A topologist's proof of the matrix-tree theorem

One of the oldest results in spectral graph theory is Kirchoff’s matrix-tree theorem, which connects the number of spanning trees of a graph with the spectrum of its Laplacian. Specifically:

Let $G$ be a connected graph on $n$ vertices with (nonnormalized) Laplacian $L$. Let $L_v$ be a principal submatrix of $L$ obtained by removing any row and its corresponding column. The number of spanning trees of $G$ is equal to $\det (L_v) = \prod_{i=2}^n \lambda_i$.

While studying for my oral exam, I came up with a proof that elides some of the tricky combinatorics by using basic algebraic topology facts. Here goes:

Choose a base vertex $v$ of the graph, and let $\partial_v$ be the boundary matrix of $G$ with the row corresponding to $v$ removed. Then $L_v = \partial_v \partial_v^T$. The Cauchy-Binet formula gives a way to compute the determinant of such a product. If we sum over subsets $S$ of $n-1$ edges in $G$, we have

$$\det(L_v) = \sum_{S \in {m \choose {n-1}}} \det( \partial_v|_S) \det((\partial_v|_S)^T) = \sum_{S \in {m \choose {n-1}}} \det(\partial_v|_S)^2.$$

Suppose that $S$ is a set of $n-1$ edges that does not form a spanning tree. Then it must contain a cycle, and hence $\ker ( \partial |\_S )$ is nontrivial, since it is the boundary matrix of a graph with nontrivial $H_1$. As a result, $\ker(\partial_v|_S) \neq 0$, and hence $\det(\partial_v|_S) = 0$. On the other hand, if $S$ is a spanning tree, $\ker(\partial|\_S)$ is trivial. Consider the reduced homology of $G_S$, the graph which has the same vertex set as $G$ but with only the edges in $S$, where the distinguished point is $v$. Since $G_S$ is a tree, $\tilde H_0(G_S) = 0$. But this implies that the map $\partial_v|_S = \pi_{G \setminus v} \circ \partial|_S : \mathbb Z^{n-1} \to \mathbb Z^n \to \mathbb Z^{n-1}$ is an isomorphism, and hence $\partial_v\mid_S$ is invertible over $\mathbb Z$, so that $\det(\partial_v|_S) = \pm 1$. Therefore, $\det(\partial_v|_S)\det((\partial_v|_S)^T) = 1$ whenever the edges in $S$ form a spanning tree and $\det(\partial_v|_S)\det((\partial_v|_S)^T) = 0$ when they do not, and hence the sum from the Cauchy-Binet formula counts the number of spanning trees.

It’s nice when all of the fiddly bits of an argument can be eliminated by appealing to general facts.