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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:
• What does $S^1$ do in many branches of math?
• 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

Ooker
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  • 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?) – Giuseppe Negro May 28 '18 at 17:04
  • Moreover, $\mathbb C^\infty$ is not a Hilbert space in a natural way. – Giuseppe Negro May 28 '18 at 17:06
  • I have rewritten the question. Is it clearer? Also, why is $\Bbb{C}^\infty$ not a Hilbert space in a natural way? Is it because of it's infinite dimensions? – Ooker May 28 '18 at 17:19
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    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. – anon May 29 '18 at 04:10
  • @anon why can't there be an interval Q which makes the basis of $L^2(Q)$ forms a group? I'm just talking about the basis not its elements. And $\Bbb C^\infty$ isn't Hilbert space? I've thought that it is the only infinite-dimensional Hilbert space? My textbook is Dym & McKean – Ooker May 29 '18 at 04:40
  • @Ooker What do you mean by "the" basis of $L^2(Q)$? Again, what does it mean for a basis of a vector space to be a group? What do you propose as a hypothetical group operation on that set? And did you read what I asked and said about $\Bbb C^{\infty}$? First I asked if you were referring to the set of all infinite sequences of complex numbers, and then I pointed out that if that's the case it can't be a Hilbert space since almost all its elements have unbounded $L^2$ norm. – anon May 29 '18 at 04:43
  • @anon I've updated my question, can you check that? I've removed the $\Bbb{C}^\infty$ part, but both Dym & McKean and Kreyszig textbook say that the space of all infinite sequences of complex numbers is a Hilbert space. – Ooker May 29 '18 at 12:50
  • I assume you mean Fourier Series and Integrals by Dym & McKean and Introductory Functional Analysis with Applications by Kreyszig. I am almost positive you're wrong about them. After skimming, nowhere did I find the notation $\Bbb C^{\infty}$, nor did I find the claim that "the space of all infinite sequences of complex numbers is a Hilbert space." Please say what page you found that claim in, or if you can't, please state what norm you think exists on $\Bbb C^{\infty}$ which makes it a Hilbert space. (It certainly isn't under the $\ell^2$ norm. Do you understand why?) – anon May 29 '18 at 18:32
  • Once again it is completely unclear to me what you're asking. What does it mean for a character or representation to be isomorphic to a group? No, a character of a group is not a point on the unit circle (Although it is true the characters of an abelian group $G$ form a basis for $L^2(G)$ with suitable inner product. If $G$ is not abelian, we can use the matrix coefficients of the irreducible representations as a basis, if everything is suitably normalized.) – anon May 29 '18 at 18:38
  • @anon Yes, those books. In D&M, p. 22, it explicitly says that, (1) "$L^2(Q)$ is isomorphic to [...] $\Bbb C^\infty=L^2(\Bbb Z^+)$, and (2) "up to isomorphism, $\Bbb C^\infty$ is the only infinite-dimensional Hilbert space there is!". In Kreyszig, example 3.1-4 (p. 131), it says that the space $\Bbb C^n$ is a Hilbert space, which is defined earlier at example 1.1-5 as the space of all ordered n-tuple of complex number. In 3.1-6, it says that the $\ell^2$ norm is generalized from $\Bbb C^n$ norm. – Ooker May 31 '18 at 09:02
  • @anon so the character or representation isn't a group by default? Is there a special case where it is? – Ooker May 31 '18 at 09:04
  • @Ooker Except on pg22 of D&M, it explicitly defines $\Bbb C^{\infty}$ to be the space of all sequences with $(\sum |c_n|^2)^{1/2}<\infty$. This is not the space of all sequences of complex numbers. (This is why I'm skeptical of the authors' choice of notation $\Bbb C^{\infty}$ for $L^2(\Bbb Z)$.) Your citation of Kreyszig is irrelevant because again it doesn't say anything about the space of all sequences of complex numbers. And no, a group is a set with a binary operation satisfying certain properties, and a representation is a function from a group satisfying certain properties. – anon May 31 '18 at 13:53
  • Indeed, the space of all sequences of complex numbers is uncountable-dimensional, while $L^2(\Bbb Z)$ has countably-infinite dimension, so they're not even isomorphic as plain old vector spaces. You seem to not be studying these subjects very systematically, because you're getting tripped up on very basic definitions (groups, characters, representations, isomorphisms, Hilbert spaces). Studying them more systematically will be easier and more natural. – anon May 31 '18 at 13:55
  • @anon thank you. So to be sure, are we just having a clash of notation, and is the space in D&M and Kreyszig the same? Is your space can be notated as $\ell^\infty$? Do you know why all $L^2(Q)$ isomorphes to each other? I've rechecked the definition of character (section 4.1.3 D&M) and turns out a character of a group G is a function, a homomorphism between G and a group of complex number of modulus 1. So for different character of G, there is a different corresponding group of complex number? My original question is about this group on the circle. – Ooker May 31 '18 at 14:27
  • And in exercise 12 - 13, it says that $\hat G$, the class of characters of G, is also a group? – Ooker May 31 '18 at 14:27
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    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. – anon May 31 '18 at 19:45
  • It's not entirely clear to me what you mean by "for different character of G, there is a different corresponding group of complex numbers." What do you mean by that? – anon May 31 '18 at 19:46
  • @anon so for example for a group $G$, through a character $\phi_1$ its corresponding group on the unit circle is $H_1$. But for $\phi_2$ the group on circle is $H_2$ and so on. I'm saying that there are many $H_n$ for $G$. Is that correct? I'm also ask that whether at least a $H_n$ is the basis of a $L^2$ space? – Ooker Jun 03 '18 at 05:13
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    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! – anon Jun 03 '18 at 05:35
  • Take, for instance, the cyclic group $\Bbb Z/n\Bbb Z$. It has $n$ characters, each of the form $x\mapsto \exp(2\pi i ax/n)$ for some $a\in \Bbb Z/n\Bbb Z$. But the image must be the group of $d$-th roots of unity for some divisor $d\mid n$, and there are only $\phi(n)$ such subgroups. So the number of images is less than the number of characters. The characters do form a basis for $L^2(\Bbb Z/n\Bbb Z)$ though, which has dimension $n$. – anon Jun 03 '18 at 05:42
  • @anon yes, the images! And yes, your second to last comment really clarifies me a lot. Thank you for patiently answering me. Can you make that an answer? Many thanks. – Ooker Jun 03 '18 at 07:36

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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).

Giuseppe Negro
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  • 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? – Ooker May 29 '18 at 03:34
  • Bases do NOT "isomorph" to circles. I don't know where you read that, but that's false. – Giuseppe Negro Jun 03 '18 at 09:39
  • 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? – Ooker Jun 04 '18 at 03:40
  • 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. – Giuseppe Negro Jun 04 '18 at 08:40
  • is it because my understanding about basic concepts is not rigorous, or because the answer to the (current) question is just no? – Ooker Jun 04 '18 at 11:45
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    Both. You are basing your question on an inexistent analogy. Just let it go. – Giuseppe Negro Jun 04 '18 at 12:55
  • Actually I want to know the relationship between harmonic analysis and projective geometry. Do you have any idea? Specific relativity deals with both projective geometry and Fourier transform, so I guess it must exist? – Ooker Jun 07 '18 at 14:46
  • There's basically no relationship to my knowledge. – Giuseppe Negro Jun 07 '18 at 14:50
  • What do you think about the Peter – Weyl theorem? – Ooker Jun 09 '18 at 04:09