A basis for Schur functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:22:45Z http://mathoverflow.net/feeds/question/120798 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120798/a-basis-for-schur-functors A basis for Schur functors Dmitry Vaintrob 2013-02-04T20:30:56Z 2013-02-05T14:47:58Z <p>Suppose $V$ is a finite-dimensional vector space (over $\mathbb{C}$) and $\lambda$ is a partition of $n$ (not necessarily the dimension). Let $S^\lambda(V)=(V^{\otimes n})_\lambda$ be the $\lambda$'th Schur functor applied to $V$ (a.k.a. the $\lambda$-component of the $S_n$ representation $V^{\otimes n}$). Pick a basis for $V$. Is there a good basis for $S^\lambda(V)$? Even better, is there a good subset of the canonical basis of $V^{\otimes n}$ which projects to a basis of $S^\lambda(V)$? (This is the case for our favorite Schur functors coming from the trivial and alternating reps). </p> <p>Further, is there a good combinatorial description for the functoriality property, i.e. $S^\lambda(M)$ where $M:V\to W$ is a matrix?</p> http://mathoverflow.net/questions/120798/a-basis-for-schur-functors/120806#120806 Answer by Y Macdisi for A basis for Schur functors Y Macdisi 2013-02-04T21:50:37Z 2013-02-04T21:50:37Z <p>You can construct "canonical" basis for the decomposition of $V^{\otimes n}$ under the symmetric group (and simultaneously under $GL(V)$) through "Young symmetrizers";</p> <p>see <a href="http://en.wikipedia.org/wiki/Young_symmetrizer" rel="nofollow">http://en.wikipedia.org/wiki/Young_symmetrizer</a> for starters..</p> <p>I'm sure the map $S^\lambda(M)$ can also be constructed "canonically". You might want to look at this software by Brian Wybourne for details on the calculations...</p> <p><a href="http://schur.sourceforge.net/" rel="nofollow">http://schur.sourceforge.net/</a></p> http://mathoverflow.net/questions/120798/a-basis-for-schur-functors/120863#120863 Answer by darij grinberg for A basis for Schur functors darij grinberg 2013-02-05T14:47:58Z 2013-02-05T14:47:58Z <p>I think this exact question answered in Theorem 1 of §8.1 of <a href="http://books.google.com/books/about/Young_Tableaux.html?id=cYA_RpBLJUkC" rel="nofollow">William Fulton's "Young Tableaux"</a>. Lazy as I am, I am unsure whether I have ever read the proof, but the answer is the following (unless I got Fulton's notations wrong): Whenever $p$ is a filling of the Young diagram of $\lambda$ <strong>with vectors in $V$</strong>, we can define an element $x_p$ of $S^{\lambda}\left(V\right)$ by taking, for each $i=1,2,...,\lambda_1$, the wedge product $w_i$ of the elements of the $i$-th column of $p$ (from top to bottom), and then taking the tensor product $w_1\otimes w_2\otimes ...\otimes w_{\lambda_1}$ of these $w_i$, and projecting this tensor product onto $S^{\lambda}\left(V\right)$. Now, letting $e_1,e_2,...,e_n$ be a basis of $V$, we can construct, for every Young tableau $T$ of shape $\lambda$ over the alphabet $\left\lbrace 1,2,...,n\right\rbrace$, an element $e_T$ of $S^{\lambda}\left(V\right)$ by $e_T = x_{p_T}$, where $p_T$ is the filling of the shape $\lambda$ in which every cell filled with a letter $j$ in $T$ is filled with the basis vector $e_j$. Then, Fulton's Theorem 1 claims that the $e_T$ with $T$ ranging over all <strong>semistandard</strong> tableaux over the alphabet $\left\lbrace 1,2,...,n\right\rbrace$ form a vector space basis of $S^{\lambda}\left(V\right)$.</p> <p>Nota bene: This is only canonical with respect to a <strong>totally ordered</strong> basis.</p> <p>Related results exist for so-called "bitableaux" (and probably contain the above result for tableaux?); see <a href="http://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/Rota2.pdf" rel="nofollow">Doubilet-Rota-Stein Foundations IX</a> and the subsequent papers by Rota, Désarménien, Kung, de Concini, Procesi and others.</p>