Is There a Simple Way to Show GL(n,C) is a Lie Group?

[ergospherical]

Homework Statement

As per title

Relevant Equations

N/A

I'd like to clarify a few things; the approach is basically just to show that $\mathrm{GL}(n,\mathbf{C})$ is isomorphic to a subgroup of $\mathrm{GL}(2n,\mathbf{R})$ which is a smooth manifold (since $\mathbf{R} \setminus \{0\}$ is an open subset of $\mathbf{R}$, so its pre-image $\mathrm{det}^{-1}(\mathbf{R} \setminus \{ 0\}) = \mathrm{GL}(2n,\mathbf{R})$ under the continuous determinant map is a smooth manifold.)

The hint is to consider replacing each $X_{ij} = a_{ij} + ib_{ij}$ with a matrix block $\begin{pmatrix} a_{ij} & -b_{ij} \\ b_{ij} & a_{ij} \end{pmatrix}$. To show this map is a group isomorphism means to show that $\Phi(XY) = \Phi(X) \Phi(Y)$, right? So I could write (summation implied over repeated suffices)\begin{align*}
(XY)_{ij} = X_{ik} Y_{kj} &= (a_{ik} + ib_{ik})(c_{kj} + id_{kj}) \\
&= a_{ik} c_{kj} - b_{ik} d_{kj} + i(a_{ik} d_{kj} + b_{ik} c_{kj})
\end{align*}I tried to write the affect of the map $\Phi$ on the elements explicitly, i.e. \begin{align*}
\Phi(X)_{ij} =
\begin{cases}
a_{\mathrm{ceil}{\frac{i}{2}} \mathrm{ceil}{\frac{j}{2}}}, & i+j \ \mathrm{even} \\
(-1)^i b_{\mathrm{ceil}{\frac{i}{2}} \mathrm{ceil}{\frac{j}{2}}}, & i+j \ \mathrm{odd}
\end{cases}
\end{align*}but this becomes a mess to work out $(\Phi(X) \Phi(Y))_{ij}$. I think it is clear, by considering e.g. a $1 \times 1$ matrix $(a_{11} + ib_{11})$, that it works, but there is surely a better approach?

 

[Office_Shredder]

For two nxn matrices (which expand to 2nx2n) you should be able to convince yourself that if you just look at the first two rows of X and the first two columns of Y, that if the multiplication works out that it will work out for the rest of the rows) columns. Then you have a much more tractable problem, just pick an arbitrary 2x2 block in the result and check that e.g. the top left entry is the sum of product of real parts of the original n complex numbers minus sum of the product of the imaginary parts.

 

[WWGD]

Part of the longer way , in showing product and inversion are smooth is using that both of these are polynomials on the entries. And for a Lie Group, a subgroup is also a Lie Group if it's a closed subgroup.

 

[fresh_42]

Multiplication are linear polynomials, hence analytic. Inversion, too, extended by the inverse of the determinants as a factor. But they are quotients of polynomials without singularity, hence analytic as well.

Here is the calculation for $T_{\operatorname{id}}(\operatorname{SL}(n)) = \mathfrak{sl}(n)$
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

It contains the derivative of the determinant.

 

[martinbn]

https://en.wikipedia.org/wiki/Block_matrix#Block_matrix_multiplication