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Given $Q = [0,1]^3$ in $\mathbb{R}^3$, how can we prove that $\partial Q$ is not smooth manifold in $\mathbb{R}^3$?

I can understand that its not, because of the connection line between $2$ sides of the cube, but what is a formal explanation for this ?

Idan Perez
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    focus on a point of some edge of the cube; there's no neighborhood of it such that here the boundary is the graph of a smooth function – 18cyclotomic Jun 17 '17 at 19:12
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    Let $f:R^2\to\partial Q\to R^3$ be the composition of a chart at one of those points with the inclusion map. This function should be smooth. Take a path $\gamma:[-1,1]\to R^2$ such that its image crossed from one facet to the other and $f\circ\gamma(0)$ is the bad point. Compute the tangent of the curve $f\circ\gamma$ at 0. Using limit form the left it points in one direction, from the right it should've point in another direction. So, there is no tangent. But there should be if it were a submanifold. – OR. Jun 17 '17 at 19:16
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    I think they need to be more specific and say that it is not a submanifold. Right? It seems to me that one can put on it a differentiable structure such that on its own it is a smooth manifold. – OR. Jun 17 '17 at 19:18
  • yes you are right, not as submanifold, I will mention that, thanks – Idan Perez Jun 17 '17 at 19:21
  • @Mla Agreed. It's homeomorphic to a sphere, you can simply transport the differentiable structure. – Matt Samuel Jun 17 '17 at 19:22

1 Answers1

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One way is to focus on one vertex of the cube, say $(0,0,0)$. There are three smooth paths starting there and lying on the boundary of the cube - the edges ( almost any parametrization should work) - whose tangent vectors at the origin are linearly independent. If this were a submanifold these vectors should lie in the tangent subspace ( of dimension $2$) at $(0,0,0)$.

Other were good ideas are in the comments above, as in fact the boundary of the cube is not a submanifold around any point on the edge.

orangeskid
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  • why so ? $\gamma:[-\delta,\delta]\to R^3$ such that $\gamma(x) = 0$ where $x \leq 0$ and lets say $\gamma(x) = (x, 0, 0)$ where x > $0$

    $\gamma'(0)$ does not exist, from left its $(0, 0, 0)$ and from right its $(1, 0, 0)$ therefor its doesn't exist @orangeskid

     – Idan Perez Jun 23 '17 at 17:45
  • @IdaPerez : Indeed, but you can take smooth paths as indicated above. – orangeskid Jun 23 '17 at 18:44
  • sorry so i do nut understand how.. do you mean to take a neighborhood of (0,0,0) such that x1 = (t1, 0, 0) x2 = (0, t2, 0), and x3 =(0, 0, t3 ) in it and for every one of them there is the tangent space

    than we get (1,0,0) $\in Tx1M $ and such and basically every one of them should be in Ker(F) but Ker(F) = 2 ( Where {F=0} = M $\bigcap$ U and such) and we got 3 independence vectors ? we are not looking on the tangent space of a single point, hence on vector in tangent space of different point at the same neighborhood

     – Idan Perez Jun 23 '17 at 19:00
  • @IdanPerez: We look at the (hypothetical) tangent space at $(0,0,0)$. – orangeskid Jun 23 '17 at 20:55
  • sorry i cant understand how do you find 3 independence vectors in the tangent space of (0, 0, 0).. i cant find even one, how you can find (hypotheticaly) 3 ?? – Idan Perez Jun 23 '17 at 21:22