Exterior Differential (and its Equivalent Differential Operator) of an Integral 0-Form

Question

I am reading Witten's 1982 paper "Supersymmetry and Morse Theory," and while I am slowly learning the material as I read through the paper, I have come across an equivalence that, while it should be rather basic, I can't seem to show rigorously.

In the paper, Witten defines a (modified) exterior derivative $\text{d}_t$ as

$$\text{d}_t = \text{d} + t\, \text{d} h$$

where

$$h(\phi) = \int_S W(\phi(x)) \, \text{d}x,$$

the region $S$ is a circle with circumference $L$ and $\phi$ is a real-valued function from $S$ to the reals, i.e. $\phi: S \rightarrow \mathbb{R}$.

Now, I am having trouble proving to myself that this exterior derivative $\text{d}_t$ is equivalent to the standard differential operator

$$\frac{d}{d \phi} + t L \frac{dW}{d\phi}$$

acting on real-valued functions of the variable $\phi$.

Now, as far as I can tell, Witten is making the identification $d \leftrightarrow \frac{d}{d \phi}$ here for this equivalency. However, given that, I can't seem to get the remaining term $t L \frac{dW}{d\phi}$.

My attempt:

We make the identification $\text{d} \leftrightarrow \frac{d}{d\phi}$, so the last term in the expression for $\text{d}_t$ is simply $\frac{dh}{d \phi}$. So, it should be as simple as evaluating

$$\frac{d}{d\phi} \int_S W(\phi(x)) \, \text{d}x.$$

However, in evaluating this, the trouble I am running in to is that $\phi$ is an arbitrary function that is not necessarily invertible, so I cannot just rewrite the integral as an integral over $\phi$.

I have also tried using the mechanics of exterior derivatives first, but then I run into the issue of evaluating

$$\text{d} \int_S W(\phi(x)) \, \text{d}x,$$ which I'm not so sure where to begin (the exterior derivative is defined on the real line, not on $S$). Possibly, we can commute the derivative, which leaves us with $$\int_S \text{d} (W(\phi(x)) \, \text{d}x).$$ Then, using one of the identities of the exterior derivative yields $$\int_S \text{d} (W(\phi(x))) \land \text{d}x,$$ at which point I think the correct next step would be $$\int_S \frac{dW}{d\phi} \,\text{d} \phi \land \text{d}x.$$

From here, though, it's not quite clear what to do. Do we integrate over the region $S$, taking the integrand (a 2-form) to a 1-form? If so, this leads me to conclude that the result be (something like) $$L \frac{dW}{d\phi} \, \text{d} \phi,$$ which, with the identification, would be $$L \frac{dW}{d\phi} \frac{d}{d\phi},$$ not what we expected.

Where is my reasoning flawed? How do I obtain the result desired?

This post imported from StackExchange Mathematics at 2016-01-19 15:54 (UTC), posted by SE-user Sam Blitz