A combinatorial proof for Cayley's identity

In a recent paper, Caracciolo, Sokal and Sportiello presented, inter alia, an algebraic/combinatorial proof for Cayley's identity. The purpose of the present paper is to give a"purely combinatorial"proof for this identity; i.e., a proof involving only combinatorial arguments together with a generalization of Laplace's Theorem, for which a"purely combinatorial"proof is already known.


Introduction
For n ∈ N, denote by [n] the set {1, 2, . . . , n} and let X = X n = (x i,j ) (i,j)∈[n]×[n] be an n × n matrix of indeterminates. For I ⊆ [n] and J ⊆ [n], we denote • the minor of X corresponding to the rows i ∈ I and the columns j ∈ J by X I,J , • the cominor of X I,J (which corresponds to the rows i ∈ I and the columns j ∈ J) by X I,J .

Combinatorial proof of Vivanti's Theorem
We may view the determinant of X as the generating function of all permutations π in S n , where the (signed) weight of a permutation π is given as ω (π) := sgn π · n i=1 x i,π(i) : det (X) = π∈Sn ω (π) .
2.1. View permutations as perfect matchings. For our considerations, it is convenient to view a permutation π ∈ S n as a perfect matching m π of the complete bipartite graph K n,n , where the vertices consist of two copies of [n] which are arranged in their natural order; see Figure 1 for an illustration of this simple idea. It is easy to see that the edges of such perfect matching can be drawn in a way such that all intersections are of precisely two (and not more) edges, and that the number of these intersections equals the number of inversions of π, whence the sign of π is sgn (π) = (−1) #(intersections in mπ ) .
This simple visualization of permutations and their inversions is already used in [1, §15, p.32]: We call it the permutation diagram. So assigning weight x i,j to the edge pointing from i to j and defining the weight ω (m π ) of the permutation diagram m π to be the product of the edges belonging to m π , we may write ω (π) = (−1) #(intersections in mπ ) · ω (m π ) .
(See Figure 2 for an illustration.)

2.2.
Action of the determinant of partial derivatives. Next we need to describe combinatorially the action of the determinant det ∂ [k],[k] of partial derivatives. Let m = (m π 1 , . . . , m πs ) be an s-tuple of permutation diagrams counted in the generating function (det (X)) s , and let τ ∈ S k : Then the summand where c τ,m is the number of ways to choose the set of k edges {(i → τ (i)) : i ∈ [k]} from all the edges in m (this number, of course, might be zero). We may visualize the action of δ τ as "erasing the edges constituting τ in m"; see Figure 3 for an illustration.
Hence we have: 2.3. Double counting. For our purposes, it is convenient to interchange the summation in (3). This application of double counting amounts here to a simple change of view: Instead of counting the ways to choose the set of edges corresponding to τ from all the edges corresponding to some fixed s-tuple m, we fix τ and consider the set of m's from which τ ' edges might be chosen. This will involve two considerations: Figure 3. Let n = 5, s = 4 and k = 3 in Corollary 1. The picture shows the four possible ways of "erasing" the edges constituting τ ∈ S 3 from the 4-tuple (m π 1 , m π 2 , m π 3 , m π 4 ) of matchings, where (π 1 , π 2 , π 3 , π 4 ) ∈ S 4 5 is ((31254) , (51324) , (14253) , (23415)). The erased edges are shown as grey dashed lines.
• In how many ways can the edges corresponding to τ be distributed on s copies of the bipartite graph K n,n ? • For each such distribution, what is the set of compatible s-tuples of permutation diagrams?
For example, if k = 3 and s = 4 (as in Figure 3), there clearly • is 1 way to distribute the three edges on a single copy of the 4 bipartite graphs (see the fourth row of pictures in Figure 3), and there are 4 ways to choose such single copy, • are 3 ways to distribute the three edges on precisely two copies of the 4 bipartite graphs (see the second and third row of pictures in Figure 3), and there are 4 · 3 ways to choose such pair of copies (whose order is relevant), • is 1 way to distribute the three edges on precisely three copies of the 4 bipartite graphs (see the first row of pictures in Figure 3), and there are 4 · 3 · 2 ways to choose such triple of copies (whose order is relevant).

Partitioned permutations.
A distribution of the edges corresponding to τ ∈ S k on s copies of the bipartite graph K n,n may be viewed (see Figure 3) • as an s-tuple of partial matchings (some of which may be empty) of K k,k • such that the union of these s partial matchings gives the perfect matching m τ of K k,k .
Clearly, to each of such partial matching corresponds a partial permutation τ i , which we may write in two-line notation as follows: • the lower line shows the domain of τ i in its natural order, • the upper line shows the image of τ i , • the ordering of the upper line represents the permutation τ i .
We say that each of these τ i is a partial permutation of τ , and that τ is a partitioned permutation. We write in short: For example, the rows of pictures in Figure 3 correspond to the partitioned permutations (written in the aforementioned two-line notation) It is straightforward to compute the number of these equivalence classes: In the language of combinatorial species (see, for instance, [2]) the s-tuples of permutation words indexing these classes correspond bijectively to the (labelled) species (Permutations) s , and since the exponential generating function of Permutations is the exponential generating function of (Permutations) s is simply s · (s + 1) · · · (s + k − 1) z k k! . Figure 4. Erase a single edge (i, π (i)) in the permutation diagram m π of some permutation π: Note that precisely the intersections with edges leading from A ⊆ domain (π) to C ⊆ image (π) and with edges leading from B ⊆ domain (π) to D ⊆ image (π) are removed by this operation.

A B C D
So the number of these equivalence classes is s · (s + 1) · · · (s + k − 1), which is precisely the factor in (2). Our proof will be complete if we manage to show that the generating functions of each of these equivalence classes are the same, namely (det (X)) s−1 · det X I,J .

2.6.
Accounting for the signs. A necessary first step for this task is to investigate how the sign of a permutation π is changed by removing a given partial permutation π ′ : We view this as erasing all the edges belonging to π ′ 's permutation diagram m π ′ from π's permutation diagram m π ; see again Figure 3. Lemma 1. Let π ∈ S n be a permutation, and let π ⋆ be the permutation corresponding to the permutation diagram m π with edge (i, π (i)) removed. Then we have sgn (π) = (−1) π(i)−i · sgn (π ⋆ ) . Assume |π −1 (C) ∩ A| = k: Then edge (i, π (i)) clearly intersects the k edges joining vertices from A to vertices from C (see again Figure 4).
The only other intersections with (i, π (i)) come from edges joining vertices from B to vertices from D: Since π is a bijection, we have Altogether, the removal of edge (i, π (i)) removes 2k +π (i)−i intersections of edges.
Proof. We proceed by induction: k = 1 simply amounts to the statement of Lemma 1.

2.7.
Sums of (signed) products of minors. Now consider a fixed equivalence class in the sense of Section 2.5, which is indexed by a partition-scheme We want to compute the generating function G [τ ] of this equivalence class: Clearly, we may concentrate on the nonempty partial permutations; so w.l.o.g. we have to consider the partition-scheme [τ 1 ⋆ τ 2 ⋆ · · · ⋆ τ m ] which consists only of nonempty partial permutations τ j for 1 ≤ j ≤ m ≤ s. For any σ ∈ S k with σ ⊆ [τ 1 ⋆ τ 2 ⋆ · · · ⋆ τ m ], such partition scheme corresponds to a unique ordered partition of the image of σ: Equation (4) gives the sign-change caused by erasing the edges corresponding to τ j (with respect to any permutation in S n which contains τ i as a partial permutation), whence we can write the generating function as This, of course, is true for m = 1. We proceed by induction on m.

(A generalization of)
Laplace's theorem. Luckily, a generalization (see [6, section 148]) of Laplace's Theorem serves as the closer for our argumentation: Theorem 2. Let a be an (m + k) × (m + k)-matrix, and let 1 ≤ i 1 < i 2 < · · · < i m ≤ m + k and 1 ≤ j 1 < j 2 < · · · < j m ≤ m + k be (the indices of ) k fixed rows and k fixed columns of a. Denote the set of these (indices of ) rows and columns by R and C, respectively. Consider some fixed set I ⊆ R. Then we have: det (a) · det a R,C = J⊆C, |J|=|I| sgn (I R) · sgn (J C) · det a R\I,C\J · det a I,J . (8) A combinatorial proof for this identity is given in [4, proof of Theorem 6]: So altogether, we achieved a "purely combinatorial" proof for (2).