## Lindström–Gessel–Viennot Lemma

Let us switch gears and describe a beautiful combinatorial result. Suppose $G$ is a graph which is directed, has no cycles, and there are only finitely many paths from a vertex to another. Given sets of n vertices:

$A =\{a_1, \ldots, a_n\}, \qquad B = \{b_1, \ldots, b_n\}$

the lemma computes the “weighted sum” of non-intersecting paths from $A$ to $B$.

To be specific, suppose each edge $e$ of the graph has a weight $w_e$ which takes values in a fixed ring. Given a directed path $P : a\to b$ from $a$ to $b,$ let $w(P)$ be the product of the edges along the path $P$. Now write $e(a, b)$ for the sum of $w(P)$ over all $P:a\to b.$

Next given $a_1, \ldots, a_n, b_1, \ldots, b_n$ above, we compute the determinant of:

$M = \begin{pmatrix} e(a_1, b_1) & e(a_1, b_2) & \cdots & e(a_1, b_n) \\e(a_2, b_1) & e(a_2, b_2) & \cdots & e(a_2, b_n) \\\vdots & \vdots & \ddots & \vdots \\e(a_n, b_1) & e(a_n, b_2) & \cdots & e(a_n, b_n)\end{pmatrix}$

On the other hand, consider a collection of paths $P = (P_1, \ldots, P_n)$ where each $P_i : a_i \to b_{\sigma(i)}$ for some permutation $\sigma = \sigma(P) \in S_n.$ Furthermore we want the paths to be pairwise non-intersecting, i.e. no two paths contain a common vertex (not even at the endpoints). Now we can state the desired result.

Lindström–Gessel–Viennot (LGV) Lemma. We have:

$\displaystyle\det(M) = \sum_{P = (P_1, \ldots, P_n)} \text{sgn}(\sigma(P)) \prod_{i=1}^n w(P_i)$

over all tuples of paths $(P_1, \ldots, P_n)$ described above.

## Proof of the LGV Lemma

First expand the determinant of $M$; each term is of the form:

$\text{sgn}(\sigma) \times e(a_1, b_{\sigma(1)}) \times e(a_2, b_{\sigma(2)}) \times \ldots \times e(a_n, b_{\sigma(n)}), \qquad \sigma \in S_n.$

Expanding each $e(a_i, b_{\sigma(i)}) = \sum_{P_i : a_i \to b_{\sigma(i)}} w(P_i)$, the above term is a sum of terms of the form

$\displaystyle\sum_{P_1, \ldots, P_n}\text{sgn}(\sigma) w(P_1) w(P_2) \ldots w(P_n),$

summed over all n-tuples $(P_i : a_i \to b_{\sigma(i)})_{i=1}^n.$ This is close to the desired result but here, the $P_i$ may intersect. Hence, we wish to show that the sum over all $(P_1, \ldots, P_n)$ for which some $P_i$ and $P_j$ intersect, vanishes.

Let $\Sigma = \{P = (P_1, \ldots, P_n)\}$ be the collection of all such paths; for each $P\in \Sigma,$ we:

• let $v$ be the first vertex occurring as an intersection between $P_i$ and $P_j$ ($i\ne j$), after taking a total ordering for the set of vertices;
• let $i$ be the smallest index such that $P_i$ contains $v$;
• let $j$ be the largest index such that $P_j$ contains $v$ (note that $j\ne i$);

and get a triplet $(i, v, j).$ Now let $P'$ be the same tuple of paths but with the tails of $P_i$ and $P_j$ switched after $v.$

This gives a map $f : \Sigma \to \Sigma$, which is an involution since the corresponding triplet for $f(P)$ is still $(i, v, j).$ However $\sigma(f(P))$ and $\sigma(P)$ have opposite signs. Denoting $w(P_1) \ldots w(P_n)$ by $T(P)$ we have $T(f(P)) = T(P)$ and so:

$\displaystyle\sum_{P\in \Sigma} \text{sgn}(\sigma(P)) T(P) = -\sum_{P\in \Sigma} \text{sgn}(\sigma(f(P))) T(f(P)) = -\sum_{P'\in \Sigma} \text{sgn}(\sigma(P')) T(P').$

So the sum is zero. ♦

## Jacobi-Trudi Identities

Here is our first application of the LGV lemma; all computations are in the formal ring $\Lambda.$

Theorem. Let $\mu = \overline \lambda$ be the transpose of $\lambda$. Then $s_\lambda$ is the determinant of either of these matrices.

$\left( h_{\lambda_i + j - i} \right)_{1 \le i, j\le d} =\begin{pmatrix} h_{\lambda_1} & h_{\lambda_1 + 1} & \ldots & h_{\lambda_1 + d - 1} \\ h_{\lambda_2 - 1} & h_{\lambda_2} & \ldots & h_{\lambda_2 + d - 2} \\ \vdots & \vdots & \ddots & \vdots \\ h_{\lambda_d - d + 1} & h_{\lambda_d - d + 2} & \ldots & h_{\lambda_d}\end{pmatrix}$

$\left( e_{\mu_i + j - i} \right)_{1 \le i, j\le d} =\begin{pmatrix} e_{\mu_1} & e_{\mu_1 + 1} & \ldots & e_{\mu_1 + d - 1} \\ e_{\mu_2 - 1} & e_{\mu_2} & \ldots & e_{\mu_2 + d - 2} \\ \vdots & \vdots & \ddots & \vdots \\ e_{\mu_d - d + 1} & e_{\mu_d - d + 2} & \ldots & e_{\mu_d}\end{pmatrix}$

where $d$ is chosen to be $\ge l(\lambda)$ and $\ge l(\mu)$ in the respective cases.

Note

One remembers the structure of the matrices by writing $\lambda$ as the subscripts for the diagonal entries, then incrementing to the right. We take $h_i = e_i = 0$ for all $i <0.$ Observe that since $h_0 = e_0 = 1$, there is no harm increasing $d$ by appending zeros to the partition.

Proof

We can obtain the second identity from the first by applying $\omega$, so let us show the first. Before we begin, observe that the Schur polynomial can be written as:

$\displaystyle s_\lambda = \sum_{\text{SSYT } T} x^T,$

where $x^T = \prod_{i\ge 1} x_i^{\text{\# of } i \text{ in } T}.$ Indeed, this follows from the proposition here.

We will use the LGV lemma: let $N$ be large; we take the following graph.

• The vertices are lattice points in $\mathbb{Z}^2$.
• The starting and ending vertices are given by:

\begin{aligned}a_d &= (0, N), a_{d-1} = (1, N), \ldots, a_1 = (d-1, N), \\ b_d &= (\lambda_d, 1), b_{d-1} = (\lambda_{d-1} + 1, 1), \ldots, b_1 = (\lambda_1 + d-1, 1).\end{aligned}

• The edges are of one unit, directed either eastwards or southwards. The southward edges are weighted 1 while the eastward edges on $y=i$ are labelled $x_i$ for $i\ge 1.$

For example if $\lambda = (4, 1, 1)$, here is an example of paths connecting $a\to b.$

Let us compute the LHS of the LGV lemma. The set of paths from $a_i = (d-i, N)$ to $b_j = (\lambda_j+ d-j, 1)$ sums up to the complete symmetric polynomial $h_{\lambda_j} + i-j$. Thus the LHS is the desired determinant expression.

On the other hand, let us consider the collection of non-intersecting paths from $a_1, \ldots, a_d$ to $b_{\sigma(1)}, \ldots, b_{\sigma(d)}$. For a path to exist, $\sigma$ must be the identity. Each collection of non-intersecting paths corresponds to a reverse SSYT of shape $\lambda$ (i.e. the rows are weakly decreasing and columns are strictly decreasing). E.g. the above diagram gives:

Hence, the RHS in the LGV lemma equals the sum of $x^T$ over all reverse SSYT $T$ of shape $\lambda.$ This is equal to the sum over all SSYT $T$ of shape $\lambda$ (we take $m, m'$ to be the minimum and maximum elements of the table respectively and replace each $u$ with $m+m'-u,$ thus obtaining an SSYT). Thus the sum is $s_\lambda.$ ♦

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