Representation theory on $\mathbb{Z}^d $, classifying invariant sub-lattices Suppose that we have a lattice $\mathbb{Z}^d$ and a subgroup $\Gamma$ of $\operatorname{SL}_d (\mathbb{Z})$ acting on it. Assuming the action on $\mathbb{R}^d$ is irreducible, does this tell us anything about the invariant sub-lattices of $\mathbb{Z}^d$ ? In particular, are they all of the form $k\mathbb{Z}^d$ for integers $k$?
 A: Irreducibility implies that the nonzero invariant sublattices have full rank (since tensoring them with $\mathbb{R}$ gives all of $\mathbb{R}^d$), but I don't think you can conclude more than that. 
Here is a method for constructing counterexamples. Let $K$ be a quadratic number field. Any algebraic integer $\alpha \in \mathcal{O}_K \cong \mathbb{Z}^2$ acts by multiplication on $\mathcal{O}_K$ in a way that preserves the sublattices of $\mathcal{O}_K$ given by any ideal $(\beta) = \beta \mathcal{O}_K$, which are not integer multiples of $\mathcal{O}_K$ as long as $\beta$ is not a unit times an integer. The determinant of $\alpha$ is the norm $N(\alpha) = \alpha \overline{\alpha}$, so $\alpha$ lies in $SL_2(\mathbb{Z})$ iff $N(\alpha) = 1$.
The remaining question is when the subgroup of $SL_2$ generated by multiplication by $\alpha$ acts irreducibly on $\mathcal{O}_K \otimes \mathbb{R}$. If $K$ is an imaginary quadratic field then $\mathcal{O}_K \otimes \mathbb{R} \cong \mathbb{C}$ and any $\alpha \in \mathcal{O}_K$ which is not real, or equivalently not an integer, generates $K \cong \mathcal{O}_K \otimes \mathbb{Q}$ as a $\mathbb{Q}$-algebra and hence generates $\mathcal{O}_K \otimes \mathbb{R}$ as an $\mathbb{R}$-algebra; consequently the action is irreducible in this case. 
So, the simplest counterexample is $K = \mathbb{Q}(i), \alpha = i$; multiplication by $\alpha$ on $\mathbb{Z}[i]$ generates a copy of $C_4$ which acts on $\mathbb{R}[i] \cong \mathbb{C}$ irreducibly, and every sublattice of the form $(\beta) = \beta \mathbb{Z}[i], \beta \in \mathbb{Z}[i]$ is invariant. Most of these sublattices are not multiples of $\mathbb{Z}[i]$, for example if $\beta = 1 + i$. We can find similar examples where $\alpha$ has infinite order. 
