Induced Metric of a 2-Sphere: Why i≠j?

[pdxautodidact]

So, by accident, while deriving the induced metric for a sphere in 3 dimensions I realized that the transpose of the jacobi matrix multiplied by the jacobi matrix (considering it as 3 row/column vectors)will work out the induced metric. Why is it that i≠j ends up being superfluous. One would have X=a 2-sphere in parameterized coordinates, and then g_ij= <X_;i,X_;j>. Thus one would compute <X_;1,X_;2> and the same for 2,3. Is this because the embedded manifold is an immersion, or is there something else? Thanks for any elucidation and best.

 

[Ben Niehoff]

Try a different parametrization and see what happens:

$$\begin{align}x &= \sin \theta \cos (\phi + \theta), \\ y &= \sin \theta \sin (\phi + \theta), \\ z &= \cos \theta.\end{align}$$

 

[pdxautodidact]

Does that, in fact, parameterize a sphere? It's not obvious to me. I included an r term in the coordinates since I want it in general, not the only the unit case. The induced metric was:
$$
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & r^2 sin^2\theta & r^2 sin^2\theta \\
0 & r^2 sin^2\theta & r^2 sin^2\theta
\end{array}
\right)
$$

I haven't had time to write it out again, so I could have made a mistake, but this looks bad to me. The matrix is singular, and thus it can't be a metric, correct? Assuming this is correct, I don't see how the Jacobian will account for i ≠ j. I'll work through it again. Thanks

 

[Ben Niehoff]

The term in the center of your matrix is wrong.

 

[pdxautodidact]

Okay, so I went through it again:

let X be the 2 sphere with your suggested parameterization

$$
X_{;\theta} = \left(
\begin{array}{c} x^1 = rcos\theta cos \left(\phi + \theta \right) - rsin \theta sin \left(\phi + \theta \right) \\
x^2 = rcos \theta sin \left( \phi + \theta \right) + rsin\theta cos \left(\phi + \theta \right) \\
x^3 = -rsin\theta
\end{array}
\right)
$$

When I dot that with itself I got the same answer again. Mathematica, with a bit of tinkering, gave me

$$ r^2 (Cos(\phi)^2 + Cos(2 \theta + \phi)^2 + sin(\theta)^2) $$

What should g_22 be?

 

[Ben Niehoff]

You should end up with

$$r^2 (1 + \sin^2 \theta)$$

Check your algebra again. You shouldn't really need Mathematica, it's pretty easy to do.

 

[pdxautodidact]

Yeah, I combined the trig terms into a 0 instead of a 1, so that's it. I see why it works for all cases now, in hindsight it's obvious. Thanks!