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The space $L^2 (\mathbb{R}^2; \mathbb{C})$ can be decomposed as $$ L^2 (\mathbb{R}^2; \mathbb{C}) = \bigoplus_{k \in \mathbb{Z}} L^2_k (\mathbb{R}^2; \mathbb{C}), $$ where $$ L^2_k (\mathbb{R}^2; \mathbb{C}) = \bigl\{ f \in L^2 (\mathbb{R}^2;\mathbb{C}) : \text{for almost every \(z \in \mathbb{R}^2 \simeq \mathbb{C}\) and \(\theta \in \mathbb{R}\), }\\ f (e^{i \theta} z) = e^{i k \theta} f (z) \bigr\}, $$ where the summands are mutualy orthogonal (see for example Stein and Weiss, Introduction to Fourier analysis on Euclidean spaces, 1971, §IV.2).

The summands are invariant under Fourier transform and the Fourier transform can there be described by Bessel functions. This decomposition appears also implicitly in separation of variables arguments.

As Stein and Weiss do not give any name to this decomposition, I was wondering whether under which name(s) this decomposition is known in the literature.

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  • $\begingroup$ Your mention of Bessel functions makes me wonder if the Euclidean motion group of the plane is lurking behind the scenes. Perhaps you could give some details of which Bessel functions appear, and how? $\endgroup$
    – Yemon Choi
    Commented Sep 22, 2015 at 15:13

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Maybe: isotypical decomposition of the representation $L^2(\mathbb{R}^2,\mathbb{C})$ of the group $U(1)$ might do.

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  • $\begingroup$ Indeed, or, if one wants to save two symbols, just "isotypic" will do, I think. $\endgroup$
    – paul garrett
    Commented Sep 22, 2015 at 22:07
  • $\begingroup$ yes. I always wondered also about combinatoric vs combinatorial... $\endgroup$
    – Abdelmalek Abdesselam
    Commented Sep 22, 2015 at 22:35
  • $\begingroup$ But do you know if this representation has some standard name? $\endgroup$
    – Yemon Choi
    Commented Sep 23, 2015 at 0:00
  • $\begingroup$ @YemonChoi, I'd call it "the/and isotypic component of the natural rotation (and reflection) group $O(n,\mathbb R)$ representation on $L^2(\mathbb R^n)$"... :) That is, srsly, just the descriptive name. $\endgroup$
    – paul garrett
    Commented Sep 23, 2015 at 0:16
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Thinking about this a bit more, here is a guess — I haven't got time right now to check the details, so comments and corrections are welcome from others.

The Euclidean motion group $E(2)$ is the group of all isometries of ${\bf R}^2$ with its usual Euclidean metric. Since $E(2)$ acts on ${\bf R}^2$ in a measure-preserving way, this action gives a complex representation $\pi$ of $E(2)$ on $L^2({\bf R}^2)$. Then the restriction of $E(2)$ to $U(1)$ should decompose in the way shown above.

If this is correct, then I guess another way to see this would be: take the trivial representation $\varepsilon$ of $U(1)$, and form the induced representaion ${\rm Ind}_{U(1)}^{E(2)} \varepsilon$. My guess is that this is once again $\pi$, with the decomposition now emerging naturally from the standard construction of "induction from a compact subgroup".

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