Unitary doesn't carry over isomorphism?

Question

I think I just heard someone say that if $\pi$ and $\rho$ are equivalent representations and if $\pi$ is unitary, then $\rho$ is not necessarily unitary.

Is that really correct?

I would think that isomorphisms between things mean that they share all properties. Does being unitary not carry over?

Answer 1

I think the problem (as is often the case) is that terminology was not picked in a perfectly logical manner a long time ago, and has stuck around.

The problem is that most of the time in math, an "adjective noun" means a noun that has the property of being adjective, but sometimes, as in this case, it means a noun together with some additional structure hinted at by adjective. To wit:

A representation of $G$ means the data of:

• a vector space $V$
• a map $\rho: G \to GL(V)$.

A unitary representation of $G$ means the data of:

• an inner product space ($V, \langle\ , \rangle$)
• a map $\rho: G \to U(V) \subset GL(V)$, where $U(V)$ is the set of linear transformations that preserve $V, \langle\ , \rangle$.

So the "forgetful functor" from unitary representations to representations "forgets structure, not just propery." That's why it's possible for two representations to be isomorphic but for one to "be unitary" and the other not, because "being unitary" isn't a property of a representation.

A simpler example of the same phenomenon: the additive groups $\mathbb{Z}[\sqrt{5}]$ and $\mathbb{Z}[\sqrt{2}]$ are isomorphic, but there is a natural ring structure on the two sides which isn't preserved by this isomorphism. That doesn't feel weird, but only because we use the word "ring" instead of the confusing phrase "ringy group."

Answer 2

It all depends on what you mean by "equivalent". If it means "linearly equivalent", then indeed there is no reason that it should preserve the property of being unitary. It could also mean "unitarily equivalent", in which case it would preserve that property.

We can look at the tautological example: the unitary group $U(n)$ has an obvious unitary representation on $\mathbb{C}^n$. Then if we conjugate by any invertible matrix $P\in GL_n(\mathbb{C})$, we get a linearly equivalent representation. But in general, unless $P$ is unitary, $PU(n)P^{-1}$ is not a subgroup of $U(n)$, so that representation is not unitary.