Let $M$ be a 2-dimensional embedded $C^1$-submanifold of $\mathbb R^3$ with a global chart$^1$ $(U,\phi)$. If $u\in U$ and $x=\phi^{-1}(u)$, let $\nu_M(x)$ denote the unique unit normal vector of $M$ with $$\det\left({\rm D}\phi(u),\nu_M(x)\right)>0\tag1.$$ Moreover, let $\sigma_M$ denote the surface measure$^2$ on the Borel $\sigma$-algebra $\mathcal B(M)$ and $$\pi:\mathbb R^3\setminus\{0\}\to S^2\;,\;\;\;x\mapsto\frac x{|x|}$$ denote the projection onto the unit 2-sphere $S^2$.

If $0\not\in B\in\mathcal B(M)$, are we able to express $\sigma_{S^2}(\pi(B))$ as an integral with respect to $\sigma_M$?

For clarity of exposition, let $$\omega_{x\to y}:=\pi(y-x)\;\;\;\text{for }x,y\in\mathbb R^3\text{ with }x\ne y.$$ There are plenty of (mathematically vaguely) references$^3$ claiming that $$\sigma_{S^2}({\rm d}\omega_{x\to y})=\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\omega_{x\to y})\rangle\right|}{|x-y|^2}\tag2,$$ which is reasonable from geometric inspection. However, I struggle to state and prove this in a measure-theoretic rigorous way.

Idea 1: Noting that $\pi=\nabla\rho$, where $\rho(x):=|x|$ for $x\in\mathbb R^3$, $(2)$ may be related to the divergence theorem. To be precise, if $K\subseteq M\setminus\{0\}$ is compact and has a $C^1$-boundary$^2$, $$\langle\nu_{\partial K},\pi\rangle=\langle\nu_{\partial K},\nabla\rho\rangle=:\frac{\partial\rho}{\partial\nu_{\partial K}}.\tag3$$ However, I'm unsure how the "outer" normal field$^5$ $\nu_{\partial K}$ and $\nu_M$ are related.

Idea 2: My guess is that $$\left(\sigma_{S^2}\circ\pi\right)(B)=\int_B\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\pi(y)\rangle\right|}{|y|^2}\tag4$$ (this would include to show that $\pi(B)\in\mathcal B(S^2)$). Since $\mathcal B(M)$ is generated by the open balls with center in $M$, it should be sufficient to prove $(4)$ for $B=B_\varepsilon(x)$ for some fixed $x\in M$ and $\varepsilon>0$. Now, $B$ is compact and has a $C^1$-boundary. Moreover, $$\nu_{\partial B}(y)=\frac{y-x}\varepsilon\;\;\;\text{for all }y\in\partial B\tag5$$ and $$\int\sigma_{\partial B}({\rm d}y)\frac{\left|\langle\nu_{\partial B}(y),\pi(y-x)\rangle\right|}{|y-x|^2}=\frac1{\varepsilon^2}\sigma_{\partial B}(\partial B)=\sigma_{S^2}(S^2)=\left(\sigma_{S^2}\circ\pi\right)(\partial B).\tag6$$ However, I don't know if $(6)$ is helpful for showing $(4)$. (But intuitively, the solid angle subtended by $\partial B$ should be the same as the solid angle subtended by $B$.)

EDIT: Isn't there a "visibility" function $v:M\times M\to\{0,1\}$ missing on the right-hand side of $(2)$ indicating whether $y$ is "visible" as seen from $x$ ($v(x,y)=1$) or is occluded by another point $z\in M$ with $\omega_{x\to y}=\omega_{x\to z}$ and $|x-z|<|x-y|$?

$^1$ i.e. $U\subseteq\mathbb R^2$ is open and $\psi:U\to M$ is an immersion and a topological embedding of $U$ onto $M$.

$^2$ Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $J_\phi$ denote the Jacobian of $\phi$ and $g_\phi:=\det J_\phi^TJ_\phi$. Then $$\sigma_M=\sqrt{g_\phi}\left.\lambda^{\otimes 2}\right|_U\circ\phi^{-1}.$$

$^3$ for example, [1, p. 14], [2, p. 53 (PDF numbering)] or [3, p. 6].

$^4$ i.e. for all $x\in\partial K$ there is an open neighborhood $V$ of $x$ and a continuously differentiable $\psi:V\to\mathbb R$ with $K\cap V=\{\psi\le0\}$ and $\nabla\psi\ne0$.

$^5$ i.e. if $\psi$ is as in footnote 4, then $$\nu_{\partial K}(x):=\frac{\nabla\psi(x)}{\left|\nabla\psi(x)\right\|}.$$

  • $\begingroup$ What would be one of the more commonly encountered lecture books which make this claim? $\endgroup$
    – Ben McKay
    Jan 17 '20 at 20:56
  • $\begingroup$ What is $\mathcal{B}(M)$? The expression $\sigma_{S^2}(d\omega_{x\to y})$ appears to apply a measure on a surface to the exterior derivative of a function on $\mathbb{R}^3\times \mathbb{R}^3 - \text{diagonal}$. Can you clarify that notation a little please? $\endgroup$
    – Ben McKay
    Jan 17 '20 at 21:00
  • $\begingroup$ I am guessing $\mathcal{B}(M)$ is the Borel sigma algebra on $M$. $\endgroup$ Jan 17 '20 at 21:27
  • $\begingroup$ I share @BenMcKay's confusion about your notation. So first, forget about $x$, which seems to just describe the center of the sphere. Let $\omega$ be the projection map $M \to \mathbb{S}^2$ where $0\not\in M$ is an embedded 2-manifold of $\mathbb{R}^3$ given by $y \mapsto y / |y|$. Are you just asking about how to relate the surface measure on $M$ and the surface measure on $\mathbb{S}^2$ via the function $\omega$? $\endgroup$ Jan 17 '20 at 21:36
  • $\begingroup$ @BenMcKay $\mathcal B(M)$ is the Borel $\sigma$-algebra on $M$. You can find the claim, for example, in this PHD thesis (on page 15): backend.orbit.dtu.dk/ws/portalfiles/portal/51112612/…. But I can give you way more references, if I search for them. $\endgroup$
    – 0xbadf00d
    Jan 18 '20 at 5:08

The density you are asking about is precisely the Jacobian of the central projection of $M$ to the unit sphere centered at the origin (as was already mentioned by Willie Wong). For an explicit expression (for instance, in the form given on p.14 in the reference you quote) it is enough to look at the picture on p.15 (which is apparently what you mean by "geometric inspection"). If you wish, one can do it in two steps by first replacing $M$ with the corresponding tangent plane (which does not change the Jacobian). I really fail to see what is the "measure-theoretic rigor" one needs in addition to this observation (which does not really seem to be at the research level). However, if you want a mathematical reference, this formula is, for instance, derived in great detail (in a somewhat different notation though) in Palamodov's book Reconstructive integral geometry, (Section 3.1, Example 5).

Talking about notation, the way you reproduce the (already somewhat confused) formula from p.14 (not p.15 as you write) of your reference is completely erroneous (which was also mentioned by Willie Wong). In that reference a point $y\in M$ is projected to the unite sphere centered at $x$, whereas in your notation it is $x\in M$ that is projected to the unite sphere centered at 0 (and the role of $y$ is completely unclear).

The right expression for the density is $|\langle x, \nu_M(x) \rangle|/|x|^3$. Geometrically, the numerator is the distance from the origin to the tangent plane to $M$ at the point $x$.

  • $\begingroup$ Thank you for your answer. You're right, there was a typo in my equation (2). However, my problem with (2) is that the left-hand side is not really well-defined, since the differential ${\rm d}\omega_{x\to y}$ does depend on $y$. So, since it is still not clear to me from your answer, is (4) the correct formula (assuming $x=0$) or is it $$\sigma_{S^2}(\Omega)=\int\sigma_M({\rm d}y)1_\Omega(\pi(y))\frac{\left|\langle\nu_M(y),\pi(y)\rangle\right|}{|y|^2}\tag7$$ for all $\Omega\in\mathcal B(S^2)$? In either case, it seems like a "visibility function" $v:M\to\{0,1\}$ is missing indicating $\endgroup$
    – 0xbadf00d
    Jan 20 '20 at 7:17
  • $\begingroup$ whether $y$ is occluded ($v(y)=0$) by another point in $M$ as seen from the origin $0$ or not ($v(y)=1$). That's what I mean with "measure-theoretic rigor". Besides that, shouldn't we be able to derive the formula without reference to the geometric picture? And most importantly: What kind of regularity assumption do we need to impose on $M$ (does it need to be the boundary of a compact set?) and how precisely is $ν_M$ (is it given by $\frac{∂_1\phi×∂_2\phi}{|∂_1ϕ×∂_2\phi|}$ (in the submanifold case) or, in the boundary case, by footnote 5?) defined? $\endgroup$
    – 0xbadf00d
    Jan 20 '20 at 7:31
  • $\begingroup$ The geometric part of your question boils dows to finding the Jacobian of the projection map. All the rest is not specific for your concrete setup and is covered by standard textbook expositions on change of variables in surface integrals; the same textbooks usually discuss normal vectors to surfaces as well. $\endgroup$
    – R W
    Jan 20 '20 at 8:50
  • $\begingroup$ You did not answer my question. Is $(4)$ or $(7)$ the correct formula? And don't we need the visibility function? $\endgroup$
    – 0xbadf00d
    Jan 20 '20 at 8:57
  • $\begingroup$ The change of variables formulas are always written locally, under the assumption that the considered transformation is locally a bijection. If this is not the case, and there are several preimages, then the image measure is the sum of several measures obtaned by applying the corresponding local bijections. See any multivariable calculus textbook for the details. If $\pi$ is not an injection, then the LHS of (4) is not defined, whereas (7) is only true if you restrict to a subset of $M$ where $\pi$ is a bijection (for instance, by using your "visibility function"). $\endgroup$
    – R W
    Jan 20 '20 at 11:26

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