The Jacobi-Trudi identities

November 20, 2009 by Qiaochu Yuan

Today I’d like to introduce a totally explicit combinatorial definition of the Schur functions. Let $\lambda \vdash n$ be a partition. A semistandard Young tableau $T$ of shape $\lambda$ is a filling of the Young diagram of $\lambda$ with positive integers that are weakly increasing along rows and strictly increasing along columns. The weight of a tableau $T$ is defined as $\mathbf{x}^T = x_1^{T_1} x_2^{T_2} ...$ where $T_i$ is the total number of times $i$ appears in the tableau.

Definition 4: $\displaystyle s_{\lambda}(x_1, x_2, ...) = \sum_T \mathbf{x}^T$

where the sum is taken over all semistandard Young tableaux of shape $\lambda$ .

As before we can readily verify that $s_{(k)} = h_k, s_{(1^k)} = e_k$ . This definition will allow us to deduce the Jacobi-Trudi identities for the Schur functions, which describe among other things the action of the fundamental involution $\omega$ . Since I’m trying to emphasize how many different ways there are to define the Schur functions, I’ll call these definitions instead of propositions.

Definition 5: $\displaystyle s_{\lambda}= \det(h_{\lambda_i+j-i})_{1 \le i, j \le n}$ .

Definition 6: $\displaystyle s_{\lambda'} = \det(e_{\lambda_i+j-i})_{1 \le i, j \le n}$ .

The proof

The following proof is one of the standard applications of the Lindstrom-Gessel-Viennot lemma, and like much of the other material in these posts it can be found (with accompanying diagrams) in both Sagan and Stanley.

We’ll work with $G = \mathbb{Z}^2$ , the acyclic plane, with edges going northward and eastward only. We’ll also add some points $(x, \infty)$ which can be reached going northward from any finite point, and we won’t weight northward steps. But we will weight eastward steps as follows:

In the e-weighting, each eastward step is weighted $x_n$ where $n$ is the sum of the coordinates of the destination vertex of the edge minus the sum of the coordinates of the initial vertex of the path. The weight of a path is the product of the weights of its eastward edges.
In the h-weighting, each eastward step is weighted $x_{n+1}$ where $n$ is the difference between the $y$ -coordinate of the destination vertex of the edge and the $y$ -coordinate of the initial vertex of the path. Again the weight of a path is the product of the weight of its eastward edges.

Choose sources $a_i = (1 - i, 0)$ and sinks $b_i = (\lambda_i - i + 1, \infty)$ . Then every path from $a_j$ to $b_i$ determines a monomial in $e_{\lambda_i + j - i}$ in the e-weighting and a monomial in $h_{\lambda_i + j - i}$ in the h-weighting, and conversely every monomial arises in this way. So by the lemma it follows that the Jacobi-Trudi determinants count non-intersecting $k$ -paths $a \to b$ .

Given such a non-intersecting $k$ -path we may construct two tableaux from it: one whose columns can be read off from the e-weighting of each path which has shape $\lambda'$ , and one whose rows can be read off from the h-weighting of each path which has shape $\lambda$ . The non-intersecting condition is then precisely the condition that rows are weakly increasing and columns are strictly increasing; intersections would correspond to a violation of one or both of those constraints. (This is much easier to see with the corresponding diagrams; my apologies.) So in fact one obtains SSYTs, and moreover it’s not hard to show that all SSYTs of the correct shapes are obtained in this way.

The fundamental involution

Since the fundamental involution $\omega : \Lambda \to \Lambda$ flips $e_n$ and $h_n$ , it follows that $\omega(s_{\lambda}) = s_{\lambda'}$ . There is a representation-theoretic interpretation of this. Recall that

$\displaystyle H(t) = \sum_{n \ge 0} h_n t^n = \exp \left( p_1 t + p_2 \frac{t^2}{2} + ... \right)$

and that

$\displaystyle E(t) = \sum_{n \ge 0} e_n t^n = \exp \left( p_1 t - p_2 \frac{t^2}{2} \pm ... \right)$ .

Applying the fundamental involution it follows that $\omega(p_n) = (-1)^{n-1} p_n$ . We should interpret $(-1)^{n-1}$ as the sign of an $n$ -cycle so that $\omega(p_{\pi}) = \text{sgn}(\pi) p_{\pi}$ . Now the identity

$\displaystyle s_{\lambda} = \frac{1}{n!} \sum_{\pi \in S_n} \chi^{\lambda}(\pi) p_{\pi}$

gives

$\displaystyle s_{\lambda'} = \frac{1}{n!} \sum_{\pi \in S_n} \chi^{\lambda}(\pi) \text{sgn}(\pi) p_{\pi}$

which has an obvious interpretation: the representation $\rho^{\lambda'}$ must come from $\rho^{\lambda}$ by tensoring with the sign representation!

Relationship to a previous definition

We are now in a position to prove that Definition 4 is equivalent to one of the other definitions we gave earlier, namely Definition 3. We follow the proof given in Sagan, p. 164. First define $e_n^{(j)}$ to be $e_n$ with all the terms containing $x_j$ removed.

Lemma: Let $\mu = (\mu_1, \mu_2, ... \mu_l)$ be a composition and define the $l \times l$ matrices $A_{\mu} = (x_j^{\mu_i}), H_{\mu} = (h_{\mu_i-l+j}), E = ((-1)^{l-i} e_{l-i}^{(J)})$ . Then $A_{\mu} = H_{\mu} E$ .

Proof. This is equivalent to the collection of identities

$\displaystyle \sum_{k=1}^{l} h_{\mu_i-l+k} (-1)^{l-k} e_{l-k}^{(j)} = x_j^{\mu_i}$

which is in turn equivalent to

$\displaystyle H(t) E^{(j)}(-t) = \frac{1}{1 - x_j t}$

which is obvious.

The equivalence of Definition 3 and Definition 4 is a simple corollary; one just picks $\mu$ so that $|A_{\mu}|$ is first $a_{\lambda}$ and then $a_{0^l}$ , and then applies the first Jacobi-Trudi identity.

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Posted in math.CO, math.RT | Tagged MaBloWriMo, representation theory of the symmetric group, symmetric functions, Young tableaux | 4 Comments

4 Responses

on December 7, 2011 at 11:26 pm | Reply anon

Thank you for this! I am currently learning about the Gessel-Viennot Lemma and the Jacobi-Trudi identities and these posts are great!
on October 20, 2012 at 2:31 pm | Reply Jacobi-Trudi identities « caozhu

[…] finally we are back to The Jacobi-Trudi identities again. This time I did see semistandard Young tableau differs from standard ones in the way that […]
on October 9, 2013 at 6:17 am | Reply Marc van Leeuwen

Maybe I am missing something, but it appears to me that the weightings used violate one fundamental rule for the Lindström-Gessel-Viennot lemma: the weight of an edge should be independent of the path of which that edge forms part. Without that requirement, the cancellation that proves the lemma does not work, as parts of paths are exchanged. And both the e-weighting and the h-weighting refer to the initial vertex of the path, vialting this rule.
- on October 9, 2013 at 4:46 pm | Reply Qiaochu Yuan
  
  Hum. So, the initial vertices all have the same $y$ -coordinates, which means you can show that the h-weighting actually doesn’t depend on the choice of initial vertex. But I’ve gone back to look at how the e-weighting is described in Sagan and Sagan’s description genuinely seems to depend on the initial vertex. I’m not sure what’s going on here.

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