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Background


I am currently writing an MSc dissertation on TQFTs (and Khovanov homology, but that is unrelated to this question).

After having read most of Kock's book on the equivalence between 2D Oriented TQFTs and Frobenius Algebras [Kock], I have started reading Lurie's seminal paper on the Cobordism Hypothesis [Lurie]. I'm currently at section 1.4, while he first states a version of the Cobordism Hypothesis at section 1.2.

These two results share both similarities and disparities alike, and I'd like to fully grasp the relationships between them. I have scoured Lurie's papers and the Internet for answers, and haven't been able to find anything enlightening enough. After bugging my (MSc dissertation's) supervisor about it a couple times, he encouraged me to try posting on here.

Apologies for the length; there are some moderately nuanced statements and I wanted to make this question as self-contained as reasonably possible. Thanks in advance to anyone who takes their time to read this mess, and even more for whichever kind soul decides to answer!

Let's first review the definitions and theorems I'm going to use, to avoid misunderstandings. Those familiar with the literature (as you likely are, if you clicked on this question!) can probably skip this section. $\def\cat#1{\mathbf{#1}}$ $\def\kk{\mathbb{k}}$ $\def\RR{\mathbb{R}}$ $\def\Sp{\mathbb{S}}$ $\def\mycat{\mathscr{C}}$

Definitions and statements


First let's state the Main Result of [Kock]. I'm also going to state its more abstract version, since the similarities are most apparent there.

Definition 1: Let $\cat{2Cob}$ be the symmetric monoidal category of oriented 2-cobordisms.

  • Objects: oriented compact 1-manifolds (that is, finite disjoint unions of circles $\Sp^1$).
  • Morphisms: oriented 2-cobordisms up to diffeomorphism (rel the boundary).
  • Composition: gluing cobordisms along the boundary.
  • Monoidal structure: disjoint union $\bigsqcup$ with unit $\emptyset$.

Definition 2: A 2D oriented TQFT is a symmetric monoidal functor $Z \colon \cat{2Cob} \to \cat{Vect}_{\kk}$.

These form a category $\cat{2TQFT}_{\kk}$, whose morphisms are monoidal natural transformations.

(I think these are also called $(1, 2)$-TQFTs).

Definition 3: A Frobenius $\kk$-algebra is a finite-dimensional $\kk$-algebra $A$ equipped with a Frobenius form $\varepsilon \colon A \to \kk$ whose kernel contains no nontrivial (left) ideals. Commutative Frobenius $\kk$-algebras form a category, $\cat{cFA}_{\kk}$.

Definition 4: A Frobenius object in a monoidal category $(V, \square, I)$ is an object $A$ equipped with four maps satisfying unit and counit axioms and the Frobenius relation. Here I think it's best just to see this screencap from [Kock], since the combinatorial diagrams are important for intuition.

About the intuition: the thing that first made me formulate this question is that the Frobenius relation can be written graphically in this way, while the "fully dualizable" condition is a higher-categorical generalization of the snake relation (also known as Zorro's lemma). You can just put on or remove caps to get from one to the other, after all!

Theorem 1a [Kock, Thm3.3.2]: There is a canonical equivalence of categories $\cat{2TQFT}_{\kk} \simeq \cat{cFA} {\kk}$.

This equivalence is obtained by mapping each TQFT $Z \colon \cat{2Cob} \to \cat{Vect} {\kk}$ to its value $Z(\Sp^1_{+})$ on the oriented circle.

Generalizing the codomain and stating it as a universal property, you get the following:

Theorem 1b [Kock, Thm3.6.19]: The skeletal cobordism category $(\cat{2Cob}, \coprod, \emptyset, T)$ is the free symmetric monoidal category generated by a (co)commutative Frobenius object.

Now, the first statement of the Cobordism Hypothesis appearing in [Lurie].

Definition 5: Let $n\cat{Bord}^{\mathrm{fr}}_n$ be the symmetric monoidal n-category of framed n-cobordisms.

  • Objects: finite disjoint unions of $n$-framed points.
  • k-morphisms: $n$-framed $k$-cobordisms between $(k-1)$-manifolds.
  • n-morphisms: framed $n$-cobordisms between $(n-1)$-manifolds, up to diffeomorphism (rel the boundary) respecting the framings.
  • Composition and monoidal structure: same as $n\cat{Cob}$.

I admit I currently don't grasp clearly enough the formal definition of "fully dualizable object" (which maybe contributes to my general confusion), but I conceptualize it as a higher-categorical analogue of the "snake relation" previously mentioned (which states the "finiteness" condition). You have a monoidal $n$-category $(\mycat, \square, I)$ with a notion of dual objects, and you consider the evaluation and coevaluation maps

$$ \mathrm{ev}_V \colon V \square V^* \to I, \qquad \mathrm{coev}_V \colon I \to V^* \square V. $$

For an object to be dualizable, it must satisfy that both ways of composing are the identity:

$$ \mathrm{id}_V \colon V \xrightarrow{\mathrm{id}_V \square \mathrm{coev}_V} V \square V^{\ast} \square V \xrightarrow{\mathrm{ev}_V \square \mathrm{id}_V} V, $$

$$ \mathrm{id}_{V^{\ast}} \colon V^{\ast} \xrightarrow{\mathrm{coev}_V \square \mathrm{id} } V^{\ast} \square V \square V^{\ast} \xrightarrow{\mathrm{ev}_V \square \mathrm{ev}_V} V^{\ast}. $$

My naïve way of higher-categorifying this would be to change every morphism to a functor, but I don't think that is quite correct. In any case, let ignore that for the time being.

Theorem 2a (Baez–Dolan Cobordism Hypothesis) [Lurie, Thm1.2.16]: The evaluation functor $Z \to Z(\cdot_{+})$ determines a bijective correspondence between (isomorphism classes of) framed fully extended $\mycat$-valued TQFTs and (isomorphism classes of) fully dualizable objects of $\mycat$.

This, too, has a corresponding universal property.

Theorem 2b: The symmetric monoidal $n$-category $n\cat{Bord}_n^{\mathrm{fr}}$ is freely generated by a fully dualizable object.

You have $(\infty, n)$-categorical versions (and a bunch of other generalizations), but I think this one should suffice.

The actual question


There are a great deal of similarities between the 1-categorical and 2-dimensional Theorem 1 and its higher-categorical, higher-dimensional analogue Theorem 2. That is no surprise, since the Cobordism Hypothesis tries to generalize the structures seen in the 2-dimensionsal case.

But how can we make this similarity explicit? More precisely: does the framed Cobordism Hypothesis (as stated) actually generalize the "classical" equivalence of Theorem 1, in the sense that it implies it? One might think that it obviously does, but there are at least three disparities which make this a far more interesting question that I would have initially expected.

  • 2D oriented TQFTs are determined by the image of the (oriented) sphere $\Sp^1_{+}$, but fully extended framed TQFTs are determined by the image of the (framed) point $\cdot_{+}$. This doesn't seem to be a problem, and in both cases the objects of the domain category are finite disjoint unions of the corresponding generator.
  • More importantly: not every (maybe extended) oriented TQFT can be made into a framed TQFT; every framing induces an orientation, but not viceversa. Disturbingly, most surfaces are not parallelizable, so the morphisms of $\cat{2Cob}$ are strictly more numerous than the 2-morphisms of $\cat{2Bord}^{\mathrm{fr}}_2$. (Although, as I will later conjecture, they are probably in some kind of bijection with 2-morphisms of $\cat{3Bord}^{\mathrm{fr}}_3$).
  • Frobenius objects and fully dualizable objects, while sharing a similar definition, aren't the same.

Of course, the 2D TQFTs-commutative Frobenius algebras correspondence is much easier to state, conceptuallize and prove than the Cobordism Hypothesis; this question arises from trying to better understand the consequences of the latter, not the former!

Some follow-up questions, too. These are a bit more rambly; I apologize.

  • Assuming everything relates nicely, how do you recover a framing (up to coherent diffeomorphism) from just an orientation? This is most likely futile in higher dimensions, but it seems doable enough in a dimension as low as 2.
  • And if the framed version doesn't suffice, what about the other (numerous) versions of the Cobordism Hypothesis stated in [Lurie]?
  • And what about framed $(1, 2)$-TQFTs, are those equivalent to the oriented ones? If they aren't, what subclass of objects of "commutative Frobenius algebras" do they correspond to?

Things I've tried


Mostly writing things out and searching. I first found a particular case of the 2D Cobordism Hypothesis.

Theorem 3 [SP, Thm3.52]: The bicategory of 2-dimensional oriented extended topological field theories with values in $\cat{Alg_2}$ is equivalent to the bicategory $\cat{Frob}$.

(See the thesis [SP] for more details). Schommer-Pries then states:

This has many consequences. The ordinary 1-categorical bordism category of closed oriented 1-manifolds and oriented 2-dimensional cobordisms sits inside $\cat{Bord}^{\mathrm{or}}_{d}$ as the endomorphism category of the empty zero manifold. The endomorphism category of the unit algebra in $\cat{Alg}_2$ is the category of $\kk$-vector spaces (or $\kk$-modules when $\kk$ is a commutative ring). Thus any extended TQFT in $\cat{Alg}_2$ gives rise to a classical 2-dimensional non-extended TQFT.

This doesn't answer my question (since it was about the framed version), but it provides a useful stepping stone and a good choice of category in which to take values. However, I think the versions characterized by universal properties could be cleaner to work with.

I'm not sure if $\cat{2Bord}^{\mathrm{fr}}_2$ is the correct category to look at, though, since most surfaces aren't parallelizable. Since every orientable surface $M$ admits a 3-framing (these can be embedded into $\RR^3$, and trivializations of $T \RR^3$ induce trivializations of $TM \oplus \underline{\RR}$), maybe the correct way to see a correspondence is to go a dimension higher. You can take the 3-category $\cat{3Bord}^{\mathrm{fr}}_3$, get rid of the objects layer by taking endomorphisms of the monoidal unit, and "truncate" it at 1-morphisms by taking the homotopy category. We conjecture this to be equivalent to $\cat{2Cob}$:

$$ \mathrm{Ho}(\mathrm{Map}_{\cat{3Bord}^{\mathrm{fr}}_3}(\emptyset, \emptyset)) \simeq \cat{2Cob}. $$

It would be very nice if that was the case (and interesting on its own), but I still don't clearly see how to get the classical equivalence from here.

References


[Kock] Joachim Kock. Frobenius Algebras and 2D Topological Quantum Field Theories. (2003). DOI: https://doi.org/10.1017/CBO9780511615443.

[Lurie] Jacob Lurie, "On the Classification of Topological Field Theories". (2009). arXiv: https://arxiv.org/abs/0905.0465

[SP] Christopher Schommer-Pries. "The Classification of Two-Dimensional Extended Topological Field Theories". (2009). arXiv: https://arxiv.org/abs/1112.1000

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  • $\begingroup$ For posterity: exploring this question eventually formed the fourth chapter of my master's thesis, "2D Topological Quantum Field Theories, Frobenius Structures, and Higher Algebra". There are some inaccuracies, and some answers raise further questions, but I tried my best to reach a satisfactory conclusion within my deadline and current capabilities — I am no expert. I have uploaded it to tiny.cc/TFM2DTQFTS, although I plan to eventually copy it to a more easily-indexable location. $\endgroup$ Commented Jun 25 at 12:19
  • $\begingroup$ I have now uploaded the text to the aforementioned more permanent location, namely hdl.handle.net/20.500.14352/105943. $\endgroup$ Commented Jul 11 at 8:08

1 Answer 1

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It turns our the most serious disparity is your first bullet point, not the second: the key difference is that the cobordism hypothesis is about fully extended TFT. A sub-issue is that in the fully extended case you have to pick a target (well, in the non extended case as well, but there is the standard choice Vect). To relate this to Frobenius algebras the natural target is as you say what [SP] call $Alg^2$, the category of algebra, bimodules and morphisms of those (which is nowadays more often denoted $Alg_1$, or more precisely $Alg_1(Vect)$). Now:

  • every object in there is 1-dualizable
  • 2-dualizable objects are finite-dimensional separable algebras, which is of course a very strong restriction.
  • the oriented version of the cobordism hypothesis tells you there is an action of $SO(2)$ on the groupoid of those (the action is trivial in that case), so that fully extended, oriented 2d TFT valued there are classified by homotopy $SO(2)$ fixed points: those are precisely separable non-necessarily commutative Frobenius algebra: this is the result in [SP] you quote (note that this important separability condition is part of what he calls Frob). See also https://arxiv.org/abs/1607.05148. The value of that TFT on the circle is then the center of your algebra, which is of course again a (now commutative) Frobenius algebra.

So I think [SP] does in fact answer your question: the punchline is that extended TFT and the cobordism hypothesis see only a very special class of Frobenius algebras (the separable ones). In other words most oriented 2d TFT won't extend down. So strictly speaking I'd say the answer of your question is no: you can't use the cobordism hypothesis to recover the classification of oriented 2d TFT.

Your comment about framing is backward: indeed every framing induces an orientation, which is why every oriented TFT (extended or not) does in fact restricts to a framed TFT.

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  • $\begingroup$ Thank you for your answer! It is true I got the framing implication backwards, I thought I had double-checked it. Oh well. The fact that most 2D TQFTs can't be extended down made me think this couldn't be possible (and the fact that Lurie didn't mention it once in his very long and detailed paper!), but the similarities were too strong to fully let it go. I'll give it a little more thought, but I suppose the problem is then "how much of the classification can you recover from the Cobordism Hypothesis?" as well as, "what does the separability condition translate to in the general-valued case"? $\endgroup$ Commented Feb 9 at 7:17
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    $\begingroup$ You might also be interested to this MO question (one of the answer is from Lurie, specifically about his paper) mathoverflow.net/questions/31154/… . Still probably for $n=2$ this isn't really easier, at least when the target is Vect. If you go all $\infty$ however this would be the way to go, e.g. AFAIK the classification of non extended oriented 2d TFT in chain complexes is an open question. Basically what makes the classical case easy is you don't have to deal with diffeomorphisms of surfaces. $\endgroup$
    – Adrien
    Commented Feb 9 at 8:30
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    $\begingroup$ Also: welcome to MO ! $\endgroup$
    – Adrien
    Commented Feb 9 at 8:30
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    $\begingroup$ Still, note the cobordism hypothesis is powerful because it does extend "easily" to the $\infty$ case, see eg section 2.1 of pages.uoregon.edu/njp/safronov.pdf . So roughly for the $\infty$ analog of separable (i.e. for dg-algebras) you do get all those coherent actions of diffeos of surfaces for free which is pretty mind blowing if you ask me. For your last question, note that separable makes sense in any (ordinary/discrete) symmetric monoidal category, since you can write this condition as the existence of a splitting of a certain map. $\endgroup$
    – Adrien
    Commented Feb 9 at 8:39
  • $\begingroup$ Thank you! I have been lurking for a couple of years, but I didn't have anything interesting enough to ask here until now :). I am familiar with that question (and Lurie's answer), but not with the notes. The characterization of "fully dualizable" in terms of admitting adjoints is pretty helpful; I looked for a similar definition while writing this question, and wasn't able to find it. $\endgroup$ Commented Feb 9 at 10:20