# Is there analog of angle in complex domain?

Given a real vector space with an inner product, i. e. a positive definite symmetric bilinear form $$\beta$$, I can define angle between any two nonzero vectors $$v_1$$, $$v_2$$ as $$\arccos\frac{\beta(v_1,v_2)}{\sqrt{\beta(v_1,v_1)\beta(v_2,v_2)}}$$. The angle will not change if I multiply the vectors by nonzero real numbers of the same sign, and will become replaced by the complementary angle if these numbers have opposite signs. The angle is also invariant under the action of the group $$\operatorname{O}(\beta)$$ of $$\beta$$-preserving linear transformations.

What are analogs of all this for complex vector spaces? I suppose that one has to take a Hermitian form for $$\beta$$ but the above expression under $$\arccos$$ will now be a complex number. Shall I take its argument? What happens under the action of the unitary group?

$$\def\eqdef{\stackrel{\text{def}}{=}}$$ Starting with the Hermitian form $$\ \beta\big(v_1,v_2\big)\eqdef v_1^\dagger v_2\$$ for $$\ v_1,v_2\in\mathbb{C}^n\$$, if we take the angle between $$\ v_1\$$ and $$\ v_2\$$ to be given by $$\arccos\frac{\mathscr{Re}(v_1^\dagger v_2)}{\|v_1\|\|v_2\|}$$ then this is the same as $$\arccos\frac{x_1^Tx_2}{\|x_ 1\|\|x_2\|}\ ,$$ where $$\ x_1,x_2\in\mathbb{R}^{2n}\$$ are given by \begin{align} x_1&=\left(\mathscr{Re}(v_{11}),\mathscr{Im}(v_{11}),\mathscr{Re}(v_{12}),\mathscr{Im}(v_{12}),\dots,\mathscr{Re}(v_{1n}),\mathscr{Im}(v_{1n})\right)\\ x_2&=\left(\mathscr{Re}(v_{21}),\mathscr{Im}(v_{21}),\mathscr{Re}(v_{22}),\mathscr{Im}(v_{22}),\dots,\mathscr{Re}(v_{2n}),\mathscr{Im}(v_{2n})\right)\ . \end{align} The most obvious analogue for a general Hermitian form $$\ \beta\$$ on a complex vector space would therefore appear to be $$\arccos\frac{\mathscr{Re}(\beta(v_1,v_2))}{\sqrt{\beta(v_1,v_1)\beta(v_2,v_2)}}\ .$$ This has the same properties of being invariant under multiplication of $$\ v_1\$$ and $$\ v_2\$$ by real numbers of the same sign (or, more generally, by non-zero complex numbers with the same phase), and under the action of the group $$\ \text{O}(\beta)\$$ on $$\ v_1\$$ and $$\ v_2\$$, and of being replaced by its complementary angle when $$\ v_1\$$ and $$\ v_2\$$ are multiplied by real numbers of opposite signs.
• This sounds logical, but then it would mean that two vectors with imaginary $\beta$ are orthogonal? Sounds unusual somehow... Mar 17 at 15:24
• Yes, that's correct! When $\ \beta:\mathbb{C}\times\mathbb{C}\rightarrow\mathbb{C}\$ is given by $\ \beta(z_1,z_2)=\bar{z}_1z_2\$, for instance, then $\ \beta(z_1,z_2)\$ imaginary implies that $\ z_1=re^{\frac{i\pi}{2}}z_2\$ or $\ z_1=re^{\frac{-i\pi}{2}}z_2\$ for some positive real number $\ r\$—that is, the representations of $\ z_1\$ and $\ z_2\$ on an Argand diagram are orthogonal. Mar 17 at 22:33
• And for the first example given in the answer, $\ \arccos\frac{\mathscr{Re}(v_1^\dagger v_2)}{\|v_1\|\|v_2\|}\$ imaginary implies that $\ x_1^tx_2=0\$ —that is, the vectors $\ x_1\$ and $\ x_2\$ are orthogonal, so the implication is not quite as odd as it might first appear. Mar 17 at 22:39