Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

One possible construction of the Specht modules goes as follows.

Given a partition $\lambda$ of $n$, we can write down Young's seminormal form for the representation of $S_n$ corresponding to $\lambda$. Writing a basis $e_1, \ldots, e_l$ in bijection with the set of standard Young tableaux $T_1, \ldots, T_l$ of shape $\lambda$, one defines Young's seminormal representation by:
$ e_i \cdot s_j = e_i$ if entries $j$ and $j+1$ are in the same row of $T_i$.
$ e_i \cdot s_j = - e_i$ if entries $j$ and $j+1$ are in the same column of $T_i$.
$s_j$ acts by the matrix $\begin{pmatrix} -1/k & 1 \\ 1-1/k^2 & 1/k \end{pmatrix}$ on the subspace with basis $(e_i, e_{i'})$, where $T_{i'}$ is obtained from $T_i$ by switching entries $j$ and $j+1$, and the entry $j$ appears higher up in $T_i$ than in $T_{i'}$. Here $k$ is the axial distance between boxes with entries $j$ and $j+1$, also equal to the difference in contents (where the content of box $(i,j)$ is $j-i$).

Using this definition, and looking at the Hasse diagram for the dominance ordering on standard Young tableaux of shape $\lambda$, one sees that this is the same as formally writing in the basis $v_i = v R(w)$, where $v$ corresponds to the Young tableau $T$ filled in along rows (so largest in the dominance ordering), $w$ is a reduced word taking $T$ to $T_i$ through the Hasse diagram (i.e. only passing through standard Young tableaux), and if $w = s_{i_1} s_{i_2} \cdots s_{i_k}$, then $R(w) = R_{i_1}(k_1) R_{i_2} (k_2) \cdots R_{i_t}(k_t)$ where` $R_i(k) = \frac{-(k+1)}{k} \cdot \frac{1}{k+1}(1- ks_i)$ and the entries $k_1, \ldots, k_t$ are given by drawing a wiring diagram for $w$ (like a braid), writing the contents of $T$ in order at the top, and putting $b-a$ at the intersection of two strands labelled $a, b$.
The Yang-Baxter equation $R_i(s) R_{i+1}(s+t) R_i(t) = R_{i+1}(t) R_i(s+t) R_{i+1}(s)$
guarantees that this is well defined, i.e. that $R(w)$ doesn't depend on which reduced word $w$ we pick to represent our tableau. (This also allows one to prove that Young's seminormal form is actually a representation, i.e. the braid relation is satisfied by the representing matrices)

Now, instead of writing in the basis $v R(w_i)$ and multiplying elements on the right and rearranging, one can instead use the basis $u_i = v w_i$, the natural basis, which defines the Specht module corresponding to $\lambda$. However, if we forget the above construction and try to write down the representation from scratch in this way, we will not be able to write down $u_i s_j$ if $s_j$ doesn't take $T_i$ to another standard tableau. The standard way around this is to use Garnir relations, but is there an interpretation possible in terms of the above $R(w)$? Ideally, one would like to make a claim like $u_i R(w) = 0$ if $w$ doesn't take $T_i$ to another standard tableau, and this would allow us to read off the $u_i w$ inductively, but these $R(w)$ run into problems (i.e. coefficients of $0$ or $-1$) precisely when $w$ doesn't take $T_i$ to another standard tableau.

As an example, consider $\lambda = (4,2)$; this gives the following picture for the Hasse diagram:

Suppose we have numbered the tableaux $T_1, \ldots, T_9$ in order, row by row from left to right.
We have $T = T_1$, and for instance $w_9 = s_4 s_5 s_3 s_4 s_2$ takes $T$ to $T_9$. Writing out the wiring diagram with the above procedure shows that
$R(w_9) = (s_4 + 1/4)(s_3+1/3)(s_4+1/2)(s_2+1/2)$
Given the seminormal representation in the basis $e_1, \ldots, e_9$, we can figure out the change of basis matrix and write down the corresponding Specht module. For instance, one finds that $u_9 \cdot s_3 = -u_1-u_2-u_3-u_4-u_5-u_6-u_7-u_8-u_9$. How does one see this only using the $R(w)$ elements, without appealing to the seminormal representation or another construction of the Specht module (say, using polytabloids and straightening)?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

You will find this explained in section 5. of:

MR1988991 (2004f:20014) Ram, Arun . Skew shape representations are irreducible. Combinatorial and geometric representation theory (Seoul, 2001), 161--189, Contemp. Math., 325, Amer. Math. Soc., Providence, RI, 2003.

share|improve this answer
arxiv.org/abs/math/0401326 This is probably more up-to-date, as it is a version 2 and the date is 2004. –  darij grinberg Jun 3 '11 at 18:27
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.