$(\infty,1)$ 2d TFTs

Question

2d topological field theories $Z : \mathrm{Cob}(2) \to \mathrm{Vect}$ are classified by commutative Frobenius algebras.

What can be said about $(\infty,1)$ 2d TFTs $Z: \mathrm{Cob}(2) \to \mathcal{S}$ with values in a symmetric monoidal $(\infty,1)$-category $\mathcal{S}$? I am interested in different targets $\mathcal{S}$.

I know that $Z(\mathbb{S}^1)$ is naturally an $E_2$-algebra in $\mathcal{S}$, but certainly it has more structure.

Please note that I do not refer to fully extended (i.e. 2-1-0-dimensional) TFTs.

Answer 1

As you point out, if you just look at the operad of bordisms with exactly one output disc, you get the framed $E_2$-operad (framed here means you can rotate the discs) and so the value of the circle is a framed $E_2$ algebra. (Aside: if we were working in categories an $E_2$-algebra is a braided monoidal category, while a framed $E_2$-algebra would also be balanced). Similarly it's a framed $E_2$-coalgebra. So it's natural to try to defined a framed Frobenius $E_2$-algebra to be a framed $E_2$-algebra and an $E_2$-coalgebra with some kind of "Frobenius" compatibility structure. The problem is that I can't really imagine what description of this Frobenius structure one could possibly hope for that won't make this statement just a tautology... That is, the definition of $E_2$-algebra is already highly geometric, and so one should expect the definition of Frobenius $E_2$-algebra to also be highly geometric, and hence the best way to define Frobenius $E_2$-algebra is to say its the value of the circle under a 2d TFT. One way one could hope to make this approach less tautological would be to say what kind of more "combinatorial" description of $E_2$-algebras you'd like the putative combinatorial description of Frobenius $E_2$-algebras to extend. For example, you could define an $E_2$ algebra to be an $A_\infty$-algebra in $A_\infty$ algebras to give a Stasheff-like combinatorial description. But you'd still need to also put in the balanced structure somehow.

Perhaps better would be to take an approach that generalizes a different definition of Frobenius algebra. Instead of saying "algebra and coalgebra with Frobenius condition" one can instead say "commutative algebra A with trace such that Tr(xy) is non-degenerate." This has a natural generalization:

Def: A framed Frobenius $E_2$-algebra is, a framed $E_2$-algebra $A$ together with an $\mathrm{SO}(2)$-invariant map $\mathrm{Tr}: A\rightarrow \mathbf{1}$ such that $X \boxtimes Y \mapsto \mathrm{Tr}(X \cdot Y)$ is the evaluation of a duality between $A$ and itself.

So now we have the following precise question:

Question: Is the map from the space of 2-dimensional TFTs valued in a symmetric monoidal $(\infty,1)$-category equivalent to the space of framed Frobenius $E_2$-algebras under the functor which evaluates the circle?

I don't know how to prove this and am not totally sure if it's true, but since duals are defined like adjoints, you can hope that the "uniqueness of coherent adjoints" lets you recover the TFT just from the framed Frobenius $E_2$-algebra structure.

I'm not even sure if this is true in the case where the target is the $(2,1)$-category of categories... My understand there from Bakalov-Kirillov is that this is asking about the classification of modular functors, which are classified by weakly rigid balanced tensor categories. But this seems to me to be saying that the trace has to be given by $\mathrm{Hom}(-,\mathbf{1})$, and I don't see how that matches up with my definition here of a Frobenius $E_2$-algebra. Maybe I'm just misunderstanding Bakalov-Kirillov's assumptions though.

Answer 2

The answer to your question is known when $\mathcal{S}$ is a symmetric monoidal $\infty$-groupoid, by work of Galatius-Madsen-Tillmann-Weiss. In other words: we understand invertible $2$-dimensional TFTs. This might seem like a somewhat silly case, but it's actually a very useful way of testing conjectures. Below I'll sketch how this implies that the answer to Noah's question is no, but let me begin by explaining the setup.

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Let's a assume that every morphism in the symmetric monoidal $(\infty,1)$-category $\mathcal{S}$ is invertible and that every object of $\mathcal{S}$ is invertible under $\otimes$. (The latter could be replaced by the assumption that every object is isomorphic to the unit $1_{\mathcal{S}}$ - it doesn't really matter since $B(\mathrm{Cob}(2))$ is connected.)

A symmetric monoidal $\infty$-groupoid is an $E_\infty$-algebra in spaces, and with the above invertiblity assumption it is an infinite loop space $\mathcal{S} = \Omega^\infty X$ of some connective spectrum $X$. Now every functor $\mathrm{Cob}(2) \to \mathcal{S}$ factors through the groupoidification $\mathrm{Cob}(2) \to (\mathrm{Cob}(2))^{\rm gp}$, which can be modeled as the classifying space of the $\infty$-category $\mathrm{Cob}(2)$. This classifying space was computed by Galatius-Madsen-Tillmann-Weiss as: $$B(\mathrm{Cob}(2)) \simeq \Omega^\infty( \Sigma^{-1} \mathrm{MTSO}_2)$$ Here $\mathrm{MTSO}_2$ is a certain Thom spectrum, which can be studied by standard methods in stable homotopy theory. (I'm happy to elaborate, if you like.) Anyways, putting this together we have that: $$ \mathrm{Fun}_\infty^\otimes(\mathrm{Cob}(2), \mathcal{S}) \simeq \mathrm{Map}_{\mathrm{Sp}}(\tau_{\ge 0} \Sigma^{-1} \mathrm{MTSO}_2, X). $$ Here $\tau_{\ge0}$ is the functor that sends a spectrum to it's connective cover. This could be dropped since $X$ is connective, but is seems more accurate to keep it around as $\mathrm{Cob}(2)$ doesn't know about $\pi_{-1} \Sigma^{-1} \mathrm{MTSO}_2 = \mathbb{Z}$.

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So why is this useful at all? I'll try to illustrate that be giving a negative answer to the question Noah posed in his post. I'll say traced $E_2^{\rm fr}$-algebra to mean a framed $E_2$-algebra $A$ equipped with an $SO_2$-equivariant trace $\tau:A \to 1$. Let $\mathcal{C}$ be the free symmetric monoidal $(\infty,1)$-category on a traced $E_2^{\rm fr}$-algebra. Since $S^1 \in \mathrm{Cob}(2)$ has this structure, there is a symmetric monoidal functor $F:\mathcal{C} \to \mathrm{Cob}(2)$ sending $A$ to $S^1$. Noah's question can now be reformulated as asking whether for all symmetric monoidal $(\infty,1)$-category $\mathcal{S}$ precomposition with $F$ induces a fully faithful functor $$ \mathrm{Fun}_\infty^\otimes(\mathrm{Cob}(2), \mathcal{S}) \longrightarrow \mathrm{Fun}_\infty^\otimes(\mathcal{C}, \mathcal{S}), \qquad \mathcal{Z} \mapsto \mathcal{Z} \circ F $$ whose image consists of those $\mathcal{Z}': \mathcal{C} \to \mathcal{S}$ such that $\mathcal{Z}'(A \otimes A \xrightarrow{ \mu } A \xrightarrow{ \tau } 1_{\mathcal{C}})$ is a non-degenerate pairing.

If $\mathcal{S}$ is an $\infty$-groupoid, then this condition is trivially satisfied, so we should have an equivalence between the functor categories. This implies that $F$ induces an equivalence $B\mathcal{C} \to B\mathrm{Cob}(2)$. However, one can show that this is not true. I'll be very brief on this. Any $E_2^{\rm fr}$-algebra in an $\infty$-groupoid can canonically be trivialised along the unit morphism $1_{\mathcal{S}} \to A$, so we basically just have to give an $\mathrm{SO}_2$-equivariant morphism $1_{\mathcal{S}} \to 1_{\mathcal{S}}$ where both sides are equipped with the trivial action. This implies that: $$ B(\mathcal{C}) \simeq \Omega^\infty \Sigma^{\infty+1} (B\mathrm{SO}_2)_+. $$ So the answer to the question is no since the connective spectra $\Sigma^{\infty+1} (B\mathrm{SO}_2)_+$ and $\tau_{\ge 0} \mathrm{MTSO}_2$ are not equivalent.

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The above argument is very roundabout and actually quite subtle: there is actually a map of spectra $$ \tau_{\ge 0} \mathrm{MTSO}_2 \to \Sigma^{\infty+1} (B\mathrm{SO}_2)_+ $$ the fiber of which is $\tau_{\ge0} \Sigma^{-1}\mathbb{S}$. (This is known as the Genauer fiber sequence and admits an interpretation on bordism categories.) So in particular this map is a rational equivalence. This might make us hopeful as it looks like we weren't that far off. However, this map goes the wrong way and I actually believe that there is no rational equivalence that goes the correct way around.

So let me try to give a more down-to-earth proof that the functor $F:\mathcal{C} \to \mathrm{Cob}(2)$ is not an equivalence on groupoidification. For a traced $E_2^{\rm fr}$-algebra $A$ in $\mathcal{S}$ one can define $\tau(1):1_{\mathcal{S}} \to A \to 1_{\mathcal{S}}$ as the composite of unit and trace. If the traced $E_2^{\rm fr}$-algebra comes from a $2$-dimesional TFT, then this is the value on the $2$-sphere. The $\mathrm{SO}_2$-action on $A$ gives us a $2$-morphism $\alpha:\tau(1) \Rightarrow \tau(1)$. (The Dehn-twist pre- and post-composed with cap and cup.) Since both $1:1_{\mathcal{S}} \to A$ and $\tau:A \to 1_{\mathcal{S}}$ are $\mathrm{SO}_2$-equivariant we have two trivialisations of $\alpha$. Together they yield a $3$-morphism $\gamma: \mathrm{id}_{\tau(1)} \Rrightarrow \alpha \Rrightarrow \mathrm{id}_{\tau(1)}$. In the above I didn't actually use that $\mathcal{S}$ is an $\infty$-groupoid; one can construct this $\gamma$-invariant for any traced $E_2^{\rm fr}$-algebra in any $(\infty,1)$-category.

If the traced $E_2^{\rm fr}$-algebra comes from a $2$-dimensional TFT $\mathcal{Z}:\mathrm{Cob}(2) \to \mathcal{S}$, then $\tau(1) = \mathcal{Z}(S^2)$ and $\alpha$ is the Dehn twist along the equator. The $3$-morphism $\gamma: \mathrm{id}_{\mathcal{Z}(S^2)} \Rrightarrow \mathrm{id}_{\mathcal{Z}(S^2)}$ probably corresponds to the $S^1$-family of rotations of $S^2$ around some axis. In particular this $3$-morphism is always of order $2$ since $\pi_1 \mathrm{Diff}^+(S^2) = \pi_1 \mathrm{SO}_3 = \mathbb{Z}/2$.

Ok, so now all that's left to show is that there are traced $E_2^{\rm fr}$-algebras where $\gamma$ is not of order $2$. For this purpose let $\mathcal{S}$ be $K(\mathbb{Z},3)$ thought of as a symmetric monoidal $3$-groupoid that has a single object, morphism, $2$-morphism, and $\mathbb{Z}$-many $3$-morphisms. Then we can take the trivial $E_2^{\rm fr}$-algebra $1_{\mathcal{S}}$ in there, equip it with the trivial trace $\tau:1_{\mathcal{S}} \to 1_{\mathcal{S}}$, and give this trace non-trivial coherence data.

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This ended up much longer than intended, but I hope it's helpful. Let me know if I should clarify anything.

I'd be very curious to see a conjectural generators and relations description of $\mathrm{Cob}(2)$ that yields both the right homotopy category and the right classifying space. (Though this seems to be very difficult as the $\Omega^{\infty} \Sigma^{-1} \mathbb{S}$ that turned up above is in some sense not finitely generated...)

Answer 3

Edited answer: As I said in the comment, I think that the structure you're looking for is that of a cyclic algebra over the framed little disc operad, which should coincide with the notion suggested by Noah, ie a framed $E_2$-algebra equipped with a non-degenerate invariant trace compatible with the structure.

Recall that a PROP is a symmetric monoidal ($\infty$-)category which is equivalent to one with has $\mathbb{N}$ as a set of object, with tensor product given by addition. An algebra over a PROP is just a symmetric monoidal functor out of it. A PROP is cyclic if the object $1$ (not to be confused with the unit !), hence every other object, is self-dual, and the evaluation is a symmetric pairing (ie formally the canonical pivotal structure you get is the identity). If $P$ is a PROP then you can look at the endomorphism operad of $1$. This operation has a left adjoint giving the free PROP on an operad: roughly just declare that $Hom(n,1)$ is the $n$th operation space of your operad, and extend the obvious way.

If the PROP is cyclic, ie $1$ is equipped with a symmetric non-degenerate pairing, then the endomorphism operad of 1 has a canonical cyclic (in fact automatically modular) structure the usual way, using the isomorphisms $$Hom(n,1) \cong Hom(n \otimes 1,0)\cong Hom(n+1,0)$$ given by this pairing to extend the $S_n$-action to an $S_{n+1}$-action. Hence you get a cyclic operad, and this operation again has a left adjoint: take the same construction as before, and then there is a unique way of adding an evaluation and coevaluation, in such a way that the induced cyclic structure on the endomorphism operad of $1$ is the cyclic structure which was given in the first place.

Remark Note that "being cyclic" is an additional structure on an operad, and the same holds for cyclic algebras. On the other hand cyclicity for PROP is "built in", a cyclic PROP is just a special kind of PROP, which is why I find those convenient.

So: the familiar category whose objects are disjoint unions of oriented discs, and morphisms smooth oriented embedding, is nothing but the PROP associated with the framed little disc operad. This operad is equivalent to the operad of genus 0 Riemann surfaces, equivalently the disc PROP/category is equivalent to the PROP with objects disjoints unions of circles, and morphisms disjoint union of genus 0 surfaces with exactly one output boundary (given a configuration of discs in a larger disc, simply cut the smaller discs out).

Now the operad of genus 0 Riemann surfaces has an obvious cyclic structure. I claim that the cyclic PROP associated to it, is nothing but $Cob(2)$: basically you just formally add the macaroni to make $S^1$ self-dual and then all possible compositions. The as in the classical case all relations formally follow from the genus 0 ones, and from properties of the trace/pairing. Note that the fact that I can switch between pairing and trace follows from the fact taht everybody in sight is unital, ie in all those PROP's I have a morphism $0 \rightarrow 1$ given either by embedding of the empty disc and by the "cap" respectively. This last bit is wrong, of course. It's true the image of circle is always a "categorified Frobenius algebra", ie a cyclic framed $E_2$-algebra, bu you need some extra conditions, perhaps extra structure to extend to the whole bordism category.

Warning I haven't actually checked this carefully. I'm fairly sure it was explained, in the slightly different language of modular operads, in an old paper of Getzler but I haven't been able to find it.