Recently, I came across a statement about random unitary transformations, but in the given context it is not clear to me whether this is a general statement or holds just in special cases. My own research in other literature did not resolve my confusion, so I hope one of you can help me!
Consider $\vec{X} = (X_1, ... , X_m)$ a vector of $\mathbb{C}^m$ and let $U$ be a random unitary transformation of the unitary group $U(m)$ drawn from the Haar measure. Define $\vec{Y} := U \vec{X}$.
Then, the vector $\vec{Y}$ is uniformly distributed on the complex sphere in $\mathbb{C}^m$ with radius $||\vec{Y}||$.
Does this statement hold for general $\vec{X}$ or do they implicitly require that $\vec{X}$ follows some distribution, e.g., the normal distribution? Has anyone seen a prove of this statement?
From my understanding of the Haar-measure in low dimensions (e.g., 2, where one can still visualize the situation), it sounds reasonable that if we choose some random rotation and apply it to our initial vector, we obtain a vector that is uniformly distributed on the sphere, independently of the input $ \vec{X}$. On the other hand, some other statements next to the one in question require standard normal distribution and, unfortunately, they are not always completely clear in pointing out all their assumptions.
Summing up, for me it is still not for 100% clear if that's a general statement or requires, for example, normally distributed inputs. Therefore, I would be grateful for hints and clarification!
Thank you in advance!
Fix a vector $x \in \mathbb{C}^m$. Let $\mu$ be the surface measure on $U(m)$, and let $dS$ be the surface measure on $S = \{z \in \mathbb{C}^m : \lVert z \rVert = \lVert x \rVert\}$, both normalized to be probability measures. Let $\nu$ be the distribution of your random variable $y$, i.e. for Borel $E \subset S$, $\nu(E) = \mu(f^{-1}(E))$, where $f(g) = gx$. We will show that $\nu = dS$.
Let $u : S \to [0, \infty]$ be an arbitrary measurable function. We have \begin{align} \int_{S}u(s)\,d\nu(s) &= \int_{U(m)}u(gx)\,dg \\ \end{align}
Now I claim that $Au(x) = \int_{U(m)}u(gx)\,dg$ is a radial function, meaning that if $y = Tx$, with $T \in U(m)$, then $Au(x) = Au(y)$. With $y = Tx$, observe that $$Au(y) = \int_{U(m)}u(gTx)\,dg.$$ Note that the map $R_T : U(m) \to U(m)$ defined by $R_Tg = gT$ is linear and is an isometry (using the operator norm) since $\lVert gT \rVert = \lVert g \rVert$. Now let $\Omega \subset U(m)$ be open and parameterized by a chart $\phi : O \to \Omega$, where $O$ is open in $\mathbb{R}^{\dim U(m)}$. Suppose $f : U(m) \to [0, \infty]$ is a measurable function supported on $\Omega$. Then $f \circ R_T$ is supported on $R_T^{-1}\Omega$ and $\psi(z) = R_T^{-1}\phi(z)$ defines a chart $\psi : O \to R_T^{-1}\Omega$. Note that $D\psi(z) = R_T^{-1}D\phi(z)$, so $D\psi(z)^tD\psi(z) = D\phi(z)^tD\phi(z)$ since $(R_T^{-1})^t(R_T^{-1}) = I$. Thus \begin{align} \int_{U(m)}f(R_Tg)\,dg &= \int_{R_T^{-1}\Omega}f(R_Tg)\,dg \\ &= \int_{O}f(R_T\psi(z))\sqrt{\det D\psi(z)^tD\psi(z) }\,dz \\ &= \int_{O}f(\phi(z))\sqrt{\det D\phi(z)^tD\phi(z)}\,dz \\ &= \int_{\Omega}f(g)\,dg \\ &= \int_{U(m)}f(g)\,dg. \end{align} Since the above holds when $f$ is supported in a single coordinate patch, it holds for all $f$ by linearity since we can cover $U(m)$ by finitely many coordinate patches since it is compact. This proves in particular that $Au(y) = Au(x)$, so $Au$ is a radial function. Thus \begin{align} \int_{S}u(s)\,d\nu(s) &= \int_{U(m)}u(gx)\,dg \\ &= \int_{S}\int_{U(m)}u(gy)\,dg\,dS(y) \\ &= \int_{U(m)}\int_{S}u(gy)\,dS(y)\,dg \end{align} Now a similar argument to before shows that $\int_{S}u(gy)\,dS(y) = \int_{S}u(y)\,dS(y)$ since $g$ is a linear isometry. Hence $$\int_{S}u(s)\,d\nu(s) = \int_{U(m)}\int_{S}u(y)\,dS(y)\,dg = \int_{S}u(y)\,dS(y).$$ Since $u : S \to [0, \infty]$ was arbitrary, this means $\nu = dS$.