Cone is a submanifold iff it is a vector subspace

Asked Jan 20 '14 at 18:32

Active Jan 20 '14 at 18:56

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Could you tell me how to prove the following?

Let $\emptyset \neq M \subset \mathbb{R}^n $ be a cone ($\forall x \in M, t \in \mathbb{R} : tx \in M $ ), $0 \le d \le n.$

Prove that $M$ is a $d$ -dimensional submanifold class $\mathcal{C}^1 \iff M$ is a $d$ -dimensional subspace of $\mathbb{R}^n $.

I will be grateful for all your help.

Thank you.

edited Jan 20 '14 at 18:56

N. Owad

asked Jan 20 '14 at 18:32

Jerry

For an idea of where to start, consider what such a cone looks like in $\mathbb{R}^3$: http://mathworld.wolfram.com/DoubleCone.html – bradhd Jan 20 '14 at 19:00
2

For another hint, consider what the charts must look like. Is there a single-chart atlas? Why or why not? – Nick Jan 20 '14 at 19:03
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For yet another hint: think about what happens around 0, and how you can reduce easily to considering a chart around 0 only. – David Hornshaw Jan 20 '14 at 19:08
Thank you.I know that $x^2+y^2=z^2$ is a submanifold only is we assume that $z>0$. It is not a submanifold if we include $(0,0,0)$. – Jerry Jan 20 '14 at 19:17
related – you can call me Al Jan 23 '14 at 05:09

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