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)?
