3
votes
2answers
166 views

A non orientable closed surface cannot be embedded into $\mathbb{R}^3$

Can someone please remind me how this goes? Here's the idea of proof I'm trying to recall: let $S$ be a closed surface (connected, compact, without boundary) embedded in $\mathbb{R}^3$. Then one can ...
3
votes
2answers
73 views

Algebraic surface as a smooth manifold

Let $S$ be the set of points $x=(x_1,x_2,\ldots,x_9)\in \mathbb{R}^9$ which satisfy the following conditions: $$x_1^{2}+x_2^{2}+x_3^{2}=x_4^{2}+x_5^{2}+x_6^{2}=x_7^{2}+x_8^{2}+x_9^{2}=1$$ ...