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Correct me if I am wrong but I believe at least conceptually (maybe even rigorously) data of a 1-dimensional TQFT and of a vector bundle with connection are equivalent.

Going into more detail (and consequently making more and more mistakes), a vector bundle with connection allows to assign to a point the fibre over that point, and to a path the monodromy (or holonomy? Yes I am that ignorant) along this path.

Three questions, but closely related:

Does there exist equally layman-ish description of going back from a TQFT to a vector bundle?

Clearly this requires the base to be at least a smooth manifold. Is there known any TQFT-like object (and maybe also connection-like object) that would work in the non-smooth context? Say, for topological manifolds, or even arbitrary finite CW-complexes?

What if one leaves manifolds alone but removes the connection? Is there known a version of TQFT that would work for vector bundles with arbitrarily severe restrictions (say, very-very nice algebraic vector bundles on very-very good algebraic varieties) but without any additional structure?

Two remarks:

Sort of minimal version of the question is whether a vector bundle $p:E\to[0,1]$ comes with any kind of map (relation? correspondence?) between $p^{-1}(0)$ and $p^{-1}(1)$. Vague association with motives comes to mind but that's all my mind offers.

Obviously I am tempted to ask the same about 2D-TQFT. But I am (almost) successfully resisting this temptation.

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1d TQFT's are in 1-1 correspondence with finite dimensional vector spaces, and the image of the circle is the dimension of that vector space.

I think what you have in mind is instead the notion of $X$-structured TQFT, aka homotopy quantum field theory. Those are defined for pairs of a topological manifold together with a map into some fixed topological manifold $X$.

So in the 1d case, you are looking for symmetric monoidal functors into Vect from the category $Bord_1^X$ which has

  • objects points together with a map into $X$, so the set of objects is just $X$.
  • morphisms bordisms between those, i.e. maps between two points of $X$ are intervals equipped with a map into $X$ in a compatible way, i.e. at the end of the day this is just a path in $X$ between your points.

Now you want to look at this up to homotopy, so the bottom line is that $Bord_1^X$ really is just the fundamental groupoid of $X$, and a 1d $X$-HQFT is thus a flat vector bundle on $X$. So not only the connection is important, but it has to be flat.

Now I'm not entirely sure, but I think given a vector bundle on $X$ with a non-necessary flat connection $A$, you get in fact an example of a 2-dimensional HQFT, where very roughly the value on a 2-dimensional surface equipped with a map into $X$ is computed by integrating the pull-back to your surface of the curvature 2-form of $A$. This is basically saying every connection is automatically "2-flat" thanks to the Bianchi identity.

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  • $\begingroup$ Many thanks for the correction, and the last part is especially interesting! So, what is a flat vector bundle over a non-smooth manifold? And, could you please indicate (say, give a reference for) how to reconstruct it from an $X$-HQFT? $\endgroup$ – მამუკა ჯიბლაძე May 29 at 9:40
  • $\begingroup$ A flat connection on a bundle is a lift of the structure group from $G$ to $G^\delta$, the group made discrete. (This is a topological notion, no smoothness needed.) This is equivalent to a representation $\pi_1(X) \to G$ which gives rise to a $G$-bundle isomorphic to your given bundle. $\endgroup$ – Mike Miller May 29 at 11:15
  • $\begingroup$ In general you can always talk about representations of the fundamental groupoid, which are the same as locally constant sheaves on $X$. $\endgroup$ – Adrien May 29 at 11:30
  • $\begingroup$ As for your last question, rephrasing what Mike said, giving an HQFT, i.e. a functor $\prod_1(X)\rightarrow Vect$, is the same as giving a vector space $V_x$ for each point $x \in X$, and an iso $V_x \cong V_y$ for every path between $x$ and $y$ which depends only on the homotopy class of that path in a way compatible with composition, so this induces a locally constant sheaf (aka locall system) on $X$ in a fairly tautological way (every choice of a contractible $U$ containing $x,y$ identifies $V_x$ and $V_y$ canonically). $\endgroup$ – Adrien May 29 at 17:26
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Going into more detail (and consequently making more and more mistakes), a vector bundle with connection allows to assign to a point the fibre over that point, and to a path the monodromy (or holonomy? Yes I am that ignorant) along this path.

Yes, this is pretty much correct. You can indeed assign holonomy to paths. Furthermore, there is an equivalence between 1-dimensional smooth TFTs over X and vector bundles with connections over X. Precise definitions with proofs can be found in the paper https://arxiv.org/abs/1501.00967.

Does there exist equally layman-ish description of going back from a TQFT to a vector bundle?

Yes. The underlying vector bundle without a connection can be recovered by evaluating the smooth TFTs at the smooth family of points given by the manifold itself. The connection is recovered by differentiating the parallel transport map.

Clearly this requires the base to be at least a smooth manifold. Is there known any TQFT-like object (and maybe also connection-like object) that would work in the non-smooth context? Say, for topological manifolds, or even arbitrary finite CW-complexes?

Yes. Replace the site of smooth manifolds with the site of topological manifolds or the site of finite CW-complexes.

What if one leaves manifolds alone but removes the connection? Is there known a version of TQFT that would work for vector bundles with arbitrarily severe restrictions (say, very-very nice algebraic vector bundles on very-very good algebraic varieties) but without any additional structure?

Yes, this is the (∞,1)-version of 1-dimensional TFTs. (“Holonomy” is now not a strict functor, but an (∞,1)-functor, which no longer produces a connection.) See, for example, the survey by Lurie and the more recent work by Chris Schommer-Pries.

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