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Can you have $\infty$-forms on infinite-dimensional manifolds or elsewhere and what are they used for?

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I know physicists speak of semi-infinite differential forms on loop spaces in some analogy to the Sato Grassmannian. I don't know how much of this is rigorous mathematics yet though. This is "used for" mirror symmetry. – Dan Petersen Sep 22 '11 at 12:41
(the word "infinity" seems a bit overloaded here... What do you mean by "infinity-forms"?) – Qfwfq Sep 22 '11 at 13:54
Please read the "how to ask" page and revise your question. – S. Carnahan Sep 23 '11 at 14:45

In the words of Bob the Builder: "Yes, we can."[1]

At the outset, I'll like to note that my thesis was called "A construction of semi-infinite de Rham cohomology".

There's a nice bit to the story and a not-so-nice bit.

The nice bit is the first bit. Let $M$ be a smooth manifold, finite or infinite dimensional. Let $E \subseteq T M$ be an integrable sub-bundle. Then it is possible to define bundles of "forms that differ from $\omega_E$ by a finite amount". Here, $\omega_E$ is a possibly hypothetical top form for $E$ (so if $E$ was locally $x^1, \dots, x^k$ then $\omega_E$ is proportional to $dx^1 \wedge \dots \wedge dx^k$).

The motivation behind the construction is as follows. One way to think of $\Lambda^k \mathbb{C}^n$ is as the space of holomorphic sections of the determinant bundle over the Grassmannian of $k$ planes in $\mathbb{C}^n$. So given a pair $V \subseteq H$ where $V$ is a (closed) subspace of $H$ we can take the Grassmannian of (closed) subspace of $H$ that differ from $V$ by a "finite" amount. This will (usually) have a determinant bundle as we can take holomorphic sections thereof to get the exterior power of things that differ from the top form on $V$ by a finite amount. (The word "finite" is in inverted commas because we don't always insist on actually being a finite difference from $V$, but "finite" is close enough for the idea.)

In nice circumstances, we can actually reinterpret that space as an honest exterior power: if $H$ is a Hilbert space we can take $\Lambda V^\perp \otimes \Lambda V^*$ (with a suitable completion). The idea here is that a finite codimension subspace of $V$ is like a finite dimensional subspace of $V$, so a form that is a bit less than the top form of $V$ is like a form that is a bit more than the bottom form of $V^*$.

So we can define the exterior powers of $T^* M$ relative to $E^*$. This is no longer an algebra, but is a module over the exterior algebra of $T^* M$.

In fact, we don't even need $E$ to be globally defined. We only need it to be defined "up to finite ambiguity". This is the secret behind so-called semi-infinite cohomology: the polarisation of a polarised manifold defines a "half-way" subspace up to "finite ambiguity".

Then the corresponding forms are simply sections of this bundle.

So for "infinite forms", we can simply take $E = T M$ and consider "finite codimension" forms.

Now comes the not-so-nice bit. Although we have forms, and a nice module structure over the exterior algebra, we cannot define the exterior derivative if $E$ is infinite dimensional. The problem is one of convergence. A typical $\infty - 1$ form will look like $f_x d\hat{x}$, where $d\hat{x}$ means "all but $d x$". Its exterior derivative will be $\sum_x \frac{\partial f_x}{\partial x} d x \wedge d\hat{x}$. But $d x \wedge d\hat{x}$ is just the top-form so all of these terms add together to give the single contribution of the top-form. There are infinitely many terms, and so we have to worry about convergence.

The simplest way around this is to only take forms for which this sum is finite. These are so-called tame forms. The easiest place to define them is on a Hilbert manifold where we have a suitable filtration of the manifold.

Unfortunately (or fortunately depending on your point of view), this ends up meaning that there isn't all that much information in this $\infty - k$ cohomology: for reasonable manifolds it ends up being isomorphic to ordinary homology (note the variance flip).

Semi-infinite cohomology ought to be more interesting, but it's also harder to work with.

Looking through the references on my thesis, the name "K.K. Mukherjea" looks one to search for. In particular, articles called: "Cohomology theory for Banach manifolds" and "The homotopy type of Fredholm manifolds". Other names to look for are Eells, Elworthy, and Palais.

[1] To any Americans reading, Bob the Builder's been using that longer.

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The question remains if this is what the OP is asking for. I guessed that it's not. Hopefully he'll come by again to clarify. – Urs Schreiber Sep 22 '11 at 19:01
Even if it isn't what the OP was after, it may still be interesting to lots of other people on mathoverflow. – euklid345 Sep 22 '11 at 19:42
I guessed that it was! If not, at least it's brought this back to the forefront of my mind and I'll stick it on my list of "thinks to write an nLab article about". Will that make you (Urs) feel better? – Loop Space Sep 23 '11 at 7:15
I vote for Andrew's interpretation, but I think the person asking the question should have spent a little more effort to make the question less ambiguous. – S. Carnahan Sep 23 '11 at 14:45

The way you phrase your question it sounds to me as if you are mixing up the $n$ in "$n$-category" with the $n$ in "$n$-dimensional manifold". While there are relations (for instance the n-catgeory of cobordisms is "built from" $n$-dimensional manifolds) in other contexts these are two entirely unrelated numbers.

So when you say "$\infty$-plectic" I am guessing that you are thinking on "n-plectic geometry" possibly as absorbed into some context of higher symplectic geometry that knows about symplectic ∞-groupoids?

But, as far as I can see, all the actual forms appearing here, representing certain de Rham classes on smooth ∞-groupoids, are of finite degree.

I wouldn't readily know what it means to have a form of infinite degree. Except maybe this: one can argue that the degree-0 elements of the BV-complex for a $dim > 1 $-dimensional field theory represent "mid $\infty$-dimensional" forms. But I don't think that's what you mean here.

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