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Suppose a continuous map $f : S^2 \rightarrow S^2$ verifies this property : $ \forall x \in S^2, f(x) - f(-x) = 2\left(f(x),x\right)x $ An equivalent expression could be : $f(x) = X(x) + u(x)x$ where $X$ is a tangent vector field on the sphere, $u$ a 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 $-1$. The question is : can we find solutions $f$ of other degree ? I particularly want to know if there exist solutions of degree $0$.

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    Is $(f(x),x)$ the dot/scalar product of $f(x)$ and $x$, as vectors in $\mathbb R^3$?2011-09-12

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Yes, it is. We can also see this symmetry such as : $ f(-x) = S_x . f(x) $ Where $S_x$ denotes the orthogonal symmetry with respect to the orthogonal plan to $x$ : $S_x(y) = y - 2(y,x)x$