17
$\begingroup$

The ring $Symm$ of symmetric functions in infinitely many variables is well-known to be a polynomial ring in the elementary symmetric functions, and has a $\mathbb Z$-basis of Schur functions $\{S_\lambda\}$, with structure constants $c_{\lambda\mu}^\nu$ called Littlewood-Richardson numbers.

I have been studying a commutative deformation of these structure constants, i.e. lifting $c_{\lambda\mu}^\nu$ to a 2-variable polynomial $c_{\lambda\mu}^\nu(a,b)$, for which I have a very explicit formula (much too complicated to describe here). However, since $Symm$ is a polynomial ring (i.e. free), any commutative deformation of it must be trivializable: there must be a way to deform each $S_\lambda$ to a function $S_\lambda(a,b)$ whose ordinary multiplication has my structure constants.

Given the $\{c_{\lambda\mu}^\nu(a,b)\}$, how can I trivialize this family, finding such deformations $S_\lambda(a,b)$?

It seems like the answer should be nonunique (hence harder to find); it's easy to imagine deformations of the basis that don't change the structure constants at all.

What extra conditions should one put on the deformation in order to make the trivialization unique?

$\endgroup$
  • $\begingroup$ Is it possible that the deformation is upper-triangular (i.e. that $S_\lambda(a,b) = S_\lambda + \textrm{lower order terms}$)? How complicated are the structure constants when $\mu$ is 1 box? $\endgroup$ – Peter Samuelson Mar 10 '15 at 14:58
  • $\begingroup$ I think $a=0$ is triangular one way, $b=0$ triangular the other way. The one-box rule is here: mathoverflow.net/questions/88569/… $\endgroup$ – Allen Knutson Mar 10 '15 at 17:40

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.