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I was reading about the appearance of Calabi-Yau manifolds in Feynman integrals, and I thought to wonder if there is such a thing as "infinite-dimensional Hodge theory". Googling the phrase turned up references to a 2010 lecture by Maxim Kontsevich called "Infinite-dimensional Hodge theory of path integrals". Does anyone know what its content was?

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The most usual Hodge theory is related to integrals, periods of algebraic differential forms on algebraic varieties, whereas the physics path integral usually involves integration of an exponential quantity. So the first thing to understand is that there is a finite dimensional Hodge theory for "exponential integrals", i.e. integrals of exponential of algebraic objects. It is sometimes called "irregular Hodge theory" (because unlike the usual Gauss-Manin connection of Hodge theory which admits regular singularities, exponential periods satisfy in general differential equations with irregular singularities). One of the numerous motivations for this theory is mirror symmetry for Fano varieties.

Some relevant names are probably: Kontsevich, Barannikov, Sabbah, Esnault, Yu, Mochizuki...

So the thing which should be more related to path integrals should be an infinite dimensional version of irregular Hodge theory. Unlike the finite dimensional case, I don't think there is yet a rigorous mathematical theory. But one can take some finite dimensional phenomenons, like the Stokes phenomenon, and see if they have some infinite dimensional counterparts (e.g. for Chern-Simons theory).

On the website of the Simons Center, you can find two video lectures by Kontsevich, with the title "Exponential integrals":

http://scgp.stonybrook.edu/video_portal/video.php?id=1798

and on the youtube channel of the IHES, you can find four video lectures with the same title:

https://www.youtube.com/watch?v=tM25X6AI5dY&index=1&list=PLx5f8IelFRgE098k6hVHriQyfMkoq--T0

In both cases, these lectures start with some finite dimensional theory and go to some infinite dimensional examples towards the end.

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    $\begingroup$ Is it the same Mochizuki as for IUTT ? $\endgroup$ – Sylvain JULIEN Oct 24 '18 at 15:15
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    $\begingroup$ I think the answer refers to Takuro Mochizuki. $\endgroup$ – j.c. Oct 24 '18 at 15:54

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