Manifold of all 2x2 Hermitian Matrices

Question

Is it true that the manifold of all $2\times 2$ Hermitian matrices is $\mathbb{R}^4$?

Answer 1

Yes, it's true. Here's a diffeomorphism from $\mathbb R^4$ to the matrix manifold: $$ (a, b, c, d) \mapsto \begin{bmatrix} a & b + ci \\ b - ci & d \end{bmatrix}. $$

The inverse map is $$ \begin{bmatrix} p & q \\ \bar{q} & s \end{bmatrix} \mapsto (p, Re(q), Im(q), s). $$

Answer 2

To give yourself a $2\times 2$ hermitian matrix amounts to give yourself its two diagonal complex coefficients that are real (by the hermitian property), that is, two real numbers, and one of its off-diagonal coefficient which (by the hermitian property) determines the, which make two more real number. In total, you need four real numbers to parametrize the real vector space of $2\times 2$ hermitian matrices. It is therefore isomorphic (i.e. diffeomorphic) to $\mathbf{R}^4$.