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If $M$ is an oriented $d$-manifold, let $D(M)$ denote the top exterior power of $H^*(M,\mathbf{C})$. Then $D(M_1 \amalg M_2) = D(M_1) \otimes D(M_2)$. Is there a good recipe for a map $D(M) \to D(N)$ induced by a cobordism from $M$ to $N$?

In some dimensions, there is a natural identification $D(M) = \mathbf{C}$ and you could take every cobordism to the identity map. But such a TQFT would not be completely trivial when $d = 4$, I think. For example it would distinguish $S^1 \times \mathbf{C}P^2$ from the mapping torus of complex conjugation.

If there's no problem defining it for $(d+1)$-manifolds, can it be "extended down" any distance, i.e. associate something to a $(d-1)$-manifold?

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    $\begingroup$ This TFT would presumably be invertible, so can in principle be classified using the cohomology of Madsen-Tillman spectra. There's a known invertible 2d TFT which assigns to a closed oriented surface the number $\lambda^{\chi(X)}$ where $\chi$ is the Euler characteristic and $\lambda \in \mathbb{C}^{\times}$, and this TFT is a categorification of that. Maybe this paper of Freed will be helpful: arxiv.org/abs/1406.7278 $\endgroup$
    – Qiaochu Yuan
    Commented Jun 9, 2016 at 20:28
  • $\begingroup$ I'm sure the answer is "yes, $D$ is part of an invertible TFT", but I'm not seeing totally immediately how to write it down. From my point of view, you want to look at the "free topological boson on $M$", which is the theory whose field is a map $\phi: M \to \mathbb C$ and whose EOM is $\mathrm d \phi = 0$. This has a "cotangent quantization" in the Costello-Gwilliam language, which up to some degree shift is $D$. If you can realize it as a fully extended TFT, you win. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 10, 2016 at 1:52
  • $\begingroup$ I would have thought that the $d$-dimensional theory (or maybe I mean $d\pm 1$) would assign $\mathrm{pt} \mapsto \mathbb C$-as-an-$E_d$-algebra. But that defines the trivial (framed) theory. Perhaps there's a nontrivial choice of how to descend from a framed theory to an oriented one, but I'm not seeing it. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 10, 2016 at 1:56

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