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I'm puzzled with how we're supposed to define $L^2(S^{n-1})$ where $S^{n-1}$ is the unit sphere in $\mathbb{R}^n$. How do we even define the inner product there? the only way that comes to mind currently is $(f,g) = \int_{S^{n-1}}fg\,d\mathcal{H}^{n-1}$. But this is not well defined since we need $f, g $ to be functions from $\mathbb{R}^n$ to $\mathbb{R}$ in order to use the Hausdorff measure. Any help or pointer to the right books would be appreciated.

hteica
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    There is a canonical Borel measure defined on the surface of the sphere. See chapter 2.7 of Folland's Real Analysis, or this thread: https://math.stackexchange.com/questions/698735/definition-of-the-surface-measure – BBBBBB Mar 04 '24 at 04:09
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    @BBBBBB instead of defining a submanifold measure and integral on the surface of the sphere, is there any way to use Hausdorff measure? – hteica Mar 04 '24 at 04:11
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    See https://mathoverflow.net/questions/134197/hausdorff-measure-on-the-sphere-is-well-defined – Eric Towers Mar 04 '24 at 04:14
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    "But this is not well defined since we need $f, g $ to be functions from $\mathbb{R}^n$ to $\mathbb{R}$ in order to use the Hausdorff measure" Why is that? It's enough that $h$ be a Borel measurable function on $S^{n-1}$ to be able to talk about integrability of it with respect to the Hausdorff measure. Why do you think you need the function to be defined on $\Bbb R^n$? – BigbearZzz Mar 04 '24 at 04:14
  • @BigbearZzz the Hausdorff measure itself is defined on $\mathbb{R}^n$, so if we have a function with domain unequal to $\mathbb{R}^n$, how do we calculate the integral? – hteica Mar 04 '24 at 04:17
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    @hteica You can simply consider $S^{n-1}$ to be the whole space and the restriction of the Hausdorff measure on it to be the measure on that space. Alternatively, you can extend any function initially defined on $S^{n-1}$ to be identically $0$ (or any other number for that matter) on $\Bbb R^n\backslash S^{n-1}$, and now you can integrate in on the whole $\Bbb R^n$ as you want. – BigbearZzz Mar 04 '24 at 04:23
  • @BigbearZzz could you please elaborate about the restriction of the hausdorff measure? and also are we allowed to extend the function in any arbitrary fashion? – hteica Mar 04 '24 at 04:51

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There are many ways to define the desired measure on $S^{n-1}$.

One way to do it is to note that $SO(n)$ acts transitively on $S^{n-1}$. Fix a point $x_0 \in S^{n-1}$, and let $H$ be the stabilizer of $x_0$. Note that $H \cong SO(n-1)$, so it is a compact subgroup of $SO(n)$. By the orbit-stabilizer theorem, we get a homeomorphism $SO(n)/H \to S^{n-1}$.

$SO(n)$ is a compact Hausdorff topological group, so it admits a canonical Haar probability measure. We push this measure forward to the quotient to obtain a probability measure on $S^{n-1}$. The fact that this measure is nonzero and invariant under the action of $SO(n)$ tells you that it is the one you want (perhaps up to scaling).

Another way is to let $\nu$ be the vector field $\nu(x) = x$ on $\mathbb{R}^n$ (where we identify the tangent spaces of $\mathbb{R}^n$ with $\mathbb{R}^n$ in the standard way).

Then let $\omega$ be the standard volume form on $\mathbb{R}^n$, i.e. $\omega = dx_1 \wedge \dots \wedge dx_n$.

Let $\iota_\nu \omega$ be the contraction of $\omega$ along $\nu$, i.e. $\iota_\nu \omega$ is the $(n-1)$-form such that for all $x \in \mathbb{R}^n$ we have $(\iota_\nu \omega)_x(v_1, \dots, v_{n-1}) = \omega_x(v_1, \dots, v_{n-1}, \nu(x))$.

Finally, let $\alpha = i^*(\iota_\nu \omega)$, where $i : S^{n-1} \to \mathbb{R}^n$ is the inclusion. Then $\alpha$ is an $(n-1)$-form on $S^{n-1}$, and it is nonzero because $\omega$ and $\nu$ are nowhere-vanishing on $S^{n-1}$.

By integrating smooth functions against $\alpha$, we obtain a nonzero, positive linear functional on the space of smooth functions $S^{n-1} \to \mathbb{R}$. This extends (uniquely) to a nonzero, positive linear functional on the space of continuous functions $S^{n-1} \to \mathbb{R}$ by (e.g.) Stone-Weierstrass + dominated convergence. Then Riesz-Markov tells us that this comes from a unique nonzero Radon measure on $S^{n-1}$, which is the measure you want (perhaps up to scaling).

diracdeltafunk
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