# 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.) – Ariyan Javanpeykar Nov 10 '16 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! – David Loeffler Nov 10 '16 at 16:48
• Aha! Thank you for your comment. :) – Ariyan Javanpeykar Nov 10 '16 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. – Alexander Premet Nov 9 '16 at 13:19