# Affine Invariant Surface Measure

In this short post I want to introduce an instance of certain objects that will be the subject of a few more posts. This particular object arises naturally in Affine Differential Geometry and turned out to have a relevant rôle in Harmonic Analysis too (in both Fourier restriction and in the theory of Radon transforms).

## 1. Affine Invariant measures

Affine Differential Geometry is the study of (differential-)geometric properties that are invariant with respect to $SL(\mathbb{R}^d)$. A very interesting object arising in Affine Geometry is the notion of an Affine Invariant Measure. Sticking to examples rather than theory (since the theory is still quite underdeveloped!), consider a hypersurface $\Sigma \subset \mathbb{R}^{d}$ sufficiently smooth to have well-defined Gaussian curvature, which we denote by $\kappa$ (a function on $\Sigma$). If we let $d\sigma$ denote the surface measure on $\Sigma$ (induced from the Lebesgue measure on the ambient space $\mathbb{R}^d$ for example, or by taking directly $d\sigma = d\mathcal{H}^{d-1}\big|_{\Sigma}$, the restriction of the $(d-1)$-dimensional Hausdorff measure to the hypersurface) then this crafty little object is called Affine Invariant Surface Measure and is given by

$\displaystyle d\Omega(\xi) = |\kappa(\xi)|^{1/(d+1)} \,d\sigma(\xi).$

It was first introduced by Blaschke for $d=3$ (finding the reference seems impossible; it’s [B] in this paper, if you feel luckier) and by Leichtweiss for general $d$. The reason this measure is so interesting is that it is (equi)affine invariant in the sense that if $\varphi(\xi) = A \xi + \eta$ is an equi-affine transformation (thus with $A \in SL(\mathbb{R}^d)$ and so volume-preserving since $\det A = \pm 1$) then, using subscripts to distinguish the two surfaces, we have

$\displaystyle \boxed{ \Omega_{\varphi(\Sigma)}(\varphi(E)) = \Omega_{\Sigma}(E) } \ \ \ \ \ \ \ (1)$

for any measurable $E \subseteq \Sigma$. We remark the following fact: that seemingly mysterious power $\frac{1}{d+1}$ in the definition of $d\Omega$ is the only exponent for which the resulting measure is (equi)affine-invariant.

Although this elementary fact is extremely well known, my lazy search could not come up with a decent reference for it (there is a slick proof when $\Sigma$ is a convex body but the fact is much more general than this), so in the rest of the post I will prove the fact in a gruesomely direct way.

## 2. Affine invariance of $\Omega$

I don’t know whether there is a clever proof of the invariance. There probably is one, but I will rather prove the fact doing some pedantic hard work.
Instead of proving just (1), we will consider more in general what happens to $\Omega$ under an affine transformation $\varphi(\xi) = A \xi$, where we omit the translation (since its action is trivially invariant) and consider the more general case of $A \in GL(\mathbb{R}^d)$ instead of just $SL(\mathbb{R}^d)$. Our claim is that

$\displaystyle \boxed{ \Omega_{\varphi(\Sigma)}(\varphi(E)) = (\det \varphi)^{\frac{d-1}{d+1}}\, \Omega_{\Sigma}(E). } \ \ \ \ \ \ \ (2)$

Before we start proving things, we list all the objects we will need.

### 2.1. Objects and notations:

We will use subscripts to denote which surface every object is referring to.
We let $\xi$ denote any point on the hypersurface $\Sigma$ and we let $\boldsymbol{\nu}_{\Sigma}(\xi)$ denote the normal at point $\xi \in \Sigma$ according to the chosen (local) orientation. The tangent space at $\xi$ is denoted by $T_{\xi}\Sigma$ – it is the space of vectors orthogonal to $\boldsymbol{\nu}_{\Sigma}(\xi)$. The shape operator $S_{\Sigma}$ is the linear operator $S_{\Sigma}(\xi) : T_{\xi}\Sigma \to T_{\boldsymbol{\nu}_{\Sigma}(\xi)} \mathbb{S}^{d-1}$ (where the latter is identified with $T_{\xi} \Sigma$ since they are parallel) defined by

$\displaystyle S_{\Sigma}(\xi) \boldsymbol{v} := D_{\boldsymbol{v}} \boldsymbol{\nu}_{\Sigma} (\xi),$

where $D_{\boldsymbol{v}}$ is the directional derivative in direction $\boldsymbol{v} \in T_{\xi} \Sigma$. The Gaussian curvature of the hypersurface $\Sigma$ at point $\xi$ is then defined to be

$\displaystyle \kappa_{\Sigma} (\xi) := \det S_{\Sigma}(\xi),$

where the determinant makes sense because we can see $S_{\Sigma}(\xi)$ as a linear map $T_{\xi} \Sigma \to T_{\xi}\Sigma$. However, it is a bit difficult to deal formally with this abstract notion of determinant – we’d rather work with the usual determinant on $d \times d$ matrices. There is a simple way to do this using projections. We let $\Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)}$ denote the orthogonal projection onto the normal $\boldsymbol{\nu}_{\Sigma}(\xi)$, that is

$\displaystyle \Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)} \boldsymbol{u} := \langle \boldsymbol{u}, \boldsymbol{\nu}_{\Sigma}(\xi) \rangle \boldsymbol{\nu}_{\Sigma}(\xi);$

similarly, we let $\Pi_{T_{\xi}\Sigma}$ denote the orthogonal projection onto the tangent space $T_{\xi} \Sigma$. Observe that $\Pi_{T_{\xi} \Sigma} + \Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)} = Id$. If we define $S'_{\Sigma}(\xi)$ to be the extension of $S_{\Sigma}(\xi)$ to the whole of $\mathbb{R}^d$ given by

$\displaystyle S'_{\Sigma}(\xi) := S_{\Sigma}(\xi)\circ \Pi_{T_{\xi} \Sigma} + \Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)},$

then the Gaussian curvature $\kappa_{\Sigma}$ can be calculated as

$\displaystyle \kappa_{\Sigma}(\xi) = \det S'_{\Sigma}(\xi),$

where now the determinant refers to the standard determinant of $d \times d$ matrices.
All the definitions above have their counterparts in the hypersurface $\varphi(\Sigma)$, clearly. To compute $d\Omega_{\varphi(\Sigma)}$ we will need to compute separately how $d\sigma_{\Sigma}$ and $\kappa_{\Sigma}$ behave under the transformation $\varphi$, which is what we will do next.

### 2.2. Change in the normal direction under $\varphi$

The first thing we need to figure out is what $\boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi))$ is. This is easy: first of all, by linearity of $\varphi$, the tangent space $T_{\varphi(\xi)} \varphi(\Sigma)$ is equal to $\varphi (T_{\xi} \Sigma)$ (as vector sub-spaces). Therefore any $\boldsymbol{w} \in T_{\varphi(\xi)} \varphi(\Sigma)$ can be written as $\boldsymbol{w} = A \boldsymbol{v}$ for some $\boldsymbol{v} \in T_{\xi} \Sigma$. Secondly, the normal $\boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi))$ is the only unit vector (up to a sign) that is orthogonal to all of $T_{\varphi(\xi)}\varphi(\Sigma)$ and so

\displaystyle \begin{aligned} \langle \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)), \boldsymbol{w} \rangle = & 0 \qquad \forall \boldsymbol{w} \in T_{\varphi(\xi)}\varphi(\Sigma) \\ \Leftrightarrow \langle \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)), A \boldsymbol{v} \rangle = & 0 \qquad \forall \boldsymbol{v} \in T_{\xi}\Sigma \\ \Leftrightarrow \langle A^{\intercal} \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)), \boldsymbol{v} \rangle = & 0 \qquad \forall \boldsymbol{v} \in T_{\xi}\Sigma , \end{aligned}

where $A^{\intercal}$ denotes the transpose of $A$. The latter can only hold (non-trivially) if $A^{\intercal} \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi))$ is parallel to $\boldsymbol{\nu}_{\Sigma}(\xi)$, and therefore we have

$\displaystyle \boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi)) = \frac{(A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)}{\|(A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|}. \ \ \ \ \ \ (3)$

It will be convenient in the following to have worked out what the corresponding projection $\Pi_{\boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi))}$ is, in terms of $\Pi_{\boldsymbol{\nu}_{\Sigma} (\xi)}$ and $\varphi$. Using (3) we have for any $\boldsymbol{u} \in \mathbb{R}^d$

\displaystyle \begin{aligned} \Pi_{\boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi))} \boldsymbol{u} = & \langle \boldsymbol{u}, \boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi)) \rangle \boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi)) \\ = & \frac{\langle \boldsymbol{u},(A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi) \rangle}{\|(A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^2} (A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi) \\ = & (A^{\intercal})^{-1} \frac{\langle A^{-1} \boldsymbol{u}, \boldsymbol{\nu}_{\Sigma}(\xi) \rangle}{\|(A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^2} \boldsymbol{\nu}_{\Sigma}(\xi), \end{aligned}

and therefore

$\displaystyle \Pi_{\boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi))} = \frac{ (A^{\intercal})^{-1} \circ \Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)} \circ A^{-1}}{\|(A^{\intercal})^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^2}. \ \ \ \ \ \ \ (4)$

### 2.3. Change in the surface area under $\varphi$

The next step is to calculate how $d\sigma_{\varphi(\Sigma)}$ is related to $d\sigma_{\Sigma}$. If we let $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_{d-1}$ denote an orthonormal basis of $T_{\xi} \Sigma$, then the density of $d\sigma_{\varphi(\Sigma)}$ with respect to $d\sigma_{\Sigma}$ is given by the ratio

$\displaystyle \frac{\begin{vmatrix} A\boldsymbol{v}_1 & \cdots & A\boldsymbol{v}_{d-1} & \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)) \end{vmatrix}}{\begin{vmatrix} \boldsymbol{v}_1 & \cdots & \boldsymbol{v}_{d-1} & \boldsymbol{\nu}_{\Sigma}(\xi) \end{vmatrix}},$

where the vertical bars denote the determinant of the $d \times d$ matrix whose columns are the vectors inbetween the bars. Notice that the denominator is equal to 1 and so it suffices to evaluate the numerator. Thanks to (3) and the properties of the determinant we can write

\displaystyle \begin{aligned} & \begin{vmatrix} A\boldsymbol{v}_1 & \cdots & A\boldsymbol{v}_{d-1} & \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)) \end{vmatrix} \\ & = \begin{vmatrix} A\boldsymbol{v}_1 & \cdots & A\boldsymbol{v}_{d-1} & A A^{-1} \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)) \end{vmatrix} \\ & = \det A \begin{vmatrix} \boldsymbol{v}_1 & \cdots & \boldsymbol{v}_{d-1} & A^{-1}\boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi)) \end{vmatrix} \\ & = \frac{\det A}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \begin{vmatrix} \boldsymbol{v}_1 & \cdots & \boldsymbol{v}_{d-1} & A^{-1} (A^\intercal)^{-1}\boldsymbol{\nu}_{\Sigma}(\xi) \end{vmatrix} \\ & = \frac{\det A}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \langle A^{-1} (A^\intercal)^{-1}\boldsymbol{\nu}_{\Sigma}(\xi), \boldsymbol{\nu}_{\Sigma}(\xi)\rangle \cdot 1 \\ & =\frac{\det A}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \langle (A^\intercal)^{-1}\boldsymbol{\nu}_{\Sigma}(\xi), (A^{-1})^\intercal \boldsymbol{\nu}_{\Sigma}(\xi)\rangle, \end{aligned}

and since $(A^{-1})^\intercal = (A^\intercal)^{-1}$ we have shown that

$\displaystyle d\sigma_{\varphi(\Sigma)}(\varphi(\xi)) = \det A \cdot \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\| \, d\sigma_{\Sigma}(\xi). \ \ \ \ \ \ \ (5)$

### 2.4. Change in the Gaussian curvature under $\varphi$

This is the most cumbersome part of the calculations. We have to evaluate the shape operator $S_{\varphi(\Sigma)}$ in terms of $S_{\Sigma}$ and $\varphi$. Taking $\boldsymbol{w} \in T_{\varphi(\xi)}\varphi(\Sigma)$ we have that $\boldsymbol{w} = A \boldsymbol{v}$ for some $\boldsymbol{v} \in T_{\xi} \Sigma$ and so by the properties of the derivative operator

\displaystyle \begin{aligned} S_{\varphi(\Sigma)}(\varphi(\xi))\boldsymbol{w} = & S_{\varphi(\Sigma)}(\varphi(\xi)) A\boldsymbol{v} \\ = & D_{A \boldsymbol{v}} \boldsymbol{\nu}_{\varphi(\Sigma)} (\varphi(\xi)) \\ = & [ D_{\boldsymbol{v}} (\boldsymbol{\nu}_{\varphi(\Sigma)} \circ A)] (A^{-1} \varphi(\xi))\\ = & [ D_{\boldsymbol{v}} (\boldsymbol{\nu}_{\varphi(\Sigma)} \circ A)] (\xi). \end{aligned}

Using (3) and linearity we have therefore

\displaystyle \begin{aligned} S_{\varphi(\Sigma)}(\varphi(\xi))\boldsymbol{w} = & \Big[ D_{\boldsymbol{v}} \Big( \frac{(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|} \Big) \Big](\xi) \\ = & (A^\intercal)^{-1} \Big[ D_{\boldsymbol{v}} \Big( \frac{ \boldsymbol{\nu}_{\Sigma}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|} \Big) \Big](\xi), \end{aligned}

which by Leibniz’s rule is equal to

$\displaystyle (A^\intercal)^{-1} \Big[ \frac{D_{\boldsymbol{v}}\boldsymbol{\nu}_{\Sigma}(\xi)}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} - \frac{D_{\boldsymbol{v}} (\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|)}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^2} \boldsymbol{\nu}_{\Sigma}(\xi) \Big].$

The numerator of the first term in square brackets we can readily recognise: it is simply $S_{\Sigma}(\xi) \boldsymbol{v}$. As for the numerator of the second term in square brackets, we can see that

\displaystyle \begin{aligned} D_{\boldsymbol{v}} (\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|) = & \frac{\langle D_{\boldsymbol{v}}[(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}](\xi),\, (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi) \rangle}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \\ = & \frac{\langle (A^\intercal)^{-1} D_{\boldsymbol{v}} \boldsymbol{\nu}_{\Sigma}(\xi), \,(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi) \rangle}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \\ = & \frac{\langle (A^\intercal)^{-1} S_{\Sigma}(\xi)\boldsymbol{v}, \, (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi) \rangle}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \\ = & \frac{ \langle A^{-1} (A^\intercal)^{-1}S_{\Sigma}(\xi)\boldsymbol{v},\, \boldsymbol{\nu}_{\Sigma}(\xi) \rangle}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \end{aligned}

(in the last line we have used that $((A^\intercal)^{-1})^\intercal = A^{-1}$). Combining this with the above we have shown that

$\displaystyle S_{\varphi(\Sigma)}(\varphi(\xi))\boldsymbol{w} = \frac{(A^\intercal)^{-1} }{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \Big[ Id - \frac{\Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)} A^{-1} (A^\intercal)^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^2} \Big] S_{\Sigma}(\xi) A^{-1} \boldsymbol{w}.$

At this point we can recognise in the above identity the projection $\Pi_{\boldsymbol{\nu}_{\varphi(\Sigma)}}$ as given by (4), so that we have

\displaystyle \begin{aligned} S_{\varphi(\Sigma)}(\varphi(\xi)) = & \frac{(A^\intercal)^{-1} S_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} - \frac{(A^\intercal)^{-1} \Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)} A^{-1} (A^\intercal)^{-1} S_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^3} \\ = & \frac{(A^\intercal)^{-1} S_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} - \Pi_{\boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi))} \frac{(A^\intercal)^{-1} S_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \\ = & \Pi_{T_{\varphi(\xi)}\varphi(\Sigma)} \circ \frac{(A^\intercal)^{-1} S_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|}. \end{aligned}

We are nearly there. At this point we remember that we want to use the extensions $S'_{\varphi(\Sigma)}$ and $S'_{\Sigma}$ in order to take advantage of the properties of the standard determinant, so we need to insert them in the expression above somehow. The first thing to notice is that we can replace $S_{\Sigma}(\xi) A^{-1}$ above with $S_{\Sigma}(\xi) \Pi_{T_{\xi}\Sigma} A^{-1}$ for free, since $A^{-1} : T_{\varphi(\xi)} \varphi(\Sigma) \to T_{\xi}\Sigma$. The second thing to notice is that we can sneak in any multiple of $\Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)}$ for free as well, because

$\displaystyle \Pi_{T_{\varphi(\xi)}\varphi(\Sigma)} (A^\intercal)^{-1} \Pi_{\boldsymbol{\nu}_{\Sigma}(\xi)} = 0.$

Indeed, this follows from the almost obvious fact that (omitting unnecessary notation)

\displaystyle \begin{aligned} \Pi_{T \varphi(\Sigma)} (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} = & (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} - \langle (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} , \boldsymbol{\nu}_{\varphi(\Sigma)}\rangle \boldsymbol{\nu}_{\varphi(\Sigma)} \\ = & (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} - \frac{\langle (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} , (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\rangle }{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|^2} (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} = 0. \end{aligned}

Therefore we can replace in the expression above $S_{\Sigma}(\xi)$ with $S'_{\Sigma}(\xi)$ for free, and as a consequence we have the beautiful identity

$\displaystyle S'_{\varphi(\Sigma)} (\varphi(\xi)) = \Pi_{T_{\varphi(\xi)}\varphi(\Sigma)} \Big[\frac{(A^\intercal)^{-1} S'_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \Big] \Pi_{T_{\varphi(\xi)}\varphi(\Sigma)} + \Pi_{\boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi))}.$

Since $\kappa_{\varphi(\Sigma)} = \det S'_{\varphi(\Sigma)}$, it remains to evaluate the determinant of the RHS. This can be done in a general way that is extremely reminiscent of what we did to calculate how the surface measure transforms under $\varphi$.
Let $N \in GL(\mathbb{R}^d)$, let $\boldsymbol{\nu}$ be an arbitrary unit vector and let $\Pi_{\boldsymbol{\nu}}$ be the orthogonal projection onto $\boldsymbol{\nu}$ and let $\Pi_{\langle \boldsymbol{\nu}\rangle^{\perp}}$ be the orthogonal projection onto the space $\langle \boldsymbol{\nu}\rangle^\perp$ orthogonal to $\boldsymbol{\nu}$. What we want is to compute

$\displaystyle \det (\Pi_{\langle \boldsymbol{\nu}\rangle^{\perp}} \, N \, \Pi_{\langle \boldsymbol{\nu}\rangle^{\perp}} + \Pi_{\boldsymbol{\nu}} ) =: \det N'$

in terms of $\det N$ and $\boldsymbol{\nu}$. If we let $\boldsymbol{u}_1, \ldots, \boldsymbol{u}_{d-1}$ be an orthonormal basis of $\langle \boldsymbol{\nu}\rangle^\perp$ we have that

$\displaystyle N' \boldsymbol{u}_j = N \boldsymbol{u}_j - \langle N \boldsymbol{u}_j, \boldsymbol{\nu} \rangle \boldsymbol{\nu}.$

Using the properties of the determinant we have therefore

\displaystyle \begin{aligned} \det N' = & \begin{vmatrix} N' \boldsymbol{u}_1 & \cdots & N' \boldsymbol{u}_{d-1} & N' \boldsymbol{\nu} \end{vmatrix} \\ = & \begin{vmatrix} N \boldsymbol{u}_1 - \langle N \boldsymbol{u}_1, \boldsymbol{\nu} \rangle \boldsymbol{\nu} & \cdots & N \boldsymbol{u}_{d-1} - \langle N \boldsymbol{u}_{d-1}, \boldsymbol{\nu} \rangle \boldsymbol{\nu} & \boldsymbol{\nu} \end{vmatrix} \\ = & \begin{vmatrix} N \boldsymbol{u}_1 & \cdots & N \boldsymbol{u}_{d-1} & \boldsymbol{\nu} \end{vmatrix} \\ = & \begin{vmatrix} N \boldsymbol{u}_1 & \cdots & N \boldsymbol{u}_{d-1} & N N^{-1} \boldsymbol{\nu} \end{vmatrix} \\ = & \det N \cdot \begin{vmatrix} \boldsymbol{u}_1 & \cdots & \boldsymbol{u}_{d-1} & N^{-1} \boldsymbol{\nu} \end{vmatrix} \\ = & \det N \cdot \langle N^{-1}\boldsymbol{\nu} , \boldsymbol{\nu} \rangle. \end{aligned}

We will apply this to $S'_{\varphi(\Sigma)}$ above, for which we have to take

$\displaystyle N = \frac{(A^\intercal)^{-1} S'_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|},$

and therefore

$\displaystyle N^{-1} = \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|\, A (S'_{\Sigma}(\xi))^{-1} A^\intercal.$

Since $\boldsymbol{\nu} = \boldsymbol{\nu}_{\varphi(\Sigma)}(\varphi(\xi))$ we have using (3) again

\displaystyle \begin{aligned} \langle N^{-1} \boldsymbol{\nu}, \boldsymbol{\nu} \rangle = & \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\| \langle A (S'_{\Sigma}(\xi))^{-1} A^\intercal \boldsymbol{\nu}_{\varphi(\Sigma)} ,\boldsymbol{\nu}_{\varphi(\Sigma)} \rangle \\ = & \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|^{-1} \langle A (S'_{\Sigma}(\xi))^{-1} A^\intercal (A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma} ,(A^\intercal)^{-1} \boldsymbol{\nu}_{\varphi(\Sigma)} \rangle \\ = & \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|^{-1} \langle (S'_{\Sigma}(\xi))^{-1} \boldsymbol{\nu}_{\Sigma} , A^\intercal (A^\intercal)^{-1}\boldsymbol{\nu}_{\varphi(\Sigma)} \rangle \\ = & \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|^{-1} \langle (S'_{\Sigma}(\xi))^{-1} \boldsymbol{\nu}_{\Sigma}, \boldsymbol{\nu}_{\Sigma} \rangle \\ = & \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}\|^{-1}, \end{aligned}

where the last line is because $S'_{\Sigma}$ acts like the identity on $\boldsymbol{\nu}_{\Sigma}$ and therefore $(S'_{\Sigma}(\xi))^{-1} \boldsymbol{\nu}_{\Sigma} = \boldsymbol{\nu}_{\Sigma}$. We have thus shown that

\displaystyle \begin{aligned} \det S'_{\varphi(\Sigma)}(\varphi(\Sigma)) = & \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^{-1} \det \Big[ \frac{(A^\intercal)^{-1} S'_{\Sigma}(\xi) A^{-1}}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|} \Big] \\ = & \frac{(\det A)^{-2} \det S'_{\Sigma}(\xi)}{\|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^{d+1}}. \end{aligned}

In other words, we have finally shown that

\displaystyle \begin{aligned} \kappa_{\varphi(\Sigma)}(\varphi(\xi)) = \frac{\kappa_{\Sigma}(\xi)}{(\det A)^2 \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi) \|^{d+1}}. \ \ \ \ \ \ (6) \end{aligned}

### 2.5. Change in $d\Omega$ under $\varphi$ – conclusion

We now have all we need to compute $d\Omega_{\varphi(\Sigma)}$. Indeed, using (5) and (6) we have for the densities

\displaystyle \begin{aligned} d\Omega_{\varphi(\Sigma)}(\varphi(\xi)) = & |\kappa_{\varphi(\Sigma)}(\varphi(\xi))|^{1/(d+1)}\, d\sigma_{\varphi(\Sigma)}(\varphi(\xi)) \\ = & (\det A)^{-2/(d+1)} \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\|^{-1} |\kappa_{\Sigma}(\xi)|^{1/(d+1)} \, \det A \|(A^\intercal)^{-1} \boldsymbol{\nu}_{\Sigma}(\xi)\| d\sigma_{\Sigma}(\xi) \\ = & (\det A)^{\frac{d-1}{d+1}} |\kappa_{\Sigma}(\xi)|^{1/(d+1)} \, d\sigma_{\Sigma}(\xi) \\ = & (\det A)^{\frac{d-1}{d+1}} d\Omega_{\Sigma}(\xi) \end{aligned}

and therefore (2) is proven.