TITLE: What are TQFTs that are multiplicative under connected sums? Do bordisms with connected sum as monoidal product exist?
QUESTION [9 upvotes]: In general, one extracts a manifold invariant from a TQFT by interpreting the closed manifold as a bordism from the empty set to the empty set. The TQFT sends this bordism to a homomorphism of the ground field, which is a number. Such invariants are always multiplicative under disjoint union, this is a consequence of the TQFT being a monoidal functor:
$$\mathcal{Z}(M_1 \sqcup M_2) = \mathcal{Z}(M_1) \otimes \mathcal{Z}(M_2) = \mathcal{Z}(M_1) \cdot \mathcal{Z}(M_2)$$
Some TQFTs, like the Crane-Yetter invariant (but not, say, the Turaev-Viro model) give manifold invariants that are multiplicative under connected sum $\#$.
One way to see this is to notice that they can be defined (for connected manifolds) with Kirby calculus: Given a manifold, choose a handle decomposition and consider its link diagram. The diagram is then labelled with morphisms from a ribbon fusion category and the whole diagram is evaluated as a morphism from the monoidal identity to itself, again a number.
Now the evaluation of the disjoint union of two link diagrams must then give the product of the evaluations of the respective diagrams, since a ribbon fusion category is monoidal.
But the disjoint union of two link diagrams of manifolds $M_1$ and $M_2$ is the link diagram of the connected sum $M_1 \# M_2$!
Which leads me to believe that this multiplicativity secretly comes from monoidality of some functor again. Is there a category of bordisms where the monoidal product of morphisms (=bordisms) is the connected sum, and not the disjoint union? Are such TQFTs actually monoidal functors from this bordism category to $\mathrm{Vect}$?
A related, noncategorical question was asked here: Monoid structure of oriented manifolds with connect sum

REPLY [6 votes]: If the dimension of $Z(S^{n-1})$ is greater than 1, then the TQFT is not even approximately multiplicative under connect sum.
If $Z(S^{n-1})$ is 1-dimensional, then a simple cut and paste argument shows that
$$ Z(M_1 \sharp\, M_2) = Z(M_1 \sqcup M_2) \,/\, Z(S^n) = Z(M_1) \cdot Z(M_2) \,/\, Z(S^n) . $$
So in this case $Z$ is multiplicative under direct sum iff $Z(S^n) = 1$.
(If $Z(S^n) \ne 1$ we might say that the theory is approximately multiplicative under connect sum.)
When $n$ is even, the Euler characteristic of $S^n$ is non-zero and we can tensor with an Euler characteristic theory to arrange that $Z(S^n) = 1$.
When $n$ is odd, the Euler characteristic of $S^n$ is zero and there is no way to tweak things so that $Z(S^n) = 1$.  Turaev-Viro and Witten-REshetikhin-Tureav theories fall into this class.

I'll also remark that if $M_i$ is not connected, then $M_1 \sharp\, M_2$ is not well-defined, even up to homeomorphism.  But nevertheless the above claims about $Z(M_1 \sharp\, M_2)$ are true, for any choice of the ambiguous connect sum.