6
$\begingroup$

Denote the Haar measure on the unitary group $U(\mathcal X)$ by $\eta$. Does this equation hold (assuming the integral exists):

$\int d\eta(U) f(U) = \int d\eta(U) f(U^\dagger)$?

Intuitively this makes sense because choosing a random $U$ seems to be the same as choosing a random $U^\dagger$, but I'm not sure how to prove this.

$\endgroup$
4
$\begingroup$

I will answer this question in a more general context. You might know that Haar's theorem tells you that on any locally compact group $G$, there is a unique left-invariant (Borel) measure $\mu$, up to a positive constant. Left-invariance means that $\mu(g A)=\mu(A)$ for any $g\in G$ and (measurable) set $A\subset G$. This is the (left-) Haar measure on $G$.

The same argument yields a unique right-invariant measure, up to a constant. Note that we can always turn a left-invariant Haar measure into a right-invariant Haar measure by taking the inverse: $\tilde\mu(A) := \mu(A^{-1})$. That is exactly what you are interested in.

For certain groups, the left- and right-invariant Haar measures coincide (unimodular groups) and the unitary group $U(d)$ is such a group. For these groups, your statement is correct. The proof is straightforward:

Unimodularity implies that the "inverse" of the normalised Haar measure $\mu$ on $U(d)$ is itself a left- and right-invariant Haar measure on $U(d)$. By uniqueness, $\tilde\mu$ can only differ by a positive constant from $\mu$. However, it is clear that both measures are normalised, thus $\tilde\mu = \mu$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.