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If $\mu$ equals Haar measure on the 3-dimensional unit sphere $S^2$, then $\hat{\mu}(\varepsilon) = \dfrac{2\sin(2\pi |\varepsilon|)}{|\varepsilon|}$.

I am not quite sure how to start this problem. First of all, I can't figure out the formula for the $\mu$ itself, let alone computing its Fourier transform. The only I can make sense of $\hat{\mu}$ is when I view $\mu$ as a distribution but even in that case I don't know where to begin with.

Any help is appreciated.

Answer: Here is my attempted solution after looking at given suggestions below. $$\hat{\mu}(\varepsilon) = \int_{S^2} e^{-i\varepsilon\cdot x}dx =4\pi \int_{S^2} e^{-2\pi i\varepsilon\cdot\gamma} d\sigma(\gamma)\, (1)$$, where $d\sigma(\gamma)$ is the surface element of $S^2$. Because $\mu$ is rotationally invariant, so is its Fourier transform. In other words, the above integral is radial in $\varepsilon$. Hence, it suffices to assume that if $|\varepsilon| = \rho$, then $\varepsilon = (0,0,\rho)$. Finally, using routine spherical coordinate transform, we see that our integral in $(1)$ equals to $${4\pi}\int_{0}^{2\pi}\int_{0}^{\pi} e^{-2\pi i\rho\cos\theta}\sin\theta d\theta d\phi = 4\pi\cdot\dfrac{1}{2}\int_{0}^{\pi}e^{-2\pi i\rho\cos\theta}\sin\theta d\theta = $$ $$2\pi\cdot\dfrac{1}{4\pi i\rho}[e^{2\pi i\rho u}]_{u=-1}^{u=1} = \dfrac{2 \sin(2\pi\rho)}{\rho}$$, as desired.

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  • $\begingroup$ Some quibbles: $S^2$ would be better referred-to as the two-dimensional unit sphere in $\mathbb R ^3$, and (since it's not a group in its own right) the measure is the $O(3,\mathbb R)$-invariant one. (Yes, there is a relation to the actual Haar measure on this orthogonal group, but maybe that's not the point here.) So: rotation-invariant distribution supported on $S^2$. It's compactly-supported, so you can meaningfully apply it to exponentials, to correctly compute its Fourier transform.. $\endgroup$ Apr 19 '16 at 22:14
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In this case the Haar measure is simply the Lebesgue measure. So you have $$ \hat \mu(\varepsilon)=\int_{S^2} e^{-i\varepsilon x}dx. $$

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  • $\begingroup$ It's not completely clear what "Lebesgue measure" on a sphere is... unless one intends exactly "group-invariant measure"... but perhaps then the questioner is asking how this could be computed. $\endgroup$ Apr 19 '16 at 22:16
  • $\begingroup$ Yes, I am interested in how exactly is this computed. But more important, I am still struggling to find a compact formula for $\mu$ itself. Or does there exist no such thing? $\endgroup$
    – dezdichado
    Apr 19 '16 at 22:18
  • $\begingroup$ I think that it is completely clear what "Lebesgue measure" is on a sphere (in fact there is a unique measure up to scaling such that translations keep the measure). On the computation, which I detailed clearly, one only needs to write down a parametrization (if one knows what a Haar measure is, for sure one knows how to parametrize a sphere...). $\endgroup$
    – John B
    Apr 19 '16 at 22:19
  • $\begingroup$ @Jonas I have written my solution based on your answer. Would you take a look at it and tell me if it has a flaw? $\endgroup$
    – dezdichado
    Apr 19 '16 at 23:16
  • $\begingroup$ @dezdichado All seems fine. $\endgroup$
    – John B
    Apr 20 '16 at 8:23

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