Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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! –  Loop Space Jun 8 '10 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 '10 at 23:23

1 Answer 1

up vote 12 down vote accepted

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 $W$.

share|improve this answer
    
Many many thanks! I see now why I didn't spot the argument myself! David Whitehouse independently pointed me to Theorem 9.0.1 of some notes by Paul Garrett called "unitary representations of topological groups", where the same proof is given. Thanks to both of you! –  Kevin Buzzard Jun 8 '10 at 23:23
    
Link to Garrett's notes: math.umn.edu/~garrett/m/v/unitary_of_top.pdf See Theorem 9.0.1. –  Kevin Buzzard Jun 9 '10 at 6:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.