compute the automorphism of Iwasawa manifold

Question

An Iwasawa manifold is a compact quotient of a 3-dimensional complex Heisenberg group by a cocompact, discrete subgroup. We can also refer to Griffiths and Harris's Principles of Algebraic Geometry p. 444 for simpler description.

I want to compute the automorphism of Iwasawa manifold,i.e.the group of biholomorphisms of X onto itself. But I don't know how to set out to do this.Is there any already known result about this queation?If not,how should I start to study this problem? Thanks in advance!

Answer 1

I'm sure this can be found in the literature, though I don't know exactly where to look. On the other hand, it is easy to calculate the automorphism group directly from the following observations: As in Griffiths--Harris, let $M = G/\Gamma$ where $G$ is the $3$-dimensional complex Heisenberg group and $\Gamma\subset G$ is a discrete, cocompact subgroup. As Griffiths and Harris point out, there is a triple $\omega_1$, $\omega_2$, $\omega_3$ of holomorphic $1$-forms defined on $M$ such that $\omega_1\wedge\omega_2\wedge\omega_3$ is nowhere vanishing, $d\omega_1 = d\omega_2 = 0$, and $d\omega_3 = \omega_1\wedge\omega_2$.

Now, clearly, any holomorphic $1$-form on $M$ must be a constant linear combination of these three $1$-forms, so if $f:M\to M$ is an automorphism, then $f^*(\omega_i)$ must be a constant linear combination of these forms for each $i$. Moreover, because $\omega_3\wedge d\omega_3$ is a nonvanishing holomorphic $3$-form $M$, any automorphism $f$ must preserve this form up to multiplication by a unit complex number. In particular, $f^*(\omega_1\wedge\omega_2) = \lambda\ \omega_1\wedge\omega_2$ for some $\lambda\in\mathbb{C}$ with $|\lambda| = 1$. Pursuing this line, one sees that $f:M\to M$ must be covered by a biholomorphism $F:G\to G$ that preserves both the right-invariant holomorphic $1$-forms (up to constant linear combinations) and the lattice $\Gamma$ (up to an appropriate translation).

This is now a purely algebraic problem (whose solution might depend on which $\Gamma$ you use). In particular, if you let $z_1$, $z_2$, $z_3$ denote the coordinates on $G$ such that the forms $\omega_i$ pull back to $dz_1$, $dz_2$, and $dz_3 + z_1\ dz_2$, respectively, on $G$, then $F$ has to be of the form $$ F(z_1,z_2,z_3) = (a_{11} z_1 {+} a_{12}z_2 {+} a_{10}, a_{21} z_1 {+} a_{22}z_2 {+} a_{20}, a_{33} (z_3{-}z_1z_2) {+} a_{31} z_1 {+} a_{32} z_2 {+} a_{30}) $$ for some constants $a_{ij}$ such that $$ a_{33} = a_{11}a_{22}-a_{21}a_{12} = \lambda $$ where $|\lambda| = 1$. It must also preserve $\Gamma$ up to an appropriate translation, which puts other restrictions on the $a_{ij}$. This will give you the automorphism group.

Answer 2

Here are some easy remarks to start with. The group $\mathrm{Aut}(X)$ is a complex Lie group, its Lie algebra is $H^0(X,T_X)$. Since $X=G/\Gamma $, the tangent bundle $T_X$ is trivial, so $H^0(X,T_X)$ has dimension 3, and is naturally identified with the Lie algebra of $G$; this implies that the neutral component of $\mathrm{Aut}(X)$ is $G$. It remains to compute the finite group $\pi _0(\mathrm{Aut}(X))$; this is probably more subtle.