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antipodal
antipodal, a. and n. (ænˈtɪpədəl) [f. antipod-es + -al1.] 1. Of or pertaining to the antipodes; situated on the opposite side of the globe.1646 Sir T. Browne Pseud. Ep. 306 The Americans are Antipodall unto the Indians. 1831 Carlyle Sart. Res. (1858) 46 The antipodal New Holland. 1877 Shields Final ...
Oxford English Dictionary
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Antipodal point
a pair of antipodal points (unlike the situation for any non-antipodal pair of points, which have a unique great circle passing through both). The antipodal map sends every point on the sphere to its antipodal point.
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en.wikipedia.org
Antipodal mapping from $S^n$ to $S^{n-1}$ Let $S^n$ denote the unit sphere in $\mathbb{R}^{n+1}$. An _antipodal mapping_ $f:A\rightarrow B$ is a continuous function such that $f(-x)=-f(x)$ for all $x\in A$. > An anti...
I would presume that "nowhere zero" simply means that the function does not take 0 as a value; i.e., for every $x\in S^n$, we have $f(x)\neq 0$. This is true here since $f(x)\in S^{n-1}$ and $0\notin S^{n-1}$.
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Show that antipodal points remain antipodal under any isometry of $S^2$ "The antipodal map of $S^2$ is the isometry $\bar m$ sending each point $P$ to its opposite point $-P$" (from textbook) Show that antipodal poin...
If $P\in S^2$, $-P$ is the only point $P'$ of $S^2$ such that $d(P,P')=2$. So, since isometries preserve distance and since $d(P,-P)=2$, $d\left(\overline m(P),\overline m(-P)\right)=2$ and therefore $\overline m(P)=\overline m(-P)$.
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Antipodal points of sphere Whenever $S^2$ is the union of three closed subsets $A_1$, $A_2$, and $A_3$, then at least one of these sets must contain a pair of antipodal points {${x,-x}$} in $S^{2}$ This is homework f...
Construct a function $f \colon S^2 \to \mathbb{R}^2$ by setting $f(x) = (d(x,A_1), d(x,A_2))$ where $d$ is the standard metric on $S^2$. Apply the Borsuk-Ulam theorem to this function. Details of the general case (the Lusternik–Schnirelmann theorem) can be found in Armstrong's book.
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Calculating an antipodal point on a sphere Consider a 3-sphere at the origin $(0,0,0)$. Let $s_1 = (x,y,z)$ be a point somewhere on the surface of the sphere. How do I calculate the antipodal point $s_1^* = (x^*,y^*,z...
For a sphere centered at the origin, the antipodal point is the one you obtain by negating all coordinates: $s_1*=(-x,-y,-z)$.
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What does it mean to say that a pair of points are antipodal in a topological sphere? A pair of points are antipodal if they are diametrically opposite to each other. This definition makes perfect sense when one think...
The meaning of "antipodal" depends on a choice of homeomorphism between an ellipsoid (or any other subset of $\mathbb{R}^n$ homeomorphic to a sphere) and
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External differential commutes with antipodal map Suppose we have a closed form $d\omega=0$ on $S^{n}$, and antipodal map $i: S^{n} \to S^{n}$ i.e $i : x \to -x$. How to see that the external differential commutes wit...
This is nothing special about $S^n$ or $i$. Let $M$, $N$ be smooth manifolds. Given any smooth map $f : M \to N$, there is an induced map $f^* : \Omega^*(N) \to \Omega^*(M)$ which respects the grading, i.e. $f^* : \Omega^k(N) \to \Omega^k(M)$ for every $k$. Most books on differential geometry would ...
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The degree of antipodal map. I am trying to solve the problem A map without fixed points - two wrong approaches. But I am not certain about the degree of antipodal map. I my thought, since the preimage of a point $y ...
The antipodal map on $S^k$ can be written as the composition of $k+1$ reflections through hyperplanes.
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Antipodal mapping $f:S^n\to S^1$ I know that " There does not exist an antipodal mapping $f:S^{n}\to S^{n-1}$" by Borsuk-Ulam. But, ¿There does not exist an antipodal mapping $f:S^{n}\to S^{1}$ (continous)?
No, embed $S^1 \hookrightarrow S^{m-1}$ equatorially so that $(x_1,x_2) \mapsto (x_1, x_2,0,\dots,0) \in S^{m-1}$ and note that such a map would provide an odd map $S^n \to S^{m-1}$. I guess if you wanted to do this from scratch: Take $f:S^n \to S^1$ an odd map. So, we can embed $i:S^1 \hookrightarr...
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Prove that if $S^k$ has a non vanishing vector field, then its antipodal map is homotopy to the identity if $k$ is even. Prove that if $S^k$ has a non vanishing vector field, then its antipodal map is homotopy to the ...
Now the homotopy between $I$ and the antipodal map of $S^n$ is the exact homotopy of those of $S^1$: Move the antipodal point $-x$ along the chosen half
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Prove that for $S^{n}$ (an n-sphere) and given antipodal map A, A is homotopic to the identity on $S^{n}$ Define an antipodal map A such that $$A: S^{n} \rightarrow S^{n}$$ by $A(x_{1},...,x_{n+1})=(-x_{1},...,-x_{n+1...
Putting these together, and viewing the antipodal map as $n+1$ reflections, we see that $deg(A)=(-1)^{n+1}$.
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Why to the points have to be non-antipodal? When coming across these two definitions(see below) I notice that it specifies that the points have to be non-antipodal points so I was wondering why the points have to be n...
But if $A$ and $C$ are antipodal, the two arcs have the same length, so the definition would still be ambiguous.
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Prove that any map $f\colon S^1 \to S^1$ mapping antipodal point to antipodal point has $\deg_2(f)=1$ by a direct computation. Without using the Borsuk-Ulam theorem. Prove that any map $f\colon S^1 \to S^1$ mapping an...
Use the $g$ you already have from the previous exercise. What do you know about $f((-1,0))$ compared to $f((1,0))$? What does this tell you about $g$? **EDIT** Further hint: First, \begin{align*} (\cos(g(t+\pi)),\sin(g(t+\pi)))&=f(\cos(t+\pi),\sin(t+\pi)) = f(-\cos t,-\sin t) = -f(\cos t,\sin t)\\\ ...
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How to show that a map without fix point from annular region to annular region is homotopic to antipodal map $\Omega=\\{x\in R^3: 1\le||x||\le2\\}$ If $L:\Omega\rightarrow \Omega $ is continuous and without fix point...
The homology of $\Omega$ is $H_0(\Omega,\Bbb{Q})=H_2(\Omega,\Bbb{Q})=\Bbb{Q}$ and $H_i(\Omega,\Bbb{Q})=0$ if $i\ne 0,2$. The Lefschetz formula tells you that a map f without fixpoints necessarily $0=Tr(H_0(f))+Tr(H_2(f)=H_0(f)+H_2(f)$. But $H_0(f)=H_0(id)=1$ for all $f$, hence in our case $H_2(f)=-1...
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