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## Injection between non-isomorphic irreducible Hilbert space reps?

I must be missing something trivial here.

Let $G$ be, say, a reductive Lie group (or more generally any locally compact Hausdorff unimodular topological group). A unitary Hilbert space representation of $G$ is a group homomorphism from $G$ to the group of unitary endomorphisms $U(H)$ of a Hilbert space $H$, with the property that for all $v\in H$ the map $G\to H$, defined by sending $g$ to $gv$, is continuous. A unitary Hilbert space representation is irreducible if $H\not=0$ and $0,H$ are the only closed $G$-invariant subspaces of $H$.

I am ploughing through the book by Jacquet and Langlands and have made it to the all-important last chapter. On page 497 they make an assertion which seems to me to be the following:

if $V$ and $W$ are irreducible Hilbert space reps of $G$, and $i:V\to W$ is a continuous $G$-equivariant map, then either $i=0$ or $V$ and $W$ are isomorphic.

But I cannot rule out the possibility that $V$ and $W$ are not isomorphic and yet $i$ is an injection with dense image.

What am I missing? Of course I might have misunderstood the assumptions being made---they are in an explicit situation with $G$ equal to $GL(2,A)$ for $A$ some adele ring---but I cannot imagine that this can make much difference. Another assumption I think one can make is that any $f\in C_c(G)$ acts on $V$ and $W$ via Hilbert-Schmidt operators. But my gut feeling is that there's a one-line observation that kills this.

I am supposed to be lecturing on this in about 14 hours so I had intended offering a big bounty! But I see from the FAQ that I have to wait 2 days before I can do so. grr

-
 I'm sure that you wouldn't overlook this, but is it possible that i is assumed to be an isometry (in the weak sense, so norm-preserving but a priori not necessarily surjective)? Or it might be that the conditions are such that this is guaranteed - for example, as the representations are irreducible it could be that the fact that the representation is unitary precisely identifies the norm (that is, given G -> Gl(H) then there is at most one inner product making this a unitary rep) (possibly up to an insignificant scalar factor). Just a couple of thoughts! – Andrew Stacey Jun 8 2010 at 21:44 @Andrew: No, i really is not known to be an isometry at this point in the argument. In fact from what I have learned in the last couple of hours I can prove it's an isometry multiplied by a scalar from what we have above. The point is that i^*i is a scalar lambda (using the spectral theorem) and so (iv,iw)=(i^*iv,w)=lambda(v,w). – Kevin Buzzard Jun 8 2010 at 23:23

No conditions are needed on the group $G$, or on the continuity of the representation, you do need the assumption that $i$ is continuous however. Since $i:V \to W$ is continuous (equivalent to being bounded) it has a continuous adjoint $i^* : W \to V$ which is also $G$ equivariant, hence $i^* i:V \to V$ is $G$ equivariant and since $G$ acts on $V$ irreducibly must be a scalar multiple of the identity (if not the spectrum would contain more than one point and you could take a spectral projection of $i^* i$ and obtain a closed non-trivial $G$-invariant subspace). This shows that $i$ is a scalar multiple of an isometry and hence the image must be closed. If $G$ also acts irreducible on $W$ then $ii^*$ is also a scalar multiple of the identity and hence if $i \not= 0$ it must be a non-zero scalar multiple of a unitary between $V$ and $W$.