= Solution
Regard complex matrices as a real vector space of dimension $2n^2$ and Hermitian matrices as a real vector space of dimension $n^2$. Set $F(A)=AA^*$. Its derivative is $DF_A(H)=HA^*+AH^*$. At a unitary $A$, any Hermitian $S$ is reached by $H=SA/2$. Hence $I$ is a <regular value>, and the <regular level set theorem> makes $\mathrm U(n)=F^{-1}(I)$ a smooth real <manifold> of dimension $\boxed{n^2}$.
The <tangent space> is the kernel of this derivative. Equivalently,
$$
\boxed{H\in T_A\mathrm U(n)\iff A^*H+H^*A=0\iff H=AK\text{ with }K^*=-K.}
$$
At $I$, therefore, $H$ is skew-Hermitian. The explicit <geodesic> with initial point $I$ and velocity $H$ is $\gamma(t)=e^{tH}$. The <matrix exponential> is unitary because $e^{tH^*}e^{tH}=I$, and $\dot\gamma(0)=H$. Its acceleration is $\gamma H^2$. Since $H^2$ is Hermitian and every tangent vector at $\gamma$ is $\gamma K$ with $K$ skew-Hermitian, the induced real inner product satisfies
$$
\langle\gamma H^2,\gamma K\rangle=\operatorname{Re}\operatorname{tr}(H^2K^*)=0.
$$
Indeed the trace of a Hermitian matrix times a skew-Hermitian matrix is purely imaginary. Its acceleration is therefore always normal to the manifold, which is the <geodesic equation> for an induced Euclidean metric. Diagonalizing $H=W\operatorname{diag}(i\lambda_j)W^*$ gives the equivalent explicit construction $W\operatorname{diag}(e^{it\lambda_j})W^*$.
Back to article page
