3
$\begingroup$

This is a bit of a followup to my previous question Intuition for the volume form - combinatorial definition?. I am looking for a certain combinatorial intuition when it comes to integrating differential forms.

Suppose, for concreteness and simplicity that $X^d$ is a compact oriented manifold with $X^d \subset \mathbb{R}^N$ and suppose that $\omega \in \Omega^d(X)$. Suppose that $\Delta_n$ is a sequence of triangulations of $X$ where the lengths of the edges (with the induced metric from Euclidean space say) of the triangulations approach $0$ as $n \to \infty$ — maybe I will just go ahead and write $\Delta_n \to X$.

I am going to write down a sequence of real numbers that I believe converges to (in an appropriate sense) $\int_X \omega$ (and if this is the case, would help my geometric understanding of $\int_X \omega$). At the risk of being overly pedantic, let me spell it all out. I apologize in advance for the indices.

Suppose that the triangulation $\Delta_n$ consists of the top-dimensional simplices $\{ \Delta_n^1,\dotsc,\Delta_n^{n_m}\}$. Let $p_n^i$ be say a preferred corner of each $\Delta_n^i$ and let $v_{p_n^i}^1,\dotsc,v_{p_n^i}^d$ be an oriented ordered orthonormal basis for the tangent space $T_{p_n^i}X$ (using the metric induced by the Euclidean metric). Finally, let $S_n = \sum_{i = 1}^{n_m} \omega_{p_n^i}(v_{p_n^i}^1,\dotsc,v_{p_n^i}^d)$.

Is it true that the $S_n$ converges to $\int_X \omega$ in the sense that the liminf of all such limits over all such sequences of such triangulations is $\int_X \omega$? (From the helpful comments in the previous question that I linked to, I learned that it does not suffice to just take an arbitrary sequence of such triangulations.)

If not, is there a way to salvage this sort of combinatorial approximation approach (maybe just in low dimensions)? I find the existence of such approaches always helps my intuition.

$\endgroup$
2
  • $\begingroup$ Trivial typo: you use $n$ both for the dimension of $X$, and an arbitrary index into your sequence of triangulations. $\endgroup$
    – LSpice
    Commented May 31, 2022 at 3:13
  • 1
    $\begingroup$ @LSpice Thank you - I think I straightened it out now. $\endgroup$
    – Sprotte
    Commented May 31, 2022 at 4:13

0

You must log in to answer this question.