Let $U(n)$ be the group of unitary $n\times n$ matrices over $\mathbb{C}$. Is there a classification of the continuous, injective group homomorphisms $U(m)\to U(n)$? If so, is there a modern account of the proof?
$\begingroup$
$\endgroup$
6
-
4$\begingroup$ You are looking at (faithful) finite dimensional rep. of a compact Lie groups, they are well studied. I guess this material is covered at Knapp's book or Fulton-Harris. (I happen to have Weyl's book in front of me here, but it is more physics-oriented). I suggest to close the topic as this is not a research level question. $\endgroup$– AsafCommented Dec 16, 2012 at 18:15
-
2$\begingroup$ This is asking about direct sums of irreducible unitary repns of $U(m)$ with total dimension at most $n$. The classification of irreducibles (they are all unitary) of $U(m)$ is by highest weights, and is well-known, well-studied, and visible in many textbooks. The question of the dimensions of the irreducibles for $U(m)$ with $m\ge 3$ is more complicated than the highest-weight classification, but Kostant's weight multiplicity formula tells the dimensions of the weight spaces, giving finer information. $\endgroup$– paul garrettCommented Dec 16, 2012 at 18:31
-
2$\begingroup$ One can find a table of dimensions of certain irreducible representations of $U(m)$ (or, which is the same, of ${\rm SL}(m)$)and also the classification by highest weights, in the book by Onishchik and Vinberg "Lie Groups and Algebraic Groups" $\endgroup$– Mikhail BorovoiCommented Dec 16, 2012 at 18:59
-
$\begingroup$ @Asaf: apologies if the question wasn't appropriate for MO; coming from a different area, I knew that such maps would be well-studied but not under which guise a classification would appear. I'm happy to vote to close but don't have the reputation at the moment. $\endgroup$– Paul McKenneyCommented Dec 16, 2012 at 21:06
-
1$\begingroup$ A couple of comments: continuous homomorphisms between compact Lie groups are analytic, so may be studied via representation theory as others have indicated. Secondly, the kernel of a non-abelian representation of $U(m), m>1$ will be a subroup of $Z(U(m))\cong U(1)$. Thus, sorting through the faithful representation might be a bit involved, but should reduce to some combinatorics. $\endgroup$– Ian AgolCommented Dec 16, 2012 at 23:15
|
Show 1 more comment