The Unapologetic Mathematician

Mathematics for the interested outsider

Standard Tableaux

So we’ve described the Specht modules, and we’ve shown that they give us a complete set of irreducible representations for the symmetric groups. But we haven’t described them very explicitly, and we certainl can’t say much about them. There’s still work to be done.

We say that a Young tableau t is “standard” if its rows and columns are all increasing sequences. In this case, we also say that the Young tabloid \{t\} and the polytabloid e_t=\kappa_t\{t\} are standard.

Recall that we had a canonical Young tableau for each shape \lambda that listed the numbers from 1 to n in each row from top to bottom, as in


It should be clear that this canonical tableau is standard, so there is always at least one standard tableau for each shape. There may be more, of course. For example:


Clearly, any two distinct standard tableaux s^\lambda and t^\lambda give rise to distinct tabloids \{s\} and \{t\}. Indeed, if \{s\}=\{t\}, then s and t would have to be row-equivalent. But only one Young tableau in any row-equivalence class has increasing rows, and only that one even has a chance to be standard. Thus if s and t are row-equivalent standard tableaux, they must be equal.

What’s not immediately clear is that the standard polytabloids e_s and e_t are distinct. Further, it turns out that the collection of standard polytabloids e_t of shape \lambda is actually independent, and furnishes a basis for the Specht module S^\lambda. This is our next major goal.

January 5, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups


  1. […] it should be made clear that this is not a standard tableau, despite the fact that the rows and columns increase. The usual line is that we imagine the tableau […]

    Pingback by Compositions « The Unapologetic Mathematician | January 6, 2011 | Reply

  2. Dear John,

    As you seem to be heading towards stating it, I was wondering if you were going prove the hook length formula. If so, have you seen Jason Bandlow’s elementary proof of it? It is by far the most accesible proof I’ve seen, requiring only the Fundamental Theorem of Algebra.



    Comment by Andrew Poulton | January 10, 2011 | Reply

  3. […] Maximality of Standard Tableaux Standard tableaux have a certain maximality property with respect to the dominance order on tabloids. Specifically, […]

    Pingback by The Maximality of Standard Tableaux « The Unapologetic Mathematician | January 13, 2011 | Reply

  4. […] are Independent Now we’re all set to show that the polytabloids that come from standard tableaux are linearly independent. This is half of showing that they form a basis of our Specht modules. […]

    Pingback by Standard Polytabloids are Independent « The Unapologetic Mathematician | January 13, 2011 | Reply

  5. […] by polytabloids of shape . 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 […]

    Pingback by Standard Polytabloids Span Specht Modules « The Unapologetic Mathematician | January 21, 2011 | Reply

  6. […] that we have a canonical basis for our Specht modules composed of standard polytabloids it gives us a matrix representation of for each . We really only need to come up with matrices for […]

    Pingback by Young’s Natural Representation « The Unapologetic Mathematician | January 26, 2011 | Reply

  7. […] as a quick use of this concept, think about how to fill a Ferrers diagram to make a standard Young tableau. It should be clear that since is the largest entry in the tableau, it must be in […]

    Pingback by Inner and Outer Corners « The Unapologetic Mathematician | January 26, 2011 | Reply

  8. […] means that we can eliminate some intertwinors from consideration by only working with things like standard tableaux. We say that a generalized tableau is semistandard if its columns strictly increase (as for […]

    Pingback by Semistandard Generalized Tableaux « The Unapologetic Mathematician | February 8, 2011 | Reply

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