Direct proof of Gelfand-Zetlin identity

Question

Denote by $D(a_1,\dots,a_n)$ the product $\prod_{j>i}(a_j-a_i)$. Assuming that $a_i$ are integers s.t. $a_1\le a_2\le\dots\le a_n$, prove that $D(a_1,...,a_n)/D(1,...,n)$ is the number of Gelfand-Zetlin triangles (that is, triangles consisting of $\frac{n(n+1)}2$ integers, s.t. each number is greater it's lower-left neighbor but not greater than lower-right neighbor) with the base $a_i$.

For example, for $n=3$ one needs to prove that number of b₁, b₂, b' s.t. $a_1\le b_1<a_2\le b_2<a_3$, $b_1\le b'<b_2$ is exactly $\frac{(a_3-a_2)(a_3-a_1)(a_2-a_1)}{3}$.

As one can guess from the name “Gelfand-Zetlin”, this fact is well-known in representation theory (namely, in LHS we count dimension of a $gl_n$-representation by Weyl formula, and in RHS we count elements in Gelfand-Zetlin basis of the same representation). But maybe someone can come with more or less direct proof? (Some kind of bijective proof, maybe.)

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Informal probabilistic argument

For simplicity, consider the case $n=3$: D(a_1,a_2,a_3) counts the number of triangles s.t. $a_1\le b_1<a_2\le b_2<a_3$, $a_1\le b'<a_3$ — and we're interested only in G-Z triangles, i.e. in triangles s.t. $b_1\le b'< b_2$. Now, mathematical expectation of the length of the interval $(b_1,b_2)$ is exactly one half of the length of the interval from which $b'$ is chosen. So one may expect that the probability that random triangle is G-Z is $1/2$ — and the answer is indeed $D(a_1,a_2,a_3)/D(1,2,3)$.

(The main problem with this computation is that we're multiplying probabilities for events that are clearly not independent. And although for $n=3$ it's not hard to transform this heuristic argument into a formal proof, even for $n=4$ I failed to do such thing.)

Answer 1

As Qiaochu points out the formula given by Gjergji Zaimi is certainly relevant. It realizes $D(a_1,\ldots,a_n)/D(1,\ldots,n)$ as the determinant of the $n$-by-$n$ matrix $M$ with $(i,j)$-entry $${a_i\choose j-1}.$$ By row operations this equals the determinant of the $(n-1)$-by-$(n-1)$ matrix $N$ with $(i,j)$-entry $${a_{i+1}\choose j}-{a_i\choose j}.$$ The $i$-th row of $n$ is the sum of vectors $v_{a_i},v_{a_i+1},\ldots,v_{a_{i+1}-1}$ where the $j$-th entry of $v_c$ is $${c\choose j-1}.$$ By the multilinearity of the determinant as a function of its rows, $N$ is the sum the determinants with rows $v_{b_1},\ldots,v_{b_{n-1}}$ where $a_1\le b_i < a_2\le b_2 < a_3\le\cdots$. These tuples $(b_1,\ldots,b_{n-1})$ are precisely the admissible penultimate rows in GZ triangles with bottom row $(a_1,\ldots,a_n)$. Hence both sides of the sought equality satisfy the same recurrence.

Answer 2

Schur polynomials proof

Observe that $D(a_1,\ldots,a_n)/D(1,\ldots,n)=s_\lambda(1,\ldots,1)$ (for $a_i=\lambda_i+i$). But by Giambelli (aka Jacobi-Trudi) formula $s_\lambda=\det h_{\lambda_i+j-i}$, so $s_\lambda(1,\ldots,1)=\det\binom{a_i}{j-1}$. Now by Lindstrom-Gessel-Viennot lemma last determinant counts non-intersecting lattice paths from the set $0,1,\ldots,n$ on Y-axis to the set $a_1,\ldots,a_n$ on X-axis. Finally, observe that any such path is characterized by x-coordinates of "steps down" — which are subject to conditions from the definition of GZ triangle.

/* Well, it's not that direct and it's more or less the proof from Robin Chapman's answer. Nevertheless, it explains something, I hope. */

Answer 3

Just to make sure: you know that what you are asking for is actually Weyl's dimension formula for the highest weight representation in the case of SL(n,C) resp. SU(n)? I suppose there is no direct combinatorial proof that does not make use of this fact.