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Let $\chi:\mathbb{R}^{3}\to\mathbb{R}^{3}$ be an orthogonal transformation such that $\det(\chi)=1$ and $\chi$ is not the identity linear transformation. Let $S \subset \mathbb{R}^{3}$, be the unit sphere. Then how do we prove that $\chi$ fixes only two points of $S$?

Any ideas of proceeding for the solution?

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Do you mean "fixes no more than $2$ points"? Cause the matrix with rows $(010, 100, 001)$ is a matrix with det = 1 fixing only one point in the unit sphere... –  Matt N. Apr 8 '13 at 9:14
@MattN. It has determinant $-1$. –  Cocopuffs Apr 8 '13 at 9:21
Idea: such a $\chi$ is a rotation in some 2D subspace $V$ (a plane) of $\mathbb{R}^3$. The fixed points are the intersection of the normal to that plane with $S$. –  Cocopuffs Apr 8 '13 at 9:25
@MattN. That's a reflection. It fixes an entire great circle of the unit sphere. –  Chris Eagle Apr 8 '13 at 9:36
Duh. Thanks. Had a moment of idiocy there. –  Matt N. Apr 8 '13 at 9:44
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2 Answers 2

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  1. First, convince yourself that $\operatorname{det}(\chi - I_3) = 0$ (hint: $\operatorname{det}(\chi - I_3) = \operatorname{det}(\chi(I_3 - \chi^T)) = \ldots$). Thus, $\dim \ker(\chi - I_3) \geq 1$, and hence $\ker(\chi - I_3)$ contains a $1$-dimensional subspace, i.e., a line through the origin, which intersects the unit sphere in two points $v_1$ and $v_2$. Thus, $\chi$ fixes at least two points.
  2. Now, suppose by contradiction that $\chi$ fixes a third point $v_3 \in S^2$, which therefore cannot line on the same line through the origin as $v_1$ and $v_2 = -v_1$. Then $\{v_1,v_2\}$ span a $2$-dimensional subspace of $\ker(\chi - I_3)$ (why?), so that $\dim \ker(\chi - I_3) \geq 2$; if $\dim \ker(\chi - I_3) = 3$ then $\chi = I_3$, which is a contradiction, so suppose instead that $\dim \ker(\chi - I_3) = 2$. Now, let $x \in \ker(\chi - I_3)^\perp$ be non-zero, so that $x$ spans $\ker(\chi - I_3)^\perp$; check that$\chi x \in \ker(\chi - I_3)^T$ (hint: for $v \in \ker(\chi - I_3)$, $v = \chi^T \chi v = \chi^T v$). Hence, $\chi x = \lambda x$ for some $\lambda \in \mathbb{R}$, so that $\chi$ is diagonalisable with eigenvalues $1$ (with multiplicity $2$) and $\lambda$ (with multiplicity $1$). Why does this then give a contradiction?

What will come out of this is that $\ker(\chi - I_3)$ is indeed $1$-dimensional, giving the axis of rotation for $\chi$; one can then show that $\chi$ restricted to $\ker(\chi - I_3)^\perp \cong \mathbb{R}^2$ is indeed a non-trivial rotation.

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An orthogonal transformation with $\det(1)$ is a rotation. The only fixed points are the one on the rotation axis. And the intersection of the rotation axis and the unit sphere are exacty two points.

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