The Cartan-Dieudonné Theorem
Reproducing and commenting a proof by Marc van Leeuwen on the Mathematics Stack Exchange.
Introduction
The Cartan-Dieudonné Theorem says that rotations in Euclidean space can be generated by reflections across hyperplanes.
Below is a mathematical formulation and proof by induction of this fact. It uses the language of modern algebra to describe the geometric scenario. My intention is to comment in what each step of the proof intends to show, while filling in details.
The advantage of this rather abstract, non-constructive proof is that it is very succint. The details are left to the reader (you and me) to figure out. My comments use the language of three-dimensional geometry for clarity, and are wrapped up in blue clickable elements.
Statement
Let \(\phi\) be an orthogonal endomorphism on an \(n\)-dimensional Euclidean vector space \(E\). Let \(d=\dim \ker (\phi - id_E)\). Then \(\phi\) can be written as a product of \(n-d\) orthogonal reflections across hyperplanes.
Show explanation
(1) An orthogonal endomorphism is a way to move points in the space so that their distance from the origin remains the same as before. Rotations and reflections have exactly this property.
(2) The identity transformation \(id_E\) doesn't change the points of \(E\). That is, \(id_E(x)=x\) for all \(x\) in \(E\).
(3) Rotations in three dimensions have an axis of rotation. A way to characterise the axis of rotation is that it constitutes the points that don't move during the rotation. Mathematically, it is the set of points \(x\) in \(E\) such that \(\phi(x)=id_E(x)\). Equivalently, \((\phi - id_E)(x)=0\). So we say that \(x\) belongs to a subspace called the kernel of the transformation \(\phi - id_E\).
(4) In three dimensions, the axis of rotation is a one-dimensional subspace of \(E\). The theorem claims that you can write a rotation as the composition of two reflections across hyperplanes.
Proof
We describe a proof by induction. For the base case, let \(n-d=0\). This is only possible if \(\phi=id_E\). Then \(\phi\) is a product of no rotations at all, so the base case is true.
Show explanation
(5) If \(d=n\) then \(\ker(\phi - id_E)\) takes up the entire space. Therefore in the base case, every point \(x\) in \(E\) stays put. We say \(\phi(x)=x\), so \(\phi=id_E\).
Now assume that \(n-d>0\), and that we have proved the statement for all values less than \(n-d\). Since \(d < n\), there exists a vector \(v\notin \ker(\phi - id_E)\). Consider the hyperplane \(H=(\phi(v)-v)^\perp\), the orthogonal reflection \(\sigma_H\) across \(H\), and the transformation \(\phi'=\sigma_H\circ \phi\). Note that we have \(\phi'(v)=v\), or \(v \in \ker(\phi'-id_E)\).
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(6) The hyperplane \(H\) cuts in between \(v\) and \(\phi(v)\). Therefore the transformation \(\phi'\) applied to \(v\) first rotates the vector and then reflects it back to its starting point. We can verify this since reflecting a vector \(b\) across a hyperplane orthogonal to a vector \(a\) is computed as $$\sigma_{a^\perp}(b)=b-2\frac{b \cdot a}{a \cdot a}a.$$ If we plug in \(a=\phi(v)-v\) and \(b=\phi(v)\), and use the fact that \(||\phi(v)||=||v||\), we can see that the numerator can be written as $$2\phi(v) \cdot (\phi(v) - v)=2||\phi(v)||^2 - 2\phi(v)\cdot v = ||\phi(v)-v||^2.$$ Therefore the reflection simplifies to \(\phi(v) - (\phi(v) - v)=v\), as required. Here, \(\ker(\phi'-id_E)\) represents the hyperplane spanned by the axis of rotation and \(v\).
Furthermore, \(\ker(\phi - id_E)\subseteq H\), since for any vector \(w\) such that \(\phi(w)=w\), we have \(w \cdot (\phi(v) - v) = 0\).
Show explanation
(7) This says that \(\phi\)'s axis of rotation is a subset of the hyperplane that cuts between \(v\) and \(\phi(v)\). It is proven by showing that a vector \(w\) in the axis of rotation (\(\phi(w)=w\)) is orthogonal to \(\phi(v)-v\). Indeed since \(\phi(w)\cdot \phi(v)=w\cdot v\) we can check that $$w \cdot (\phi(v) - v) = \phi(w)\cdot \phi(v) - w \cdot v = 0.$$
Therefore on one hand we have $$\ker(\phi - id_E) \subset \ker(\phi + id_E) + \text{span}\{v\} \subseteq \ker(\phi' - id_E).$$ And on the other hand we have $$\ker(\phi' - id_E) \cap H \subseteq \ker(\phi - id_E).$$ So \(\dim \ker (\phi' - id_E) = \dim \ker (\phi - id_E) + 1 = d+1\).
Show explanation
(8) What a mouthful! But this is the crucial argument. The way the story goes is that the axis of rotation is on one hand contained in a strictly larger subspace. But on the other hand the axis of rotation contains that subspace intersected with a hyperplane, cutting its dimension down by one. Therefore, the difference in dimension between the axis of rotation and that larger subspace must be exactly one.
Then the induction hypothesis applies to \(\phi'\), and we may express it as a product of \(n-d-1\) reflections. Then \(\phi=\sigma_H\circ \phi'\) can be written as a product of \(n-d\) reflections. This concludes the proof.
Last edited 25/11/2024