Let $\mathit{Bord}_d$ be the symmetric monoidal category of $(d-1)$-manifolds and bordisms between them. Let $\mathcal{C}$ be the symmetric monoidal category of $k$-modules. Then, for two symmetric monoidal functors $F,G:\mathit{Bord}_d \to \mathcal{C}$ their tensor product is defined by 'point-wise product': $$ (F\otimes G)(M) = F(M) \otimes G(M) \quad \text{and} \quad (F\otimes G)(X:M \to N) = F(X) \otimes G(X). $$ We can also define their direct sum, using the fact that any manifold $M$ decomposes into connected components $M = \amalg_{i \in \pi_0 M} M_i$: \begin{align} (F \oplus G)(M) &:= \bigoplus_{\pi_0 M = A \amalg B} F(\amalg_{a\in A} M_a) \otimes G(\amalg_{b\in B} M_b) \\ &\cong \bigoplus_{\pi_0 M = A \amalg B} \otimes_{a\in A} F(M_a) \otimes \otimes_{b\in B} G(M_b) \end{align} (Defining this on morphisms is a bit more subtle and uses the fact that $\mathcal{C}$ is an additive category.)

One can show that both operations yield symmetric monodial functors. And I suppose it is not to hard to show that they induce the structure of a 'rig-category' on the category (actually, groupoid) of symmetric monoidal functors from $\mathit{Bord}_d$ to $\mathcal{C}$. (Here 'rig'='ring without inverses', ie. two symmetric monoidal structures that are compatible in a distributive way.)

Note that this direct-sum operation on TQFTs has been studied from the mathematical physics side, here a random example, but does not seem to appear in the mathematical literature. Any references are very much welcome.

My question now is how to set up these definitions in the $\infty$-categorial world.

More precisely, if $\mathit{Bord}_d$ is the $(\infty,1)$-category of bordisms and $\mathcal{C}$ is a rig-$(\infty,1)$-category$({}^*)$, how can we define a rig-structure on the $(\infty,1)$-category $$ TQFT_d(\mathcal{C}) := \mathit{Fun}_{(\infty,1)}^\otimes(\mathit{Bord}_d,\mathcal{C}) $$ such that the evaluation at any $(d-1)$-manifold $M$: $$ ev_M: TQFT_d(\mathcal{C}) \to \mathcal{C},\quad F \mapsto F(M) $$ is a rig-homomorphism for $M$ connected, and symmetric monoidal wrt $\otimes$ for general $M$?

$({}^*)$ Maybe this only makes sense when $\mathcal{C}$ is a stable $(\infty,1)$-category with bilinear symmetric monoidal structure and the rig-structure is the one coming from direct sum and symmetric monoidal product.

My attempts so far are as follows:

One easily defines the multiplicative part of the rig-structure using the $\otimes$ on $\mathcal{C}$.

The additive $F\oplus G$ is not the Day convolution $F*G$ as discussed in this MO question. (This would be for the case that $\oplus$ on $\mathcal{C}$ is the categorical coproduct. However, this approach fails as $F*G$ is lax monoidal, but not always strong monoidal.)

One might try to give a construction similar to the Day convolution by taking a Kan-extension among symmetric monoidal functors, but this and similar universal constructions are unlikely to work as all natural transformations between TQFTs are invertible.

It seems to me as if the fact that $\mathit{Bord}_d$ has a unique 'prime decomposition' wrt $\amalg$ is essential to the construction. So I would be surprised (but positively surprised) to see a solution that does not make use of having this specific domain.

  • $\begingroup$ $\mathit{Bord}_d$ as it is usually defined is not an $(\infty,1)$-category in general, but rather an $(\infty, d)$-category; for example, the pair-of-pants cobordism is a noninvertible 2-morphism in $\mathit{Bord}_2$. Alternatively, one could define an $(\infty,1)$-category of bordisms by not extending to lower dimensions, but including diffeomorphisms, isotopies, etc.; is this what you had in mind for $\mathit{Bord}_d$? $\endgroup$ Aug 6, 2018 at 2:27
  • 1
    $\begingroup$ Yes, exactly. I meant non-extended TQFTs in dimension $d$. So objects are $d-1$ manifolds, morphisms are $d$-bordisms and all higher morphisms come from diffeomorphisms and isotopies between them. I suppose this is also sometimes denoted as $\mathit{Bord}_{\langle d-1, d \rangle}$. I made this assumption for simplicity, if there is an answer in the fully extended case I would also be interested. $\endgroup$ Aug 6, 2018 at 6:08


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.