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Lattices in the sense of discrete subgroups of Euclidean spaces, as used in number theory, discrete geometry, Lie groups, etc. (Not to be confused with lattice theory or lattices as used in physics! For lattices (ordered sets), use the tag: [lattice-theory])
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can minimal volume rational subspaces in a lattice be arbitrarily 'close'.
Let $\Gamma$ be a cocompact lattice in $\mathbb{R}^n$, eg. $\Gamma = A \mathbb{Z}^n$ for some $A \in SL_n \mathbb{R}$. Then any $k$-dimensional subspace $P$ which is rational in $\Gamma$ has a volume: …
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Accepted
Lorentz quotient and orientation
This particular question (``consistently orienting the reductions") and all those arising from your above link are interesting.
However, the answer to this question seems `easy' -- and i'm wondering …
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Linear symmetric spaces are spaces with ''orthogonal complements''?
The collection of marked unimodular lattices in the euclidean $n$-space corresponds to the symmetric space $S_n:=SO_n(\mathbb{R}) \backslash SL_n(\mathbb{R})$. … The collection of (marked) nondegenerate lattices in $n$-space corresponds to the collection $P_n$ of positive definite $n\times n$ symmetric real matrices. …
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Linear symmetric spaces are spaces with ''orthogonal complements''?
Finally, i understand the answer! It is Hermann's Convexity Theorem -- and Harish-Chandra's canonical embedding, which proves that EVERY symmetric space (of noncompact type) is conformally equivalent …
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Comparing the volume of a rational lagrangian under a linear symplectomorphism.
The following problem arises in computations with symplectic lattices: Suppose $W$ is a rational lagrangian subspace in the symplectic lattice $\Lambda$, and suppose we have a basis $w_1, w_2, \ldots$ …
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List of Automorphism groups of Abelian Varieties for Dummies
(%Edited after abx comment%)
I seek explicit linear integral representations $\rho: Aut(X,\omega) \to Sp_{2g}(\mathbb{Z})$, when $(X,\omega)$ is complex $g$-dimensional PPAV. I prefer explicit matri …
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When does a subgroup of $\operatorname{GL}(n, \mathbb Q)$ have a bounded fundamental domain ...
After reading the comments, I think the underlying algebraic question is: " if $G=U$ is a unipotent linear algebraic group defined over $\mathbb{Q}$, then is the arithmetic unipotent subgroup $U(\math …