$GL_n\mathbb{C}$ as subgroup of $GL_{2n}\mathbb{R}$

differential-geometrygeneral-linear-groupgroup-theorylie-groups

I am confused about the following paragraph in Fulton W., Harris J. "Representation Theory: A First Course":

Of course, the group $GL_n\mathbb{C}$ of complex linear automorphisms
of a complex vector space $V=\mathbb{C}^n$ can be viewed as subgroup
of the general linear group $GL_{2n}\mathbb{R};$ it is, thus, a real
Lie group as well, as is the subgroup $SL_n\mathbb{C}$. Similarly, the
subgroups $SO_n\mathbb{C} \subset SL_n\mathbb{C}$ and
$Sp_{2n}\mathbb{C} \subset SL_{2n}\mathbb{C}$ are real as well as
complex Lie subgroups. Note that since all nondegenerate bilinear
symmetric forms on a complex vector space are isomorphic (in
particular, there is no such thing as a signature), there is one
complex orthogonal subgroup $SO_n\mathbb{C} \subset SL_n\mathbb{C}$ up
to conjugation; there are no analogs of the groups
$SO_{k,l}\mathbb{R}.$

The things I don't understand:

  1. How do we get the subgroup structure on $GL_n\mathbb{C}$? It seems, that according to the definition of a subgroup (it is also a closed submanifold), we should construct a map $f: GL_n\mathbb{C} \rightarrow GL_{2n}\mathbb{R}$. Should $f$ also be a homomorphism of groups? If it is the case, then I can hardly imagine its coordinate representation.
  2. How do we prove that all nondegenerate bilinear symmetric forms are isomorphic?
  3. I have never encountered such a designation $SO_n\mathbb{C}$, it was always $SO_n$. And now I don't understand why we ignore the field.

Also, I want to emphasize that before this paragraph, the structure of a complex manifold was not defined.

Best Answer

  1. Consider, for instance, the case $n=2$. An element of $GL_2\Bbb C$ is a matrix of the form$$\begin{bmatrix}z_1&z_2\\z_3&z_4\end{bmatrix}$$with $z_1z_4-z_2z_3\ne0$. But you can see it as$$\begin{bmatrix}\operatorname{Re}z_1&-\operatorname{Im}z_1&\operatorname{Re}z_2&-\operatorname{Im}z_2\\\operatorname{Im}z_1&\operatorname{Re}z_1&\operatorname{Im}z_2&\operatorname{Re}z_2\\\operatorname{Re}z_3&-\operatorname{Im}z_3&\operatorname{Re}z_4&-\operatorname{Im}z_4\\\operatorname{Im}z_3&\operatorname{Re}z_3&\operatorname{Im}z_4&\operatorname{Re}z_4\end{bmatrix}\in GL_4\Bbb R.$$This works in any dimension.
  2. You will find an answer here.
  3. The group $SO_n$ is the group of those matrices $A\in GL_n\Bbb R$ such that $A^TA=\operatorname{Id}$ and that $\det A=1$. The condition $A^TA=\operatorname{Id}$ is very inportant over the reals, since it means that $A$ is the matrix of an isometry. But over $\Bbb C$ the same condition has no particular geometric meaning.
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