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Every unitary representation is a direct sum of cyclic representations. it can be proved without the Zorn's Lemma ?

user147972
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1 Answers1

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I don't think you need Zorn's Lemma. Let $V$ be a unitary representation of the group $G$ and let $U$ be the subrepresentation that is the direct sum of all cyclic subrepresentations. Suppose for contradiction that $U \not = V$. Then, $$V = U \oplus U^{\perp}$$ with $U^{\perp}$ nonzero. So, pick a vector $v \in U^{\perp}$. Then, $\text{span}\{G\cdot v\}$ is a cyclic subrepresentation of $V$ contained in $U^{\perp}$ and hence not contained in $U$, a contradiction. Hence, $U = V$ and hence every unitary representation is a direct sum of its cyclic subrepresentations.

Actually, note that in the above argument, unitary isn't necessary. All you need is complete reducibility i.e. every subrepresentation of $V$ has a complementary subrepresentation.

Siddharth Venkatesh
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  • thank you very much.it's good.why V=U + U⊥ ?and why U⊥ is nonzero? – user147972 May 12 '14 at 05:31
  • In any unitary representation $V$, is you have a subrepresentation $U$, then $V$ is the direct sum of $U$ and its orthogonal complement as vector spaces, and the orthogonal complement is also a subrepresentation because the group preserves the inner product and hence preserves orthogonality. $U^{\perp}$ is nonzero because $V = U \oplus U^{\perp}$ and I assumed for contradiction that $V \not= U$. – Siddharth Venkatesh May 12 '14 at 07:22
  • Thank you for your help., I've already asked two questions. Someone has replied to them. Would you help me?The two posts below. – user147972 May 12 '14 at 08:28
  • http://math.stackexchange.com/questions/783268/convolution-of-f-and-g-is-in-lp-where-f-has-compact-support – user147972 May 12 '14 at 08:29
  • http://math.stackexchange.com/questions/783643/suppose-g-is-unimodular-if-f-is-in-lpg-and-g-is-in-lqg-where-1-p-q-an – user147972 May 12 '14 at 08:29
  • I'm not that good at harmonic analysis sorry. The reason I could answer this question is that I'm interested and have studied representation theory. I'll take a look at those problems tomorrow but I can't guarantee being able to solve them. – Siddharth Venkatesh May 12 '14 at 10:23
  • ok . thank you very much – user147972 May 12 '14 at 12:45
  • @ Siddharth Venkatesh:why V≠∅? – user147972 May 26 '14 at 06:19
  • I'm sorry, I don't understand the question. Are you asking why $V$ is nonempty? – Siddharth Venkatesh May 27 '14 at 06:53
  • ooh,I'm sorry, why U is nonempty? – user147972 May 30 '14 at 07:38
  • Vector spaces cannot be empty. If you are asking why it is not zero, I have not made that assumption but in that case $U^{\perp}$ becomes all of $V$ and you can still get a contradiction. – Siddharth Venkatesh May 30 '14 at 18:29