# Ruling out a lone eigenvalue of -1 to complete this proof?

Euler's rotation theorem (very informally stated) says that an rotation of a sphere in three dimensional space always has two fixed points. I believe this claim generalizes to (n-1)-hyperspheres in n-dimensional space for odd n, but I am unable to patch the final gap in the proof. My strategy so far is as follows; any ideas on how to proceed would be greatly appreciated.

First, note that an arbitrary rotation of our hypersphere can be characterized by rotations about n-1 different axes. Since a rotation by theta in $\mathbb{R}^2$ is characterized by $\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{array}\right]$, it is not hard to see that a rotation about n-1 different axes can be characterized by the matrix product

$\left[\begin{array}{ccccccc} \cos\theta_{1} & -\sin\theta_{1} & 0 & \cdots & & & 0\\ \sin\theta_{1} & \cos\theta_{1} & 0 & \cdots & & & 0\\ 0 & 0 & 1\\ \vdots & \vdots & & & & & \vdots\\ & & & & \ddots\\ & & & & & & 0\\ 0 & 0 & & \cdots & & 0 & 1\end{array}\right]\left[\begin{array}{ccccccc} 1 & 0 & 0 & \cdots & & & 0\\ 0 & \cos\theta_{2} & -\sin\theta_{2} & \cdots & & & 0\\ 0 & \sin\theta_{2} & \cos\theta_{2} & \cdots & & & 0\\ \vdots & \vdots & \vdots & 1 & & & \vdots\\ & & & & \ddots\\ & & & & & & 0\\ 0 & 0 & 0 & \cdots & & 0 & 1\end{array}\right]$

$\cdots\left[\begin{array}{ccccccc} \cos\theta_{n-1} & 0 & & \cdots & & 0 & -\sin\theta_{n-1}\\ 0 & 1 & & \cdots & & & 0\\ \\\vdots & \vdots & & \ddots & & & \vdots\\ \\0 & & & & & 1 & 0\\ \sin\theta_{n-1} & 0 & & \cdots & & 0 & \cos\theta_{n-1}\end{array}\right]$.

Then notice that we have two fixed points if the matrix product just described has an eigenvalue of 1. The eigenvalues of a matrix are given by the roots of the characteristic polynomial; the characteristic polynomial here has real coefficents; therefore any complex roots of the characteristic polynomial must come in conjugate pairs; therefore if n is odd, we must have a real eigenvalue. But since rotations preserve length, any real eigenvalues must have an absolute value of one.

So we've shown that for odd n, there must be an eigenvalue of 1 or -1, but to complete the proof, we need to show that there can't be a lone eigenvalue of -1, and I haven't been able to figure out how to do that.

I thought I might be able to glean some intuition from the three-dimensional case, where a half-rotation about one axis gives us two eigenvalues of -1. Perhaps eigenvalues of -1 must also come in pairs? That seems plausible---dare I say even obvious?---on intuitive and physical grounds, but how might one prove it?

This excursion was inspired by an exercise in Bretscher Linear Algebra.

• @Zach M. Davis: The product of the eigenvalues equals the determinant, and the determinant of a product is the product of the determinants. If you have a lone eigenvalue of $-1$, then the determinant of the matrix is $-1$, isn't it? – Arturo Magidin Dec 13 '10 at 4:35
• I deleted my comment because I thought it wasn't enough, but in fact it is; as Arturo says you only need to observe that rotations have determinant 1. Also, as far as I can tell, the first bit of your argument is not used at all in the second. – Qiaochu Yuan Dec 13 '10 at 4:40
• Not that it changes the argument, but one has to rule out an odd number of $-1$ eigenvalues, not just a single such eigenvalue. – Matt E Dec 13 '10 at 4:41

You need to show that the determinant of your transformation is equal to $1$ (or at least is positive; but then it will necessarily equal $1$).

Given this: Recall that this determinant is the product of the eigenvalues. Now the (possibly complex) eigenvalues are all of absolute value 1, and there are an odd number of them. The non-real eigenvalues come in conjugate pairs, which have product 1, so there are also an odd number of real eigenvalues, which are $\pm 1$, and whose product equals $1$. It follows that at least one of them equals $1$, as desired.

I'll let you work out the determinant fact.

[Added: Several people wrote down the same argument in comments, reflecting (I think) the fact that this is the standard approach to this quesion.]

• Of course one can always use the Poincare-Hopf theorem...! – Qiaochu Yuan Dec 13 '10 at 4:48
• @Qiaochu: Dear Qiaochu, True! – Matt E Dec 13 '10 at 4:52

Here is an approach which gives a stronger result. Let $g$ be a rotation of $\mathbb{R}^n$ and suppose that $g$ has a complex eigenvalue $\lambda$. By the spectral theorem, there is a complex eigenvector $v$ associated to $\lambda$, hence a conjugate eigenvector $\bar{v}$ associated to $\bar{\lambda}$. It follows that on the real subspace $V$ of $\mathbb{R}^n$ spanned by $v + \bar{v}$ and $\frac{v - \bar{v}}{i}$, $g$ acts as a 2-dimensional rotation, and we can break $\mathbb{R}^n$ up into a direct sum $V \oplus V^{\perp}$, and this respects the action of $g$ (so $g$ acts as a rotation on $V^{\perp} \simeq \mathbb{R}^{n-2}$).

By induction, we obtain a structure theorem for rotations: any rotation is a direct sum of 2-dimensional rotations and 1-dimensional rotations (multiplication by $\pm 1$). The determinant condition then tells us that the multiplications by $-1$ must occur in pairs, hence must pair up as 2-dimensional rotations by $\pi$. So we can say more: any rotation has an invariant subspace $V$ of even codimension, and on $V^{\perp}$ it acts as a direct sum of nontrivial 2-dimensional rotations.