If $H$ is a one-dimensional Hilbert space then the zero representation of a C*-algebra on $H$ is irreducible.

Question

It says on page 143 of Murphy's $C^*$-algebras and operator theory that if $H$ is a one-dimensional Hilbert space then the zero representation of any C*-algebra on H is irreducible.

What is the zero representation?

Why is the zero representation of any C*-algebra on $H$ irreducible?

Answer 1

The zero representation of an algebra on a Hilbert space $H$ is the map that sends every element of the algebra to the zero operator on $H$.

Murphy's book gives the following definitions:

If $A$ is a C*-subalgebra of $B(H)$, it is said to be irreducible, or to act irreducibly on $H$, if the only closed vector subspaces of $H$ that are invariant for $A$ are $0$ and $H$ [page 58].

A representation $(H,\varphi)$ of a C*-algebra $A$ is irreducible if the algebra $\varphi(A)$ acts irreducibly on $H$ [page 143].

If $\varphi$ is the zero representation, then $\varphi(A)$ contains only the zero operator. If $H$ is one dimensional, then the only subspaces it has are $0$ and $H$, so $\varphi(A)$ trivially acts irreducibly according to Murphy's definition. However, it is not nondegenerate (so it would be degenerate?).

The subject index was useful for finding the definitions.

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Aside: There is a common convention, apparently not used by Murphy here, to define away this example, and state that part of the definition of an irreducible representation is that it be nonzero. One motivation for doing so is that the zero representation does not provide any information. The kernels of irreducible representations play a role in C*-algebras analogous to prime ideals in rings, and a familiar convention is that the entire ring is not a prime ideal, by definition. Concretely, the set of irreducible representations of $C(X)$ can be identified with $X$ if $X$ is a compact Hausdorff space, as long as we don't include the zero representation.