The Unapologetic Mathematician

Mathematics for the interested outsider

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.


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


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

    Pingback by Standard Polytabloids are Independent : : standardy | January 16, 2011 | Reply

  2. […] 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

  3. […] 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

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: