# the group of all biholomorphic group automorphisms of complex tori

My background is complex geometry, but when I confront complex tori, I feel it is easier to consider it as a compact connected complex Lie group although I just know the definition of Lie group.

Let $X=\mathbb{C}^n/\Lambda$,where $\Lambda$ is the discrete subgroup of maximal rank in $\mathbb{C^n}$ whose entries $(x_1,\ldots,x_n)$ are of the form $x_i=a_i+b_i\sqrt{-1}(a_i,b_i\in\mathbb{Z})$. I want to calculate the group of all biholomorphic group automorphisms $Aut(X)$.

Geometrically, it is the groups of isomorphisms as complex manifolds fixing zero. When $n=1$, it is intuitive that $Aut(X)=\mathbb{Z}/4\mathbb{Z}$. But when the dimension becomes higher，I feel difficult to set out to calculate $Aut(X)$.

I wonder if this problem is easier from the viewpoint of Lie group. Complex geometrical approach is also welcome! Thanks in advance!

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The automorphism group you describe is always equal to the group of automorphisms of the lattice - that is, the group of linear maps $\mathbb C^n \to \mathbb C^n$ that send $\Lambda$ to $\Lambda$. in your case, this group is $GL_n(\mathbb Z[i])$, the group of invertible matrices whose coefficients are Gaussian integers.
no,say $n=1$,$Aut(X)$ should be the cyclic group generated by $i$.But according to your answer,$Aut(X)=\mathbb{Z}[i]\setminus \{0\}$,so you are wrong. – user108005 Jan 11 '14 at 4:57
No, he is right. $GL_1(\Bbb{Z}[i]$ is $\Bbb{Z}[i]^*=\{\pm 1,\pm i\}$. – abx Jan 11 '14 at 5:45