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I am reading Theorem 1 on page 110 of Fulton's Young Tableaux and have several questions on it. Let $E$ be a free module on $e_1,\ldots,e_m$ (for our purposes $E$ being a finite dimensional complex vector space will do) and consider the module $E^\lambda$ as constructed earlier in the chapter. If $e_T$ is the vector as defined in the paragraph before Lemma 1 of page 107, the Theorem states:

Theorem 1 (pg 110): The module $E^\lambda$ is free on $e_T$ as $T$ varies over the tableaux on $\lambda$ with entries in $\{1,\ldots,m\}$.

Now Fulton begins by arguing that all such $e_T$ have to generate $E^\lambda$ as follows. He uses the presentation $E^\lambda = F/Q$ of Lemma 1 and orders the tableau as follows: $T' > T$ if in the right most column where they differ, the entry in $T'$ is larger than $T$. Now to show that the $e_{T'}$ with $T'$ a tableau generate $E^\lambda$, we need to show that if $T$ is not a tableau then we can write $e_T$ as a linear combination of $e_S$ where $e_S$ are tableau. Using the relations of Lemma 1, we can assume that the columns of $T$ are strictly increasing. Now suppose the $k$ -th entry of the $j$ -th column of $T$ is strictly larger than the $k$ - th entry of the $j+1$ - th column.

What I don't understand: He says in this case we can write $e_T = \sum e_S$ the sum over all $S$ obtained from $T$ by exchanging the top $k$ entries of the $j+1$ - th column of $T$ with $k$ entries of the $j$ -th column. How is this possible? If I consider the specific case of $\begin{array}{|c|c|c|} \hline 1 &2 & 3 \\ \hline 5 & 4 & 6 \\\hline \end{array}$ then by selecting the two middle boxes, he is saying that $$\begin{array}{|c|c|c|} \hline 1 &2 & 3 \\ \hline 5 & 4 & 6 \\\hline \end{array} = \begin{array}{|c|c|c|} \hline 2 &1 & 3 \\ \hline 4 & 5 & 6 \\\hline \end{array} ?$$ How does this make sense? Also I tried to use the available relation to write by filling of $(3,3)$ in terms of a sum of tableaux, but nothing has worked out. What am I not understanding here?

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  • $\begingroup$ Note that your definition of the ordering is imprecise. It should be: $T' > T$ if in the rightmost column in which $T'$ and $T$ are different, the lowermost cell which has a different entry in $T'$ and $T$ has a greater entry in $T'$ than in $T$. $\endgroup$ Aug 15, 2013 at 9:35

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Yes, he is saying that

$$e_{\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline 5 & 4 & 6 \\\hline \end{array}} \equiv e_{\begin{array}{|c|c|c|} \hline 2 & 1 & 3 \\ \hline 4 & 5 & 6 \\\hline \end{array}} \mod Q$$

(please don't write this as $\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline 5 & 4 & 6 \\\hline \end{array} = \begin{array}{|c|c|c|} \hline 2 & 1 & 3 \\ \hline 4 & 5 & 6 \\\hline \end{array}$).

Generally, if $S$ is any filling of a Young diagram, and $T$ is a filling obtained from $S$ by interchanging two columns of equal length, then $e_S \equiv e_T \mod Q$. This is a very simple case of the exchange relation (iii) in Lemma 1 (at least when the two columns are adjacent; otherwise, iterate). When the Young diagram $\lambda$ has just one line, this shows that $e_S$ depends only on the multiset of entries of $S$; this agrees with the fact that $E^{\lambda} \cong S^{\left|\lambda\right|} E$ in this case.

What might have confused you is why such a simple exchange of columns was a "step forward" for the proof. But it is; just check that $\begin{array}{|c|c|c|} \hline 2 & 1 & 3 \\ \hline 4 & 5 & 6 \\\hline \end{array} > \begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline 5 & 4 & 6 \\\hline \end{array}$.

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