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How to solve the following:

Let $K\subset \mathbb{R}^3$ be a convex, compact set with smooth boundary $C=\partial K$ and let $\vec{u}$ be any vector. Show that there exist points $x\neq y$, $x,y\in C$ such that vector $\vec{xy}$ is colinear with $\vec{u}$ and tangent planes $T_{x}C$, $T_{y}C$ are parallel.

Any hint would be helpful, because I don't have any idea how to start. Thanks in advance.

Seirios
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alans
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    @wordsthatendinGRY: I don't see why that should be the case...For the OP: this is an interesting question! Is this for a particular class? I'm curious as to what tools you have at your disposal. – Phillip Andreae May 30 '14 at 19:18
  • @PhillipAndreae Oh, I overlooked the constraint on $\vec u$. –  May 30 '14 at 19:19
  • @PhillipAndreae This problem was for my topology class, but I didn't know how to prove it. – alans May 30 '14 at 20:37

1 Answers1

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It seems I find solution. Let $\phi: C \rightarrow C$ is the involution, which maps point x to its opposite point y by vector u.

Let's consider the following vector field on C: $$v(x) = pr_{T_x} [u, n(x) + n(\phi(x))]$$ where n(x) is the normal vector for C in point x with length one. By the Hairy ball theorem it's exists the point x for which $v(x)=0$. This is the required point.

To proof this we need to check that if $v(x) = pr_{T_x} [u, n(x) + n(\phi(x))]=0$ then $n(x) + n(\phi(x)) = 0$.

Let $a = n(x)$, $b = n(\phi(x))$.

Suppose that $pr_{T_x}[u, a+b] = 0$. Then vector $v = [u, a+b]$ is parallel to vector a, and if v is nonzero then vector a is orthogonal to vector u, because v is orthogonal to u. But if vector a is orthogonal to u then u belongs to the tangent space $T_x$ of surface C. In this case a = b and $pr_{T_x}[u, a+b] \neq 0$. We have contradiction. Thus $v = [u, a+b] = 0$, so $a + b = 0$.

DenisMath
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    What do you mean by "opposite point y by vector u"? I also do not understand the notation $[u, n(x) + ...]$. Do you mean "cross product"? – Moishe Kohan May 30 '14 at 21:26
  • Yes, of course, it`s cross product. "Opposite point y by vector u" means that one takes point x on the surface C and draw a straight line L passing through point x which is parallel to vector u. $phi(x)$ is the intersection of L with the surface. – DenisMath May 30 '14 at 21:32
  • So you mean to say that $\phi$ is an involution, not that it is idempotent? –  May 30 '14 at 21:35
  • Yes, of course, it's an involution :) I am edit text. – DenisMath May 30 '14 at 21:37
  • The implication "Suppose that $pr_{T_x}[u,a+b]=0$. Then vector $v=[u,a+b]$ is parallel to vector $a$" is false for general vectors $a$ and $b$. As a general suggestion, you should not rush but proofread what you write before posting. – Moishe Kohan May 30 '14 at 21:51
  • Ok, thank you for suggestion! But previously I define vector a as n(x), so $pr_{T_x}=pr_a$ is a projection on the tangent plane which has normal vector a. – DenisMath May 30 '14 at 22:00
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    Oh, I did not see it. Then the question is why $v=0$ implies $a+b=0$? All what I see is that $a+b$ is collinear to $u$. – Moishe Kohan May 30 '14 at 22:27
  • It's a very good question! First, $|a|=|b|=1$. So if $[u,a+b]=0$ then $pr_u(a+b)=0$ where $pr_u$ is a projection to a plane which is orthogonal to vector u, and we have exactly two variants to restore a and b from their projections. But it`s permissible only one of them because normal vectors a and b look outside the body. – DenisMath May 30 '14 at 22:38
  • I have no idea what the last argument meant. I think, one has to use convexity to rule out this possibility. – Moishe Kohan May 30 '14 at 23:39
  • The body is the convex, compact set K. – DenisMath May 31 '14 at 07:51
  • @Denis: If you need help completing your argument, let me know. Right now, it is in limbo: Neither correct, nor false (but incomplete). – Moishe Kohan May 31 '14 at 20:18
  • Let's consider convex compact "body" K and two points $x$ and $y=\phi(x)$ on its surface C. Segment $xy$ is parallel to vector u. For simplicity suppose that $u=xy$. Vectors $a$ and $b$ are outer normal vectors to body K in points $x$ and $y$ correspondingly, $|a|=|b|=1$. Proof that if $[u,a+b]=0$ then $a+b = 0$. – DenisMath Jun 01 '14 at 10:15
  • Let $a=a_1 + a_2$, $b=b_1+b_2$, where $a_2$ and $b_2$ are orthogonal to u, $a_1$ and $b_1$ are parallel to u. From $[u,a+b]=0$ we have that $a_2+b_2=0$. Hence $|a_1|^2 = 1 - |a_2|^2 = 1 - |b_2|^2 = |b_1|^2 \Rightarrow |a_1|=|b_1|$. Vectors $a_1$ and $b_1$ are parallel to u. So there are only two variants: $a_1 = b_1$ or $a_1 = - b_1$. But because $a$ and $b$ are outer normal vectors to body K we have that $a_1 = - b_1$. Thus, $a+b=0$. – DenisMath Jun 01 '14 at 10:16