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For instance, the cyclic group $\Bbb Z/n\Bbb Z$ has n characters in the form $x\mapsto \exp(2\pi i ax/n)$ for some $a\in \Bbb Z/n\Bbb Z$. The characters form a basis for $L^2(\Bbb Z/n\Bbb Z)$, which has dimension n (the Peter–Weyl theorem).

Are there some projective groups having the same property of $\Bbb Z/n\Bbb Z$? My ultimate goal is to see the link between projective geometry and harmonic analysis.


Related:
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How is orthonormal basis $\{e^{2πikt}\}_{k=-\infty}^\infty$ with index going to $-\infty$ constructed in $L^2(\Bbb S^1)$?
How does projective geometry related with Fourier transform via special relativity? in Physics

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  • $\begingroup$ As is, this question is borderline incomprehensible, and I am afraid that it will attract close votes. What kind of "homomorphism" have you in mind? And what do you mean by "basis of a Hilbert space"? (I suppose you are thinking of orthonormal bases, but then, how can a set of bases be homomorphic to a group?) $\endgroup$ Commented May 28, 2018 at 17:04
  • $\begingroup$ Moreover, $\mathbb C^\infty$ is not a Hilbert space in a natural way. $\endgroup$ Commented May 28, 2018 at 17:06
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    $\begingroup$ What would it even mean for a basis of a Hilbert space to be isomorphic to a group or a representation? No, a basis of $L^2(S^1)$ is not simply a point on $S^1$. Nor is an element of $L^2(S^1)$ simply a point on $S^1$. Nor is a basis of $\Bbb C^{\infty}$ "the circle itself." (Also, is $\Bbb C^{\infty}$ supposed to just be the space of all infinite sequences of complex numbers? If so, it's not a Hilbert space; almost all of its elements have unbounded $L^2$ norm.) Nothing you're saying or asking makes any sense at all to me. $\endgroup$
    – anon
    Commented May 29, 2018 at 4:10
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    $\begingroup$ Yes the Hilbert space can be denoted $\ell^{\infty}$. All $L^2(Q)$ are isomorphic as Hilbert spaces, where $Q$ denotes an interval, since they're all isomorphic to $\ell^2$ (they have a canonical basis: complex exponentials, shifted and scaled appropriately). One definition of character is a homomorphism from $G$ to $\Bbb C^{\times}$, in which case the set of all characters is itself a group. A more general definition of character is the trace of a representation, in which case a character is not a homomorphism. $\endgroup$
    – anon
    Commented May 31, 2018 at 19:45
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    $\begingroup$ What do you mean by "the corresponding group on the unit circle"? What does that mean? Please do not expect me to read your mind. Do you mean the image (aka "range") of the homomorphism $\phi$? If so, then do not treat a homomorphism the same as its image. Different homomorphisms can have the same image, after all. And it doesn't make sense to ask if the images form a basis for an $L^2$-space of functions, because an image of a character is a set of numbers and elements of $L^2$ are functions. They're different things! $\endgroup$
    – anon
    Commented Jun 3, 2018 at 5:35

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There is no relation. Don't read too much into these informal analogies. There is no natural group structure on the set of all bases of a vector space, and so, it makes no sense to speak of "isomorphism" with a projective group. (The same with the set of all orthonormal bases of a Hilbert space).

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  • $\begingroup$ No? But why is that? I think if anything isomorphes to a common thing like circle, then it would be easy to find its other isomorphisms? $\endgroup$
    – Ooker
    Commented May 29, 2018 at 3:34
  • $\begingroup$ Bases do NOT "isomorph" to circles. I don't know where you read that, but that's false. $\endgroup$ Commented Jun 3, 2018 at 9:39
  • $\begingroup$ I'm sorry, my knowledge is screwed up. I mean the base of a $L^2$ space is all function, and so are characters. So is there a connection between them in projective group? $\endgroup$
    – Ooker
    Commented Jun 4, 2018 at 3:40
  • $\begingroup$ In the original question, I see a long thread of comment by anon, trying to convince you that this train of thoughts is not leading you anywhere. I agree with these comments. I think that the analogy you are trying hard to see simply does not exist. $\endgroup$ Commented Jun 4, 2018 at 8:40
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    $\begingroup$ Both. You are basing your question on an inexistent analogy. Just let it go. $\endgroup$ Commented Jun 4, 2018 at 12:55

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