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Let $u = (u_1,u_2,u_3)$, $v = (v_1,v_2,v_3)$ and

$$X = \{(u,v) \in \mathbb{R^3} \times \mathbb{R^3} \mid u_1^2+u_2^2+u_3^3=1, v_1^2+v_2^2-v_3^2=1, u \cdot v=0 \}$$

Then, is $X$ a smooth manifold?

What I have in mind is try to apply regular value theorem.

So, let $F:\mathbb{R^6} \rightarrow \mathbb{R^3}$ define by

$$F(u,v) = (u_1^2+u_2^2+u_3^3,v_1^2+v_2^2-v_3^2,u_1v_1+u_2v_2+u_3v_3)$$

$$DF = \left( \begin{array}{ccc} 2u_1 & 2u_2 & 2u_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2v_1 & 2v_2 & -2v_3 \\ v_1&v_2&v_3&u_1&u_2&u_3 \end{array} \right) $$

I know that $F(u',v')$ will be a regular value when $DF|_{(u',v')}$ has full rank (i.e. $\text{rank}(DF) = 3$). But I don't know how to show that. Isn't as long as none of the row has all zero entries, the matrix will have full rank? (Which mean as long as $u \neq 0$ or $v \neq 0$)

Michael Hardy
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SamC
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    Please note my edit. ${(u,v)\in\mathbb R^3\times\mathbb R^3\mid u_1^2+\cdots\cdots$ looks different from ${(u,v)\in\mathbb R^3\times\mathbb R^3 | u_1^2+\cdots\cdots$. Seeing people use the latter incorrect form every day here, I might someday begin to suspect, far-fetched though is seems, that the difference is to some people not conspicuous. ${}\qquad{}$ – Michael Hardy Mar 28 '15 at 13:49
  • Thanks for the edit. I'll keep in mind that those two lines are different. – SamC Mar 28 '15 at 14:05

2 Answers2

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For the matrix to have full rank, it is necessary that no row have all zero entries, but ordinarily that is not sufficient. For a $3 \times 6$ matrix like this one, what is necessary and sufficient is that the rows be linearly independent, meaning that if $c_1,c_2,c_3$ are constants and if $$c_1 (\text{Row 1}) + c_2 (\text{Row 2}) + c_3 (\text{Row 3})=(0,0,0,0,0,0) $$ then $c_1=c_2=c_3=0$.

So can you show this to be true (using, of course, the assumption that $F(u,v)=(1,1,0)$ )?

Lee Mosher
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$Y: = \{ (u,v)\in \mathbb{R}^3\times \mathbb{R}^3 \mid |u|=1,\ v_1^2+v_2^2-v_3^2 =1\} $

Clearly $ Y$ is a smooth manifold, i.e., $S^2(1)$-bundle over $ H:= \{ v_1^2+v_2^2-v_3^2 =1 \}$.

$ v\in H\Rightarrow v\neq 0$ That is, condition $u\cdot v=0$ implies that $X$ is $S^1(1)$-bundle over $H$.

Michael Hardy
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HK Lee
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