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I have a smooth mapping $v \colon [0,1] \to S^{n-1}$ such that for any $u \in S^{n-1}$ exists $t \in [0,1]: v(t)\cdot u = 0$ and $n \geq 3$. So a have an assumption that such a mapping $v(\cdot)$ doesn't exist.


I tried to consider a function $f(t,s) = v(t)\cdot v(s)$, then I selected a function $s(t)$: $v(t)\cdot v(s(t)) = 0$ for any $t$ and I tried to show that $s(t)$ can be choosen to be contunious. If it is true, then a mapping $s(t)$ has a fixed point and then $|v(t)|^2=0$ for some $t$. Maybe contunious $s(t)$ doesn't exists at all.

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    What does this mean: "So a have an assumption that such a mapping $v(\cdot)$ doesn't exist"?2011-11-24
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    I think that function $v(\cdot)$ with specified properties doesn't exist.2011-11-24
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    So your question is this: Prove that there does not exist a smooth mapping $v:[0,1]\rightarrow\mathbb{S}^{n-1}$ where $n\geq 3$ with the following property: for any $u\in\mathbb{S}^{n-1}$ there exists $t\in[0,1]$ such that $v(t)\cdot u=0$. Moreover, $v(t)\cdot u=0$ means the dot product in $\mathbb{R}^n$ by considering $\mathbb{S}^{n-1}\subset\mathbb{R}^n$. Am I correct?2011-11-24
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    You are correct :)2011-11-24
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    So yet another translation is: prove that there are no smooth mappings $v : [0,1] \to S^{n-1}$ ($n\geq 3$) intersecting every great circle. I must say I find that rather unlikely.2011-11-24
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    For $n=3$, any smooth parametrisation of the equator is such a mapping.2011-11-24
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    And unless I'm mistaken, if you consider the equator $E \subset S^2$ as a smooth curve in $S^{n-1}$ via the standard inclusion $S^2 \subset S^{n-1}$, it still works for every $n \geq 3$.2011-11-24
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    Yes, it works. Thank you.2011-11-24

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