9

Let $U(n)$ denote the group of $n\times n$ unitary complex matrices. Is it possible to characterize the set of continuous group homomorphisms between $U(n)$ and $U(m)$?

Some simple homomorphisms are:

  • $m=1$: for an integer $k$, $A\mapsto \det(A)^k$
  • $m=n$: for a unitary matrix $V$, $A\mapsto VAV^\dagger$ where $V^\dagger$ is the conjugate transpose
  • $m=n$: $A\mapsto A^*$, conjugating each element of $A$.
  • $m>n$: $A\mapsto\left(\begin{array}{cc}A&\\&I_{m-n}\end{array}\right)$ where $I_{m-n}$ is the $(m-n)\times (m-n)$ identity matrix.

We can of course compose homomorphisms of the above form as well. Lastly, for $m=n+n'$, there is the trivial embedding of $U(n)\otimes U(n')$ into $U(m)$ by $(A,A')\mapsto \left(\begin{array}{cc}A&\\&A'\end{array}\right)$, which when composed with the above homomorphisms gives even more possibilities.

Are there any other continuous group homomorphisms from $U(n)$ to $U(m)$ that are not some combination of these?

AAA
  • 857

1 Answers1

6

This is more or less equivalent to asking for a classification of the (finite-dimensional, complex) linear representations of each unitary group $U(n)$; given such a representation $W$ of dimension $m$ we get an induced homomorphism $U(n) \to GL_m(\mathbb{C})$, then we can pick an arbitrary inner product on $W$ and average it over the action of $U(n)$ (Weyl's unitary trick) to get an invariant inner product. After further choosing an orthonormal basis of $W$ this produces a homomorphism $U(n) \to U(m)$, and choosing another orthonormal basis of $W$ conjugates this homomorphism (corresponding to your second bullet). This inner product is not necessarily unique, but it is unique up to scale if $W$ is irreducible by Schur's lemma.

The simplest representation to understand is the defining representation $V = \mathbb{C}^n$; this corresponds to your second bullet ($V$ doesn't matter up to conjugacy / isomorphism). From here we can build other representations. The top exterior power $\Lambda^n(V)$ corresponds to $A \mapsto \det(A)$, and then taking tensor powers of this representation corresponds to your first bullet $A \mapsto \det(A)^k$.

Your third bullet corresponds to taking the complex conjugate representation $\overline{V}$, which is the same representation but with scalar multiplication twisted by complex conjugate. By choosing an invariant inner product as above we can show that this representation is isomorphic to the dual $V^{\ast}$. Your fourth bullet corresponds to taking the direct sum with copies of the trivial representation, and I'm not sure what you have in mind wrt the stuff after the fourth bullet but it could be interpreted as taking the direct sum of representations in general.

There are many representations which are not obtained by composing the above homomorphisms, such as the other exterior powers $\Lambda^k(V)$ and the symmetric powers $S^k(V)$. The list of all irreducible representations is known and can be described in terms of either highest weight theory or Schur-Weyl duality / Schur functors, which generalize the symmetric and exterior powers. And finite-dimensional representations are completely reducible, so all finite-dimensional representations are finite direct sums of irreducibles.

The simplest interesting representation not on your list is probably the symmetric square $S^2(\mathbb{C}^2)$ of the defining representation of $U(2)$; this corresponds explicitly to a homomorphism $U(2) \to U(3)$ which is given by

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto \begin{bmatrix} a^2 & \sqrt{2} ab & b^2 \\ \sqrt{2} ac & ad + bc & \sqrt{2} bd \\ c^2 & \sqrt{2} cd & d^2 \end{bmatrix}$$

(the funny $\sqrt{2}$ factors are there to preserve unitarity).

Qiaochu Yuan
  • 468,795