I would appreciate your insight on how to prove that a complex matrix group is a Lie Group.
Let $G =\left\{ U(\alpha, \beta) = \left( \begin{array}{cc} \alpha & \beta \\ \beta^* & \alpha* \end{array} \right)\; | \; det \, U(\alpha, \beta) = 1 \right\}$ be a group (I am able to show this).
To prove that this is in fact a Lie Group, I need to prove that it is indeed a manifold, and then show that the group operations multiplication and inverse are smooth.
First, I construct $\phi : G \rightarrow \mathbb R^4 - \{x, y, z, t \; | \; x^2 + y^2 = z^2 + t^2\}$ (open set of $\mathbb R^4$) defined by $\phi \left(U(\alpha, \beta)\right) = (Re(\alpha), Im(\alpha), Re(\beta), Im(\beta))$. $\phi$ is a bijective map. There is only one transition chart which is the identity, which is smooth. Is that enough to show that G is a manifold?
The elements of the product of two matrices being polynomials of the elements of the matrices, the multiplication map is smooth. Same for the inverse since the matrices have determinant 1.
Does this prove that $G$ is a Lie Group or am I missing something?
Thank you and have a good day!