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Put $G= \mathbb{C} \rtimes_{\phi} \mathbb{C} \setminus \{0\}$ where $\phi_{a} (z)= az$ for $a \in \mathbb{C} \setminus \{0\}$. $G$ is a real $4$ dimensional Lie group; then it has a unique left invariant metric which restricts to the standard Euclidean metric at the neutral element. Let $I(G)$ be the group of isometries of $G$ with respect to this invariant metric.

On the other hand $G$ is an open subset of the complex Euclidean space. The group of holomorphic automorphisms of $G$ is denoted by $Aut(G)$.

We constructed $G$ as an obvious generalization of the $2$ dimensional Lie group $H= \mathbb{R} \rtimes \mathbb{R} \setminus \{0\}$, the Poincare upper half plane. We know that the group of holomorphic automorphisms of $H$ preserves the left invariant metric of $H$.

Now a natural question is: What is the structure of $Aut(G)$? What geometric structure on $G$ is preserved by $Aut(G)$?

Are there any relations between $I(G)$ and $Aut(G)$?

Is there a precise description for these two groups?

Considering $G$ as an open set in $\mathbb{C}^2$, is it true to say that all geodesics are either closed or perpendicular to the boundary? (Perpendicular with respect to the usual geometry of $\mathbb{C}^2 = \mathbb{R}^4$?

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    $\begingroup$ $G$ is isomorphic to the group of upper triangular matrices in $\mathrm{PGL}_2(\mathbf{C}$; it acts transitively by isometries (fixing a boundary point) on the 3-dimensional hyperbolic space and thus can probably be seen inside its unit tangent bundle. Computing the automorphism group is an exercise using the Lie algebra; in particular its unit component is made of the inner automorphisms. $\endgroup$
    – YCor
    Commented May 13, 2017 at 21:42
  • $\begingroup$ @YCor I think G is isomorphic to all matrices with (a, b,0, a). A on diagonal. But could you please explain on remaining part of your comment?(or please give a reference) $\endgroup$ Commented May 13, 2017 at 22:35
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    $\begingroup$ No, the group you mention is commutative. $\endgroup$
    – YCor
    Commented May 13, 2017 at 22:39
  • $\begingroup$ @YCor yes i was incorrect $\endgroup$ Commented May 13, 2017 at 22:50
  • $\begingroup$ @YCor regarding the automorphism group, are you saying every bi holomirphic map on G is an inner group automorphism provided it is homotopic to identity? $\endgroup$ Commented May 13, 2017 at 22:52

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