# Degree of maps on the sphere with a property of symmetry

Suppose a continuous map $f : S^2 \rightarrow S^2$ verifies : $$f(x) - f(-x) = 2 \left(f(x),x\right)x$$ where $\left(x,y\right)$ is the scalar product in $R^3$. An equivalent way of expressing $f$ could be : $$f(x) = X(x) + u(x)x$$ where $X$ is a tangent vector field of $S^2$ and $u$ a real-valued function with : $$X(-x)=X(x) \hspace{15mm} u(-x) = u(x) \hspace{15mm} \left|X\right|^2+u^2 = 1$$ For instance, the identity and antipodal maps are solutions of degree 1 and -1. My question is :

Do there exist solutions $f$ of degree $0$ ?

Edit : The answer is no, solutions have odd degree. What about the result for dimension n>2 ?

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Any motivation? –  Ben McKay Sep 13 '11 at 9:44

First, for any $x\in S^2$ we have an endomorphism $A(x)$ of $\mathbb{R}^3$ given by $A(x)(w)=x\times w$. More generally, we have an orthogonal matrix $B(t,x)=\exp(t A(x))$, which is a rotation through angle $t$ around $x$. When $w$ is perpendicular to $x$ we have $B(\pi/2,x)(w)=A(x)(w)$.
Let $F$ be the space of maps $f$ as in the question, and let $G$ be the space of maps $g:S^2\to S^2$ satisfying $g(-x)=-g(x)$ for all $x$. Given a function $f(x)=X(x)+u(x)x$, put $\phi(f)(x)=A(x)(X(x))+u(x)x=x\times X(x)+u(x)x$. This gives a homeomorphism $\phi:F\to G$. Using the maps $x\mapsto B(t,x)(X(x))+u(x)x$ (for $0 \leq t\leq \pi/2$) we see that $\phi(f)$ is homotopic to $f$ and so has the same degree. It is fairly standard that maps in $G$ have odd degree, and it follows that maps in $F$ have odd degree.
UPDATE: Romain asks about a generalisation for $n>2$. This seems harder. The analogous space $F$ is the space of sections of the bundle $W=n+2-L$ over $\mathbb{R}P^n$, where $L$ is the tautological bundle. By thinking about exterior algebras, one can produce an isomorphism $2^nL\simeq 2^n$. In the case $n=2$ this gives $4L\simeq 4$ so $W=4-L=4L-L=3L$, and by working out explicit formulae for this identification one is led to the proof above. If $n>2$ then $2^n>n+2$ so this approach will not work without some additional ideas.
Thanks a lot ! It is more elementary than I expected. This method works thanks to the vectorial product in $R^3$, but I failed to find a canonical way to do a quarter turn of $X$ homotopically for all $x$ in higher dimensions. Can we generalize this result to higher dimensions ? –  Romain Sep 13 '11 at 17:45
There is a small typo in line 3 of the second paragraph. it should be $... x\times X(x)$. –  Gjergji Zaimi Sep 13 '11 at 19:06
I am sorry to ask it, but I am not an algebraic topologist. I just guessed that F is the space of sections of a sphere bundle over $RP^n$. I do not understand the terminology. What do "1", "+" and above all "-" stand for in the formula $W=n+2-L$ ? I just need a reference to be able to understand what you said. Thank you again, Neil Strickland, for the time you have dedicated to this question. –  Romain Sep 16 '11 at 20:59