5
$\begingroup$

Schur functions are irreducible characters of the unitary group $U(N)$. This implies $$ \int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu},$$ where the overline means complex conjugation. My question is what is the result of the same integral performed over $SU(N)$ instead, $$ \int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=?$$

$\endgroup$
3
  • $\begingroup$ isn't the answer the same for $U(N)$ and $SU(N)$? $\endgroup$ Aug 8 '19 at 18:55
  • 1
    $\begingroup$ @CarloBeenakker Why should it be the same? (I don't think they are the same) $\endgroup$
    – thedude
    Aug 8 '19 at 18:57
  • 1
    $\begingroup$ At least for $N=2$ it's explictly known how a $U(N)$ irreducible repsentation reduces upon restricting to $SU(N)$ (see e.g. here math.stackexchange.com/questions/2284576/…); the integral counts the number of irreducibles, if I'm not led astray by analogy with finite groups... $\endgroup$ Aug 8 '19 at 21:38
6
$\begingroup$

The desired integral is given in equation (13) of arXiv:1812.06069:

$$\int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\sum_{q=-\infty}^\infty\prod_{i=1}^N\delta_{\lambda_i,\mu_i+q},$$ where $\lambda=(\lambda_1,\lambda_2,\ldots\lambda_N)$ and $|\lambda|=\sum_{i}\lambda_i$, with $\lambda_1\geq\lambda_2\cdots\geq 0$. This is still an orthogonality relation, because the Schur functions $s_\lambda$ and $s_\mu$ are identical in $SU(N)$ iff $\lambda=\mu+(q,q,\ldots q)$ for some integer $q$.

The $U(N)$ integral $\int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu}$ corresponds to the $q=0$ term in the sum over $q$. It follows that the integrals over $SU(N)$ and over $U(N)$ are the same if $|\lambda|,|\mu|<N$, because then only the $q=0$ term contributes.

$\endgroup$

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.