# Integral lattices in Lie group representations

Let $G$ be a split semisimple algebraic group scheme over $\mathbf{Z}$ (I'm mostly interested in the case $G = Sp_4$).

Let $V$ be an irreducible representation of the generic fibre $G_{\mathbf{Q}}$, of some highest weight $\lambda$, and $v$ a choice of highest-weight vector. Then there is a notion of an "admissible $\mathbf{Z}$-lattice" in $V$: it's a $\mathbf{Z}$-lattice $V_{\mathbf{Z}}$ in $V$ for which the structure map $G_{\mathbf{Q}} \to GL(V)$ extends to a map of $\mathbf{Z}$-group schemes $G_\mathbf{Z} \to GL(V_\mathbf{Z})$, and such that the intersection of $V_{\mathbf{Z}}$ with the highest-weight space $\mathbf{Q} \cdot v$ is $\mathbf{Z} \cdot v$. There are finitely many of these, and in particular there's a maximal and a minimal one.

My question is this: let $P$ be a parabolic subgroup (containing our fixed choice of Borel) and $M$ its Levi factor. It's easy to see that $V$ breaks up as a direct sum of eigenspaces uder $Z(M)$, and the highest-weight eigenspace is the weight $\lambda$ representation $W$ of $M$, with $v$ as its highest weight vector. Moreover, the image in $W$ of an admissible lattice in $V$ is an admissible lattice in $W$.

Is the projection to $W$ of the maximal (or minimal) admissible lattice in $V$ equal to the maximal (resp. minimal) admissible lattice in $W$?

(If this works for the minimal lattice it works for the maximal one, and vice versa, because the minimal and maximal lattices are dual to each other.)

• Why are there only finitely many of "these"? I assume you work up to conjugation by some group, but which group? $GL(V_{\mathbb Z})(\mathbb Z)$ or $GL(V_{\mathbb Z})(\mathbb Q)$? Or some other group? (I know this is not what your question is about, but your statement sparked my curiosity.) Nov 10, 2016 at 13:44
• Once you fix the highest-weight vector $v$ there are literally only finitely many admissible $V_{\mathbf{Z}}$'s -- up to conjugation by the trivial group! Nov 10, 2016 at 16:48
• Aha! Thank you for your comment. :) Nov 10, 2016 at 17:41

Let $$\mathfrak{n}_-$$ be the Lie algebra of the derived subgroup $$N_-$$ of $$B_{-}$$ where $$B_{-}=TN_-$$ is a Borel subgroup of $$G$$ opposite to a Borel subgroup $$B_+\subset P$$ (here $$T$$ is a split maximal torus of $$G$$ defined over $$\mathbb{Z}$$). If $$V_\mathbb{Z}$$ is a minimal admissible $$\mathbb{Z}$$-lattice then there is a highest weight vector $$v\in V$$ (for $$B_+$$) such that $$V_{\mathbb{Z}}=U^{-}_{\mathbb Z}\cdot v$$ where $$U^{-}_{\mathbb Z}$$ is the Kostant $$\mathbb{Z}$$-form of the universal enveloping algebra $$U(\mathfrak{n}_-)$$. It is spanned over $$\mathbb{Z}$$ by the PBW-monomials in $$e_\alpha^n/n!$$ where $$\alpha$$ is a negative root of $$G$$ with respect to $$T$$ and $$n\in \mathbb{Z}_{\ge 0}$$ (the root vectors $$e_\alpha$$ should come from a Chevalley basis of $$\mathfrak{g}={\rm Lie}(G)$$). It follows that $$V_\mathbb{Z}\cap W$$ is spanned over $$\mathbb{Z}$$ by the elements of the form $$X\cdot v$$ where $$X$$ is a PBW monomial in $$e_\alpha^n/n!$$ and $$\alpha$$ is a negative root of $$P$$ with respect to $$T$$. This shows that $$V_\mathbb{Z}\cap W$$ is a minimal admissible lattice for a Levi subgroup of $$P$$.
• Most of what I used can be found in Lectures on Chevalley groups '' by Robert Steinberg. The book is easy to find on the web and it can be downloaded from several sources. Nov 9, 2016 at 13:19