0
$\begingroup$

I am trying to learn about differentiable manifolds. I came across this example but I am not too sure how to solve it.

Problem:

Consider S a set consisting of a pair ($\textbf{a}, \textbf{b}$) of vectors in $\mathbb{R}^3$ where $\textbf{a}.\textbf{a}=1$, $\textbf{b}.\textbf{b}=1$, $\textbf{a}.\textbf{b}=0$

Check the system has constant rank. Conclude it specifies a smooth manifold. Find the dimension.

I get rank of the system = 3, and dimension 3

Using conditions stipulated in problem on $\textbf{a}$ and $\textbf{b}$

System of equations is:

$f^1(x,y,z,r,s,t) = x^2 + y^2 + z^2 - 1$

$f^2(x,y,z,r,s,t) = r^2 + s^2 + t^2 - 1$

$f^3(x,y,z,r,s,t) = xr + ys + zt$

Calculate Jacobi and see it has constant rank $= 3$

Then by constant rank theorem solutions set S to system of equation has dimension k= 6-3 = 3

Is this correct?

Thanks, J

  • 0
    Where does it have constant rank $3$? Not everywhere... – 2017-01-10
  • 0
    constant rank for all points of the solutions set. – 2017-01-10
  • 1
    Yea that's pretty much it. – 2017-01-10
  • 0
    No problem----- – 2017-01-10

0 Answers 0