Representation theory of finite symmetric groups

Young operator

Let be a Young tableau with boxes, be the subgroup of row permutations, and be the subgroup of column permutations. The row symmetrizer is given by

and the column antisymmetrizer by

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

Notation

The partition subscripts will typically be written out as a sum for the purpose of these notes, e.g. . 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

  1. and are subgroups of with . Thus .
  2. and are total Symmetrizer and antisymmetrizer elements for the subgroups and .
  3. and are essentially idempotent but in general not primitive.
  4. The young operators are essentially idempotent and primitive.
  5. The irreps generated by and are equivalent iff , regardless of and . 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

for the term we get , and for all other terms either

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

hence , but since and thus . Hence is essentially idempotent. Moreover the above argument already demonstrates the Idempotent primitivity criterion, hence is primitive.

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

Now since , it follows that , hence we have equivalence by the Equivalence of irreps on left ideals criterion.

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