Representation theory of finite symmetric groups

Young operator

Let $Θ 𝑝 𝜆$ be a Young tableau with $𝑛$ boxes, $𝐻 𝑝 𝜆$ be the subgroup of row permutations, and $𝑉 𝑝 𝜆$ be the subgroup of column permutations. The row symmetrizer $𝔰 𝑝 𝜆 \in ℂ [𝑆 𝑛]$ is given by

𝔰 𝑝 𝜆 = \sum ℎ \in 𝐻 𝑝 𝜆 𝛿 ℎ

and the column antisymmetrizer $𝔞 𝑝 𝜆 \in ℂ [𝑆 𝑛]$ by

𝔞 𝑝 𝜆 = \sum 𝑣 \in 𝑉 𝑝 𝜆 s g n (𝑣) 𝛿 𝑣

then the Young operator is given by #m/def/rep/sym

𝔢 𝑝 𝜆 = 𝔰 𝑝 𝜆 𝔞 𝑝 𝜆 = \sum 𝑣 \in 𝑉 𝑝 𝜆 \sum ℎ \in 𝐻 𝑝 𝜆 s g n (𝑣) 𝛿 ℎ 𝑣

Notation

The partition subscripts will typically be written out as a sum for the purpose of these notes, e.g. $𝔢 (12) 3 + 2 + 2 = 𝛿 (12) * 𝔢 3 + 2 + 2 * 𝛿 (12)$ . In the literature, it is common to use an entire Young diagram as a subscript.

A practical way to do pen-and-paper calculations is with a Birdtrack notation. If single-box symmetrizers and antisymmetrizers are drawn, each line passes through exactly one symmetrizer and exactly one antisymmetrizer. Each (anti)symmetrizer corresponds to a different row (column), with the number of lines passing through given by the number of boxes therein.

$Birdtrack diagram for a Young operator$

Properties

$𝐻 𝑝 𝜆 = 𝑝 𝐻 𝜆 𝑝 - 1$ and $𝑉 𝑝 𝜆 = 𝑝 𝑉 𝜆 𝑝 - 1$ are subgroups of $𝑆 𝑛$ with $𝐻 𝑝 𝜆 \cap 𝑉 𝑝 𝜆 = {𝑒}$ . Thus $𝔢 𝑝 𝜆 = 𝛿 𝑝 * 𝔢 𝜆 * 𝛿 𝑝 - 1$ .
$𝔰 𝑝 𝜆$ and $𝔞 𝑝 𝜆$ are total Symmetrizer and antisymmetrizer elements for the subgroups $𝐻 𝑝 𝜆$ and $𝑉 𝑝 𝜆$ .
$𝔰 𝑝 𝜆$ and $𝔞 𝑝 𝜆$ are essentially idempotent but in general not primitive.
The young operators $𝔢 𝑝 𝜆$ are essentially idempotent and primitive.
The irreps generated by $𝔢 𝑝 𝜆$ and $𝔢 𝑞 𝜇$ are equivalent iff $𝜆 = 𝜇$ , regardless of $𝑝$ and $𝑞$ . Thus, the young operators for standard tableaux generate minimal left ideals for every non-equivalent irrep. #m/thm/rep/sym

Proof

The proof is most intuitive with birdtrack arguments.

For 4., we expands the convolution as follows

𝔢 𝑝 𝜆 𝔢 𝑝 𝜆 = 𝔞 𝑝 𝜆 𝔰 𝑝 𝜆 𝔞 𝑝 𝜆 𝔰 𝑝 𝜆 = \sum 𝑞 \in 𝑆 𝑛 [𝔰 𝑝 𝜆 𝔞 𝑝 𝜆] (𝑞) 𝔞 𝑝 𝜆 𝛿 𝑞 𝔰 𝑝 𝜆

for the term $𝑞 = 𝑒$ we get $𝔢 𝑝 𝜆$ , and for all other terms either

$𝑞$ produces a zero connection (i.e. two lines intersect the same symmetrizer and antisymmetrizer)
$𝑞$ switches lines connected to the same symmetrizer, giving $𝔢 𝑝 𝜆$
$𝑞$ switches two lines connected to the same antisymmetrizer, giving $- 𝔢 𝑝 𝜆$
A combination of 2. and 3. gives at most a sign change

hence $𝔢 𝑝 𝜆 𝔢 𝑝 𝜆 = 𝜂 𝜆 𝔢 𝑝 𝜆$ , but $𝜂 𝜆 \neq 0$ since $𝔢 𝑝 𝜆 (𝑒) \neq 0$ and thus $t r P (𝔢 𝑝 𝜆) \neq 0$ . Hence $𝔢 𝑝 𝜆$ is essentially idempotent. Moreover the above argument already demonstrates the Idempotent primitivity criterion, hence $1 𝜂 𝜆 𝔢 𝑝 𝜆$ is primitive.

For 5. we will show that if $𝜆 \neq 𝜇$ then $𝔢 𝜆 𝑞 𝔢 𝜇 = 0$ for all $𝑞 \in ℂ [𝐺]$ , i.e. the Equivalence of irreps on left ideals criterion. Consider $𝔢 𝜆 = 𝔰 𝜆 𝔞 𝜆$ and $𝔢 𝜇 = 𝔰 𝜇 𝔢 𝜇$ in birdtracks, with the symmetrizers and antisymmetrizers ordered from top to bottom from longest to shortest. Then the first symmetrizer of $𝔰 𝜆$ has $𝜆 1$ lines, each of which must enter a different one of $𝔞 𝜇$ 's $𝜇 1$ antisymmetrizers if $𝔰 𝜆 𝛿 𝑝 𝔞 𝜇 \neq 0$ , which by the Pigeonhole principle is impossible if $𝜆 1 > 𝜇 1$ . If $𝜆 1 = 𝜇 1$ , only antisymmetrizers of $𝔞 𝜇$ with at least two lines are available, of which there are $𝜇 2$ . For a nonzero connection, each of the second symmetrizers $𝜆 2$ lines must connect to a different one of these, which is impossible if $𝜆 2 > 𝜇 2$ . Continuing this argument we see a nonzero connection of $𝔰 𝜆$ to $𝔞 𝜇$ is impossible if $𝜆 > 𝜇$ . The same goes for connecting $𝔞 𝜆$ to $𝔰 𝜇$ if $𝜇 > 𝜆$ . Thus by lineärity, if $𝜆 \neq 𝜇$ then $𝔢 𝜆 𝑞 𝔢 𝜇 = 0$ for all $𝑞 \in ℂ [𝐺]$ .

Now since $𝔢 𝑝 𝜆 = 𝛿 𝑝 𝔢 𝜆 𝛿 𝑝 - 1$ , it follows that $𝔢 𝑝 𝜆 𝛿 𝑝 𝔢 𝜆 = 𝛿 𝑝 𝔢 𝜆 𝔢 𝜆 = 𝛿 𝑝 𝜂 𝜆 𝔢 𝜆 \neq 0$ , hence we have equivalence by the Equivalence of irreps on left ideals criterion.

#state/tidy | #lang/en | #SemBr