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Let's say I want to integrate over a sphere $S^2$.

Take $f \in (L^1(S^2),dS)$, then we have that $$\int_{S^2} |f| dS = \int_{S^2} |f| \sin^2 (\theta) d \theta d \phi < \infty,$$ right?

Now, assuming that $f$ is the solution to a PDE on $S^2$, like $\Delta_{\theta,\phi}f=0$. Does this mean now that if I substitute for example this $$f = g_1(\phi)g_2(\theta) \frac{1}{\sin(\theta)}$$ in the PDE , that my new spaces for the two ODEs with respect to $g_1$ and $g_2$ are given by $L^1([0,\pi], \sin(\theta) d\theta)$ and $L^1 ([0,2\pi],d \phi)$? ( And if I had substituted ($f = g_1(\phi)g_2(\theta) $) it would have been $L^1([0,\pi], \sin^2(\theta) d\theta)$ and $L^1 ([0,2\pi],d \phi)$?)

By the way: Is the notion of Substitution correct, that every diffeomorphism $\Phi$ gives rise to a unitary map $L^1 ( \Omega, d\lambda) \rightarrow L^1(\Phi(\Omega),|\det \Phi'| d \lambda)$?

Sorry, is it so confusing what I want to ask?

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  • $\begingroup$ Your substitution is problematic as $\sin \theta$ may get the value 0 (say for $\theta = 0$). The problem is that a shpere is not diffeomorphic to a plane globally, it is only locally. $\endgroup$ – LinAlgMan Jul 22 '14 at 13:36
  • $\begingroup$ yes, but these are actually only two points 0 and pi, so you take the open interval $(0,\pi)$ and you should be fine, or am I wrong? $\endgroup$ – user159356 Jul 22 '14 at 13:37
  • $\begingroup$ You still don't cover all the sphere. $\endgroup$ – LinAlgMan Jul 22 '14 at 13:40
  • $\begingroup$ ah sorry, forgot the squares, did you mean this? $\endgroup$ – user159356 Jul 22 '14 at 13:42
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Spherical coordinates: $$ x = r\sin\theta\cos\phi,\;\; y =r\sin\theta\sin\phi,\;\;z=r\cos\theta. $$ Restrict to $r=1$. The tangent vectors $$ \frac{\partial}{\partial\theta}(x,y,z)=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta)\\ \frac{\partial}{\partial\phi}(x,y,z) = (-\sin\theta\sin\phi,\sin\theta\cos\phi,0). $$ The tangent vectors are orthogonal. So the area element, which is the magnitude of the cross product, is the product of the lengths of these gradient vectors, which is $$ 1\cdot\sin\theta. $$ So the scalar area element is $\sin\theta\,d\theta\,d\phi$. You have an extra $\sin\theta$ factor running around.

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