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Consider the irreducible representation $V$ in the symmetric group $S_5$ corresponding to the Young diagram (these are meant to be boxes): $$[\;\;][\;\;] \\ [\;\;][\;\;] \\ [\;\;]\;\;\;\;$$

(a) List all standard Young tableaux of the given shape (that is, list all the possible fillings of the boxes with $\{1,2,3,4,5\}$ such that the numbers increase along the rows and down the columns).

(b) Give the dimension of $V$.

(c) For any standard tableau $\Omega$ of the given shape, let $e_{\Omega}$ be the standard basis element of the representation $V$. Evaluate the action of the permutation $g=(23)(45)(23)(34)(23) \in S_5$ on the vector $e_{\Omega}$ where $\Omega =$ $$[\;1\;][\;4\;] \\ [\;2\;][\;5\;] \\ [\;3\;]\;\;\;\;\;\,$$

My answers: (a) There are 5 standard Young tableaux of the given shape. (b) So dim(V)=5. (c) The basis vector $e_{\Omega} = P_{\Omega}Q_{\Omega}$ where $P_\Omega \in \mathbb{C}S_5$ and $Q_{\Omega}\in\mathbb{C}S_5$ are the row symmetrizer and column antisymmetrizer of $\Omega$: $$P_{\Omega} = [e+(14)][e+(25)] = e + (25) + (14) + (14)(25) \\ Q_{\Omega} = [e - (12) - (13) - (23) + (123) + (132)][e-(45)]$$

The given permutation $g=(23)(45)(23)(34)(23)=(2543)$. Is there a quicker way to evaluate the action of this permutation on $e_{\Omega}$ then to work out $(2543)P_{\Omega}Q_{\Omega}$?

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I don't think there's anything quicker than working out $(2543)P_{\Omega}Q_{\Omega}$; I don't know of any shortcut to $P_{\Omega}Q_{\Omega}$, and the multiplication by $(2543)$ isn't exactly hard to compute. As for the answers, they look right. – darij grinberg Jan 27 '13 at 14:39

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