# Cone is a submanifold iff it is a vector subspace

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.

• For an idea of where to start, consider what such a cone looks like in $\mathbb{R}^3$: mathworld.wolfram.com/DoubleCone.html Jan 20 '14 at 19:00
• 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
• For yet another hint: think about what happens around 0, and how you can reduce easily to considering a chart around 0 only. 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 $(0,0,0)$. Jan 20 '14 at 19:17
• related Jan 23 '14 at 5:09