## Correction

Todd Trimble pointed out to me a mistake I made in saying that internalizations commute. I glossed over a number of important subtleties and he set me straight.

Now I’m just getting back from a visit to Bourbon Street, since the real intarwobs and television won’t be reaching my apartment until Wednesday morning. I’ve skimmed his points and they check out, but I invite him to post them here as a comment so I don’t mangle them. Basically, I think I’m “morally” right (and I’m sure I’ve read John Baez saying somthing along these lines), but I’ve messed something up in my presentation as I attempt to get back in the saddle. I’d be glad of any clarification that can be made to see where I’ve gone Pete Tong.

These last two posts are sort of the tail end of internalization anyhow, and I was planning on moving on to something new on Monday. If they’re a little messy, so be it.

Ok. I had written John a longish email on this, as one math wonk to another. I agree, he’s morally right: there is this body of results along the lines of “a group in the category of categories is the same as a category in the category of groups”, as in his following post “Internal Categories”, and this is an important theme (as well as being kind of fun). But I complain that the general result he is shooting for here doesn’t work. I’m actually not sure what the general result should be!

Here’s a somewhat edited version of my email:

I’m writing with regard to your blog entry, “Internalizations Commute.” I think there are problems with it, but that it would be more appropriate to point them out privately than in the Comments section. [Edit: Well, now that the cat’s out of the bag…]

I take it that things like Th(Mon), Th(Ab), etc. are supposed to refer to Lawvere algebraic theories, where the models in a category C with finite products are product-preserving functors Th –> C. Of course, there are various ways we could interpret “theory” and “product” — in the case of Mon for example, the word “product” could be interpreted as just a monoidal product, and the word “theory” as just a ‘prop’ or a ‘pro’ — this is the point of view you take in your entry on the simplicial category. In other cases like Grp or Ab, you need diagonal maps (i.e. cartesian products) to describe the theory. In order for the notation ‘Th’ to carry a uniform meaning for the purposes of the blog entry, I take it to refer to cartesian products and Lawvere theories.

I say all this because there are points in the entry where the use of the word “product” looks weasel-y. For example, where you say

“We’ve taken abelian groups and built free monoid objects on them, and we’ve taken monoids and built free abelian group objects on them. And the rings we get from each case are isomorphic to each other.”

That’s a bit sloppy; I’m guessing you mean something like the free monoid on a free abelian group yields a ring which is isomorphic to the ring obtained as the free abelian group on a free monoid. Careful! I think you are trying to say that the free (unital) ring on a set X is obtainable in two ways:

Z[Sum_n X^n] = free abelian group on free monoid on X

= Sum_n (X.Z)^{\otimes n} [free monoid in (Ab, \otimes)]

(where X.Z denotes a coproduct of X copies of Z in Ab, aka the free abelian group on X). That’s correct, but it involves some sleight-of-hand, since you’ve switched between two notions of product, one cartesian and the other one a non-cartesian tensor product, \otimes.

It’s just not true that all internalizations commute. The case of rings you’ve picked is an interesting one, because here is a case where the relevant monad is a composite TS of two monads, where T = free abelian group and S = free monoid. It doesn’t work the other way (i.e., ST), and it’s relevant to say that what makes this example tick is the presence of a “distributive law” ST –> TS. I won’t go into this too much because I expect you have some familiarity with this notion (or if you don’t, you can easily look it up, say in Toposes, Triples and Theories) — I’ll just make the prosaic remarks that the map ST –> TS captures the idea that the S-operations in the composite TS distribute over the T-operations, and that the multiplication on the monad TS comes about as

TSTS –> TTSS –> TS

where the first arrow uses the distributive law and the second the multiplications of the monads T, S.

[As an aside, there’s a cute little result, that the only

algebraic theory which “commutes” with all others is the algebraic theory (the monad) for compact Hausdorff spaces! (Edit: the operations for this infinitary theory are given by ultrafilters; see the book Algebraic Theories by Manes

for more information on this lovely result.)]

Your entry proposes a proof of commutation, based on the fact that Cat is cartesian closed. So the question is: where does the proof break down?

Well, we’re not really using Cat, are we? We’re using product-preserving functors after all. More to the point: your last paragraph seems to suggest that the category of product-preserving functors,

Prod(Th(S) x Th(T), C)

is equivalent to

Prod(Th(S), Prod(Th(T), C))

but that’s not true. This is sort of akin to the difference,

in the world of abelian groups, between (joint) linearity

A x B –> C and (separate) bilinearity A x B –> C. As a test case, you can substitute the category 2 = {0 –> 1} for Th(S) and Th(T) and see what happens. [Edit: namely, product-preserving functors 2 –> C are essentially the same as subobjects of the terminal object 1 in C; call this category Sub(1). Then

Prod(2, Prod(2, C)) ~ Sub(Sub(1)) ~ Sub(1),

whereas product-preserving functors 2 x 2 –> C are essentially given by *pairs* of subobjects of 1.]

BTW, there is a tensor product of theories S \otimes T, where you force perfect interchange (i.e. all S-operations

are T-algebra homomorphisms, and vice-versa), but the result can be a pretty degenerate theory, as I’m sure you can imagine. There is on the other hand a coproduct of theories S + T, where the S and T structures have nothing to do with one another. I’m not sure off-hand what S x T gives you!

Comment by Todd Trimble | August 12, 2007 |