# Why do we define TQFTs as functors to vector spaces instead of Hilbert spaces?

Let

• $$\mathrm{Cob}_n$$ be the category with objects closed oriented $$n-1$$-manifolds and morphisms being cobordisms identified upto boundary preserving diffeomorphism
• $$\mathrm{Vect}_\mathbb C$$ be the category of complex vector spaces and linear maps
• $$\mathrm{Hilb}_\mathbb C$$ be the category of complex Hilbert spaces and continuous linear maps
• $$\mathrm{Hilb}_\mathbb C^\mathrm{unit}$$ be the category of complex Hilbert spaces and unitary maps

Then my questions are:

• Why do we define a TQFT to be a functor $$Z\colon\mathrm{Cob}_n\to\mathrm{Vect}_\mathbb C$$? Isn't it closer to quantum mechanics to define $$Z\colon\mathrm{Cob}_n\to\mathrm{Hilb}_\mathbb C^\mathrm{unit}$$?
• Even if we want to consider nonunitary theories, then shouldn't we consider $$Z\colon\to\mathrm{Cob}_n\to\mathrm{Hilb}_\mathbb C$$?

What are the reasons we don't choose the other two categories?

Why do we define a TQFT to be a functor $$Z:\mathrm{Cob}_n→\mathrm{Vect}_\mathbb C$$? Isn't it closer to quantum mechanics to define $$Z:\mathrm{Cob}_n→\mathrm{Hilb}^\mathrm{unit}_\mathbb C$$?

Quantum mechanics is not a TQFT, but merely a functorial field theory.

TQFTs have fully dualizable images, which in the case of Hilbert spaces amounts to being finite-dimensional.

One way to encode time evolution in quantum mechanics is via a one-parameter semigroup of (say) unitary operators on a Hilbert space (possibly infinite-dimensional).

In terms of functorial field theories, this is encoded by a functor from the category of 1-dimensional manifolds equipped with a metric (i.e., length) to the category of Hilbert spaces and unitary maps. We send the point to the Hilbert space under consideration, and to an interval of length T we assign the value of the one-parameter semigroup at T. This assignment is functorial precisely because of the semigroup condition. Such a field theory is clearly nontopological, since its values depend on lengths of intervals.

Even if we want to consider nonunitary theories, then shouldn't we consider $$Z:\mathrm{Cob}_n→\mathrm{Hilb}_\mathbb C$$?

If nonunitary field theories are your goal, then certainly Hilbert spaces and bounded linear maps is a legitimate target category.

• 'Quantum mechanics is not a TQFT, but merely a functorial field theory.' However, a TQFT is a generalisation of quantum mechanics (to higher dimensions) and also a simplification (to the topological case instead of Riemannian). So why do we change our target category from $\mathrm{Hilb}_\mathbb C$ to $\mathrm{Vect}_\mathbb C$? – Chetan Vuppulury Jun 6 at 17:12
• @ChetanVuppulury: TQFT is not a generalization of quantum mechanics (it cannot encode time evolution or infinite-dimensional spaces). But you can certainly consider TQFTs valued in finite-dimensional Hilbert spaces, with unitary or arbitrary linear maps. The resulting theory is not substantially different. In fact, the value of a nonoriented 1-dimensional TQFT valued in finite-dimensional real vector spaces is automatically equipped with a canonical metric, which provides you with a way to introduce metrics using the structure of a TQFT. – Dmitri Pavlov Jun 6 at 17:29
• @Dmiti, in physics TQFT arises as a quantum mechanical system. So I should be able to view it as both a generalisation and simplification of quantum mechanics in the sense of my previous comment. We can think of the cobordisms to represent 'time evolution' though 'time' doesn't really mean anything in a topological theory as there is no metric. That's why I said it's a simplification. Please do correct me if I am wrong. Maybe I should rephrase my question as: What difference does it make if I look at functors $Z\colon\mathrm{Cob}_n\to\mathrm{Hilb}_\mathbb{C}^\mathrm{unit}$ rather ...continued – Chetan Vuppulury Jun 6 at 18:34
• ... than into $\mathrm{Vect}_\mathbb{C}$. Is there a physical reason to make that replacement. Of course, as a mathematician I would be interested in functors into any good category, but I am looking for a more physical motivation. – Chetan Vuppulury Jun 6 at 18:35
• @ChetanVuppulury: As I mentioned in my comment above, you consider either category as a target. However, unitary operators as a target essentially do not allow for any interesting TQFTs because any unitary operator is an isomorphism of underlying vector spaces, which forces the dimension of the Hilbert space assigned to any object to be equal to 1, making the theory trivial. – Dmitri Pavlov Jun 6 at 20:13