A complex orthonormal basis induces a real orthonormal basis

Question

Let $V$ be a real inner product space with an orthogonal almost complex structure $J:V\to V$. Then we can view $V$ as a complex vector space using $J$. Choose a complex basis $\{e_1,\dots,e_n\}$ for $V$ with $\langle e_i,e_j\rangle=\delta_{ij}$. Then it is easily seen that $\mathcal{B}=\{e_1,Je_1,\dots,e_n,Je_n\}$ is a real basis for $V$. Is it necessarily true that $\mathcal{B}$ is an orthonormal basis? We have $\langle e_i,Je_i\rangle= \langle Je_i,J^2e_i\rangle=\langle Je_i,-e_i\rangle$, so $Je_i$ is orthogonal to $e_i$. But I can't see whether $\langle e_i,Je_j\rangle=0$ for $i\neq j$.

Answer 1

Real-orthormality of a complex basis does not guarantee orthonormality of the "extended" real basis. For example, identify $\mathbf{R}^{4}$ with $\mathbf{C}^{2}$ equipped with the natural complex structure. Let $a$, $b$, and $c$ be positive real numbers such that $a^{2} + b^{2} + c^{2} = 1$, and consider the vectors \begin{align*} e_{1} &= (1, 0, 0, 0) \sim (1, 0), & e_{2} &= (0, a, b, c) \sim (ai, b + ci), \\ Je_{1} &= (0, 1, 0, 0) \sim (i, 0), & Je_{2} &= (-a, 0, -c, b) \sim (-a, -c + bi). \end{align*} The set $\{e_{1}, Je_{1}, e_{2}, Je_{2}\}$ is a real basis, so the real-orthogonal set $\{e_{1}, e_{2}\}$ is a complex basis, but $e_{1}$ and $Je_{2}$ are not orthogonal.

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But wait: This example contradicts a "well-known phenomenon"...! The reconciling details are hidden in

1. The term complex basis, which refers to defining complex scalar multiplication by $$ (a + bi)v = av + bJv,\quad\text{$a$, $b$ real,} $$
2. The Hermitian structure $h$ used to define orthonormality of the complex basis.

Since $J$ is orthogonal with respect to an inner product $g$, the formula $$ \omega(u, v) = g(Ju, v) $$ defines a skew-symmetric $2$-tensor on $V$. (Proof: apply $J$ to each argument on the right and use symmetry of $g$, and $J^{2} = -I$.) Consequently, $$ h(u, v) = g(u, v) + i\omega(u, v) $$ defines an Hermitian (conjugate-symmetric) inner product on $V$.

When we say $\langle e_{i}, e_{j}\rangle = \delta_{ij}$ in complex geometry, we usually mean $h(e_{i}, e_{j}) = \delta_{ij}$. The vanishing of the imaginary part is precisely the condition $\langle Je_{i}, e_{j}\rangle = 0$ for all $i$ and $j$. (In the example above, $h(e_{1}, e_{2})$ is pure imaginary, but not zero.)

Geometrically, to say a complex basis is orthonormal (or unitary) is to say the complex lines spanned by the basis elements are orthogonal as real $2$-dimensional subspaces.