## Independence of Intertwinors from Semistandard Tableaux

Let’s start with the semistandard generalized tableaux and use them to construct intertwinors . I say that this collection is linearly independent.

Indeed, let’s index the semistandard generalized tableaux as . We will take our reference tableau and show that the vectors are independent. This will show that the are independent, since any linear dependence between the operators would immediately give a linear dependence between the for all .

Anyway, we have

Since we assumed to be semistandard, we know that for all summands . Now the permutations in do not change column equivalence classes, so this still holds: for all summands . And further all the are distinct since no column equivalence class can contain more than one semistandard tableau.

But now we can go back to the lemma we used when showing that the standard polytabloids were independent! The are a collection of vectors in . For each one, we can pick a basis vector which is maximum among all those having a nonzero coefficient in the vector, and these selected maximum basis elements are all distinct. We conclude that our collection of vectors in independent, and then it follows that the intertwinors are independent.

[...] that we’ve shown the intertwinors that come from semistandard tableaux are independent, we want to show that they span the space . This is a bit fidgety, but should somewhat resemble the [...]

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[...] the space of all intertwinors from the Specht module to the Young tabloid module . We also know that they’re linearly independent, and so they form a basis of the space of intertwinors [...]

Pingback by Kostka Numbers « The Unapologetic Mathematician | February 17, 2011 |