User melanie matchett wood - MathOverflow

Computer package for representation theory of the symmetric group

2012-05-11T20:29:27Z

Is there a computer algebra package in which I can compute the following for representations of a specific symmetric group (e.g. $S_7$):

(1) $V \otimes W$

(2) $S_\lambda V$, where $S_\lambda$ is a Schur functor, or even just $\wedge^s V$,

where $V$ and $W$ are input as sums of irreducible representations, i.e. by partitions with coefficients, and output in the same format?

Answer by Melanie Matchett Wood for Symmetric Power and Gauss quadratic form

2011-08-15T16:12:10Z

Forms $ax^2+bxy+cy^2$ for which $a,b,c,\in \mathbb Z$ and $b$ is even correspond to the symmetric submodule of $(\mathbb Z^2)^{\otimes 2}$, i.e. the $S_2$ fixed elements of $(\mathbb Z x\oplus \mathbb Z y)^{\otimes 2}$.

Forms $ax^2+bxy+cy^2$ for which $a,b,c,\in \mathbb Z$ correspond to the symmetric quotient of $(\mathbb Z^2)^{\otimes 2}$, i.e. the quotient of $(\mathbb Z x\oplus \mathbb Z y)^{\otimes 2}$ by elements $a-\sigma(a)$ where $\sigma$ is the transposition of what is left and right of the tensor.

Answer by Melanie Matchett Wood for Binary Quadratic Forms in Characteristic 2

2011-08-15T16:04:10Z

Classes of binary quadratic forms over any commutative ring R (with no conditions on characteristic, PID, etc.) correspond to certain modules over quadratic extensions of R. This theory is completely general also in the forms it covers, i.e. covers non-primitive and discriminant zero forms. We can then restrict our attention to the cases of interest.

For example, when R is an integral domain and the binary quadratic form has non-zero discriminant, the modules associated to the forms are exactly the ideal classes of the quadratic extension.

The construction is as follows, simplied for your case of interest. From a form $ax^2+bxy+cy^2$, we form a quadratic $R$-algebra $Q:=R[\tau]/(\tau^2+b\tau+ac)$ and a module $M =Rx\oplus Ry$ with $\tau x=-cy-bx$ and $\tau y=ax$. Given a quadratic algebra and ideal, we can pick an $R$ basis $x,y$ of the ideal and then shift a generator $\tau$ of the quadratic algebra as necessary so that $\tau y$ is a multiple of $x$. Then you can simply read off the inverse map.

What I've given above is the case when the binary quadratic form, quadratic algebra, and module are free over $R$. For general $R$, they will only be locally free, and the above gives a construction locally on free patches. There are also global, coordinate-free descriptions of the correspondence between binary quadratic forms and modules over quadratic algebras. For the details I refer you to my paper "Gauss composition over an arbitrary base" http://www.sciencedirect.com/science/article/pii/S0001870810003257

Hypercohomology of a dg-algebra

2009-12-03T00:33:02Z

Can someone give me a reference (note I am looking for a reference and not a proof) for the following:

If a complex $C$ has a dg-algebra structure, then the hypercohomology $H^0R\pi_*C$ has an algebra structure, and if $M$ is a dg-module for $C$, then $H^0R\pi_*M$ is a module under $H^0R\pi_*C$. (Here I am thinking of $C$ as a complex of sheaves on some scheme $S$ with $\pi : S \rightarrow T$, but this should just be a fact of homological algebra.)

Comment by Melanie Matchett Wood

2012-05-11T21:30:36Z

I just looked at this, and it seems "itensor" will do (1), but I am not sure if there is a function there for (2) (but there could be and I am missing it because I don't know how to translate schur functors, or even exterior products, into symmetric polynomial language).

Comment by Melanie Matchett Wood

2011-08-24T16:16:32Z

Example: $a=t$ and $b=2us+2t$ and $c=2us+t$. The discriminant is $4u^2s^2$. This form is not congruent to a square modulo $u^2$

Comment by Melanie Matchett Wood

2011-08-11T19:02:04Z

If we take Jordan's example $a = s^2 + pust$, $b = 2st + put^2$, $c = t^2$, then the discriminant is $p^2u^2t^4$, which has $u$-adic valuation $2$ (which is the relevant one---the $2$ or $p$ is a red herring).