Standard Polytabloids 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. We’ll actually use a lemma that applies to any vector space $V$ with an ordered basis $e_\alpha$ . Here $\alpha$ indexes some set $B$ of basis vectors which has some partial order $\preceq$ .

So, let $v_1,\dots,v_m$ be vectors in $V$ , and suppose that for each $v_i$ we can pick some basis vector $e_{\alpha_i}$ which shows up with a nonzero coefficient in $v_i$ subject to the following two conditions. First, for each $i$ the basis element $e_{\alpha_i}$ should be the maximum of all the basis vectors having nonzero coefficients in $v_i$ . Second, the $e_{\alpha_i}$ are all distinct.

We should note that the first of these conditions actually places some restrictions on what vectors the $v_i$ can be in the first place. For each one, the collection of basis vectors with nonzero coefficients must have a maximum. That is, there must be some basis vector in the collection which is actually bigger (according to the partial order $\preceq$ ) than all the others in the collection. It’s not sufficient for $e_{\alpha_i}$ to be maximal, which only means that there is no larger index in the collection. The difference is similar to that between local maxima and a global maximum for a real-valued function.

This distinction should be kept in mind, since now we’re going to shuffle the order of the $v_i$ so that $e_{\alpha_1}$ is maximal among the basis elements $e_{\alpha_i}$ . That is, none of the other $e_{\alpha_i}$ should be bigger than $e_{\alpha_1}$ , although some may be incomparable with it. Now I say that $e_{\alpha_i}$ cannot have a nonzero coefficient in any other of the $v_i$ . Indeed, if it had a nonzero coefficient in, say, $v_2$ , then by assumption we would have $e_{\alpha_1}\prec e_{\alpha_2}$ , which contradicts the maximality of $e_{\alpha_1}$ . Thus in any linear combination

$\displaystyle c_1v_1+\dots+c_mv_m=0$

we must have $c_1=0$ , since there is no other way to cancel off all the occurrences of $e_{\alpha_1}$ . Removing $v_1$ from the collection, we can repeat the reasoning with the remaining vectors until we get down to a single one, which is trivially independent.

So in the case we care about the space is the Young tabloid module $M^\lambda$ , with the basis of Young tabloids having the dominance ordering. In particular, we consider for our $v_i$ the collection of polytabloids $e_t$ where $t$ is a standard tableau. In this case, we know that $\{t\}$ is the maximum of all the tabloids showing up as summands in $e_t$ . And these standard tabloids are all distinct, since they arise from distinct standard tableaux. Thus our lemma shows that not only are the standard polytabloids $e_t$ distinct, they are actually linearly independent vectors in $M^\lambda$ .

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January 13, 2011 - Posted by John Armstrong | Algebra, Representation Theory, Representations of Symmetric Groups

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[...] Przeczytaj artykuł: Standard Polytabloids are Independent [...]

Pingback by Standard Polytabloids are Independent : : standardy | January 16, 2011 | Reply
[...] 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 the Specht [...]

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

Pingback by Independence of Intertwinors from Semistandard Tableaux « The Unapologetic Mathematician | February 9, 2011 | Reply

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