I think it’s even underappreciated that the tensor product of commutative rings is their coproduct. It’s not an obscure fact, but nevertheless I’ve known it to take people by surprise.

True. Of course the right way to think of it is that $Rings$ is really $Aff^{op}$ and the tensor product of two rings of functions on $X$ and $Y$ is the ring of functions on the product $X \times Y$.

This becomes even more “impressive” for more general Lawvere theories: for instance the $C^\infty$- completed tensor product of two rings of smooth functions $C^\infty(X)$ and $C^\infty(Y)$ is just their coproduct as $C^\infty$-rings.

I have seen differential geometers who first learned about this quite impressed (myself once included, thanks to initiation by Todd). Of course when one thinks about it a little, it is a triviality. But that’s the point, right: making that which is trivial be trivially trivial. ;-)

Peter’s proof uses slightly more than:

• Any object in a category with finite coproducts becomes a monoid in a unique way.

and its dual; he really uses that:

• Any object in a category with finite products becomes a comonoid in a unique way, and every map between them a map of comonoids

and its dual. In fact, you can do without this. For any monoidal category $\mathcal{V}$, the forgetful functor $\mathbf{Comon}(\mathcal{V}, \otimes) \to \mathcal{V}$ creates colimits. In particular, for any $\mathcal{E}$ with biproducts, the forgetful functor $\mathrm{Comon}(\mathcal{E}, \times) \to \mathcal{E}$ creates coproducts. Now each $X \in \mathcal{E}$ is uniquely a $\times$-comonoid, and the unique lifting of $X$ to $\mathbf{Comon}(\mathcal{E}, \times)$ is itself a $+$-comonoid in a unique way; whence $X$ is uniquely a bimonoid for $\times = +$. Of course, there is a dual version.

Note that, in fact, we can do without the assumption that $\times = +$ here, so long as we are prepared to interpret the notion of bimonoid in a more general sense. Here is that sense. Suppose that $\mathcal{V}$ is a category equipped with two monoidal structures $\otimes_1$ and $\otimes_2$ such that the second one is lax monoidal with respect to the first. This is sometimes called a two-fold monoidal category. This means in particular that there are (non-invertible) interchange maps

$$\sigma_{X,Y,W,Z} \colon (X \otimes_2 Y) \otimes_1 (W \otimes_2 Z) \to (X \otimes_1 W) \otimes_2 (Y \otimes_1 Z)$$

satisfying various axioms. Then we can define a $\otimes_1$-$\otimes_2$-bimonoid to be an object of $X \in \mathcal{V}$ which is a $\otimes_2$-comonoid, an $\otimes_1$-monoid and which satisfies, amongst other things, the bialgebra law that

$$X \otimes_1 X \stackrel{\delta \otimes_1 \delta}{\to} (X \otimes_2 X) \otimes_1 (X \otimes_2 X) \stackrel{\sigma_{X,X,X,X}}{\to} (X \otimes_1 X) \otimes_2 (X \otimes_1 X) \stackrel{\mu \otimes_2 \mu}{\to} X \otimes_2 X$$

should equal

$$X \otimes_1 X \stackrel{\mu}{\to} X \stackrel{\delta}{\to} X \otimes_2 X.$$

More abstractly, we have that $\otimes_2$, being lax monoidal with respect to $\otimes_1$, lifts to the category of $\otimes_1$-monoids; so that we may consider $\otimes_2$-comonoids in there, and these are precisely the bimonoids defined above. Of course you could also first take $\otimes_2$-comonoids and then $\otimes_1$-monoids, but you’d get the same answer. To summarise, we have that:

$$\mathbf{Bimon}(\mathcal{V}, \otimes_1, \otimes_2) \cong\mathbf{Comon}(\mathbf{Mon}(\mathcal{V}, \otimes_1), \otimes_2) \cong\mathbf{Mon}(\mathbf{Comon}(\mathcal{V}, \otimes_2), \otimes_1)$$

Now for any category $\mathcal{E}$ with both products and coproducts, the product structure is lax monoidal with respect to the coproduct structure (because any functor is lax monoidal with respect to coproducts), and so we have a two-fold monoidal category $(\mathcal{E}, +, \times)$. The above argument now shows that any object of $\mathcal{E}$ is a $+$-$\times$-bimonoid in a unique way.

If the green blob is the diagonal, then the equation holds no matter what morphism the red dot stands for.

So you made us think that the problem has something to do with addition, when obviously the result is $(A (x, y), A (x, y)) = (A (x, y), A (x, y))$ for any linear map $A\colon V \oplus V \to V$. So it has nothing to do with addition at all, only duplication. How sneaky!

If the red blob is the codiagonal, then the equation holds no matter what morphism the green dot stands for.

So you made us think that the problem has something to do with duplication, when obviously the result is $(A_1 x + A_1 y, A_2 x + A_2 y) = (A_1 (x + y), A_2 (x + y))$ for any linear map $A\colon V \to V \oplus V$. So it has nothing to do with duplication at all, only addition. How sneaky!