The Unapologetic Mathematician

Mathematics for the interested outsider

Standard Polytabloids Span Specht Modules

We defined the Specht module S^\lambda as the subspace of the Young tabloid module M^\lambda spanned by polytabloids of shape \lambda. But these polytabloids are not independent. We’ve seen that standard polytabloids are independent, and it turns out that they also span. That is, they provide an explicit basis for the Specht module S^\lambda.

Anyway, first off we can take care of all the e_t where the columns of t don’t increase down the column. Indeed, if \sigma\in C_t stabilizes the columns of t, then

\displaystyle e_{\sigma t}=\sigma e_t=\mathrm{sgn}(\sigma)e_t

where we’ve used the sign lemma. So any two polytabloids coming from tableaux in the same column equivalence class are scalar multiples of each other. We’ve just cut our spanning set down from one element for each tableau to one for each column equivalence class of tableaux.

To deal with column equivalence classes, start with the tableau t_0 that we get by filling in the shape with the numbers 1 to n in order down the first column, then the second, and so on. This is the maximum element in the column dominance order, and it’s standard. Given any other tableau t, we assume that every tableau s with [s]\triangleright[t] is already in the span of the standard polytabloids. This is an inductive hypothesis, and the base case is taken care of by the maximum tabloid t_0.

If t is itself standard, we’re done, since it’s obviously in the span of the standard polytabloids. If not, there must be a row descent — we’ve ruled out column descents already — and so we can pick our Garnir element to write e_t as the sum of a bunch of other polytabloids e_{\pi t}, where [\pi t]\triangleright[t] in the column dominance order. But by our inductive hypothesis, all these e_{\pi t} are in the span of the standard polytabloids, and thus e_t is as well.

As an immediate consequence, we conclude that \dim(S^\lambda)=f^\lambda, where f^\lambda is the number of standard tableaux of shape \lambda. Further, since we know from our decomposition of the left regular representation that each irreducible representation of S_n shows up in \mathbb{C}[S_n] with a multiplicity equal to its dimension, we can write

\displaystyle\mathbb{C}[S_n]=\bigoplus\limits_{\lambda\vdash n}f^\lambda S^\lambda

Taking dimensions on both sides we find

\displaystyle \sum\limits_{\lambda\vdash n}\left(f^\lambda\right)^2=\sum\limits_{\lambda\vdash n}f^\lambda\dim(S^\lambda)=\lvert S_n\rvert=n!

January 21, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups

3 Comments »

  1. […] look at one example of “straightening” out a polytabloid to show it’s in the span of the standard polytabloids, using the Garnir […]

    Pingback by “Straightening” a Polytabloid « The Unapologetic Mathematician | January 25, 2011 | Reply

  2. […] be clear that an obvious choice for the objects is to replace with the Specht module , since we’ve seen that . But what category are they in? On the left side, is an -module, but on the right side all […]

    Pingback by The Branching Rule, Part 1 « The Unapologetic Mathematician | January 27, 2011 | Reply

  3. […] want to show that they span the space . This is a bit fidgety, but should somewhat resemble the way we showed that standard polytabloids span Specht […]

    Pingback by Intertwinors from Semistandard Tableaux Span, part 1 « The Unapologetic Mathematician | February 11, 2011 | Reply


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