The Unapologetic Mathematician

Well, that was disappointing

I leave for two whole days and nobody has anything to talk about? Hm.

I’m not really up to a proper post right now. Maybe later tonight, or maybe not until tomorrow. Then I shall be back on my regular schedule. For now, though, I leave you with the mental image of a Portuguese woman walking through a very touristy area in Faro wearing a Vote For Pedro shirt.

Seriously, I nearly collapsed onto the lovely mosaic street from laughing too hard once she was past me.

July 10, 2007 Posted by | Uncategorized | 1 Comment

My Macintosh’s brick just blew, so with the last few minutes of power I have I’ll let you know that I won’t be posting until Tuesday night (I hope) when I’ll be back in Maryland and I’ll be able to get to an Apple store. Until then, discuss amongst yourselves.

Categories with Duals

Now we’ve got monoidal categories to categorify the notion of a monoid, we should consider what the proper analogue of an inverse is. What we’ll do is define a “dual object”, which is sort of halfway a left inverse and halfway a right inverse.

So, we say a monoidal category $\mathcal{M}$ “has duals” if for every object $M$ there is a “dual object” $M^*$ and natural transformations $\eta_M:\mathbf{1}\rightarrow M^*\otimes M$ and $\epsilon_M:M\otimes M^*\rightarrow\mathbf{1}$. We do not ask that these be isomorphisms, since for the best examples they’re actually not.

Instead we ask that the transformations be compatible in the sense that the following equations hold:
$\lambda_M\circ(\epsilon_M\otimes1_M)\circ\alpha_{M,M^*,M}^{-1}\circ(1_M\otimes\eta_M)\circ\rho_M^{-1}=1_M$
$\rho_{M^*}\circ(1_{M^*}\otimes\epsilon_M)\circ\alpha_{M^*,M,M^*}\circ(\eta_M\otimes1_{M^*})\circ\lambda_M^{-1}=1_{M^*}$
The first one means (reading right to left) that if we create a copy of the identity on the right with $\rho^{-1}$, then hit it with $\eta$ to create $M^*\otimes M$, then associate, then hit the $M\otimes M^*$ on the left with $\epsilon$ to get the identity, then remove it with $\lambda$, it’s just the same as doing nothing at all! And there’s a similar interpretation of the other equation. I can hear your furiously scratching your heads and saying “huh?” right now, but I’ll come back to this with another viewpoint later.

Anyhow, we also want this duality to be a functor — a contravariant functor even — which you should have guessed when I called $\eta$ and $\epsilon$ natural transformations. So we need for every $f:M\rightarrow N$ and arrow $f^*:N^*\rightarrow M^*$ so that $(f\circ g)^*=g^*\circ f^*$ and $1_M^*=1_{M^*}$. There are also a naturality conditions for $\eta$ and $\epsilon$ that you should be able to write down if you think of this as a covariant functor $(\underline{\hphantom{X}})^*:\mathcal{M}^\mathrm{op}\rightarrow\mathcal{M}$. The functor should also be monoidal, in the sense that $(M\otimes N)^*\cong N^*\otimes M^*$ (why this form? write out the condition explicitly).

The motivating example here is the category $\mathbf{FinVect}_k$ of finite-dimensional vector spaces over a field $k$. Here we know that tensor product over $k$ gives a monoidal structure, and we’ll use the dual module of linear functionals as our functor. Indeed, if we set $M^*=\hom_{\mathbf{Vect}_k}(M,k)$ we clearly have a contravariant functor. We can verify that it’s monoidal in the proper sense, and that the pairing gives a natural transformation $M\otimes M^*\rightarrow\mathbf{1}=k$. The other natural transformation required is also there, but we don’t yet have the tools needed (unless you’ve taken some linear algebra on your own). Still, you can keep this example in mind.

Now one thing that should be bugging you is that we have arrows $M\otimes M^*\rightarrow\mathbf{1}$ and $\mathbf{1}\rightarrow M^*\otimes M\rightarrow$, but there are two more it seems we should have. To handle this, we’ll insist that $(M^*)^*\cong M$ — that duality “is its own inverse”, so that we can just use $\eta_{M^*}$ and $\epsilon_{M^*}$ as the other natural transformations we need. This does indeed hold in our motivating example, but we again won’t prove it completely until later.

Note that there’s no reason to assume that $\epsilon_{M^*}\circ\eta_M=1_\mathbf{1}$ or $\epsilon_{M}\circ\eta_{M_*}=1_\mathbf{1}$, so these still aren’t isomorphisms. If they are, though, then we really do have natural isomorphisms to replace left and right inverse rules in the definition of a group. If you’re feeling up to it, try to state (prove?) a coherence theorem that says what diagrams must commute to ensure they all do.

July 7, 2007 Posted by | Category theory | 10 Comments

The Braided Coherence Theorem

I find myself up earlier than expected this morning, so I thought I’d bang out the proof I’d promised of the “coherence theorem” for braided monoidal categories.

Recall that we’re considering a braided monoidal category $\mathcal{M}$ with underlying “ordinary category” $\mathcal{M}_0$ — the same category, just forgetting that it’s braided and monoidal. Then the statement is that the category of braided monoidal functors from $\mathcal{B}raid$ to $\mathcal{M}$ is equivalent to $\mathcal{M}_0$.

To make our lives a bit easier, we’ll use the strictification theorem to pass from the (weak) monoidal category $\mathcal{M}$ to a monoidally equivalent strict monoidal category $\mathcal{S}$, with underlying category $\mathcal{S}_0$. Now the braiding on $\mathcal{M}$ induces (via the equivalence) one on $\mathcal{S}$. The new statement is that the category of strictly monoidal functors from $\mathcal{B}raid$ to $\mathcal{S}$ is isomorphic to $\mathcal{S}_0$. Then the result we really want (about $\mathcal{M}$) will follow.

One direction is easy: we take a functor $F$ and evaluate it at the object $1$ to get an object $F(1)\in\mathcal{S}_0$. A monoidal natural transformation $\eta:F\rightarrow G$ has, in particular, the component $\eta_1:F(1)\rightarrow G(1)$, which makes “evaluate at $1$” a functor.

For the other direction, we’ll pick an object $S$ of $\mathcal{S}_0$ and use it to construct a strictly braided monoidal functor from $\mathcal{B}raid$ to $\mathcal{S}$ that sends $1$ to $S$, and then show that an arrow $f:S\rightarrow S'$ in $\mathcal{S}_0$ induces a natural transformation between the functors built on $S$ and $S'$. We’ll see that everything in sight is unique, so this construction actually provides an “on the nose” inverse functor to “evaluate at $1$“.

So, we’ve got an object $S$ and we set $F_S(1)=S$. Then for $F$ to be monoidal we must set $F(n)=S^{\otimes n}$, since $n$ is $1^{\otimes n}$ (remember that the “tensor power” is defined just like regular exponentiation: take the tensor product of $n$ copies of the object in question). This completely defines the functor $F_S$ on objects.

Now any morphism in $\mathcal{B}raid$ is an element of some braid group $B_n$. Let’s look at simply crossing the left strand over the right in $B_2$: this is $\beta_{1,1}$, so to have a strictly braided functor we must set $F_S(\beta_{1,1})=\beta_{S,S}$. In $B_n$, the crossings of one strand over the one to its right generate the whole group. If we cross strand $i$ over strand $i+1$ we’ll call the generator $\sigma_i$. In terms of the category $\mathcal{B}raid$, we can write $\sigma_i$ as $1_{i-1}\otimes\beta_{1,1}\otimes1_{n-i-1}$, so monoidality forces us to set $F_S(\sigma_i)=1_{S^{\otimes i-1}}\otimes\beta_{S,S}\otimes1_{S^{\otimes n-i-1}}$.

We do have to check that the relations of the braid group are satisfied. But if $|i-j|>1$ then we have $F_S(\sigma_i\sigma_j)=F_S(\sigma_j\sigma_i)$ because the monoidal product in $\mathcal{S}$ is a functor. And we see that $F_S(\sigma_i\sigma_{i+1}\sigma_i)=F_S(\sigma_{i+1}\sigma_i\sigma_{i+1})$ because this is exactly an instance of the triangle relation for the braiding on $\mathcal{S}$!

So once we pick the single object $S=F_S(1)$ everything else about $F_S$ is uniquely determined. If we have an arrow $f:S\rightarrow S'$ then we must set the component $\eta_1$ of the induced natural transformation to be $f:F_S(1)\rightarrow F_{S'}(1)$. Then monoidality forces $\eta_n:F_S(n)\rightarrow F_{S'}(n)$ to be $f^{\otimes n}$. The construction is indeed a functor sending $\mathcal{S}_0$ to the category of strictly monoidal functors from $\mathcal{B}raid$ to $\mathcal{S}$, and it is clearly inverse to evaluation at $1$.

July 6, 2007

Second Day in Faro

I think my talk went well enough, and I’m glad to have it done so early so I can relax a bit more.

I’ll wait on deep comments until I return stateside and can scan in some of my notes. I did feel, though, that some of the major speakers were dragging their feet a bit so they could fill all three hours. Still, their previews of “coming attractions” look interesting enough. Maybe I’m still recovering from lost sleep a bit, but at least I didn’t nod off.

Oh, I did run into a reader today. I won’t out him myself, knowing how some value their privacy, but if he wants to say hello…

First Day in Faro: Random Thoughts

Well, it was an adventure getting from my flight arriving in Lisbon through customs (even with nothing to declare and no checked baggage) and on to my next flight. On the way, Scott Morrison got to see me grumbling (I partly chalk that up to such a long time without indulging my addiction) and as I got to the connection I ran into Peter Ozsvath, which was a nice little surprise.

He and I ran into each other again a little while ago while wandering around. I was killing time until I could check in to my hotel and he was just trying to stay awake. So we had lunch. It was good, a little more than I’d have liked to pay, but there’s time to find the little holes-in-the-walls later. I’m also hoping his general reactions were due to being tired rather than me being insanely boring. It’s probably a little of both.

Anyhow, I’ve started to bump into a few more people, and we’re starting to clump a bit. I like to think I’m rather cosmopolitan, but culture shock is still a real thing. And it’s easier to manage in a small group, ’cause someone won’t be a complete idiot in a given situation. On the other hand, before running into Oszvath (the second time) I negotiated my way around a bit. The lemon gelado here is good for cutting through the heat, and it scours your palate with a refreshing sourness that lingers on.

Another random thought: in the Philadelphia airport I randomly ran into a guy who’d just graduated high school and is on his way to Cornell. I mention this because while he was still undecided last fall he sat in on my Calculus 3 class on a campus visit. One hour he saw me — almost a full year ago, no less — and he still recgonized me and wanted to say hi and that he’d liked that one lecture. Of course, since it’s a compliment I said thianks (I’ve manged to learn to do that much at least), wished him good luck, and told him to watch out for that high suicide rate in Ithaca. Yeah, I’m a bad person.

More on the mathematical front, I bought Douglas Hofstadter’s I Am a Strange Loop for this trip. It’s a good read so far, but that’s to be expected from him by now. Particularly nice is the way he approaches (reapproaches) epiphenomena, and specifically how we talk about them. He notes that at a certain point it’s not only unhelpful to describe a macroscopic system explicitly in terms of its microscopic pieces, but it’s flat-out harmful.

Sure, an engineer could write down untold numbers of equations describing the position and momentum of every single molecule of gasoline and air in a car’s cylinder, but “pressure” isn’t about this molecule or that one. It’s an epiphenomenon that a very large number of different microstates give rise to. If we talk about statistical mechanics instead of thermodynamics we’re missing the forest and the trees alike for the leaves

I’ve been saying much the same thing. In particular, with regard to the natural numbers: we don’t care how we build them, we just want them to satisfy the universal (epiphenomenal) property and we use that property to prove everything else without caring about the petty details of how our model satisfies the condition.

But this all then raises a natural question: we’re looking at microstates (statistical mechanics) and declare examples of them to be “the same” if their macrostates are the same. But we don’t like “the same” — we like “isomorphic. So, is the passage from statistical mechanics to thermodymanics a decategorification?

Anyhow, I’m going to also push through until the local sleep time. In a bout an hour I’m meeting up with someone and we’ll try to figure out how to get from here to the university. This ought to be fun.

July 4, 2007 Posted by | Uncategorized | 3 Comments

The Category of Braids

Before I jet off for a week of knot homology — where I’m looking forward to seeing Scott Morrison, Jacob Rasmussen, and Ben Webster of the Secret Blogging Seminar — I thought I’d finally start tying in (sorry) this category theory stuff with knot theory.

As I said yesterday, the analogue of commutativity for a monoidal category is called a braiding. The name comes from a deep connection between braided monoidal categories and braid groups. Specifically, there is a (strict) braided monoidal category of braids: $\mathcal{B}raid$. The objects of this category are the natural numbers, and $\hom_{\mathcal{B}raid}(m,n)$ is the group $B_n$ if $m=n$ and empty otherwise. We think of a braid with $n$ strands as taking a collection of $n$ points and shuffling them around, and we draw a diagram like this:

As a side note, I always intend diagrams like these to be read from bottom to top, though other authors go the other way.

First, the monoidal structure. We don’t need any of the structural isomorphisms because we’re making a strict monoidal structure, but we do need a bifunctor $\underline{\hphantom{X}}\otimes\underline{\hphantom{X}}$. On the objects it’s just addition. For morphisms we need a way of taking a braid with $m$ strands and one with $n$ strands and making a braid with $m+n$ strands. We’ll just stand both of our braids next to each other like this:

Now we need a braiding. That is, for each pair of objects $(m,n)$ we need a morphism (a braid) from $m+n$ to $n+m$ that “commutes” with all the other morphisms. We’ll just pass the $m$ strands from the left over the $n$ strands from the right. Here’s what I mean for $m=3$ and $n=2$:

This is natural because if we stick on a braid below the leftmost three strands on the bottom, the Reidemeister moves will let us pull it up and over the other two strands until it’s sitting on top of the three strands at the right on the top. Similarly, a braid can be pulled along the undercrossing strands. Since our monoidal structure is strict, the hexagon identities degenerate into triangles, and the proof is just the following diagram:

So we indeed have a strict braided monoidal category. It turns out to be extremely important because of the following theorem of Joyal and Street:

If $\mathcal{M}$ is a braided monoidal category with underlying category $\mathcal{M}_0$ (forget the monoidal structure and the braiding to just have a regular old category), then the category of braided monoidal functors from $\mathcal{B}raid$ to $\mathcal{M}$ is equivalent to $\mathcal{M}_0$.

This is sort of like a coherence theorem for braided monoidal categories: for each natural isomorphism built from $\alpha$, $\lambda$, $\rho$, and $\beta$ there is an “underlying braid”, and two such isomorphisms are equal if and only if they have the same underlying braid. I’ll defer the proof of this for now, but you should think about it a bit. The details aren’t that bad, but the basic idea just leaps out at you after a bit.

July 3, 2007

Braidings and Symmetries

So we’ve got monoidal categories that categorify the notion of monoids. In building up, we had to weaken things, refusing to talk about “equalities” of objects. Instead we replaced them with (natural) isomorphisms and then talked about equalities of morphisms. Now we’re ready to specialize a bit.

Many monoids we’re interested in are commutative: $ab=ba$ for all elements $a$ and $b$ of our monoid $M$. So what does this look like up at the category level? We learned our lesson with associativity, so we already know we should look for a natural isomorphism with components $\beta_{A,B}:A\otimes B\rightarrow B\otimes A$ to swap objects around the monoidal product.

Of course, there’s a coherence condition:

If we start with $(A\otimes B)\otimes C$ and want to get to $B\otimes (C\otimes A)$, we could pull $A$ past $B$, associate, and then pull $A$ past $C$. We could also associate, pull $A$ past $B\otimes C$, and associate again. And, naturally, we want these two to be the same. There’s another coherence condition we’ll need, but first let’s notice something.

Since $\beta_{A,B}:A\otimes B\rightarrow B\otimes A$ is an isomorphism, there’s an inverse $\beta_{A,B}^{-1}:B\otimes A\rightarrow A\otimes B$. Of course, we also have $\beta_{B,A}:B\otimes A\rightarrow A\otimes B$ sitting around. Now the obvious thing is for these two to be the same, but actually that’s not what we want. In fact, there are very good reasons to allow them to be different. So in general we’ll have two different ways of pulling $A$ past $B$, either with a $\beta$ or with a $\beta^{-1}$, and both of them should satisfy the hexagon identity shown above. That’s our second coherence condition.

We call such a natural transformation satisfying (both forms of) the hexagon identity a “braiding”, and a monoidal category equipped with such a “braided monoidal category”, for a very good reason I’ll talk about tomorrow (hint). Now if by chance the braiding $\beta$ is its own inverse, we call it a “symmetry”, and call the category a “symmetric monoidal category”.

We say that a monoidal functor $F:\mathcal{C}\rightarrow\mathcal{D}$ is braided (symmetric) if its monoidal structure plays well with both braidings. That is, if $F_{(B,A)}\circ\beta_{F(A),F(B)}=F(\beta_{A,B})\circ F_{(A,B)}$. Notice that if $F$ is strictly monoidal this just says that $F(\beta_{A,B})=\beta_{F(A),F(B)}$.

Both of these kinds of categories give back commutative monoids when we decategorify. This tells us that even though commutativity is relatively straightforward down at the level of sets, categorification lets us tease apart a subtle distinction and get a better insight into what’s “really going on”.

We know that any category with finite products (or coproducts) can use them as a monoidal structure. As an exercise, use the universal properties of products to come up with a braiding, then show that it’s symmetric.

Another exercise: show that any braiding satisfies $\rho_A=\lambda_A\circ\beta_{A,I}$ and $\lambda_A=\rho_A\circ\beta_{I,A}$. That is, when we have a braiding the left and right units are determined by each other and the braiding.

July 2, 2007 Posted by | Category theory | 6 Comments

The “Strictification” Theorem

I want to get in another post in the current line before I head off to Portugal on Tuesday. This theorem is another heavy one to prove, so if the proof of the Coherence Theorem was overwhelming you might just want to skim this one and remember the result: Any monoidal category $\mathcal{M}$ is monoidally equivalent to a strict monoidal category $\mathcal{S}$ — one whose objects form an honest monoid without any “up to isomorphism” dodging about. This will justify the usual laxity with which we treat associativity and unit laws in monoidal categories, writing $A\otimes B\otimes C\otimes\cdots$ without regard to parentheses and such.

When we naively write down a monoidal product like this we’re treating it as a list of objects of $\mathcal{M}$, and we take the product of two lists just by concatenation. So let’s start by defining the collection of objects of $\mathcal{S}$ to be the free monoid on the objects of $\mathcal{M}$, which is exactly this collection of lists. We’ll write a typical list as $A=[A_1,A_2,...,A_m]$. Then we have the concatenation product $\left[A_1,...A_m\right]\cdot\left[B_1,...B_n\right]=\left[A_1,...A_m,B_1,...,B_n\right]$ and the empty list $\left[\right]$ for an identity object. All the monoid rules hold “on the nose” already, so we don’t need to worry about associators and all.

We don’t have any arrows in our category yet, but let’s press on a bit first to define a map $F:\mathcal{S}\rightarrow \mathcal{M}$. If this has any hope of being the object function of a monoidal functor we’ll need $F([])=\mathbf{1}$. We also need to figure out how $F$ should deal with a monoidal product. What we’re going to do is just choose to make the result have all its parentheses on the right. We define $F([A_1,A_2,...])=A_1\otimes F([A_2,...])$ to get (for example) $F([A_1,A_2,A_3,A_4])=A_1\otimes(A_2\otimes(A_3\otimes A_4))$.

Now we’ll go back and add arrows to $\mathcal{S}$ in just such a way as to make $F$ into a monoidal functor. Given two lists $A$ and $B$ in $\mathcal{S}$ I’ll define $\hom_\mathcal{S}(A,B)=\hom_\mathcal{M}(F(A),F(B))$. Notice that this makes the functor $F$ fully faithful by definition, and that every object $M$ in $\mathcal{M}$ is $F([M])$, so we’re well on our way to having an equivalence of categories!

What’s missing now? Well, we still don’t know quite how to take the monoidal product of arrows in $\mathcal{S}$, so it’s still not quite a monoidal category yet. We can use the isomorphisms of hom-sets we just set up to do this. Given $f:A\rightarrow C$ and $g:B\rightarrow D$ in $\mathcal{S}$ their monoidal product $f\cdot g:A\cdot B\rightarrow C\cdot D$ is an arrow in $\hom_\mathcal{M}(F(A\cdot B),F(C\cdot B))$. We define it to be the composite:
$F(A\cdot B)\rightarrow F(A)\otimes F(B)\rightarrow F(C)\otimes F(D)\rightarrow F(C\cdot D)$
Here the middle arrow is the monoidal product of $f$ and $g$ in $\mathcal{M}$, and the outer two arrows are the (unique!) ways to change parentheses as needed. The uniqueness follows from the Coherence Theorem, so we have a well-defined monoidal product on $\mathcal{S}$.

So now $F$ is a monoidal functor. We take $F_*$ to be the identity $F([])=\mathbf{1}$, and we let $F_{(A,B)}:F(A)\otimes F(B)\rightarrow F(A\cdot B)$ be the (unique!) arrow moving all the parentheses to the right. The Coherence Theorem gives us this uniqueness, and also handles the diagrams required for a monoidal functor.

Now for the other direction we’ll just go ahead and define $G:\mathcal{M}\rightarrow\mathcal{S}$ by setting $G(A)=[A]$ and $G(f)=f$. We also define $G_*:[]\rightarrow[\mathbf{1}]$ by noticing that $G_*\in\hom_\mathcal{M}(\mathbf{1},\mathbf{1}$, so we can just use the identity. We define $G_{(A,B)}:[A,B]\rightarrow[A\otimes B]$ similarly.

I’ll leave the cleanup to you: $G$ does satisfy coherence conditions for a monoidal functor, $F\circ G:\mathcal{M}\rightarrow\mathcal{M}$ is the identity functor, and there is a monoidal natural isomorphism from $G\circ F$ to the identity functor on $\mathcal{S}$. It’s all pretty striaghtforward, though you’ll have to appeal to the Coherence Theorem a couple times like we have above.

July 1, 2007 Posted by | Category theory | 3 Comments

Things Change

I am no longer a lecturer in Yale’s mathematics department. I am now an assistant professor in Tulane’s.

July 1, 2007 Posted by | Uncategorized | 3 Comments