Categorifying Fractional Euler Characteristics

December 16, 2011

Categorifying Fractional Euler Characteristics

Posted by John Baez

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Mathematicians should know how to count. Most of us do. But still there are some mysteries left when we try to count things in a way that gives a negative, fractional, irrational or complex answer.

Luckily, we’ve been making lots of progress. The Euler characteristic of a space can be a negative integer. The cardinality of a groupoid can be a nonnegative real number. The Euler characteristic of a category can even be a negative real number! And here’s yet another approach:

Igor Frenkel, Catharina Stroppel, Joshua Sussan, Categorifying fractional Euler characteristics, the Jones-Wenzl projector and 3j-symbols.

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The fun starts around page 17, when they look for a kind of entity whose Euler characteristic can be a function like this:

$\frac{1}{1 + q^2}$

Their first trick is to expand it as a power series:

$\frac{1}{1 + q^2} = 1 - q^2 + q^4 - \cdots$

Then they use a standard trick in this game, namely to define the ‘ $q$ -dimension’ of a graded vector space in such a way that the $q$ -dimension of a $d$ -dimensional space of grade $k$ is $d q^k$ . This allows them to define the Euler characteristic of a chain complex of graded vector spaces: it’s the alternating sum of their $q$ -dimensions.

So, suppose we have a chain complex of graded vector spaces

$V_0 \leftarrow V_1 \leftarrow V_2 \leftarrow \cdots$

where $V_k$ is 1-dimensional and of grade $2k$ . Then its Euler characteristic is

$1 - q^2 + q^4 - \cdots$

as desired!

This is just the beginning of the game; it quickly gets a lot more deep. But this is nice to see.

(By the way, I have tried to give their paper the title it deserves, not the one it has.)

Posted at December 16, 2011 7:20 AM UTC

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Re: Categorifying Fractional Euler Characteristics

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Then they use a standard trick in this game, namely to define the ‘ $q$ -dimension’ of a graded vector space in such a way that the $q$ -dimension of a $d$ -dimensional space of grade $k$ is $d q^k$ .

This is the part that drives me crazy — not because I don’t like it, but because I don’t feel like I understand it fully. This is clearly a great trick, but it always seems totally ad hoc. Is there some abstract context that it comes out of?

In particular, there is a general definition of the dimension/Euler-characteristic of a dualizable object in a symmetric monoidal category: the trace of its identity map. I would really like to know whether this sort of ‘dimension’ be realized in such a way.

Perhaps there is an answer to this in the paper; I didn’t see it, but I only read the small portion that looked like it would be comprehensible to me. (-:

Posted by: Mike Shulman on December 16, 2011 4:32 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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This is clearly a great trick, but it always seems totally ad hoc.

Yes! I’d love to understand why this is a natural move.

Posted by: Tim Silverman on December 16, 2011 9:36 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

I believe the motivation for this is the identification of the Grothendieck group of graded vector spaces/chain complexes (or representations of the multiplicative group in vector spaces/chain complexes) with Laurent polynomials in one variable, which we call q.

Posted by: David Ben-Zvi on December 17, 2011 1:14 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Thanks — but I’m not looking for motivation, I’m looking for an abstract context. They’re not the same thing; in fact, some might say they’re polar opposites. (-:

Posted by: Mike Shulman on December 17, 2011 7:06 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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I think David Ben-Zvi almost gave away what I’d call the abstract context.

Finite dimensional $\mathbb{Z}$ -graded vector spaces form a kind of ‘symmetric 2-rig’. To be precise, they form a symmetric monoidal category with finite colimits, where the tensor product distributes over these colimits. If we decategorify any such symmetric 2-rig we get a commutative rig. In this particular example, the commutative rig we get is isomorphic to $\mathbb{N}[q,q^{-1}]$ : that is, Laurent polynomials in one variable, with natural number coefficients.

If we want $\mathbb{Z}[q,q^{-1}]$ we have to work harder, since we need things that act like negative-dimensional vector spaces. We can do this using chain complexes instead of vector spaces, and taking the derived category.

If we want Laurent series instead of Laurent polynomials, we can use graded vector spaces that are allowed to be nonzero in infinitely many grades.

I’m not sure what part of this “drives you crazy”, so I’ll stop here and let you say!

Posted by: John Baez on December 17, 2011 7:33 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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It sounds like what you’re saying is that you consider graded vector spaces or chain complexes, choose some notion of “finiteness” (either total finite dimensionality, or finite dimensionality in each degree), decategorify the “finite” objects, and define the “dimension” of a “finite” object to be its image under the decategorification. Is that right?

I’ll grant that that’s an abstract context. But it’s not what I’m looking for, because (1) it depends on the input of a choice of some notion of “finiteness”, and (2) it’s not clear to me what sort of decategorified ri(n)gs you get in other examples. With vector spaces, everything is easy because all vector spaces are free, but what about, say, graded abelian groups, or chain complexes thereof? Spectra? Perhaps we have to start talking about algebraic $K$ -theory — which is certainly an interesting direction, but it’s not what I want.

What I would really like is a symmetric monoidal category in which these “ $q$ -dimensions” are traces of identity maps of dualizable objects. I like that context because (1) the notion of “finiteness” (dualizability) comes out of the abstract context, instead of being put into it, and (2) in any example it is easy to identify the ring where the traces live (the endomorphisms of the unit object), and also (3) it generalizes to Lefschetz numbers and twisted traces and (4) it supports nice additivity and multiplicativity theorems.

Posted by: Mike Shulman on December 17, 2011 11:22 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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I’m missing what the issue here is - the construction seems as tautological and abstract as anything I know (maybe that’s saying more about me though). We are given a ( $\infty$ -)category, namely (chain complexes of) graded vector spaces (it’s also symmetric monoidal but that’s less relevant for the basic construction). There are two tautological targets of “character” constructions on dualizable objects. Each such defines a class (or homotopy point) in $K_0$ (or the K-theory spectrum), which is the universal additive invariant of the object. There is also a universal trace map, landing in Hochschild homology (or a version of cyclic if you want to keep track of cyclic symmetry). In particular the trace of the identity of the object is a class (its character) in Hochschild homology. So either one of these characters is a form of the universal notion of dimension of the object (and of course they’re related by the relevant form of Chern character).

In our case then we just need to describe the relevant K-groups or Hochschild homology groups (or spectra if you prefer). Our ( $\infty$ -)category in question is symmetric monoidal, in fact it’s just representations of the multiplicative group, and so its Hochschild homology is easy to describe, as functions on the multiplicative group. (There are similar easy descriptions for HH of the category of sheaves on any nice stack, in terms of the loop space of the stack). Calling the coordinate $q$ we get the desired Laurent polynomials in $q$ .

Is the issue you want to get functions of $q$ without calculating what K-theory or HH of this category actually is? If you prefer, rather than calculating the naive trace of the identity, the tautological Hochschild trace here amounts to the function that attaches to each group element the trace of the corresponding operator on our representation. So yes this uses that our symmetric monoidal category happens to be representations of a group or groupoid or sheaves on a stack, but that’s fairly typical (and probably close to universal given suitable hypotheses - ie we can attach a “spectrum” stack to any symmetric monoidal category and I’d be surprised if the HH doesn’t factor through this construction..)

Posted by: David Ben-Zvi on December 18, 2011 1:53 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Maybe this is just too tautological for my taste. Or possibly I’m just not comfortable enough with the stuff you’re talking about.

Here’s my level of understanding. Let $C$ be any category; we can define a trace on $C$ to be an operation from endomorphisms in $C$ to some fixed target, which satisfies a cyclicity property $tr(g f) = tr(f g)$ . Obviously any category has a “universal” trace map, whose target is the quotient of $\coprod_x C(x,x)$ by $f g \sim g f$ . If we work additively and derived, then that quotient is the Hochschild homology of the category. And since the construction of $\coprod_x C(x,x) / \sim$ is invariant under Morita equivalence, if the category is something like dualizable modules over a ring, which is the Cauchy completion of that ring itself, then its Hochschild homology will be that of the ring. Hence, every endomorphism yields a universal “trace” lying in Hochschild homology.

I have no idea why there would be two such traces — can you explain further? Also, I have no idea how you calculated the HH of the category of graded vector spaces (or dualizable graded vector spaces?) — in what sense is it representations of something? How would I calculate the HH (or the trace without the HH) of other related things like the category of spectra, or the category of spectra parametrized over a given space?

Can this approach give us Laurent series in addition to Laurent polynomials? Can it tell us that the Euler characteristic of the homology of a topological space or spectrum is equal to some geometric invariant? And can it tell us why $B G$ has Euler characteristic $1/{|G|}$ ?

Posted by: Mike Shulman on December 19, 2011 7:48 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Let me try to elaborate. First the category of graded vector spaces is canonically equivalent to the category of representations of the multiplicative group - this is a nice illustration of Tannaka duality: the multiplicative group arises as automorphisms of the forgetful functor to plain vector spaces (equivalently through the corresponding monad on vector spaces). The same is true on the $\infty$ -level if you prefer to work with chain complexes of graded vector spaces.

Next, we agree on the universal notion of trace - the point about two traces is of course you can calculate the trace of an underlying vector space (ie apply the above forgetful functor), but there is more information in the universal trace on the original category, which gives a class in HH. So one just needs to calculate this HH. For this there are explicit complexes or general methods. e.g. for the $\infty$ -category of spectra (modules for the sphere spectrum) it’s fairly direct from the definition (ie self-Tor of the sphere as a bimodule over itself) that the HH is just the sphere itself (the trace map if I’m not mistaken just assigns to any finite spectrum its Euler characteristic, considered as an element of $\pi_0$ of the sphere).

I haven’t thought about parametrized spectra over general spaces, but if you’re interested in quasicoherent sheaves of spectra over a nice derived stack the answer is functions [global sections of the structure sheaf] on the loop space of the stack [self-intersection of the diagonal]. This setting includes both the example of the stable category itself (the stack being just a point), the example of modules over any connective $E_\infty$ -ring spectrum, and the example of graded vector spaces (the stack being the classifying stack of the multiplicative group). In the case of modules over an $E_\infty$ spectrum this is just the definition of THH, not a calculation, though in characteristic zero you can calculate this as the symmetric algebra of the shifted cotangent complex. Finally in the case of graded vector spaces, the loop space is just the multiplicative group itself (modulo itself acting trivially) and the derived functions are just Laurent polynomials in one variable (which we can call “q”). More generally for any affine algebraic group $G$ the HH of the category of modules (quasicoherent sheaves on the classifying stack $BG$ ) is class functions on the group (functions on the loop space $G/G$ ), as you might expect for the target of a universal trace or character map.

[The proofs of these assertions are easy modulo the $\infty$ -categorical machinery. You just need to check that endofunctors on sheaves on a reasonable stack $X$ are given by sheaves on $X\times X$ , so that trace of the identity functor is represented geometrically by functions on self-intersection of the diagonal, aka free loop space. I hesitate to give a reference since everything here is old and well known but my paper with Francis and Nadler on the arXiv does discuss these issues in the language I’m using.]

Unfortunately I haven’t thought about the questions in the last paragraph so don’t have anything to add (naively I’d guess that to get Laurent series you’d probably want to take the ind category of graded vector spaces with a one-sided bound on the grading, which is a completion of the usual category of graded vector spaces).

Posted by: David Ben-Zvi on December 19, 2011 9:54 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Unfortunately, if you want me to really understand, you’ll have to use smaller words. I don’t yet speak algebraic-geometer; to me a “multiplicative group” still means a group whose operation is written $x y$ instead of $x+y$ . (-: But in any case, I think that this:

for the ∞-category of spectra… the HH is just the sphere itself (the trace map if I’m not mistaken just assigns to any finite spectrum its Euler characteristic

means that this approach can’t possibly give what I want. The graded vector spaces that I see come from the homology of spaces or spectra. Functoriality of trace means that the Euler characteristic of a space can equivalently be calculated on its homology. But I don’t want just the plain old Euler characteristic of a space; I want to understand examples like $\chi(RP^\infty) = \frac{1}{2}$ that you get by taking the “ $q$ -Euler characteristic” of the homology and evaluating it (after analytic continuation or whatever) at $q=-1$ . If this approach to trace just gives the ordinary Euler characteristic of a space/spectrum, then it seems as though it can’t say anything interesting about that. Unless I’m missing something?

Posted by: Mike Shulman on December 20, 2011 1:18 AM | Permalink | Reply to this

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to me a “multiplicative group” still means a group whose operation is written $x y$ instead of $x + y$

See the $n$ Lab entry affine line in the section multiplicative group: it is the $Spec$ of the Laurent polynomial ring

$\mathbb{G}_m := \mathbb{A}^1 - \{0\} \simeq Spec\; k[t,t^{-1}] \,.$

It’s called “multiplicative group” because it is the group of multiplicatively invertible elements in $\mathbb{A}^1$ . Just as $\mathbb{C}^\times = \mathbb{C} - \{0\}$ is the mutliplicative group inside $\mathbb{C}$ .

So that’s why global sections of the structure sheaf on it are Laurent polynomials

$\mathcal{O}( \mathbb{G}_m ) = k[t,t^{-1}] \,.$

The rest that David Ben-Zvi hints at is the stuff in the article here.

As you know, I once tried to spell out this perspective on Hochschild (co)homology in the $n$ Lab entry Hochschild cohomology (and, to be frank, tried to fill in some technical details).

The upshot is that for whatever suitably nice behaved notion of “geometric $n$ -functions” $\mathcal{O}$ you have, the HH of $A$ is functions (geometric $n$ -functions) $\mathcal{O}( \mathcal{L} Spec A )$ on the derived free loop space object of $X = Spec A$ .

For the case at hand, $X = \mathbf{B}\mathbb{G}_m$ is the delooping stack of the multiplicative group, so that graded vector spaces are 2-functions (quasicoherent sheaves of modules) on $\mathbf{B}\mathbb{G}_m$ . And the free loop space object of that is

$\mathcal{L} \mathbf{B}\mathbb{G}_m = \mathbb{G}_m \times ( * /\!/ \mathbb{G}_m) \,.$

Posted by: Urs Schreiber on December 20, 2011 7:51 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

The intent was certainly not to throw around big words, but to address your request for a setting that might apply both to things like spectra and parametrized spectra and to graded vector spaces. Of course if one wants to calculate traces or HH of graded vector spaces one doesn’t need any of the general nonsense, you just need to know what class functions on a group are and that representations have traces which are such.

One caveat: I think the grading I’m talking about and the one you’re talking about are different.. a chain complex of graded vector spaces (the kind of thing I’m talking about as are the authors I believe) has homology which is a BIgraded vector space, ie there’s a homological grading and an additional grading. This is the category that relates to representations of $C^*$ or the circle. I would be careful trying to compare this directly with graded vector spaces coming from cohomology of spaces (i.e. in your language evaluating $q$ to $-1$). I think this is a source of some of the confusion in the discussions.

Finally just a comment on parametrized spectra: I was trying to find some category of parametrized spectra whose HH I knew. But one can say something much more naive: given a parametrized spectrum over a space $X$, we get a “parametrized trace” which is a homotopy class of maps from $X$ to the sphere. Before we considered $X$ to be a point, so we got just an element of $\pi_0(S)=Z$, the Euler characteristic, but for more interesting $X$ these will be more interesting invariants.. (same for $X$-parametrized objects of some more general $\infty$-category, we’ll get a homotopy class of maps to $THH$ of that category..)

Posted by: David Ben-Zvi on December 20, 2011 10:28 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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I didn’t mean to imply that you were throwing around big words for its own sake; just to ask you to dumb it down a bit for me. (-: Your comments about bigradings and parametrized traces seem also closely in line with what John said below. Thanks for all your help; I think I understand a bit better now.

Posted by: Mike Shulman on December 22, 2011 10:55 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Mike wrote:

Unfortunately, if you want me to really understand, you’ll have to use smaller words.

Hear hear! Me too. Well, it’s not actually smaller words that I need, it’s more words per sentence whose meaning I can parse without significant effort. As it is, I read a few sentences and collapse from the exertion.

I don’t yet speak algebraic-geometer; to me a “multiplicative group” still means a group whose operation is written $x y$ instead of $x+y$ . (-:

That’s on the right track. In algebraic geometry, the multiplicative group is the group of invertible elements of your favorite field. In this group, the operation is indeed written $x y$ instead of $x+y$ .

Or if you’re fancier, it means the group of invertible elements of your commutative ring… or if you’re even fancier, it’s something like the affine scheme that summarizes all these multiplicative groups in a single package! I guess that’s the affine scheme corresponding to the commutative ring generated by $x$ and $y$ together with a relation saying that $y$ is the inverse of $x$ . Jim Dolan might call it the ‘walking invertible element of a commutative ring’.

But since I’m not fancy at all, and I like the complex numbers, I often interpret ‘the multiplicative group’ as the group $\mathbb{C}^\times$ of nonzero complex numbers.

And then it’s a famous fact that the category of (algebraic or holomorphic) representations of $\mathbb{C}^\times$ on complex vector spaces is equivalent to the category of (continuous or smooth) representations of its maximal compact subgroup, $\mathrm{U}(1)$ , on complex vector space.

And this category is in turn equivalent to the category of $\mathbb{Z}$ -graded vector spaces.

There are lots of fun ways to see this, or at least see why it’s cool: David mentioned Tannaka–Krein duality, but it’s also a famous special case of Pontryagin duality.

Namely: any compact abelian group $A$ has a Pontryagin dual:

$\widehat{A} = hom(A,\mathrm{U}(1))$

such that the category of continuous representations of $A$ on complex vector spaces is equivalent to the category of $\widehat{A}$ -graded complex vector spaces. And

$\widehat{U(1)} = \mathbb{Z}$

As I mentioned, the closed string has $\mathrm{U}(1)$ as symmetries, because a closed string is a circle and you can turn a circle around. So every complex vector space associated to the closed string naturally gets to be a representation of $\mathrm{U}(1)$ , or if you prefer, a $\mathbb{Z}$ -graded vector space.

So, when we ‘stringify’ mathematics, by thinking of ordinary math as being about points and then replacing those points with little circles, a lot of vector spaces become $\mathbb{Z}$ -graded, so their dimensions switch from being integers to being Laurent series: elements of $\mathbf{Z}[[q,q^{-1}]]$ .

But ‘stringification’ can also be seen as a form of ‘categorification’, as Urs pointed out in a paper once: the worldline of a particle is a 1-morphism, but the worldsheet of a string is a 2-morphism. And all the scary stuff David is saying about ‘loop spaces of stacks’ is really somehow about strings moving around in various high-powered spaces (namely stacks).

So, there’s a lot of interesting stuff going on here. Unfortunately I will never fully understand it, because I’m too busy doing other things, and I’ve never managed to persuade anyone to explain this stuff in a way I could easily understand. But it should, ultimately, be rather simple.

Posted by: John Baez on December 20, 2011 3:51 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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As it is, I read a few sentences and collapse from the exertion.

Thanks; you always make me feel better about being publicly ignorant. (-:

Thanks also for the explanation about the multiplicative group.

any compact abelian group $A$ has a Pontryagin dual… such that the category of continuous representations of $A$ on complex vector spaces is equivalent to the category of $\hat{A}$ -graded complex vector spaces.

I’ve encountered Pontryagin duality, and the fact that $\widehat{U(1)}=\mathbb{Z}$ . But I haven’t heard this fact about representations before. What is the functor between those two categories? Is it some sort of eigenspace decomposition?

And, I wonder, is it a monoidal equivalence?

Posted by: Mike Shulman on December 20, 2011 6:44 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Thanks; you [John] always make me feel better about being publicly ignorant. (-:

Mike, many people would aspire to be as ignorant as you.

This “multiplicative group” thing is, to me, an excellent illustration of how it can be simpler to think of schemes as functors $Ring \to Set$ rather than as ringed spaces.

It’s just a matter of personal reaction, but if someone says to me “the functor $Ring \to Set$ sending a ring to its multiplicative group”, nothing could be easier. (At least, once I’m sure I know that by “multiplicative group” they mean “group of units”.)

On the other hand, if someone says to me “ $Spec(\mathbb{Z}[x, x^{-1}])$ ”, I kind of block. I think “hmm, I don’t know much about Laurent polynomials”, and then “oops, what are the prime ideals of $\mathbb{Z}[x, x^{-1}]$ ?”, then “yikes, what’s its Zariski topology?”, and finally, progressing to stronger expletives, “what’s its structure sheaf?”, before wondering whether I should really be doing something else.

Yes, the two things are the same: if you view a scheme as a functor $Ring \to Set$ rather than a ringed space, then $Spec(R)$ just means $Ring(R, -)$ (for a ring $R$ ). So $Spec(\mathbb{Z}[x, x^{-1}]$ just means the functor represented by $\mathbb{Z}[x, x^{-1}]$ . But $\mathbb{Z}[x, x^{-1}]$ is the free ring containing an invertible element, so this functor is the one sending a ring to its set of invertible elements — a.k.a. its multiplicative group.

Still, although the two descriptions are equivalent, one appeals to my intuition a lot more directly than the other.

Posted by: Tom Leinster on December 20, 2011 9:31 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Thanks, Tom. How about this as a third alternative? Let $FPRing$ denote the category of finitely presented rings; then the topos $[FPRing, Set]$ is the classifying topos for rings. Thus in particular it contains a generic ring object, say $R$ . (As a functor $FPRing \to Set$ , $R$ is the underlying-set functor.) The multiplicative group, regarded as a functor restricted to $FPRing$ , is the group of units of $R$ qua ring object in $[FPRing,Set]$ — the generic group of units of a ring.

I just thought of that right now, and I like it, but I’m not sure about that finitely-presented business. What does the topos $[Ring,Set]$ (or $[Ring_\kappa,Set]$ , I suppose, to make it small) classify?

Posted by: Mike Shulman on December 21, 2011 12:11 AM | Permalink | Reply to this

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What does the topos $[Ring, Set]$ (or $[Ring_\kappa, Set]$ , I suppose, to make it small) classify?

I feel the need to restate the question in more elementary terms:

What are the finite limit preserving functors $Ring^op \to Set$ ?

For that matter, what are the finite limit preserving functors $Set^op \to Set$ ? Do I know this? Is this part of the Boolean algebra/Stone-Čech compactification story?

Posted by: Tom Leinster on December 21, 2011 1:06 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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Do I know this?

I don’t think I know it.

Posted by: Mike Shulman on December 21, 2011 7:47 AM | Permalink | Reply to this

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Tom wrote:

It’s just a matter of personal reaction, but if someone says to me “the functor $Ring \to Set$ sending a ring to its multiplicative group”, nothing could be easier. (At least, once I’m sure I know that by “multiplicative group” they mean “group of units”.)

On the other hand, if someone says to me “ $Spec(\mathbb{Z}[x, x^{-1}])$ ”, I kind of block. I think “hmm, I don’t know much about Laurent polynomials”, and then “oops, what are the prime ideals of $\mathbb{Z}[x, x^{-1}]$ ?”, then “yikes, what’s its Zariski topology?”, and finally, progressing to stronger expletives, “what’s its structure sheaf?”, before wondering whether I should really be doing something else.

… and then you start thinking better thoughts. Since I never studied algebraic geometry the usual way, these preliminary bad thoughts don’t infest my mind. Here’s what I think. It amounts to the same thing you think, with different words:

If someone says “give me the free commutative ring on an element” I say “ $\mathbb{Z}[x]$ ”. A homomorphism from this to any commutative ring $R$ picks out an element of $R$ , namely the image of $x$ .

But if someone says “give me the free commutative ring on an invertible element”, I know I have to give $x$ an inverse. So I say “ $\mathbb{Z}[x, x^{-1}]$ ”. A homomorphism from this to $R$ picks out an invertible element of $R$ .

And $Spec$ , to me, is just a name for “turn around the arrows”. It turns commutative rings into things called affine schemes, which are simply objects of the opposite category. So a map

$Spec(R) \to Spec(\mathbb{Z}[x, x^{-1}])$

is just a homomorphism

$\mathbb{Z}[x, x^{-1}] \to R$

which is an invertible element of $R$ .

So when I see $Spec(\mathbb{Z}[x, x^{-1}])$ , I think “the walking invertible element”, with the proviso that $Spec$ turns around arrows and makes it “walk backwards”. So if someone says “what’s the multiplicative group?”, I look for a group object in affine schemes that acts like “the walking invertible element”, and I say “ $Spec(\mathbb{Z}[x, x^{-1}])$ ”.

Posted by: John Baez on December 21, 2011 1:42 AM | Permalink | Reply to this

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Right — we’re in total agreement here, and saying the same thing in very nearly the same way.

The viewpoint of schemes as (special) functors $Ring \to Set$ makes this kind of thing very easy. The viewpoint of schemes as (special) ringed spaces makes it harder, at least for me.

That’s not to say there’s no occasion when viewing schemes as ringed spaces is useful. I think there was a thread about exactly that on a blog a while ago — maybe the Secret Blogging Seminar?

Posted by: Tom Leinster on December 21, 2011 1:53 AM | Permalink | Reply to this

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Tom wrote:

The viewpoint of schemes as (special) functors $Ring \to Set$ makes this kind of thing very easy.

Right, I like that. But you’ll notice that in my little description I was deliberately taking an even less strenuous viewpoint: I was only working with affine schemes. For those you can just think ‘commutative rings, but write all the arrows backwards’.

Affine algebraic geometry is a kind of cozy little harbor that’s good to practice sailing your sailboat around in before exploring deeper waters.

Posted by: John Baez on December 21, 2011 2:09 AM | Permalink | Reply to this

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I was only working with affine schemes. For those you can just think ‘commutative rings, but write all the arrows backwards’.

I was thinking that this was actually a substantive difference between our comments, and that I have some reservations about the point of view you’re presenting. I’m certain you understand all of what I’m about to say perfectly well, but since we’re talking here about the helpfulness of different points of view rather than black-and-white mathematics, I’ll say it anyway.

We start from the fact that the category $Aff$ of affine schemes is dual to the category $Ring$ of commutative rings. For the purposes of this comment, let’s define $Aff = Ring^{op}$ (which is what you were doing too). Let’s also regard the category $Sch$ of all schemes as a full subcategory of $[Ring, Set]$ . By the Yoneda lemma, $Aff$ is equivalent to the full subcategory of $Sch$ consisting of the representables. The scheme represented by a ring $R$ is traditionally called $Spec(R)$ . And of course, $Spec$ is contravariant.

However, I’m uneasy with saying that $Spec$ is

just a name for “turn around the arrows”.

I’m uneasy because colimits in $Sch$ are different from limits in $Ring$ . Take a diagram of affine schemes — let’s say one of the form

$\begin{matrix} \cdot &\to &\cdot\\ \downarrow& & \\ \cdot& & \\ \end{matrix}$

for the sake of argument. Since $Aff = Ring^{op}$ , and $Ring$ has all limits, $Aff$ has all colimits. So you can take its pushout in $Aff$ (which is a pullback in $Ring$ ). But on the other hand, you could regard $Aff$ as embedded in the larger category $Sch$ of schemes. And the pushout in $Sch$ is different from the pushout in $Aff$ , in general. For example, consider gluing together two affine lines to get a projective line.

Categorically, this phenomenon is to do with the total failure of the Yoneda embedding to preserve colimits, but geometrically it’s something central to the idea of scheme: that gluing together affine schemes can give you a non-affine scheme. Indeed, gluing together affine schemes can give you any scheme you desire.

So I view $Spec$ as more than just turning the arrows round. I view it as turning the arrows round and moving to a larger geometric context.

To some extent this is simply a matter of notation. We have functors

$Ring^{op} \stackrel{\simeq}{\to} Aff \hookrightarrow Sch$

(where the first one might be the identity, according to taste). It seems that you want to use ‘ $Spec$ ’ as a name for the first functor, and I want to use it as a name for the composite functor. I would have thought that the latter usage was more common. And it does matter somewhat, because that embedding $Aff \hookrightarrow Sch$ doesn’t preserve colimits.

Posted by: Tom Leinster on December 21, 2011 10:40 AM | Permalink | Reply to this

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Tom wrote:

However, I’m uneasy with saying that $Spec$ is

just a name for “turn around the arrows”.

I’m uneasy because colimits in $Sch$ are different from limits in $Ring$ .

Right. When I wrote that I was thinking of $Spec$ as a very tautologous sort of contravariant functor

$Spec: CommRing \to AffSch^{op}$

but probably nobody ever does that except me; probably everyone always uses

$Spec: CommRing \to Sch^{op}$

and the point is then that colimits in $Sch$ are more geometrically nice than in $AffSch$ .

Posted by: John Baez on December 21, 2011 11:17 AM | Permalink | Reply to this

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I have an unrelated and probably stupid question. Does a homomorphism $\mathbb{Z}[x,x^{-1}] \to R$ pick out an invertible element, or just an invertible element up to inverses?

The problem is that there’s no way to distinguish $x$ and $x^{-1}$ . This is the same as the “indistinguishability” of $i$ and $-i$ in $\mathbb{C}$ . Of course if you have $r$ and $r^{-1}$ in $R$ you can tell they’re different, you just can’t tell which is which.

This is something that’s puzzled me since undergraduate days.

Posted by: Tom Ellis on December 21, 2011 7:33 AM | Permalink | Reply to this

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Tom wrote:

I have an unrelated and probably stupid question. Does a homomorphism $\mathbb{Z}[x,x^{-1}] \to R$ pick out an invertible element, or just an invertible element up to inverses?

It picks out a specific invertible element, namely the image of $x$ .

Of course it also picks out a bunch of other specific invertible elements, namely the images of all the integer powers of $x$ . But that does not lessen the truth of my first sentence.

The problem is that there’s no way to distinguish $x$ and $x^{-1}$ .

You’re not going to like this, but actually there is. One looks like this:

$x$

while the other looks like this:

$x^{-1}$

These are purely syntactic entities (symbols on the page), and they’re very easy to distinguish.

I think what you should say, not that they’re ‘indistinguishable’, but that there’s a symmetry that interchanges them. That’s true: the ring $\mathbb{Z}[x,x^{-1}]$ has precisely two automorphisms, the identity and the homomorphism that switches $x$ and $x^{-1}$ .

I think you’re suffering from some blurring of the concepts of equality and isomorphism.

Posted by: John Baez on December 21, 2011 8:00 AM | Permalink | Reply to this

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It picks out a specific invertible element, namely the image of $x$ .

Ah I see. $\mathbb{Z}[x, x^{-1}]$ is a particular object which happens to be an abelian group. I suppose I thought that notation meant “the isomorphim class of the free abelian group on one generator”. For some reason I have got into the habit of subconsciously identifying everything that’s isomorphic.

Posted by: Tom Ellis on December 21, 2011 2:58 PM | Permalink | Reply to this

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Tom wrote:

$\mathbb{Z}[x, x^{-1}]$ is a particular object which happens to be an abelian group.

It’s a particular object, which happens to be a commutative ring.

I suppose I thought that notation meant “the isomorphism class of the free abelian group on one generator”.

No, that notation means “the free commutative ring on one generator $x$ which has a multiplicative inverse $x^{-1}$ ”. This commutative ring is called the ring of Laurent polynomials in the variable $x$ . Here’s a typical element:

$6x^7 + 4x^5 - 3x^2 - 42 + 24x^{-3} - 248 x^{-49}$

This ring is related in a certain interesting way to the free abelian group on one generator: you can think of the monomials $x^n$ as elements of the free abelian group on one generator $x$ . But it’s different.

For some reason I have got into the habit of subconsciously identifying everything that’s isomorphic.

Posted by: John Baez on December 22, 2011 9:28 AM | Permalink | Reply to this

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I think this is a somewhat subtle point, though. It’s related to the question of whether “the symmetric group on three letters” is well-defined up to unique isomorphism, or only up to conjugacy. The answer is that when we say “ $G$ is the symmetric group on three letters” we mean that it comes with the structure of a particular set of three letters and an action of $G$ as the symmetries of that set. This structure is then well-defined up to unique isomorphism.

Likewise, when we say “the free group on one generator” we mean a particular generator. A sentence like “ $\mathbb{Z}$ is the free group on one generator” is strictly speaking incomplete; we should say that $\mathbb{Z}$ is the free group on the single generator $1$ . Of course, $\mathbb{Z}$ is also the free group on the single generator $-1$ . Since the free group on one generator is unique up to unique isomorphism, these two incarnations of it are isomorphic—by the isomorphism $n\mapsto -n$ .

Finally, “the free ring on an invertible element” means a particular invertible element. $\mathbb{Z}[x,x^{-1}]$ is the free ring on the single invertible element $x$ … and also the free ring on the single invertible element $x^{-1}$ .

In more abstract language, we’re talking about representations of the functor $R\mapsto R^\times$ which takes a ring to its group of units. But remember that a representation of a functor $F$ consists of an object $Z$ together with an isomorphism $F \cong Hom(Z,-)$ . Composing such an isomorphism with an automorphism of $Z$ will give you a different representation of the same functor (albeit one with the same underlying object $Z$ ). Bringing that back down to earth, we could say “a homomorphism $\mathbb{Z}[x,x^{-1}]\to R$ picks out an invertible element (the image of $x$ )”, or we could say “a homomorphism $\mathbb{Z}[x,x^{-1}]\to R$ picks out an invertible element (the image of $x^{-1}$ )”, and we would have two different true statements.

I think it’s easy to get confused because we often use langage like “ $Z$ represents the functor $F$ ” as if it were a property of the object $Z$ , whereas actually a representation is a structure that we’re implicitly giving it.

Posted by: Mike Shulman on December 21, 2011 9:06 AM | Permalink | PGP Sig | Reply to this

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nothing could be easier.

By the way, since elsewhere I see the word “scary” thrown around so much:

it is precisely this easiness that, one categorical level up, makes it feel good and easy to speak of the stack $\mathbf{B}\mathbb{G}_m$ instead of a “locally ringed higher space/topos” (which one could do as well).

Because this means to think of $\mathbf{B}\mathbb{G}_m$ as the assignment $Ring \to Groupoid$ that sends every ring $R$ to the groupoid with a single object and the multiplicative group $R^\times$ as its morphisms.

Again, nothing could be easier. Don’t let anyone scare you of this concept ;-)

Posted by: Urs Schreiber on December 20, 2011 10:04 PM | Permalink | Reply to this

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I did realize when I was writing my comment that I was sweeping something small under the carpet. Sending a ring to its multiplicative group isn’t literally speaking a functor $Ring \to Set$ ; one really ought to say “underlying set of the multiplicative group”. If you stick with the group itself, you get a functor $Ring \to Group$ (or $Ring \to Groupoid$ , if you like), giving you a group scheme or groupoid scheme. But that was peripheral to the main point I wanted to make, as well as kind of obvious, so I didn’t say it.

Posted by: Tom Leinster on December 20, 2011 10:37 PM | Permalink | Reply to this

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If you stick with the group itself, you get a functor $Ring \to Group$ (or $Ring \to Groupoid$ , if you like),

No, I don’t like that. The first is $\mathbb{G}_m$ . The second is $\mathbf{B}\mathbb{G}_m$ . By no means to be conflated.

Posted by: Urs Schreiber on December 20, 2011 10:58 PM | Permalink | Reply to this

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I presume you mean that the functor $Ring\to Group$ is $\mathbb{G}_m$ (with its group structure) while the functor $Ring\to Groupoid$ is $\mathbf{B}\mathbb{G}_m$ ?

Posted by: Mike Shulman on December 20, 2011 11:57 PM | Permalink | Reply to this

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Whoa, I don’t think I’m “conflating” anything. “If you like” just means that you have the choice.

Posted by: Tom Leinster on December 20, 2011 11:58 PM | Permalink | Reply to this

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Urs wrote:

By the way, since elsewhere I see the word “scary” thrown around so much…

By the way, whenever I say something is “scary” I’m not trying to make people scared of something; quite the opposite. I’m guessing that certain people are scared of it, and by bringing this fear out in the open, I’m trying to start dealing with it. These certain people might include me… or they might not. Often include an earlier version of me: an earlier self whom I pity and want to help.

I find that a little sympathy helps when explaining concepts. If I act like nothing is ever scary, the people who are scared will feel inferior and… leave. But if I admit to my negative emotions, I’ll look human and they’ll think “he’s human, whatever he’s doing can’t be all that hard”.

Also, as soon as someone says something in mathematics is “scary”, it becomes a bit of a joke—it won’t bite you, after all! And so, it automatically starts becoming less scary.

I once wanted to write a book called Scary Concepts in Mathematics, which would explain all sorts of concepts like groupoids and categories and sheaves and schemes and stacks and so on: concepts that various people have claimed are ‘too abstract’ and thus ‘scary’. I think such a book could be a lot of fun—especially when people saw you reading it!

Posted by: John Baez on December 21, 2011 1:52 AM | Permalink | Reply to this

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Mike wrote:

Thanks; you always make me feel better about being publicly ignorant. (-:

Thanks. Maybe someday I’ll be elected president of APIM: the Association of Publicly Ignorant Mathematicians. That’ll look great on my resumé.

[…] I haven’t heard this fact about representations before. What is the functor between those two categories? Is it some sort of eigenspace decomposition?

And, I wonder, is it a monoidal equivalence?

You answered your first question, and the answers to the other questions are both ‘yes’.

But I’ll go over it here just so everyone can see how easy it is to see the category of continuous representations of $\mathrm{U}(1)$ on finite-dimensional complex vector spaces is equivalent to the category of $\mathbb{Z}$ -graded finite-dimensional complex vector spaces.

I inserted ‘finite-dimensionality’ in the list of assumptions because it makes the topology on the complex vector space obvious, and we can use a standard fact: every continuous rep of a compact group on a finite-dimensional complex vector space is a direct sum of irreducible representations. I’ll stop saying ‘continuous’, ‘finite-dimensional’ and ‘complex’—they’ll be implicit from now on.

What are the irreducible representations of $\mathrm{U}(1)$ ? By Schur’s Lemma, in such a representation each element $q \in \mathrm{U}(1)$ must act as a scalar, say $\rho(q)$ . Since we need

$\rho(qq') = \rho(q) \rho(q')$

$\rho$ must be a (continuous) group homomorphism from $\mathrm{U}(1)$ to the invertible complex numbers. The only options — checking this is a fun little puzzle — are the obvious ones:

$\rho(q) = q^n$

for $n \in \mathbb{Z}$ .

So, there’s one irreducible representation of $\mathrm{U}(1)$ for each $n \in \mathbb{Z}$ , and the process of breaking a representation into its irreducible pieces is precisely what turns it into a $\mathrm{Z}$ -graded vector space.

Conversely, it’s obvious how to turn a $\mathrm{Z}$ -graded vector space into a representation of $\mathrm{U}(1)$ : let $q \in \mathrm{U}(1)$ act as multiplication by $q^n$ on the $n$ th grade.

And it’s easy to see that tensoring the representation

$\rho(q) = q^n$

with the representation

$\rho'(q) = (q)^{n'}$

gives

$(\rho \otimes \rho')(q) = q^{n+n'}$

so the equivalence of categories is a monoidal equivalence.

In physics, $\mathrm{U}(1)$ is the symmetry group whose corresponding conserved quantity is electric charge. Every elementary particle is, among other things, an irreducible representation of $\mathrm{U}(1)$ , and the number $n$ is called its charge. When you stick a bunch of particles together you tensor them, and the above equation says that the electric charges of the pieces add to give the electric charge of the whole thing! So this monoidal equivalence of categories is fundamental in the study of electromagnetism.

Posted by: John Baez on December 21, 2011 1:19 AM | Permalink | Reply to this

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Maybe someday I’ll be elected president of APIM: the Association of Publicly Ignorant Mathematicians.

A natural progression from you current appointment as the Charles Sanders Peirce Professor of Miserable Ignorance.

Posted by: David Corfield on December 21, 2011 10:26 AM | Permalink | Reply to this

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Mike wrote:

What I would really like is a symmetric monoidal category in which these “q-dimensions” are traces of identity maps of dualizable objects.

Of course in categories of representations of quantum groups, the dualizable objects have ‘dimensions’, defined as traces of their identity maps, that are Laurent polynomials in a formal variable $q$ . This is the context where the term ‘quantum dimension’ first showed up. But these are braided monoidal categories.

The goal of Khovanov and his colleagues is to ‘categorify’ these Laurent polynomials by reinterpreting them as actual things. That is, actual objects in a 2-rig whose decategorification is the rig of Laurent polynomials. And an obvious choice of such a 2-rig is the category of finite-dimensional $\mathbb{Z}$ -graded vector spaces, so that’s what they use. Then they proceed to try to categorify the whole theory of quantum groups, along with all its applications to knot theory.

Their recipe for categorifying Laurent polynomials could be looked on unfavorably as a ‘trick’, but since it’s giving lots of new results in knot theory it’s certainly not a ‘mere trick’. It seems to be something good that we need to understand better.

It’s worth noting that in string theory the group $S^1$ tends to act on everything, since you can turn a string around. The category of finite-dimensional complex vector spaces equipped with a continuous representation of $S^1$ is equivalent to the category of finite-dimensional $\mathbb{Z}$ -graded complex vector spaces. So that’s one reason why this category shows up, and Laurent polynomials show up.

Anyway, maybe you know all this, but I thought some people would enjoy hearing the context surrounding this ‘ $q$ -dimension’ business. Your goal of finding a symmetric monoidal category in which the dualizable objects have dimensions that are Laurent polynomials seems new to me.

Posted by: John Baez on December 18, 2011 12:08 AM | Permalink | Reply to this

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Their recipe for categorifying Laurent polynomials could be looked on unfavorably as a ‘trick’, but since it’s giving lots of new results in knot theory it’s certainly not a ‘mere trick’. It seems to be something good that we need to understand better.

This seems to be the case for many things that go under the name “categorification” in knot theory and related subjects. In category theory, we generally categorify things conceptually, by thinking about what the corresponding notion would be at higher category levels. But they just seem to say, how can we cook up some category whose decategorification is the thing we started with? And somehow, out comes interesting stuff! It’s amazing. (-: And it gives me the same feeling that “there must be something going on in the background here that we don’t understand yet, explaining why this works so well.”

In any case, I don’t know hardly anything about quantum groups — are you telling me that there’s a way to regard any graded module as a representation of some quantum group, in such a way that you recover its $q$ -dimension as the trace of its identity map? That would make me pretty happy; I’d be willing to take braiding if that’s what’s required for it to work (and maybe I’d go learn some quantum group theory). You mention:

The category of finite-dimensional complex vector spaces equipped with a continuous representation of $S^1$ is equivalent to the category of finite-dimensional ℤ-graded complex vector spaces.

But $S^1$ isn’t a quantum group… or is it? Can the $q$ -dimension of a graded vector space be computed via this equivalence?

Ben Cooper’s comment suggests that it is still a big deal to find one thing in the world of quantum groups whose Euler characteristic is something we expect to find in the world of graded modules.

Posted by: Mike Shulman on December 19, 2011 7:04 PM | Permalink | Reply to this

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Mike wrote:

In category theory, we generally categorify things conceptually, by thinking about what the corresponding notion would be at higher category levels. But they just seem to say, how can we cook up some category whose decategorification is the thing we started with? And somehow, out comes interesting stuff! It’s amazing. (-:

Well, you guys are trying to categorify frameworks, while those guys are trying to categorify interesting examples of things that live in frameworks.

And the problem is that if all you have is an example of thing that lives in some framework, the framework isn’t uniquely determined. So, it’s hard to know which categorified framework the categorified example will live in. So, you just have to try stuff and see what happens. Usually nothing interesting, but often enough something amazing.

But this is typical of physics: you do experiments and see what happens. As Arnol’d said, mathematics is just the branch of physics where experiments are very cheap.

In any case, I don’t know hardly anything about quantum groups — are you telling me that there’s a way to regard any graded module as a representation of some quantum group, in such a way that you recover its $q$ -dimension as the trace of its identity map?

No, I’m not saying that.

For now, we can say that a quantum group is a certain kind of Hopf algebra over the commutative ring $\mathbb{Z}[q,q^{-1}]$ , equipped with enough structure so that its category of representations is braided. There’s one quantum group for each simple Lie algebra. When you set $q = 1$ , to make the Laurent polynomials reduce to natural numbers, everything about quantum groups reduces to known stuff about simple Lie algebras.

In particular, the representation of a quantum group will have a ‘dimension’, the trace of its identity map, which will be an element of $\mathbb{N}[q,q^{-1}]$ . But when you set $q = 1$ , this element evaluates to a natural number, which is the dimension of the corresponding representation of the simple Lie algebra.

But of course elements of $\mathbb{Z}[q,q^{-1}]$ show up throughout the theory of quantum groups, because they are Hopf algebras over this ring.

What Khovanov and company are trying to do is categorify everything I just told you. So in particular, wherever they see an element of $\mathbb{Z}[q,q^{-1}]$ , they try to replace it with a $\mathbb{Z}$ -graded chain complex whose ‘ $q$ -dimension’, or better ‘ $q$ -Euler characteristic’, is that element.

So you see that ‘dimensions’ are entering in two different ways here. Maybe that’s part of how I managed to confuse you.

But $S^1$ isn’t a quantum group…or is it?

It’s just an ordinary group, but I see why you’re asking that: yes, it too has the funny feature that its representations can be assigned ‘dimensions’ that are elements of $\mathbb{N}[q,q^{-1}]$ .

If you ask me what the deep significance of this fact is, I’ll have to make something up. It might be fun.

Can the $q$ -dimension of a graded vector space be computed via this equivalence?

Yes! Suppose you take $q$ to be a complex number with $|q| = 1$ , so $q \in \mathrm{U}(1) \cong S^1$ . It acts as an operator on any representation of $S^1$ , so we can try to take the trace of that operator.

If our representation is finite-dimensional this trace will be well-defined. It’ll depend on $q$ , of course. It will be a Laurent polynomial in $q$ — and that’s just the $q$ -dimension we’re talking about!

If our representation is infinite-dimensional taking the trace is more iffy, but sometimes we’ll get a Laurent series in $q$ .

Posted by: John Baez on December 20, 2011 4:32 AM | Permalink | Reply to this

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So you see that ‘dimensions’ are entering in two different ways here. Maybe that’s part of how I managed to confuse you.

Ah, yes. I was looking for a context in which the $q$ -dimension of a graded vector space was the trace of its identity map. (Maybe I didn’t say that fully explicitly.) So when you said “categories of representations of quantum groups, the dualizable objects have ‘dimensions’, defined as traces of their identity maps, that are Laurent polynomials in a formal variable $q$ ” I assumed you were presenting that as an answer to what I was looking for.

If you ask me what the deep significance of this fact is, I’ll have to make something up. It might be fun.

What’s the deep significance of that fact?

Seriously, I’m starting to wonder whether the answer to my question might lie in that direction. For instance:

Suppose you take $q$ to be a complex number with ${|q|}=1$ , so $q\in U(1)\cong S^1$ . It acts as an operator on any representation of $S^1$ , so we can try to take the trace of that operator.

This reminds me of bicategorical traces. If $R$ and $S$ are not-necessarily-commutative rings and $M$ is an $R$ - $S$ -bimodule, which is dualizable as a right $S$ -module, then any endomorphism of $M$ has a trace in the bicategory of rings and bimodules. If $R$ and $S$ are commutative, then this trace is an abelian group map $R\to S$ which is computed as follows: given $r\in R$ , consider the endomorphism $x\mapsto r\cdot f(x)$ of $M$ , and take its trace regarding $M$ as a dualizable $S$ -module. (If $R$ and $S$ are not commutative, then the trace is a map $HH(R)\to HH(S)$ instead.)

In particular, such an $M$ has a “bicategorical Euler characteristic” — the trace of its identity map — which is the map $R\to S$ sending each $r\in R$ to the trace of $x\mapsto r\cdot x$ on $M$ as an $S$ -module. That sounds kind of like what’s going on here, with $S^1$ -representations regarded as bimodules from $S^1$ to $\mathbb{C}$ .

It will be a Laurent polynomial in $q$ — and that’s just the $q$ -dimension we’re talking about!

I guess that tells me the answer to “what is the functor between these two categories”. We have to take a graded vector space $(V_n)$ to $\bigoplus_n V_n$ , with $z\in S^1$ acting by $x\mapsto z^n x$ on each piece $V_n$ , right?

So now I have another question. If $X$ is a topological space (or a spectrum), then it has an associated graded vector space, namely its homology. That graded vector space then has an associated $S^1$ -representation. Does that composite functor have any nice direct description? Does the equivalence of graded vector spaces with $S^1$ -representations have any topological analogue?

Posted by: Mike Shulman on December 20, 2011 7:17 PM | Permalink | PGP Sig | Reply to this

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John wrote approximately:

$S^1$ is just an ordinary group, but I see why you’re asking that: yes, it too, like any quantum group, has the funny feature that its representations can be assigned ‘dimensions’ that are elements of $\mathbb{N}[q,q^{-1}]$ .

Mike wrote:

What’s the deep significance of that fact?

Right now I’m thinking it has something to do with how quantum groups are a ‘stringification’ of ordinary groups: the category of representations of the quantum group associated to a simple Lie group $G$ can also be seen as the category of (certain) representations of the universal central extension of loop group $L G = Maps(S^1, G)$ . So there’s a kind of ‘hidden $S^1$ action’ lurking in the theory of quantum groups.

The stuff about Hochschild homology and free loop spaces, which some folks here have also been mentioning, should fit into this somehow.

But that’s all I can say right now. Probably some people understand all this a lot better than I do; the problem will be getting them to explain it in a way you can understand.

(Even if we had an infinitely knowledgeable mathematician in our midst, we’d still be stuck on most of the problems we face, because we couldn’t understand most of their answers to our questions!)

If X is a topological space (or a spectrum), then it has an associated graded vector space, namely its homology. That graded vector space then has an associated $S^1$ -representation. Does that composite functor have any nice direct description?

I don’t know—good puzzle! But you’re reminding me of a related thing, which I mentioned in “week182”:

The main thing to remember is that when your manifold is complex, the cohomology becomes "bigraded": instead of just

H^{^p}(M)

you get

H^{^i,j}(M).

So now, what’s nonabelian Hodge theory?

The basic idea is simple: instead of askng what extra structure the homology groups get when M is a complex manifold, we ask what extra structure the homotopy type of M gets when M is a complex manifold. The homotopy type includes invariants like the homotopy groups, but also more. How are these constrained by the fact that M is complex?

Unfortunately, to describe the answer - even a little teeny part of the answer - I need to turn up the math level a notch.

For starters we can consider the fundamental group π₁(M). But this is hard to relate to differential geometry, so we will immediately water it down by picking an algebraic group G and looking at homomorphisms of π₁(M) into G. These are basically the same thing as flat G-bundles over M, so it’s easier to see how M being a complex manifold affects things. We can even be sneaky and study this for all G at once by forming a group Π₁(M) called the "proalgebraic completion" of π₁(M). This is a proalgebraic group - an inverse limit of algebraic groups - which contains π₁(M) and has the property that any homomorphism from π₁(M) into an algebraic group G extends uniquely to a proalgebraic group homomorphism from Π₁(M) to G.

It’s nice to ask what extra structure Π₁(M) gets when M is a complex manifold, because this question has a nice answer.

To get ready for how nice the answer is, first go back to plain old abelian Hodge theory. Note that making the cohomology of M bigraded gives an obvious way for the algebraic group C*, the nonzero complex numbers, to act on the cohomology. The reason is that for each integer there’s a representation of C* where the number z acts as multiplication by zⁿ, so gradings are just another way of talking about C* actions. Since the cohomology of M is automatically graded, putting another grading on it amounts to letting C* act on it.

So in plain old Hodge theory, the answer to "What extra structure does the cohomology of M get when M is complex?" is:

"It gets an action of C*!"

And it turns out that in nonabelian Hodge theory, the answer to "What extra structure does Π₁(M) get when M is complex?" is:

"It gets an action of C*!"

This is incredibly cool, but the story goes a lot further. The fundamental group is just the beginning; you can do something similar for the higher homotopy groups - but it’s a lot more subtle. In fact, you can do something similar directly to the homotopy type of M! When M is a compact complex manifold, there’s a homotopy type called the "schematization of M" whose fundamental group is Π₁(M) - and there’s an action of C* on this homotopy type!

By the way, when M is a compact Kaehler manifold the action of C* on its cohomology extends to a natural action of SL(2,C), as explained in Wells’ book. I wonder if SL(2,C) acts on the schematization of M?

I’m still curious about my question here. But you’re going back to the more basic question of whether there’s a C* action or $S^1$ action lurking around somewhere that makes the cohomology of every space graded in the first place!

Posted by: John Baez on December 21, 2011 3:09 AM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

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That’s all very interesting, thanks. Lots to think about. I feel like maybe I’m one step closer to understanding $q$ -Euler characteristics.

Posted by: Mike Shulman on December 22, 2011 10:53 PM | Permalink | Reply to this

Re: Categorifying Fractional Euler Characteristics

Interested readers should see my ongoing work with Krushkal (link). Actually, this blog post resembles the introduction to that paper!

Concerning Michael Shulman’s comments, one beautiful observation in the Frenkel-Stroppel-Sussan paper is that modules with the correct graded dimension for [n], essentially H^*(CP^n), appear naturally within the Bernstein-Frenkel-Khovanov category in which they are working. The chain complex for 1/[n] can obtained by constructing a resolution which starts with the augmentation map. Once this is done then it is possible to show that the usual definitions correspond to the ones being considered here.

Posted by: Ben Cooper on December 19, 2011 4:24 PM | Permalink | Reply to this

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December 16, 2011