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Concider a lattice in $\mathbb R^{n}$ (i.e. all $\mathbb Z$-linear combinations of a chosen basis). The coordinates of all vectors in this lattice with respect to its basis, is given by $\mathbb Z^{n}$. Am I right in stating that the automorphism group of a lattice consists of all permutations ($n!$) and sign changes ($2^n$) of $\mathbb Z^{n}$? This would mean that the order of the automorphism group is $2^nn!$, regardless of which lattice I'm talking about (e.g. in $\mathbb R^{2}$ you have the square, hexagonal, rectagonal, ... lattice).

I suspect this only goes for the square lattice (or cubic lattice in $\mathbb R^{3}$), but I can't understand why. Choosing the right basis for any lattice makes it isomorphic to $\mathbb Z^{n}$ right?

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    You're right in reducing your search to the case of a square lattice. An automorphism of the lattice is a linear transformation in $\mathbb{R}^n$, and if you write its matrix in the "chosen basis", then it will be the matrix of an automorphism of $\mathbb{Z}^n$. But, the group of automorphisms of $\mathbb{Z}^n$ is much bigger than just signed permutations, it's infinite (for $n \ge 2$) and consists of all integer matrices whose determinant is $1$ or $-1$ (which we denote $\textrm{GL}_n(\mathbb{Z})$).2011-10-07
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    The automorphism group of $\mathbb{Z}^n$ is not finite when $n\geq 2$, since it consists of all $n\times n$ matrices with integer coefficients and determinant 1 or -1.2011-10-07
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    @Joel - Shouldn't an automorphism of a lattice be an orthogonal transformation (i.e. linear transformation that preserves the inner product), which is represented by a (not necessarily orthogonal) matrix from $GL(n,\mathbb Z)$ with respect to the lattice basis? Since there are only a finite number of those, the automorphism group is a finite subgroup of $GL(n,\mathbb Z)$. How do I relate this to the fact that (as I understand from your comment) $GL(n,\mathbb Z)$ is the automorphism group of $\mathbb Z^n$ while lattice and $\mathbb Z^n$ are isomorphic?2011-10-07
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    @Wox : The expression of the inner product will change if you change the basis. When your lattice is generated by an orthonormal basis (the square lattice), then the matrices will be orthogonal and in $\textrm{GL}_n(\mathbb{Z})$, so it's a signed permutation. If your lattice is generated by an arbitrary basis, denote $A$ its Gramian matrix, automorphisms that preserve the inner product will be matrices in $M \in \textrm{GL}_n(\mathbb{Z})$ that satisfy $M^t A M = A$.2011-10-07
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    For example if you take $A = \textrm{diag}(2,1)$ (a rectangular lattice), you find only four automorphisms which are $\textrm{diag}(e_1,e_2)$ where $e_1$ and $e_2$ are either $1$ or $-1$ (I invite to check that using $M^t A M = A$). Geometrically it means that because you have a rectangle (and not a square), you can't swap the coordinates.2011-10-07
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    @Joel - I understand it when looking at examples, but I don't understand it theoretically. If a lattice is isomorphic (as free $\mathbb Z$-module) with $\mathbb Z^n$ then why aren't all lattice automorphism groups the same? This is obviously not true so I'm missing something very basic here...2011-10-10
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    @Wox : If you require automorphisms to preserve an inner product, it means you're looking at the lattice as $\mathbb{Z}$-module **and** a metric space. From that point of view, square and rectangular lattices are **not** the same (they are not isometric). So it's not a big surprise that their automorphism group turn out to be different. Remember that the expression of the inner product changes with the basis. So choosing the "right" basis does indeed make the lattice isomorphic to $\mathbb{Z}^n$ as a $\mathbb{Z}$-module, but messes up the inner product doing so.2011-10-10

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