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I have been reading the online lecture notes by Fiona Murnaghan

http://www.math.toronto.edu/murnaghan/courses/mat1197/notes.pdf

The first lemma in p.35 says that every unitary representation of locally compact group $G$ is semisimple. In her notes, she defines a semisimple representation to be a representation which is a direct sum of irreducible representations.

On the other hand, people often say that the right regular representation of $G$ on $L^2(G)$, which is unitary, does not decompose into a direct sum of irreducible representations but it is a direct integral.

But if the above lemma by Murnaghan is correct, it seems that $L^2(G)$ must be a direct sum of irreducible representations.

I read the proof of the lemma in her note carefully, and noticed that even though she stated the lemma under the assumption that the group $G$ is $p$-adic and the representation is smooth, the proof goes through for any locally compact group and any uniterizable representation. Indeed, the proof only uses the property that any subrepresentation has the orthogonal complement and Zorn's lemma.

What am I missing?

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  • $\begingroup$ I've added a tag for p-adic groups, but it's also a good idea to indicate this restriction in your header and/or question to avoid confusion. $\endgroup$
    – Jim Humphreys
    Commented Nov 22, 2013 at 14:25
  • $\begingroup$ Thanks Jim. But actually, if you look at the proof of the lemma in Murnaphan's note, you can see that the proof seems to go through any locally compact topological group. That's why I didn't put any restriction on the group $G$ to begin with. So I changed the question itsself accordingly $\endgroup$
    – Windi
    Commented Nov 22, 2013 at 15:59
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    $\begingroup$ As the various answers and comments suggest, it's risky to take lecture notes at face value. Yhey may be useful but are seldom complete or authoritative. While "semisimple" has a standard meaning in representation theory, "unitary representation" definitely requires more context. $\endgroup$
    – Jim Humphreys
    Commented Nov 22, 2013 at 19:50

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Probably the version of "semi-simplicity" relevant in the context in those notes refers to a repn of $G$ that is admissible with respect to a compact subgroup $K$, in the sense that it is assumed to decompose with finite multiplicities over $K$. This assumption holds for repns induced from cuspidal repns on the compact itself, for example, as observed by Jacquet already in 1971. In such a context, the otherwise-too-glib remarks about complete-reducibility still reach a correct conclusion, even without completeness in a Hilbert-space sense.

The more down-to-earth situations, like $L^2(\mathbb R)$, or subspaces generated by rough functions on the circle, do not satisfy suitable "admissibility" conditions, perhaps-oddly.

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$L^2(\mathbb{R})$ acted on by translation does decompose into a direct integral of one-dimensional representations. So in general you have a direct integral if $G$ is not compact. If $G=SL_2(Q_p)$, you have the same issue. So no direct sum, but a direct integral.

You should look up type 1 groups, for various more complicated issues. The decomposition might not be unique otherwise.

I haven't time to read the full script. I seems that she is using it in the context of supercuspidal representations, which are in fact semisimple in the sense that they decompose into a direct sum (actually a finite sum). Here it is okay.

Murnagahan is certainly wrong if she means unitary = unitarizabile and smooth,e.g. it fails for $C_c^\infty(G)$. Unitary (being a unitary Hilbertspace representation and smooth) implies finite-dimensional, so isn't really interesting.

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  • $\begingroup$ Actually you can look at $C_c^\infty(Q_p)$ already;) $\endgroup$
    – Marc Palm
    Commented Nov 22, 2013 at 15:45
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In the lemmas on semisimplicity, the representation is assumed to be smooth, which $L^2(G)$ is not.

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    $\begingroup$ A representation which is infinite-dimensional for SL_2(Q_p) and smooth, can't be unitary, only unitarizable. For example, the right translation on $C_c^\infty(G)$ is certainly smooth, its unitarisation is $L^2(G)$, but it can't be decomposable into a direct sum, but only a direct integral. $\endgroup$
    – Marc Palm
    Commented Nov 22, 2013 at 15:26
  • $\begingroup$ Thanks, user43112. But it seems to me that the proof of the lemma in Murnaghan's note does't use the smoothness of the representation, even though the lemma assumes the representation is smooth. So I edited the question itself accordingly. $\endgroup$
    – Windi
    Commented Nov 22, 2013 at 16:00
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The dimension. Every finite-dimensional unitary representation is a direct sum of irreducible representations (given an irreducible sub-representation, take the orthogonal, and go on…), but this is not true of an infinite-dimensional unitary representation -- you cannot go on forever, so you need a more elaborate notion of direct sum.

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  • $\begingroup$ Thanks for your quick reply. But of course I'm asking about the infinite dimensional case. In her note (top of p. 13), she explicitly defines a semisimple representation (including the infinite dimensional case) to be a direct sum of irreducible representations. And the above mentioned lemma in p.35 includes the infinite dimensional case. $\endgroup$
    – Windi
    Commented Nov 22, 2013 at 7:28
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    $\begingroup$ Oops, sorry, I had not realized this is over a $p$-adic field! (at least the group $G$ is totally disconnected). What you refer to about $L^2(G)$ is for the classical case. $\endgroup$
    – abx
    Commented Nov 22, 2013 at 7:34
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The proof does not go through. As far as I can tell, it is just false that $V = W \oplus W^{\perp}$ with only the hypotheses Murnaghan is using; the standard hypotheses are that $V$ is complete and that $W$ is closed. $W$ closed is necessary since the above implies $W = (W^{\perp})^{\perp}$, and on the other hand the earlier lemma to which the proof alludes needs complements for arbitrary $W$ to get semisimplicity.

For an easy counterexample let $V = L^2(\mathbb{Z})$ be the regular representation of $\mathbb{Z}$ (locally compact, Hausdorff, and totally disconnected as required) and let $W$ be the subspace of compactly supported functions. Then $W^{\perp} = 0$.

Edit: Noncompactness isn't even the issue here. For a compact counterexample let $V = L^2(S^1)$ be the regular representation of $S^1$ and let $W$ be the subspace spanned by translates of, say, a sawtooth wave. Here we don't get a direct sum of irreducibles but a Hilbert space direct sum of irreducibles. In general an infinite-dimensional Hilbert space can never be the orthogonal direct sum of any collection of its finite-dimensional subspaces.

I know nothing about $p$-adic representation theory but it seems like that whole section is disregarding the issue of topologies on $V$ and other hypotheses concerning only the structure of $V$. Is this standard?

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