2

$GL_n(C)$ is isomorphic to a lie subgroup of $GL_{2n}(R)$.

I see some posts concerning this (not the same claim):

$GL(n, \mathbb{C})$ is isomorphic to a subgroup of $GL(2n, \mathbb{R})$

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

$GL(n, \mathbb{C})$ is a properly embedded Lie subgroup of $GL(2n, \mathbb{R})$

My definitions of a lie group and a ie subgroup are as follows:

Lie group is a manifold $G$ together with a binary operation $m : G×G→G$ that is a smooth map converting $G$ into a group, so that the map $G→G$ carrying $x ∈ G$ to $x^{−1}$, is smooth.

A Lie subgroup of a Lie group $G$ is a subgroup $H$ that is an immersed submanifold.

A subset $M$ of a manifold $N$ is an immersed submanifold if $M$ is endowed with a structure of manifold such that the embedding $M → N$ is an immersion.

A smooth map $f : M → N$ is called immersion if, for any $x ∈M$, the tangent map $T_x(f) : T_xM → T_f(x)N$ is injective.

Can you kindly explain-provide details for this claim.

Mat999
  • 537
  • Why does your final link not answer your question? Are you in doubt as to why it is sufficient to prove that it is just a topological embedding? – SomeCallMeTim Mar 23 '23 at 16:18
  • 1
    I will adress your comment here. I think any proof will basically boil down to showing the second theorem of the latter linked post, so let's just assume it. By applying this, you only need to check that the map in the link is injective, smooth, and a group homomorphism. For smoothness, just note that it is linear. – SomeCallMeTim Mar 23 '23 at 17:42
  • Where I can find a proof of this theorem? – Mat999 Mar 23 '23 at 19:02
  • 1
    As is stated in the link, Lee's book. – SomeCallMeTim Mar 24 '23 at 13:32
  • Is this map smooth since it is a map between subspaces of the euclidean space so it is linear? And to include that that $GL_n(C)$ is isomorphic to the image of this map, can I use the isomorphism theorem for groups? – Mat999 Mar 25 '23 at 16:21
  • 1
    @Matt99 The map is smooth as it is a linear map between euclidian spaces. And yes, you can use the isomorphism theorem to prove isomorphisms of groups. – SomeCallMeTim Mar 27 '23 at 10:14

1 Answers1

2

A non-zero complex number $z = x+iy$ can be mapped to a $2\times 2$ real matrix $$ A_z = \left( \begin{matrix} x & -y \\ y & x \end{matrix} \right). $$ Since $\det A_z = x^2+y^2 \neq 0$, then $A_z \in \mathrm{GL}(2,\mathbb R)$. This establishes a correspondence $\mathrm{GL}(1,\mathbb C) \to \mathrm{GL}(2,\mathbb R)$. Can you generalize it?

Gibbs
  • 8,230
  • This and more general statements have been proven in the links OP provided. He is really asking why the topological group isomorphism is an immersion. – SomeCallMeTim Mar 23 '23 at 16:05
  • 1
    The OP is asking why $\mathrm{GL}(n,\mathbb C)$ is isomorphic to a Lie subgroup of $\mathrm{GL}(2n,\mathbb R)$. The case $n=1$ provides a very explicit isomorphism which is easily generalized. – Gibbs Mar 23 '23 at 16:08
  • 1
    Sure, but you haven't proven why $\mathcal{A}:=\lbrace A_z\rbrace$ is a Lie subgroup of $\text{GL}(2,\mathbb{R})$, nor why the implied isomorphism $\mathbb{C}^\times \xrightarrow{\sim} \mathcal{A}$ is a Lie isomorphism, and this is where the meat of his question lies. – SomeCallMeTim Mar 23 '23 at 16:14
  • 1
    The subgroup ${A_z} \subset \mathrm{GL}(2,\mathbb R)$ is a Lie subgroup as it is closed under multiplication of matrices of the form $A_z$, plus multiplication and inverse operation are smooth. – Gibbs Mar 23 '23 at 17:22
  • @Mat999 what is exactly your question? – Gibbs Mar 23 '23 at 18:09
  • @Gibbs About defining the loe isomorphism $GL_n(C)\to GL_n(R)$ is the general way ..what is the best way to show that it is indeed a lie isomorphism? – Mat999 Mar 23 '23 at 19:04
  • You should just use the definition of Lie isomorphism... – Gibbs Mar 24 '23 at 20:03
  • @Gibbs Yes, but is what you said about ${A_z}$ being a lie subgroup of $GL(2,R)$ enough? Do not we have to say that it is a smoot submanifold? – Mat999 Mar 25 '23 at 16:26