# Norms and angles of vectors

In an inner product space, the norm (AKA length) of a vector is defined as $${\left\Vert v\right\Vert\equiv\sqrt{\left\langle v,v\right\rangle }}$$, leading to common relations such as the Cauchy-Schwarz inequality $${\left|\left\langle v,w\right\rangle \right|\leq\left\Vert v\right\Vert \left\Vert w\right\Vert}$$ and the triangle inequality (see below).

Alternatively, we can define a norm on a vector space as a non-negative real function that only vanishes for the zero vector and satisfies $${\left\Vert av\right\Vert =\left|a\right|\left\Vert v\right\Vert}$$ as well as the triangle inequality $${\left\Vert v+w\right\Vert \leq\left\Vert v\right\Vert +\left\Vert w\right\Vert}$$. This makes the space a normed vector space. Existence of a norm does not in general imply the existence of an inner product, but if a norm satisfies the parallelogram identity

$$\displaystyle \left\Vert v+w\right\Vert ^{2}+\left\Vert v-w\right\Vert {}^{2}=2\left(\left\Vert v\right\Vert {}^{2}+\left\Vert w\right\Vert {}^{2}\right),$$

an inner product can be obtained using the polarization identity, which for a real vector space is

\begin{aligned} \left\langle v,w\right\rangle &=\frac{1}{4}\left(\left\Vert v+w\right\Vert {}^{2}-\left\Vert v-w\right\Vert {}^{2}\right)\\&=\frac{1}{2}\left(\left\Vert v\right\Vert {}^{2}+\left\Vert w\right\Vert {}^{2}-\left\Vert v-w\right\Vert {}^{2}\right)\\&=\frac{1}{2}\left(\left\Vert v+w\right\Vert {}^{2}-\left\Vert v\right\Vert {}^{2}-\left\Vert w\right\Vert {}^{2}\right),\end{aligned}

and for a complex vector space is

\begin{aligned}\left\langle v,w\right\rangle & =\frac{1}{4}\left(\left\Vert v+w\right\Vert {}^{2}-\left\Vert v-w\right\Vert {}^{2}+i\left\Vert v-iw\right\Vert {}^{2}-i\left\Vert v+iw\right\Vert {}^{2}\right).\end{aligned}

Note that the parallelogram and polarization identities, which only involve the squared norm $${\left\Vert v\right\Vert ^{2}=\left\langle v,v\right\rangle}$$, also hold for a pseudo inner product.

Comparing the polarization identities immediately identifies the real part of the complex inner product as just the real inner product defined by the orthonormal basis in the decomplexification $${V^{\mathbb{R}}}$$, i.e.

\begin{aligned}\mathrm{Re}\left(\left\langle v,w\right\rangle _{\mathbb{C}}\right)=\left\langle v,w\right\rangle _{\mathbb{R}}.\end{aligned}

The imaginary part is then the inner product of vectors

\begin{aligned}\mathrm{Im}\left(\left\langle v,w\right\rangle _{\mathbb{C}}\right) & =\left\langle v,-iw\right\rangle _{\mathbb{R}}\\
& =\left\langle iv,w\right\rangle _{\mathbb{R}}\end{aligned}

in $${V^{\mathbb{R}}}$$. Note that both parts vanish for vectors in different complex planes, but for vectors in the same complex plane $${\mathrm{Im}\left(\left\langle v,w\right\rangle \right)}$$ acts as a sort of “inverse” inner product on the vectors in $${V^{\mathbb{R}}}$$: it vanishes if they are a real multiple of each other (parallel in $${V^{\mathbb{R}}}$$), and is the squared norm for vectors related by an imaginary factor (orthogonal in the decomplexified complex plane), i.e. $${\mathrm{Im}\left(\left\langle iv,v\right\rangle _{\mathbb{C}}\right)=\left\langle v,v\right\rangle _{\mathbb{R}}}$$. As we will see, $${\mathrm{Im}\left(\left\langle v,w\right\rangle \right)}$$ is a symplectic form on $${V^{\mathbb{R}}}$$.

In a real inner product space, we can define the angle between vectors by $${\cos\theta\equiv\left\langle v,w\right\rangle /\left(\left\Vert v\right\Vert \left\Vert w\right\Vert \right)}$$. If we then decompose $${v}$$ into components parallel and orthogonal to $${w}$$, the parallel component is called the orthogonal projection of $${v}$$ onto $${w}$$, and has length $${v_{\parallel w}=\left\langle v,w\right\rangle /\left\Vert w\right\Vert =\left\Vert v\right\Vert \cos\theta}$$. In a complex vector space $${V}$$, taking the real part of the cosine defines the Euclidean angle $${\cos\theta_{E}\equiv\mathrm{Re}\left(\left\langle v,w\right\rangle \right)/\left(\left\Vert v\right\Vert \left\Vert w\right\Vert \right)}$$, which is the angle between the vectors in the decomplexification $${V^{\mathbb{R}}}$$; the orthogonal projection of $${v}$$ onto $${w}$$ in $${V^{\mathbb{R}}}$$ is then $${v_{\parallel w}=\mathrm{Re}\left(\left\langle v,w\right\rangle \right)/\left\Vert w\right\Vert =\left\Vert v\right\Vert \cos\theta_{E}}$$.

For two vectors in a single complex line in $${V}$$, called a holomorphic plane in $${V^{\mathbb{R}}}$$, we may write $${\sin\theta_{\mathrm{E}}\equiv\mathrm{Im}\left(\left\langle z,u\right\rangle \right)/\left(\left\Vert z\right\Vert \left\Vert u\right\Vert \right)}$$, so that the component of $${z}$$ orthogonal to $${u}$$ is $${z_{\perp u}=\mathrm{Im}\left(\left\langle z,u\right\rangle \right)/\left\Vert u\right\Vert}$$. In the more general case, two arbitrary complex vectors (assumed to be non-parallel in $${V^{\mathbb{R}}}$$) have a Kähler angle defined by $${\cos\theta_{\mathrm{K}}\sin\theta_{\mathrm{E}}\equiv\mathrm{Im}\left(\left\langle v,w\right\rangle \right)/\left(\left\Vert v\right\Vert \left\Vert w\right\Vert \right)}$$. $${\cos\theta_{\mathrm{K}}}$$ is dependent only upon the plane in $${V^{\mathbb{R}}}$$ defined by the two vectors, and is unity if this plane is holomorphic, while vanishing for vectors in orthogonal complex lines.

The above depicts how in the holomorphic plane defined by $${e_{1}^{\mathbb{R}}}$$ and $${e_{2}^{\mathbb{R}}}$$, the real part of the complex inner product $${\left\langle z,u\right\rangle}$$ determines the parallel component $${z_{\parallel u}}$$ (the orthogonal projection), while the imaginary part determines the orthogonal component $${z_{\perp u}}$$; these can also be expressed in terms of the Euclidean angle as $${z_{\parallel u}=\left\Vert z\right\Vert \cos\varphi_{\mathrm{E}}}$$ and $${z_{\perp u}=\left\Vert z\right\Vert \sin\varphi_{\mathrm{E}}}$$. For two vectors $${v}$$ and $${w}$$ with an extra component in a direction orthogonal to the holomorphic plane, the Kähler angle is the angle between the $${vw}$$ plane and the holomorphic plane. We can see this in the figure by noting that since the $${e_{3}^{\mathbb{R}}}$$ components of the vectors are parallel, the imaginary part of $${\left\langle v,w\right\rangle}$$ does not include any contribution from this component. In the holomorphic plane, $${v}$$ has no imaginary component, so we have $${\mathrm{Im}\left\langle v,w\right\rangle =v_{1}w_{2}=\left\Vert v\right\Vert \cos\theta_{\mathrm{K}}\left\Vert w\right\Vert \sin\theta_{\mathrm{E}}}$$, the defining relation for the Kähler angle. This geometric view of the Kähler angle as the angle between planes remains valid if the line of intersection between the planes is not along the imaginary axis, or if $${v}$$ has an imaginary component; but for vectors with components in multiple holomorphic planes, the situation is more complicated.

Finally, taking the modulus of the cosine defines the Hermitian angle $${\cos\theta_{\mathrm{H}}\equiv\left|\left\langle v,w\right\rangle \right|/\left(\left\Vert v\right\Vert \left\Vert w\right\Vert \right)}$$, where $${v_{\parallel w}^{\mathbb{C}}=\left\Vert v\right\Vert \cos\theta_{\mathrm{H}}}$$ is the (complex) orthogonal projection of $${v}$$ onto $${w}$$. The pseudo-angle is then defined by $${\left\langle v,w\right\rangle \equiv\left|\left\langle v,w\right\rangle \right|e^{i\theta_{\mathrm{P}}}}$$. Like the Kähler angle, the Hermitian angle vanishes for vectors in orthogonal complex lines; for two vectors in the same holomorphic plane in $${V^{\mathbb{R}}}$$, the Hermitian angle is unity and the pseudo-angle is just the Euclidean angle.

 Δ It is important to remember that a Euclidean angle of $${\pi/2}$$ does not ensure a vanishing complex inner product, and that parallel vectors in $${V}$$ may be orthogonal using the corresponding real inner product in $${V^{\mathbb{R}}}$$.