Why numerical invariants of n-manifolds are secretly n-categories

So, in another thread Scott C. questioned why I would say that TQFT’s produce all numerical invariants of n-manifolds. There is a good reason for this, though maybe not good enough.

The reason is that there is a construction for starting with a numerical invariant, and extending it to a TQFT:

• Start with a numerical invariant $n(M)$ of n-manifolds.
• Now consider your favorite n-1-manifold X.
• Consider the vector space $\tilde V(X)$ spanned by n-manifolds with isomorphisms of their boundary to X.
• The space $\tilde V(X)$ has an inner product, given by gluing two n-manifolds along X, and summing their invariants. Quotient by the kernel of this and call this $V(X)$.

Congratulations. You’ve got your TQFT. This construction is a little weird; in particular, it doesn’t seem to have to be monoidal, which is bad. Still, it tells you TQFT’s are in some sense easier to construct than you probably thought.

Presumably, one can extend this further by giving an n-2-manifold the category whose objects are formal sums of manifolds bounding it, with morphism spaces given by gluing, etc. Anyways, this is what I had in mind when writing the infamous sentence in question.

7 thoughts on “Why numerical invariants of n-manifolds are secretly n-categories”

1. Jan says:

If I believe the wiki page on TQFT’s, there still is some problem relating the Atiyah-Segal axioms to Witten type TQFT’s, in particular the Seiberg-Witten invariants.
Is this true? Or can you give some explanation for a person not so well-versed in TQFT’s why your argument works here (I especially have some problems visualizing $\tilde{V}(X)$, what are the vectors? )

2. 1. What I wrote above doesn’t necessarily satisfy the Atiyah-Segal axioms (it might not be monoidal). Also, its hopelessly uncomputable in general.

2. $\tilde V(X)$ is just a huge infinite-dimensional vector space, with basis given by every way of making X isomorphic to the boundary of a n-manifold. It’s way too big to visualize.

3. martin says:

@jan:
In the case of Seiberg-Witten invariants, one reason is the following:
Seiberg-Witten invariants of 4-manifolds are only defined under assumtions on $b_2^+$, everything is fine for $b_2^+\geq 2$ and for $b_2^+=1$ the so called wall-crossing appears.
Therefore it’s not possible to consider the whole cobordism category as in the axiomatic approach.

4. Hi Ben,

While this construction produces something, I have a hard time viewing it as a TFT.

I like examples, so here is a good one. Fix your favorite n-manifold X. My favorite numerical invariant of closed n-manifolds assigns to a manifold M the number one if M is isomorphic to X, and zero otherwise. The thing you get is totally bizarre and practically uncomputable. It doesn’t respect tensor products, as you pointed out.

In fact, if n=2, and your favorite manifold X is the torus, then you can work out what happens explicitly. We have

$dim \tilde{V}( \emptyset )= 1$

$dim \tilde{V} (S^1) = 2$

and

$dim \tilde{V} (S^1 \cup S^1) = 4$

but the linear maps assigned to bordisms are crazy! If you take the disk and view it as a bordism from the empty set to the circle, then it induces the zero map! Viewed the other way around, it induces a surjective map.

5. Ben Cooper says:

Monoidality is a strong requirement.

6. Charles Frohman says:

Whether the construction you gave leads to a TQFT is relatively (heh heh) easy to check. In comparison, checking that it gives an extended TQFT is monstrously hard to check. In fact, I am pretty skeptical about the existence of many extended TQFTs. For instance in Chris Schommers-Preis Thesis he really limited which 1+1 dimensional TQFT give rise to extended ones. Building down… is hard.

7. Charlie Frohman says:

Here is a disturbing example.

The Frobenius extension underlying the $sl(3$)-knot homology is $A=Z[x]/(x^3)$ with $\epsilon:A\rightarrow \mathbb{Z}$, $\epsilon(1)=\epsilon(x)=0$ and $\epislon(x^2)=-1$.

The corresponding invariant of surfaces sends any surface not of genus one to 0, and a genus one surface to 3.

Starting with just the invariant of surfaces, try to build down to get A.