Let $\lambda = (n-1,1)$ be the partition of $n$. I am trying to compute the Specht module $S^{(n-1,1)}$ which is a certain submodule of the free module $M^{(n-1,1)} = \mathbb{C}\left\{\underset{\scriptstyle\textstyle\underline{i}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{i} \cdots n}}} : i=1,2\cdots,n\right\}$ where $\mathbf{i} :=\underset{\scriptstyle\textstyle\underline{i}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{i} \cdots n}}}$ is a tabloid (an equivalence class of row equivalent Young Tableaux) and $\hat{i}$ denotes omission.
According to the definition of $S^{(n-1,1)}$, it is the submodule spanned by the polytabloids $e_t = \sum_{\pi \in C_t}\text{sgn}(\pi) \{t\}$ where $t$ is any Young Tableau and $\{t\}$ is the tabloid containing $t$, and $C_t$ is the subgroup of $S_n$ that fixes the columns of $t$ (as sets).
So let $1\leq i \leq n$ and lets compute $e_t$ for $t\in \mathbf{i} = \underset{\scriptstyle\textstyle\underline{i}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{i} \cdots n}}}$. Suppose the first entry in the first row of $t$ is $j\neq i$, then we have $C_t = \{1,(i,j)\}$ so that
\begin{align*}e_t & = \left(\sum_{\pi \in C_t}\text{sgn}(\pi)\pi \right) \underset{\scriptstyle\textstyle\underline{i}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{i} \cdots n}}}\\
& = (1 – (i,j))\underset{\scriptstyle\textstyle\underline{i}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{i} \cdots n}}}\\
& = \underset{\scriptstyle\textstyle\underline{i}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{i} \cdots n}}} – \underset{\scriptstyle\textstyle\underline{j}\phantom{aaaaaaaaa}}{\underline{\overline{1\ 2 \cdots \hat{j} \cdots n}}}\\
& = \mathbf{i} – \mathbf{j}.
\end{align*}
So $S^{(n-1,1)}$ is the span of all such vectors. According to the book I am reading, we get that
$$S^{(n-1,1)} = \left\{\sum_{i=1}^n c_i \mathbf{i} : \sum_{i=1}^n c_i = 0\right\}.$$
My question: How did they compute this last step?
Best Answer
Instead of $\mathbf{i}$ I'll use $v_i$. This just amounts to showing that $$\operatorname{Span}_{\mathbb C}(v_i-v_j:1\leq i<j\leq n)=\left\{\sum_{i=1}^n c_iv_i:\sum_{i=1}^n c_i=0\right\}.$$
The forward inclusion is clear since $v_i-v_j$ has coefficients of the $v_k$'s which sum to zero for each $1\leq i<j\leq n$. Now try to write an arbitrary element $\sum_{i=1}^n c_iv_i$ with $\sum_{i=1}^n c_i=0$ as a sum of $v_i-v_j$ terms. For example, you can use $c_n=-c_1-\cdots-c_{n-1}$ so that $$\sum_{i=1}^n c_iv_i=c_1(v_1-v_n)+\cdots+c_{n-1}(v_{n-1}-v_n).$$