Uniqueness of the Degree of Faithful Irreducible Representations

Question

Can a group have two faithful irreducible unitary representations with distinct degrees? For example, if you were to find a faithful irreducible unitary representation of a group $G$ of dimension $\aleph_0,$ could you immediately deduce that any other faithful irreducible unitary representation must also have this dimension?

Answer 1

Qiaochu already provided a counterexample for finite-dimensional cases in the comments. Here is an infinite-dimensional counterexample, constructed using $C^\ast$-algebra and von Neumann algebra machinery. There should be much more simpler constructions, but this is the only one I came up with.

Fix a separable II$_1$ factor $M$. Let $U(M)$ be its unitary group. Then its center is $U(1)$, the group of scalar operators. Let,

$$G = (U(M) \times U(M))/\{(z, z): z \in U(1)\}$$

Now, consider the following two representations of the group:

1. Let $H = L^2(M)$, the standard representation of $M$, which comes equipped with both left and right actions of $M$. Let $\pi: G \to U(H)$ be given by,

$$\pi([(g, h)])\xi = g\xi h^{-1}, \forall \xi \in H$$

Note that, for any $z \in U(1)$, $z\xi z^{-1} = \xi$, so this is indeed a well-defined homomorphism. Furthermore, we note that it is faithful. Indeed, if $g\xi h^{-1} = \xi$ for all $\xi \in H$, then in particular, $ghh^{-1} = h \Rightarrow g = h$. But then, when $g = h$, $g\xi g^{-1} = \xi$ for all $\xi \in H$ implies $g \in Z(M)$, so $g \in U(1)$, i.e., $(g, h) \in \{(z, z): z \in U(1)\}$.

We also note that it is irreducible. Indeed, $\pi([(g, 1)])$ gives all unitaries in the left action of $M$ while $\pi([(1, h)])$ gives all unitaries in the right action of $M$. The left and right actions together generate the entirety of $\mathbb{B}(L^2(M))$, so the representation is irreducible. Since $M$ is separable, $H = L^2(M)$ is a separable Hilbert space. Thus, $G$ has a faithful irreducible representation of dimension $\aleph_0$.

2. By Krein-Milman, $M$ has a pure state and thus an irreducible representation $\phi: M \to \mathbb{B}(H)$ for some Hilbert space $H$. It is known that, if $H$ is separable, then $\phi$ must be normal. (See Theorem V.5.1 in Takesaki’s Theory of Operator Algebras I.) But if $\phi$ is normal, then as $M$ is of type II$_1$, $\phi(M)$ and thus $\phi(M)’$ must be of type II. However, the representation is irreducible, so $\phi(M)’ = \mathbb{C}$, which is a contradiction, so $H$ must be non-separable. Note also that, as II$_1$ factors are simple as $C^\ast$-algebras, $\phi$ must be faithful.

Now, let $\pi: G \to U(H \otimes \overline{H})$ be defined by,

$$\pi([(g, h)]) = \phi(g) \otimes \overline{\phi(h)}$$

Since $z \otimes \overline{z} = 1$ for any scalar $z \in U(1)$, we again have $\pi$ is a well-defined homomorphism. $\pi$ is also faithful, as $\phi(g) \otimes \overline{\phi(h)} = 1$ implies both $\phi(g)$ and $\phi(h)$ are scalars, in which case it is easy to check that $\phi(g) \in U(1)$ and $\phi(h) = \phi(g)$. But $\phi$ is faithful, so,

$$(g, h) = (\phi(g), \phi(g)) \in \{(z, z): z \in U(1)\}$$

This verifies that $\pi$ is faithful. Finally, it is irreducible as $\pi([(g, 1)])$ generates $\phi(M) \otimes 1$ and thus generates $\mathbb{B}(H) \otimes 1$, as $\phi$ is irreducible. Similarly, $\pi([(1, h)])$ generates $1 \otimes \mathbb{B}(\overline{H})$, so $\pi(G)$ generates the entirety of $\mathbb{B}(H \otimes \overline{H})$, i.e., $\pi$ is irreducible. Hence, $G$ also has a faithful irreducible representation on a non-separable Hilbert space.