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!


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.

  • $\begingroup$ 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. $\endgroup$ – user108005 Jan 11 '14 at 4:57
  • 2
    $\begingroup$ No, he is right. $GL_1(\Bbb{Z}[i]$ is $\Bbb{Z}[i]^*=\{\pm 1,\pm i\}$. $\endgroup$ – abx Jan 11 '14 at 5:45

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