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Quantum Topology and Classifying Manifolds October 11, 2007

Posted by Noah Snyder in low-dimensional topology, quantum algebra, topology, tqft.
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In this post I want to explain an old idea of (our frequent commenter) Greg Kuperberg on classifying low-dimensional manifolds using quantum algebra. In particular I want to discuss a 2-dimensional analogue of it that I’ve thought about a lot, but have recently mostly given up on.

Here’s the essence of Kuperberg’s idea:

  • To every Heegaard splitting of a 3-manifold there is a corresponding expression in the structure constants of a Hopf algebra.
  • The allowable moves of Heegaard splittings correspond to identities which follow from the axioms of an involutory (S^2=1) Hopf algebra.
  • Conversely, if you have an identity that follows from the axioms of an involutory Hopf algebra, then you can do the corresponding move on the Heegaard splitting.

Kupperberg then conjectures that if an identity holds for any finite dimensional involutory Hopf algebra, than it must follow from the axioms. This would mean that his invariants distinguish 3-manifolds!

Now let me turn my attention to the 2-dimensional case, which is conceptually the same, but the details are much clearer. For the rest of the paper I’m going to assume that you understand the basics of lattice 2-d tqfts. The best way to understand these is to read John Baez’s lecture notes from the quantum gravity seminar in Winter 2001, track 1, weeks 16 and 17. For a little more detail and some references see my recent paperlet, and for a completely thorough discussion see Lauda-Pfeiffer.

The basic idea of lattice 2-d tqfts is that to every triangulated oriented closed surface you assign an expression in the structure constants of a symmetric (but not necessarily commutative) suitably normalized Frobenius algebra. The standard moves for relating different triangulations translate directly over to identities in symmetric normalized Frobenius algebras. Just as in Kuperberg’s case, the converse is also true. Namely an algebraic identity follows from the axioms if and only if the corresponding topological move is allowed.

In this case, however, because of the classification of surfaces we know that the analogue of Kuperberg’s conjecture is true! Namely the only invariant of oriented surfaces is Euler characteristic, and there are finite dimensional Frobenius algebras (namely matrix algebras) which distinguish them. However, this proof is unsatsifactory, what we would like to do is turn it on its head and use quantum topology to prove the classification of surfaces! (Yes, I know, this is the sort of thing that only I care about.) But now we’re stuck in the same bind Kuperberg was stuck in, how do we know that any identity that holds for finite dimensional algebras follows from the axioms?

Now it turns out this is difficult, but I think I at least understand why it is difficult now. Suppose we had a similar question about, say, groups. We would just take the free group on countably many generators (where the only relations are those coming from the axioms) and show that this group is residually finite.

Why won’t this idea work for us? Because algebraic structures studied in quantum topology typically don’t have a notion of free algebra. What is a free involutory Hopf algebra? What is a free symmetric normalized Frobenius algebra? They don’t behave well under infinite products. (Also see this comment on our blog from Simon Willerton.)

The way I want to think about Kuperberg’s idea is as follows. There is a universal quantum invariant which distinguishes all manifolds. This universal invariant comes from a free construction. Since the relevant free algebra is residually finite, the actual known invariants distinguish everything. To some extent this is still ok, however the universal invariant doesn’t come from an actual free Hopf algebra, it comes from a Hopf algebra object in some diagram category, and hence is much much more difficult to deal with.

If any of you have any ideas about how to complete this idea in 2-dimensions please contact me, as I think this is a very beautiful idea, but I just have been stuck too long to keep thining about a new way to prove such an old result.

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Comments

1. Charlie Stromeyer Jr - October 11, 2007

Since I don’t know much about 2-d tqfts, I would like to ask the question: Doesn’t a Frobenius algebra have to be commutative to determine uniquely (up to isomorphism) a 2-d tqft?

At least this is what I seem to deduce from looking at the paper “Homotopy algebras and noncommutative geometry” by A. Hamilton and A. Lazarev which introduces an oo_generalization of a Frobenius algebra, available in the arXiv as math/0410621, or am I missing something here?

I will try to think about the rest of what you say.

2. Noah Snyder - October 11, 2007

Charlie, excellent question and a source of constant confusion.

There are two ways of building up 2-d TQFT from frobenius algebras. In one of them the building block is a pair of pants, in the other the building block is a triangle. Any closed 2-manifold can be built by gluing together pants and cups and caps, and any closed 2-manifold can be built by gluing together triangles. So, both of these approaches should give 2-manifold invariants. A priori the pants approach is more general, but the triangles approach is more local.

The pants approach takes a commutative frobenius algebra as its input. The triangles approach takes a symmetric normalized frobenius algebra as its input. The pants approach classifies all TQFT, whereas the triangles approach is closer in spirit to extended TQFTs, and in particular doesn’t let you see all TQFTs.

3. Noah Snyder - October 11, 2007

Oh, by the way, the reason I’m using the triangles approach here is that you know automatically that these invariants make sense for an arbitrary PL-surface. However, to know that an arbitrary surface has a pants decomposition you’re already most of the way to the usual proof of the classification of surfaces, so why bother.

4. Charlie Stromeyer Jr - October 12, 2007

Thanks for the clarification, Noah. I was not able to even come close to imagining such a thing as a free Hopf algebra whose antipode is an involution.

For anyone who might be interested, in the case of a commutative Frobenius algebra, M. Hyland introduces the concept of a free symmetric monoidal category generated by a Frobenius algebra. See Section 2.2 on page 12 of “Computer science logic: 18th international workshop, CSL 2004″.

5. davidspeyer - October 12, 2007

I’m ignorant but willing to think. Why isn’t there a free Hopf algebra on a set X, over a field k? Take all k-linear combinations of grammatical expressions involving elements of X, multiplication, comulitplication and S, and mod out by the equivalence relations generated by the axioms of a Hopf algebra. (Oh, you should also include a symbol for the identity.) I think this should define a Hopf algebra.

I assume what you mean is that you aren’t as good at thinking about free Hopf algebras as you are at thinking about free algebras. So let’s start with bialgebras for practice and see if we can come up with a basis for the free bialgera on an n-element set (x_1, …, x_n). A spanning set is “computation networks” which are directed graphs with one sink, some number of sources, each labelled by one of x_1, …, x_N, and in which every other vertex either has outdegree one or indegree one. (Corresponding to multiplying n-elements, or comultiplying one element up to A^n.) The in edges (resp. out edges) of a multiplication (resp. comultiplication) vertex are ordered. There is now only one “basic move”, corresponding to the axiom that comulatiplication is a map of algebras. Is there no canonical form theorem for these guys?

6. Todd Trimble - October 12, 2007

David, one sense in which free bialgebras or free Hopf algebras don’t exist is that there is no adjoint pair of functors

Free –| U = Underlying: Hopf –> Set

where the free functor is left adjoint to the underlying-set or forgetful functor. So if “free” is in the sense of a map (a function)

X –> UFree(X)

which is universal (given a function X –> H into a Hopf algebra over k, there exists a unique Hopf algebra map Free(X) –> H which extends this function in the sense of an obvious commutative triangle), then there is no free Hopf algebra.

To see this, note that the ground field k is terminal among Hopf algebras: for any H there is a unique Hopf algebra map H –> k given by the counit. But there are many functions X –> k, so how can you extend these to Hopf algebra maps Free(X) –> k?

A less formal way to think about this is that “usually” you can form a free object in the way you say, by taking equivalence classes of syntactic expressions, if your theory T is described by operations. Indeed the free object F[n] on n elements is the set of n-ary operations of the theory, in the sense that there is a tautological action F[n] x X^n –> X on any model X of the theory, which interprets F[n] as n-ary operations on X, and in fact F[n] is the set of natural operations on the “generic model” U: T-Alg –> Set,

UF[n] = Nat(U^n, U),

by an application of the Yoneda lemma. But in the case of Hopf algebras or bialgebras, the theory is described by both operations and co-operations (in k-Mod), and this method isn’t quite available. Somehow the “operations” or syntactic expressions can’t be contained or valued in a single free object H — you need all the tensor powers of H. I think that’s roughly the problem.

7. Charlie Stromeyer Jr - October 12, 2007

Todd, it turns out that if you google the phrase “free hopf algebra takeuchi” then you will see various examples of free Hopf algebras, however, I am not right now near an academic library so I have not been able to look at these papers.

Further, I can’t tell just from using google whether the notion of a free Hopf algebra was first introduced by Mitsuhiro Takeuchi or by David Radford. It looks like one would have to look at both their papers from 1971 to see who first introduced the concept (albeit separate concepts of what a free Hopf algebra is).

8. Todd Trimble - October 12, 2007

Charlie, as far as I can tell, Takeuchi was discussing free Hopf algebras over (generated by) coalgebras, which unquestionably exist. But that’s a far cry from what I was discussing in response to David’s comment, which was about free Hopf algebras generated by sets.

Please let me know if you see something wrong with what I actually said.

9. Charlie Stromeyer Jr - October 12, 2007

Todd, I don’t see anything wrong with what you actually wrote above, but I say this with the caveat that I have never even looked at any of the various papers (including the more recent ones) about free Hopf algebras.

10. Noah Snyder - October 12, 2007

Todd already explained some of this, but David let me try to expand a bit on his answer.

An example where you can build a free algebra, but it’s a little tricky, is algebras with trace. However, you have to change your idea of trace. Instead of trace being a map from A to k, you instead have trace be a map from A to A, and then require that its image lie in the center. The reason for this is that when you add some formal variable x, then you want to have tr(x) be something arbitrary. But if you have to pick some element of k then it’s not all that arbitrary is it? You want the base field to grow to be k(tr(x)) or something. So you have to tweak things a bit so that trace is a 1-ary operation on A.

The problem with Hopf algebras is that Delta(x) lives in A(x)A, not in A. So it isn’t so clear how you quotient out by some relations *just in A* in order to force the axioms to hold.

11. Noah Snyder - October 12, 2007

Oh yeah, for free algebras with trace you also need some more axioms like tr(tr(x)y) = tr(x)tr(y), but I’m sure you can figure them out.

12. Scott Carter - October 12, 2007

See my post on the previous thread. It addresses a freeness question. The free-est Hopf thing is the cat of diagrams mod the relations that define a Hopf alg.

13. Bruce Bartlett - October 12, 2007

This is an interesting idea Noah, I’m trying to get my head around it. So let me throw out a confusion I have right at the outset :

“Kupperberg then conjectures that if an identity holds for any finite dimensional involutory Hopf algebra, than it must follow from the axioms.”

Doesn’t Gödel’s completeness theorem, or something like it, come into play at this point? As I understand it, it says roughly “if a certain equation holds for all concrete models of a bunch of axioms, then that equation must follow from the axioms”.

14. Greg Kuperberg - October 13, 2007

Hi Noah et al. Thanks a lot for the credit in this post. My own thinking about my conjecture follows exactly the lines that you say. The analogous conjecture is true for Frobenius algebras and surfaces, but the framework of Frobenius algebras hardly helps to prove the result. Meanwhile, thanks to Perelman, “my” conjecture is at least nearly true for closed, irreducible 3-manifolds. The reason is that geometrizable manifolds are residually finite and can probably be distinguished by finite-group invariants. It would be even closer to true if it were the analogous result for non-involutory Hopf algebras, because that would take care of the technicality of inequivalent lens spaces with the same fundamental group. But of course, Perelman’s work does not really have anything to do with Hopf algebras or with me. The real remark, so far, is that the classification of 3-manifolds has an interesting corollary for finite-dimensional Hopf algebras.

Since the Hopf algebras in question have to be finite-dimensional, there is no free example of such an algebra. What there is instead is the free Hopf PROP with traces. (I am not sure if this terminology is exactly correct, but the idea is to make a universal Hopf object, with traces, in a symmetric tensor category built for it.) My theorem says that an equivalence class of scalar words in this PROP corresponds to an oriented, compact 3-manifold M with two colors of boundary which is minimal in a natural sense. The minimality conditions are: M has no essential 2-spheres, including no such boundary components; M need not be connected but has no 3-sphere components; all boundary components are incompressible; and there are no essential annuli whose circles use both colors. So in particular M could be a closed and irreducible 3-manifold.

Some basic examples: The empty 3-manifold is the trivial word 1. RP^3 is Tr(S), the trace of the antipode. S^2 x S^1 is the trace of the identity operator on the Hopf algebra (although in the non-involutory case it would be Tr(S^2)).

Also to answer this question:

Doesn’t Gödel’s completeness theorem, or something like it, come into play at this point?

No, because the axioms are much too weak for Gödel’s theorem. Gödel considered axioms that described a full theory of logic, such as the Peano axioms, rather than axioms that just describe an algebraic object like a group or a Hopf algebra. Indeed, there are relations that do not hold in the abstract PROP for finite groups, but do hold in every finite group. (The finite-group PROP is subtly different from the traced Hopf PROP, because group algebras must also be cocommutative. The geometric model for the finite-group PROP is a class of 2-complexes rather than 3-manifolds.)

15. Charlie Stromeyer Jr - October 14, 2007

Greg, unless I am mistaken (which is certainly possible), the correct term is “properad” when talking about bialgebras, Lie bialgebras or infinitesimal Hopf algebras.

At least this is what it says on the top of page 2 of the paper “A Koszul duality for PROPS” by Bruno Vallette in the arXiv as math/0411542

By the way, I still can’t figure out an answer to Scott Carter’s question in the previous thread.

16. Scott Carter - October 14, 2007

Charlie,

Essentially, I was trying to describe a Hopf
Properad, BUT I only want trivalent vertices, units and counits.

Brooms and cobrooms (which are part of an operadic like structure) can be inductively defined using associativity and coassociativity. A cobroom
would correspond to the Sweedler
x(1) ox x(2) ox \ldots ox x(n).

I am hopeful that the nature of the relations to be imposed are contextually clear since I can’t draw the pictures here.

17. Greg Kuperberg - October 15, 2007

I looked at the definitions of props and properads in Vallette’s paper. I got lost. I do not know if this terminology applies to the object that I studied. So let me stick to terminology that I know is valid.

You can define a Hopf object in any symmetric tensor category. In particular, there is a free Hopf symmetric tensor category that is presented by the Hopf axioms. There is also a pivotal symmetric tensor category which is freely involutory Hopf. Both of these categories have a canonical Hopf object. My result is a description of End(1) in the latter category.

Without the pivotal structure, End(1) is trivially the one-element semigroup. But it would still be worthwhile to figure out the other Hom spaces, with and without the pivotal structure.

It would also be worthwhile to figure out the Hom spaces of the free non-involutory Hopf categories, again with and without the pivotal structure.

18. Bruno Vallette - October 16, 2007

Hello everybody and thank you for your interest in properads. Maybe I can try to clarify the first questions that arose here (but unfortunately not the very last ones).

First, if you have an algebraic structure whose definition only involves products (eg associative algebras, Lie algebras, Gerstenhaber algebras), that is operations with several inputs but one output, you can model it by an operad. In this case, the notion of free algebra exists and is equivalent to a presentation of the operad (for more details about the “presentation” of an operad, see below). Now, if you consider an algebraic structure defined by products AND coproducts (like bialgebras, Lie bialgebras, Frobenius (bi)algebras), we need to use a properad to model these operations (instead of an operad). The graphical representation of the composite of all these products-coproducts involve directed graphs (trees for operads). As long as you have to deal with algebraic structures whose relations can be depicted with connected graphs, the related properad faithfully encode the whole data. On the contrary, you will have to use a prop and non-connected graphs which are much bigger objects. In the case of a type of (bial)gebra modeled by a properad P, there is no notion of free P-gebra, that is the forgetful functor does not admit a left adjoint (the category of P-gebra has no coproduct). This is the source of many “problems” in this theory. Some constructions can not be generalized from operads to properads (eg. the bar and cobar constructions of P-(al)gebras).

But maybe your question is not related to a free gebra over a properad but to the notion of free properad and/or properad defined by generators and relations. One should always be aware that there are two levels when working with operads/properads/props and (bial)gebras. The first level is the level of operads/properads/props itself and the second level is the level of gebras over them (modules over an operad/properad/prop). There are properads and, for instance, there is A properad which models the category of bialgebras. So you know the notion of semi-group or associative algebras. An operad is a “multiple input” generalization of an associative algebras. I like to think at elements of an associative algebra as operations with one input and one output (a labeled vertex with one input and one output). Then, the tensor product of two elements is represented by the grafting of these two vertices one above the other. The image under the composition map is a vertex with one input and one output labeled by the product of the two elements. An operad has the same flavor. Elements of an operad are (formal, abstract) operations that can be represented by corollas (vertices with several inputs and one output). Now, you can graft them (some formal composition). The (concrete) composition map of an operad tells you how these operations really compose : it associates to any tree whose vertices are labeled by operations of the operad one corolla labeled by one operation of your operad with the same number of inputs. Such a composition map has to be associative : if you “compose” (contract) one subtree of such a tree and then the rest of the tree, this is independent of the subtree chosen. (This is the cotriple definition of an operad). Hence, an associative algebra is a particular operad (the operations have only one input). You can continue and consider corollas with multiple inputs and multiple outputs and with graphs of any genus instead of trees. If you only allow to compose connected graphs, you define the notion of properad and if you allow any kind of graphs, you get the notion of props. Finally, an associative algebra is an operad and an operad is a properad.

At the beginning of this thread, you raised the very good question of a presentation of a group defined by generators and relations, which is not an easy problem. Instead of groups, you can consider semi-groups or associative algebras and of course now operad and properads. The easiest example is the following one : consider the free non-symmetric operad (forgetful the action of the symmetric group and work with planar trees only) generated by one binary operation. Then take the quotient by the ideal generated by the associative relation (a*b)*c=a*(b*c) (imagine trees of course). You can give a very simple presentation of this operad : the free non-symmetric operad on one binary generator is equal to the set of planar binary trees and if you identify left ((a*b)*c) and right (a*(b*c)) compositions, there is just one class for any trees with the same number of inputs. That is, there is just one n-mutlilinear operation on an associative algebra. (This also explains the shape of the free associative algebra given by the tensor module.) Now, you can play this game with more operations. In the case of Frobenius (bi)algebras, you will have to consider a generating binary (commutative or not) product (a corolla with 2 inputs and 1 output), a generating coproduct (1 input and 2 outputs). Then you have to create all connected directed graphs with trivalent vertices. Finally, you quotient by the relations of a Frobenius algebra. And the result is …. (The orbits are in one-to-one correspondence with the number of inputs, the number of outputs and the genus of the graph in the commutative, cocommutative case. It is a good exercise.).
It seems that the “identity from the axioms” you are looking for can come from this setting.

(If you want pictures, there is short survey on operads/properads/props on Page 34 of arxiv.org/abs/0707.0889. It is explained there that in each case, there are monoids in a particular monoidal category. This definition is different from the one given above but equivalent. Therefore, it makes sense to consider this “presentation” problem.) If you have to work with traces, you will have to use wheeled properad as they were defined in arxiv.org/abs/math/0610683.

I hope this helped but if something remains unclear, feel free to ask.

P.S.1: There is a properad which codes Hopf algebras but the notion of Hopf properad is something different (namely, it is a properad in the category of coalgebras).

P.S.2: Greg, do you have any reference for the constructions you mention in your last message (17) ? Since I do not know that “pivotal” means I can hardly understand.

19. Charlie Stromeyer Jr - October 16, 2007

Bruno, in your short survey on operads/properads/props you say that a semi Hopf algebra can only be represented by a prop. What is a semi Hopf algebra?

Also, I don’t want to speak for Greg but just in case he is busy with something else you can see what he means on page 17 of his paper “Non-involutory Hopf algebras and 3 manifold invariants” as arxiv.org/abs/q-alg/9712047

For more on what Greg means by pivotal also see page 9 of the paper to which Greg refers by Barrett and Westbury called “Spherical categories” as arxiv.org/abs/hep-th/9310164

20. Dylan Thurston - October 16, 2007

Vogel has a similar negative result for Lie algebras. Consider finite-dimensional Lie algebras with an invariant, non-degenerate bilinear form. There’s a small list of axioms you can right down for such an object, that are enough to guarantee that you get a sequence of invariants of 3-manifolds. (These are finite-type invariants.) You can also add in a finite-dimensional representation to get invariants of knots in 3-manifolds. Vogel showed that there are some expressions in these operations that are not consequences of the axioms and yet vanish for every finite-dimensional Lie algebra and representation of it (this part is fairly easy), and even for every finite-dimensional Lie super-algebra and representation of it. (I think the question is still open if you drop the representation, but I’m not sure about that.)

Does this relate to the conjecture on Hopf algebras? I’m guessing not; maybe the Lie algebra construction is in some sense coming from derivatives of Hopf algebras, and finite-dimensional Hopf algebras like those coming from finite groups, which I guess is involved in Greg’s construction from (14).

21. Scott Carter - October 16, 2007

Bruno,

Thanks, very much. In relation to PS1, is the Properad that encodes Hopf algebras different than the category of Hopf algebras — more general than modules thereover?
Scott Carter

22. Greg Kuperberg - October 16, 2007

Bruno: A pivotal category is a particular kind of monoidal category. (A monoidal category is also called a tensor category.) In a pivotal category, objects have left and right duals and preduals and you can take traces. The concepts are explained pretty well in an old paper by Barrett and Westbury, arXiv:hep-th/9310164. The upshot is that in a pivotal category, any planar tensor diagram has a well-defined and unique tensor value. Without the pivotal condition, only acyclic diagrams can be evaluated. If a tensor category is symmetric, then A tensor B is canonically isomorphic to B tensor A. In this case the tensor diagrams need not be planar; they are just graphs.

It all looks very similar in some ways to the formalism in your paper. One difference is that the edges are colored by objects in the category instead of all having the same color.

Anyway a category, or a tensor category, or a pivotal symmetric tensor category, can have a presentation with generators and relations, just as a group or a ring can. In particular, you can use the Hopf axioms as a presentation of a symmetric tensor category whose set of objects is a free semigroup with one generator. The pivotal version of this is a bit more complicated; the set of (isomorphism classes of) objects is a free abelian semigroup with two dual generators. The generators are the universal Hopf object H and its dual H^*.

23. Greg Kuperberg - October 16, 2007

Dylan, Scott, and Charlie: We crossed paths when I was writing comment #22, so my posting is somewhat redundant. To restate Dylan’s remark in my language: There is a universal Frobenius Lie object in its enveloping complex-linear category. Vogel showed that there are elements that vanish in every Frobenius Lie superalgebra, but not in the universal object. It would be hard to relate this to the analogous Hopf algebra conjecture, at least in the non-involutory case, given that I think that the representability conjecture is actually true in that case.

But it is possible that involutory Hopf algebras have weaknesses so that they do not distinguish homotopy-equivalent lens spaces, for example. They may be too close finite groups.

Well, maybe I am wrong to be optimistic in the non-involutory case. There is a recent paper in the arXiv that says that the Hennings invariant for a truncated quantum group equals the Reshetikhin-Turaev invariant times some power of the determinant of the 3-manifold. (Namely, the cardinality of H_1(M) if it is finite and zero otherwise.) Something similar is probably true for my invariant, if use the quantized enveloping algebra of a Borel subalgebra.

If that is a way to defeat my old conjecture, then it would be just because my invariant in arXiv:q-alg/9712047 has some annoying factors of zero when the manifold has rational homology and the Hopf algebra isn’t involutory. Certainly if you take all TQFTs, you would be hard-pressed to find two closed 3-manifolds that they don’t distinguish. But if you take two homotopy-equivalent lens spaces, for example L(7,1) and L(7,2), you would be hard-pressed to find two involutory Hopf algebras that do distinguish them. (If you’re not hard-pressed on that last one, then you would surprise me and you might have a very nice new result.)

24. Charlie Stromeyer Jr - October 27, 2007

Hello, everyone. Could someone please explain to me if Greg Kuperberg’s invariant is related to SU(2)/SO(3) quantum invariants? I am asking because these quantum invariants do distinguish at least some lens spaces of the same homotopy type.

There are various papers about these quantum invariants. See, for example, B.H. Li and T.J. Li, Generalized Gaussian sum and Chern-Simons-Witten-Jones invariants of lens spaces, Journal of Knot Theory and its Ramifications, vol. 5(2) (1996), pp.183-224. Thanks.

25. Greg Kuperberg - November 1, 2007

My invariant for U_q(sl(2)^+) probably does distinguish lens spaces and so on — but that is the non-involutory invariant. And it was recently shown that the closely related Hennings invariant for U_q(sl(2)) is Reshetikhin-Turaev times the determinant of the 3-manifold. (This quantity is the cardinality of the first homology, if finite, and zero otherwise.)


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