Suppose $f$ is a diffeomorphism from the standard two sphere $S^2$ to itself.

Given any three points $a,b,c$ on a geodesic curve on the sphere $S^2$ and $b$ is the middle points of the geodesic arc $\widehat{ac}$,

$f$ satisfies the following two conditions:

(1) $f(a), f(b), f(c)$ are also on a geodesic.

(2) $f(b)$ is the middle points of the geodesic arc $\widehat{f(a)f(c)}$;

I want to know whether $f$ is an isometry on $S^2$ or not.

  • $\begingroup$ Great question. Can you please cite the boss you read to learn about this terminology and particular question? I would like to be able to follow, and catch up. Thanks. :-) $\endgroup$ – Jack Maddington Oct 24 '16 at 16:33

Yes, $f$ is an isometry; you only need to know that $f$ is a homeomorpphism, diffeomorphism is not needed. You first show that (by continuity of $f$) that $f$ sends geodesics to geodesics. From this, you conclude that $f$ is the lift of a projective transformation $g: RP^2\to RP^2$. ($f$ preserves anipodality since antipodal points are intersections of two great circles; then use von Staudt's theorem about characterization of elements of $PGL(3,R)$.) By composing with a random isometry of $S^2$, you can assume that $g$ has three fixed points $x, y, z$ in general position on $RP^2$. If $f$ is not an isometry, you get a contradiction by taking $a, c$ to be lifts of two of the fixed points, say, $x, z$, and $b$ the midpoint of $a, c$ and observing that $$ \lim_{n\to\infty} g^n(p)\in \{x, z\} $$ for every $p$ on the projective line $xz$.

Edit. Here are the details. Let $SL_{\pm}(n,R)$ denote the subgroup of $GL(n,R)$ consisting to matrices with determinant $\pm 1$. This subgroup projects onto $PGL(n,R)$.

Definition. An element $g\in SL_\pm(n,R)$ is a transvection if it is diagonalizable with real positive eigenvalues. A transvection is nontrivial if it is different from the identity matrix.

Examples of transvections used in the proof will come from the following. Let $sl(n,R)= o(n) \oplus {\mathfrak p}$ be the Cartan decomposition of the Lie algebra of $SL_{\pm}(n,R)$; the subspace ${\mathfrak p}$ consists of traceless symmetric matrices. Then $exp({\mathfrak p})$ consists of transactions. (Actually, every transvection is conjugate to one of these.)

Proposition. If $G< SL_\pm(n,R)$ is a closed subgroup strictly containing $O(n)$ then $G$ contains a nontrivial transvection. (Working more one can show that $G=SL_\pm(n,R)$ but I will not need this.)


Lemma. $O(n)$ preserves unique up to scalar multiple positive definite quadratic form, namely the form $q_0$ with the identity Gramm matrix.

Corollary. The normalizer of $O(n)$ in $SL_\pm(n,R)$ equals $O(n)$.

Proof. If $g\in SL_\pm(n,R)$ normalizes $O(n)$, it sends $q_0$ to an $O(n)$-invariant quadratic form, hence, $g^*(q_0)=q_0$, hence $g\in O(n)$. qed

Let $g\in G - O(n)$. Then $g O(n) g^{-1}\ne O(n)$ and is contained in $G$. The subset $g O(n) g^{-1}$is connected and passes through $1\in G$. Hence, the identity component of $1$ in $G$ is strictly larger than $SO(n)$. Since $G$ is a closed subgroup of $SL_\pm(n,R)$, it is a Lie subgroup (by Chevaliey’s theorem). Thus, the Lie algebra ${\mathfrak g}$ of $G$ is strictly larger than $o(n)$, hence, it has nonzero intersection with ${\mathfrak p}$. Taking a nonzero element $\xi$ of this intersection, $\exp(\xi)\in G$ is a nontrivial transvection. qed

Lemma. If $g\in SL(n,R)$ is a transvection, it cannot send midpoints to midpoints in $RP^{n-1}$.

Proof. Let $v_1,…,v_n$ be a basis of eigenvectors of $g$. There exist two vectors in this basis, say, $v_1, v_2$ which correspond to distinct eigenvalues $\lambda_1, \lambda_2$, say, $\lambda_1> \lambda_2$. Let $V\subset R^n$ be the span of $v_1, v_2$ and let $L\subset RP^{n-1}$ be the projectivization of $V$, a projective line in $RP^{n-1}$. Considering the action of $g$ on the projective space, we see that it preserves the line $L$ and has exactly two fixed points in this line, $p_1, p_2$, which are projections of the eigenvectors $v_1, v_2$. Since $\lambda_1> \lambda_2$, the fixed point $p_1$ is attractive and $p_2$ is the reclusive fixed point for the action of $g$ on $L$. Let $p\in L$ denote the midpoint of $p_1, p_2$. Since $g$ is not the identity on $L$, $g(p)\ne p$ (otherwise $g$ has three fixed points in $L$). But this implies that $g$ does not preserve midpoints in $RP^{n-1}$. A contradiction. qed

We conclude that $O(n)$ is the maximal subgroup of $SL_\pm(n,R)$ which sends midpoint to midpoints in $RP^{n-1}$.

| cite | improve this answer | |
  • 1
    $\begingroup$ By Hilbert's characterisation of the projective structure of the real projective plane, (Hartshorne, Foundations of projective geometry, theorem 8.4), we don't need to assume that $f$ is even a homeomorphism. $\endgroup$ – Ben McKay Oct 24 '16 at 14:01
  • $\begingroup$ @BenMcKay: I am worried about the continuity argument in going from midpoints to sending geodesics to geodesics. Once this is established, it indeed suffices to have a bijection (indeed, this is Hartshorne's theorem 8.4, which is von Staudt's fundamental theorem of projective geometry, from 1850, either 1st or 2nd, I always get confused). $\endgroup$ – Misha Oct 24 '16 at 16:23
  • $\begingroup$ You mean $f$ could be, say, some non-continuous lift of the identity of $RP^2$? $\endgroup$ – მამუკა ჯიბლაძე Oct 24 '16 at 17:03
  • $\begingroup$ @მამუკაჯიბლაძე: For instance; but, I am also worried about existence of bijective discontinuous maps of the projective plane to itself which preserve midpoints. $\endgroup$ – Misha Oct 24 '16 at 17:14
  • $\begingroup$ @Misha i am confuse by the two assertions: (1) By composing with a random isometry of $S^2$, $g$ has three fixed point $x, y, z$ in general position on $RP^2$. (2) if $f$ is not isometry, we can get a contradiction, what is the contradiction? Could you explain them in more details? i will appreciate it . $\endgroup$ – Jacob.Z.Lee Oct 25 '16 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.