# Definition The unitary group is the [[group]] of all complex $n \times n$ [[Unitary Operator|unitary matrices]]. This [[Lie group]] has [[dimension]] $n^2$. $U(n)$ is an [[Isometry Group]] over complex [[vector space|vector spaces]] equipped with a [[Non-Degeneracy (Quadratic Forms)|non-degenerate]] positive-definite [[Hermitian form]]. See [[Isometry Group#Example 2 U n|here]] for a discussion. # Lie Algebra $\mathfrak{u}(n)$ Since $U(n) \cong \text{Isom}(V)$, element $X$ of its [[Lie Algebra]] $\mathfrak{u}(n)$ must obey: $ \braket{X \pmb{v} | \pmb{w}} = - \braket{\pmb{v}|X\pmb{w}} \quad \forall \pmb{v}, \pmb{w} \in V $ Now take $V = \mathbb{C}^n$ with a positive definite [[Non-Degeneracy (Quadratic Forms)|non-degenerate]] [[Hermitian form]], and pick an [[orthogonality|orthonormal]] [[basis]]. Then we have: $ \begin{align} (X [\pmb{v}])^\dagger [\pmb{w}] &= -[\pmb{v}]^\dagger X [\pmb{w}]\\ [\pmb{v}]^\dagger X^\dagger [\pmb{w}] &= -[\pmb{v}]^\dagger X [\pmb{w}]\\ X^\dagger &= -X \end{align} $ Thus, $\mathfrak{u}(n)$ is the set of $n\times n$ [[Hermitian Operator|anti-Hermitian matrices]]. %% $ \mathfrak{u}(n) = \{X \in M_n(\mathbb{C}) | X = - X^\dagger\} $ %% The Lie algebra has the same dimension as its Lie group, $n^2$, as one can easily verify.