sp2n-fundamental-tilting-charaters

You can find a (short and hopefully sweet) Mathematica notebook to compute the Weyl characters for fundamental tilting modules that you can download on this page.

This uses the below formula from https://arxiv.org/abs/2303.04264

References for the recalled background below can also be found in that paper.

Background

Fix some $n>0$. Let $\mathcal{A}=\mathbb{Z}[q,q^{-1}]$ for a formal parameter $q$. We work with the divided power quantum group for $\mathfrak{sp}_{2n}$.

Fix any field $\mathbb{K}$ and any $\xi\in\mathbb{K}\setminus{0}$, we can specialize this quantum algebra to $U_{\xi}(\mathfrak{sp}_{2n})$ by sending $q\mapsto\xi$. We have no restrictions on $\mathbb{K}$ or $\xi$.

The quantum group $U_{\xi}(\mathfrak{sp}_{2n})$ has four types of natural modules, all of which are indexed by dominant integral weights $\lambda\in X^{+}$:

• The Weyl modules $\Delta(\lambda)$ that play the role of standard modules.
• The dual Weyl modules $\nabla(\lambda)$ that play the role of costandard modules.
• The simple modules $L(\lambda)$ that play the role of atoms. These are not addressed in this notebook.
• The indecomposable tilting modules $T(\lambda)$ that play the role of projective modules.

As usual in representation theory, we would like to understand the characters of these modules. For the first two this is easy: $\Delta(\lambda)$ and $\nabla(\lambda)$ have the same characters as the simple (quantum) $\mathfrak{sp}_{2n}$-modules in the semisimple case which are given by (the quantum version of) Weyl's character formula. In other words, they are known.

Let $\varpi_{i}$ be the ith fundamental weight of the symplectic group where we use the convention that the double edge of the Dynkin diagram of $\mathfrak{sp}_{2n}$ is from $n-1$ to $n$. Then the tilting characters (we denote these using square brackets) are given as follows.

First, the initial conditions are $[T_{\mathbb{K}}(\varpi_{0})]=[\Delta_{\mathbb{K}}(\varpi_{0})]$ (the trivial modules) and $[T_{\mathbb{K}}(\varpi_{1})]=[\Delta_{\mathbb{K}}(\varpi_{1})]$. Then, for $k>1$: $$[T_{\mathbb{K}}(\varpi_{k})]=\sum_{i\geq 0,k-2i\geq 0}[\Delta_{\mathbb{K}}(\varpi_{k-2i})]-\sum_{i\geq 1,k-2i\geq 0,\binom{n-k}{i}\neq 0}[T_{\mathbb{K}}(\varpi_{k-2i})].$$ (Here $\binom{n-k}{i}$ is the quantum binomial.) Note that this determines them as the Weyl characters are known.

The Mathematica notebook computes these characters.

Running the notebook

There are only a few functions in the notebook:

qbin[n_, k_, q_] is the quantum binomial $\binom{n}{k}$ (or rather a shift of it); with input n,k, and $\xi\leftrightsquigarrow q$

WeylCharSp[n_, k_, p_, q_] computes the character of $T(\varpi_{k})$ for $\mathfrak{sp}_{2n}$. Here $p$ is the characteristic of $\mathbb{K}$ (take $p\gg 0$ to simulate characteristic zero) and $\xi\leftrightsquigarrow q$

XX = Table[WeylCharSp[78, k, 7, 2], {k, 0, 78}]; MM = Table[If[Count[XX[[j]], i] == 1, 1, 0], {i, 0, 78}, {j, 0, 78}]; MatrixPlot[MM] then outputs the whole base change matrix as in the example below

Example

For example, for $n=78$ and quantum characteristic $p=7,\ell=3$ and $p=3,\ell=2$, respectively, we get:

The columns are the fundamental tilting modules $T(\varpi_{i-1})$ and the rows the fundamental Weyl modules $\Delta(\varpi_{i-1})$, both starting to count at one.

Contact

In case of questions drop us an email: elijah.bodish@gmail.com dtubbenhauer@gmail.com