# What is weight lattice modulo coroot lattice?

In Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov, the $S$ matrix (expression 3.3.7) is expressed in the form $\vert P/kQ^\vee \vert^{-1/2}\times(\cdots)$, where $k\in \mathbb{N}$ is the shifted level, $P$ is the weight lattice and $Q^\vee$ is the coroot lattice. This expression seems to suggest that $Q^\vee$ can be interpreted as a sublattice of $P$, but this is giving me a confusion because $P\subset \mathfrak{h}^*$ while $Q^\vee\subset\mathfrak{h}$. What is the natural way to embed $Q^\vee$ inside $P$?

For instance, in case of $G_2$, let's say $\alpha$ and $\beta$ are simple short and long roots, respectively. The fundamental weights are then $w_1 = 2\alpha+\beta$, $w_2 = 3\alpha+2\beta$. Hence the weight lattice in this case is the same as the root lattice; $P=Q$. Now the coroots are $\alpha^\vee = \alpha$, $\beta^\vee = \frac{1}{3}\beta$, and they don't sit inside $P$.

Should I just assume that $Q^\vee$ is the sublattice of $P$ generated by long roots, or is there a better way to interpret $\vert P/Q^\vee \vert$?

## 1 Answer

In Theorem 3.3.6 the authors of said book mention that $Q^\vee$ is embedded into $\mathfrak h^*$ via the form $\langle , \rangle$. Here $\langle, \rangle$ denotes an invariant bilinear form on $\mathfrak g$ such that $\langle \alpha,\alpha\rangle = 2$ for long roots (see the beginning of section 1.4).

In this way $kQ^\vee$ is viewed as a sublattice of $P$. Actually, the authors state in the proof of Theorem 3.3.20 that $W^a = W\ltimes kQ^\vee$ acts on $P \subseteq \mathfrak h^*$; as such this means that $P/kQ^\vee$ is the set of orbits of the action of $kQ^\vee$ on $P$ (which is the same thing).

• Thanks for the answer. That makes sense. I read section 1.3 but not 1.4, so I wasn't aware that they are using two different invariant bilinear forms on $\mathfrak{g}$, one normalized so that $\langle\langle\alpha,\alpha\rangle\rangle=2$ for short roots while the other normalized so that $\langle \alpha,\alpha \rangle$ for long roots. After all, $Q^\vee$ is the sublattice of $Q$ generated by long roots, as I expected. Jul 27, 2018 at 17:25