Integration with a map having zero determinant

Question

I am trying to compute an integral of a $d$-manifold $\mathcal{M}$, projected onto the unit (hyper-)sphere $S^d = \{\vec{\omega} \in \mathbb{R}^{d+1} \,:\, \|\vec{\omega}\|=1\}$, both embedded in $\mathbb{R}^{d+1}$. Let $\vec{x} : [0,1]^d \rightarrow \mathbb{R}^{d+1}$ be a parametrization of $\mathcal{M} = \{\vec{x}(\vec{t}) \in \mathbb{R}^{d+1} \,:\, \vec{t} \in [0,1]^d\}$ and let it be sufficiently smooth. Also let a function $f:S^d\rightarrow \mathbb{R}$ be given. The radial projection from $\mathcal{M}$ onto $S^d$ is:

$$P: \mathbb{R}^{d+1} \rightarrow \mathbb{R}^{d+1}, \,P(\vec{x}) = \frac{\vec{x}}{\|\vec{x}\|}$$

The integral then reads (taking into consideration Shifrin's comment, this is where I made a mistake):

$$\int_{S^d}f(\vec{\omega})\,d\vec{\omega} = \int_{S^d}f(P(\vec{x}))\,dP(\vec{x}) = \int_{\mathcal{M}}f(P(\vec{x}))\left|det[J_P(\vec{x})]\right|\,d\vec{x} \\ J_P(\vec{x}) = \frac{\partial P}{\partial \vec{x}}(\vec{x})$$

However, unsurprisingly the determinant of the Jacobian is zero, since it has a zero eigenvalue along $\vec{x}$:

$$J_{P}(\vec{x}) = \frac{1}{\|\vec{x}\|}\left(I- \frac{\vec{x}\vec{x}^T}{\|\vec{x}\|^2}\right)$$

I have seen elsewhere (e.g. wikipedia, the page on solid angle for $d=2$) that one gets a term $\frac{1}{\|x\|^2}\left(\frac{\vec{x}}{\|\vec{x}\|} \cdot n_{\vec{x}}\right)$ instead (where $n_{\vec{x}}$ is the normal of $\mathcal{M}$ at $\vec{x}$), i.e.:

$$\int_{S^d}f(\vec{\omega})\,d\vec{\omega} = \int_{\mathcal{M}}f(P(\vec{x}))\frac{1}{\|x\|^2}\left(\frac{\vec{x}}{\|\vec{x}\|} \cdot n_{\vec{x}}\right)\,d\vec{x}$$

After this I can use:

$$\int_{\mathcal{M}}f(P(\vec{x}))\frac{1}{\|x\|^2}\left(\frac{\vec{x}}{\|\vec{x}\|} \cdot n_{\vec{x}}\right)\,d\vec{x} = \int_{[0,1]^d}f(P(\vec{x}(\vec{t})))\frac{1}{\|x\|^2}\left(\frac{\vec{x}}{\|\vec{x}\|} \cdot n_{\vec{x}}\right)\sqrt{det[g(\vec{t})]}\,d\vec{t} \\ g(\vec{t}) = \left(\frac{\partial \vec{x}}{\partial \vec{t}}(\vec{t})\right)^T\left(\frac{\partial \vec{x}}{\partial \vec{t}}(\vec{t})\right),$$

to compute the integral.

Provided that the above is not wrong, my question is how is the term $\frac{1}{\|x\|^2}\left(\frac{\vec{x}}{\|\vec{x}\|} \cdot n_{\vec{x}}\right)$ derived in the first place, and how would I go about deriving a similar term for other projections (the current one projects on the unit sphere).

Edit: Taking into account Shifrin's comment I realized that I cannot split the two transformations as I have done above. Namely, when I was computing the metric tensor's determinant, I had used a property which doesn't hold for non-square matrices:

$$det[AB] = det[BA] = det[A]det[B],$$

in order to get:

$$det \left[\left(\frac{\partial P}{\partial \vec{t}}\right)^T\left(\frac{\partial P}{\partial \vec{t}}\right)\right] = det \left[\left(\frac{\partial P}{\partial \vec{x}}\frac{\partial \vec{x}}{\partial \vec{t}}\right)^T\left(\frac{\partial P}{\partial \vec{x}}\frac{\partial \vec{x}}{\partial \vec{t}}\right)\right] \\ \ne det \left[\left(\frac{\partial P}{\partial \vec{x}}\right)^T\left(\frac{\partial P}{\partial \vec{x}}\right)\right] det \left[\left(\frac{\partial \vec{x}}{\partial \vec{t}}\right)^T\left(\frac{\partial \vec{x}}{\partial \vec{t}}\right)\right].$$

Nevertheless, it seems like this can be factored in 3D (following the solid angle article in wikipedia) in the following manner instead:

$$\sqrt{det \left[\left(\frac{\partial P}{\partial \vec{t}}\right)^T\left(\frac{\partial P}{\partial \vec{t}}\right)\right]} = \frac{1}{\|\vec{x}\|^2}\left|\vec{n}_{\vec{x}} \cdot \frac{\vec{x}}{\|\vec{x}\|}\right| \sqrt{det\left[\left(\frac{\partial \vec{x}}{\partial \vec{t}}\right)^T\left(\frac{\partial \vec{x}}{\partial \vec{t}}\right)\right]}.$$

I am assuming that after many algebraic transformations the above can be proven at least in 3D. I am not aware whether there's a more general result for $d$ dimensions or how/whether this generalizes to other maps except $P$ (e.g. $Q:\mathcal{M} \rightarrow \mathcal{N}$, $Q$ sufficiently smooth, $\mathcal{N}$ being a different manifold). I have the suspicion that this can be derived somehow from the divergence theorem.

Note that in my comments below my question, $S$ corresponds to $P$ (I cannot edit the comments).

Answer 1

As I said in the comments, you cannot apply the change of variables theorem unless you look at the mapping from the hypersurface to the sphere. So your "zero determinant" is a complete red herring.

The most powerful way to do such computations is with differential forms. You can see the details of one such computation here. Of course, the normal $\vec n$ to your hypersurface $\mathcal M$ has to appear, since the element of "surface area" on an oriented hypersurface is $$d\vec\sigma = \iota_{\vec n}dt_1\wedge\dots\wedge dt_{d-1} = \sum_{i=1}^{d+1} (-1)^{i-1} n_i dt_1\wedge\dots dt_{i-1}\wedge dt_{i+1}\wedge\dots dt_{d+1}.$$ In particular, the element of "surface area" on the unit sphere is $$d\vec\omega = \sum_{i=1}^{d+1} (-1)^{i-1} t_i dt_1\wedge\dots dt_{i-1}\wedge dt_{i+1}\wedge\dots dt_{d+1}.$$ As the computation in that linked answer shows for $d=2$, solid angle appears here quite naturally — but be careful about the exponent on $r=\|\vec t\|$: If $\pi\colon\Bbb R^{d+1}-\{0\}\to S^d$ is the radial projection, $\pi(\vec t) = \dfrac{\vec t}r$, then the pullback $$\pi^*d\vec\omega = d\vec\Omega = \frac1{r^d}\sum (-1)^{i-1} t_i dt_1\wedge\dots dt_{i-1}\wedge dt_{i+1}\wedge\dots dt_{d+1}.$$

If we now restrict this solid angle form to $\mathcal M$, you want to see how it is related to $d\vec\sigma$. This is quite analogous to the flux computations that appear in multivariable calculus (see, e.g., my YouTube lecture). In particular, to get the flux of $\vec F = (F_1,\dots,F_{d+1})$ across the hypersurface $\mathcal M$, we want $$\int_{\mathcal M} \vec F\cdot\vec n\, d\vec\sigma = \int_{\mathcal M} \sum (-1)^{i-1} F_i\,dt_1\wedge\dots dt_{i-1}\wedge dt_{i+1}\wedge\dots\wedge dt_{d+1}.$$ As I explained in that YouTube lecture, you can see that this magic formula is correct by evaluating on an oriented orthonormal basis for the tangent plane of $\vec M$, noting that putting $\vec n$ at the head of that list gives an oriented orthonormal basis for $\Bbb R^{d+1}$; only the normal component of $\vec F$ appears, since any tangential component gives a $0$ term in the determinant.

In our situation, restricting $d\vec\Omega$ to $\mathcal M$ is finding the flux of $\vec F = \dfrac{\vec t}{r^d}$, which is in turn the integral of $\vec F\cdot\vec n\,d\vec\sigma$, as you desired.

I would urge you to learn the basics of differential form computations and get rid of Gram determinants and the awkward vector computations. :)