Notion of a distribution as acting on tangent spaces

Question

I'm reading a paper that uses distributions in a way I'm not entirely comfortable with. To be precise, I'm not sure what definitions the author is working with and can't find any natural way to fill in the gaps.

I'll first ask the question in greatest generality, avoiding the context of the paper. I'll then restate the question in more detail, should the context become important.

Put generally, for functions: Consider a continuous map $f: M \to N$ between smooth manifolds. One would like to take the usual derivative $df$, but $f$ may not have sufficient regularity. However, if we view $f$ as a distribution then $df$ exists in the distribution sense, by passing the differential operator to the test function. That being said, is there still a notion of $df$ as a map between tangent spaces if we view it as a distribution? I cannot think of a natural way to define this.

Slightly more in the direction of what I am reading, say that $v$ is a vector field on $M$. We want to apply some operator to $v$ that turns it into a $(1,1)$-tensor. This operator might only give us a distributional object. Can we still think of it as giving us endomorphisms of the tangent space?

Most specifically: I am reading some notes on quasiconformal deformations of hyperbolic structure. Let $\mathbb{H}^n$ be hyperbolic $n$-space for $n \geq 3$, and consider $S_{\infty}^{n-1}$, the boundary at infinity. A deformation is a quasiconformal vector field $v$ on $S_{\infty}^{n-1}$. That is, a vector field whose associated strain tensor $Sv$ is in $L^{\infty}$ as a distribution. To me, that means that the integral of $Sv$ against a smooth, compactly supported $(1,1)$-tensor in the $L^{\infty}$ unit ball is uniformly bounded.

The author then goes on to consider the eigenspaces of this tensor as determining an invariant plane field on $S_{\infty}^{n-1}$. This is where my confusion arises: doesn't such an argument implicitly view the tensor as acting by endomorphisms on tangent spaces? How can this be reconciled with the fact that the tensor is defined distributionally?

Edit: The paper is McMullen's Renormalization and 3-manifolds which Fiber over the circle, specifically page 23 of the paper (page 29 of the PDF), although this shows up in other places.

Thanks for any insight you can share!

Answer 1

I won't give a complete answer to this (for an obvious reason), but here are some possibly helpful clues.

The reason why distributions allow you to differentiate non-differentiable functions is because of the "integration by parts" formula. In other words, if you multiply your non-differentiable real-valued function $f$ by a very smooth test function $\phi$ with compact support, then the integral of $f'\phi$ is the same as the integral of $-f\phi'$, or something like that, when $f$ is differentiable. So when $f$ is not differentiable, you just generalise the derivative by saying that it is defined by its action on $\phi$ when integrated in this way. (I am assuming that you know all about distributions. So I am just summarising the relevant bits very colloquially so that I can get to the next step. I haven't used distributions for many years.)

Now suppose your function $f$ is valued in $R^n$. You could regard this as $n$ real-valued functions. In fact, I guess you can do this for any Banach-space-valued function. (There are many mentions of vector-valued distributions in a certain popular search engine.)

Now here is my main point. Since the very simplest kind of target manifold is going to be a perfectly flat Euclidean space, you need to be able to handle at least that case. This looks very much like a vector-valued distribution question. Then one may ask how to extend target Euclidean spaces to target manifolds. Unfortunately, the tangent space on the target manifold is a vector bundle, not a vector space, and the manifold itself is not even a vector bundle.

The main clue, I think, is to try to find some analog of the integration by parts for functions valued in a Euclidean space, and then extend this to functions valued in a manifold. The problem is to define some notion of "average". When you integrate the product of a function $f$ multiplied by a test function $\phi$, you are averaging $f$ by smoothing it. The integral of a real-valued function on a manifold works for distribution theory because $R$ is a linear (or affine) space. What does it mean to average out or smooth out the values of a function $f:M\to N$? For example, what is the average of two points $p_1,p_2\in N$? If that doesn't mean anything, then nor does the integral of $f\phi$ over a region of $M$. One answer could be to use geodesics to interpolate between $p_1$ and $p_2$. But this lacks the kind of commutativity properties required for an average or integral. The order in which you do the sums in, say, a Riemann integral affects the answer.

Conclusion:
If you can't say what the weighted average of a set of points in $N$ is, then you can't really define $N$-valued distributions in an obvious way.