This is in relation to Kodaira's Complex Manifolds and Deformation Complex Structures Chpt 2, Sec 2.

$W$ is a complex Lie group. A discrete subgroup $G\leq W$ gives properly discontinuous and fixed point free action on $W$. Thus $W/G$ makes sense as a complex manifold. Now the book says $W/G$ is a complex Lie group as well without mentioning $G$ normal.

$\textbf{Q:}$ Why does $W/G$ inherits a group structure? Note that I need $W\times W\xrightarrow{\cdot} W$ descends to the quotient level map. From standard group theory, $W/G$ is group iff $G$ is normal by considering $wG\cdot 1G=wG$. The natural procedure is to assume $G$ is normal which forces $G\leq Z(W)$ if $W$ is connected where $Z(W)$ is the center. However, the book did not mention $G$ being normal. Where does normality coming from then or have I missed something here?

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    $\begingroup$ In a connected group every discrete normal subgroup is central. Hence discrete subgroups are non-normal as soon as they are not contained in the center. So there are plenty of examples of non-normal discrete subgroups. $\endgroup$ – YCor Jun 2 at 20:07
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    $\begingroup$ I checked Kodaira's book, the statement occurs on page 48 and is indeed plain false. Moreover, if $G$ is non-abelian discrete and $W$ is simply-connected then $W/G$ cannot even be homeomorphic to a Lie groups, as Lie groups have abelian fundamental groups. As Yves said, there are many examples of nonabelian discrete subgroups of complex Lie groups. For instance, take the permutation group $S_n$ as a subgroup of $GL(n, {\mathbb C})$. $\endgroup$ – Moishe Kohan Jun 2 at 22:39
  • $\begingroup$ @MoisheKohan That is unfortunate. Thank. $\endgroup$ – user45765 Jun 2 at 23:10

I can't comment on the book, since I don't have access to it now. But I can tell you that the statement is false. Just take, for instance$$\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right\},$$which is a non-normal subgroup of $GL(2,\mathbb C)$.

  • $\begingroup$ So you mean without normality, it is not group. I just checked your example. Denote off diagonal matrix as $A=a_{ij}$ with $a_{ii}=0$ and $a_{12}=1,a_{21}=1$ for $A$. Similarly for $B=b_{ij}$ with $b_{ii}=0,b_{12}=-1,b_{21}=1$. Set $G$ the group to be quotient out in your example. Then clearly $AG=BG$. So $A^2G=B^2G$ if there is group structure. One compute to get $A^2G=IG, B^2G=-IG$ which is impossible. $\endgroup$ – user45765 Jun 2 at 18:51
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    $\begingroup$ I don't really understand what you were computing. But I know this: if $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]\in GL(2,\mathbb C)$, then$$\begin{bmatrix}a&b\\c&d\end{bmatrix}.\begin{bmatrix}1&0\\0&-1\end{bmatrix}.\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\begin{bmatrix}\frac{b c+a d}{a d-b c} & \frac{2 a b}{b c-a d} \\ \frac{2 c d}{a d-b c} & \frac{b c+a d}{b c-a d}\end{bmatrix},$$which will seldom be equal to $\left[\begin{smallmatrix}1&0\\0&\pm1\end{smallmatrix}\right]$. Therefore, my subgroup is not normal. $\endgroup$ – José Carlos Santos Jun 2 at 19:09
  • $\begingroup$ I was trying to check whether there is group structure endowed on $GL(2,C)/G$. And I checked the specific elements for multiplication property on the coset to see the quotient not being group which deduces non-normality. $\endgroup$ – user45765 Jun 2 at 19:16

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