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I'm going over some GR from more of a differential geometry perspective and had a quick question about a simple calculation - my differential geometry background isn't too strong so I apologise if any of the terminology is incorrect, but I'd be grateful for any clarification.

I'm following an example in Appendix A of Sean Carroll's Introduction to GR, where there is a map $\phi:M \to N$, given by $$ \phi(\theta,\phi) = (\sin\theta \cos \phi , \sin \theta \sin \phi, \cos \phi) , $$ and $M=S^2$ is a submanifold of $N =\mathbb{R}^3$ (i.e. the two-sphere embedded in $\mathbb{R}^3$). The coordinates on the manifolds are $x^{\mu} = (\theta,\phi)$ on $M$, and $y^{\alpha} = (x,y,z)$ on $N$. The induced metric on $M$ is just the pullback of the flat-space metric $\phi^*g$, which is given by the formula

$$ (\phi^*g)_{\mu \nu} = \frac{\partial y^{\alpha} }{\partial x^{\mu}} \frac{\partial y^{\beta} }{\partial x^{\nu}} g_{\alpha \beta}. $$
I understand how to calculate the individual Jacobian matrices of partial derivatives $\frac{\partial y^{\alpha} }{\partial x^{\mu}}$ (e.g. just using $y^1 = \sin\theta \cos \phi $, $y^2 = \sin \theta \sin \phi$ and $y^3 = \cos \phi$ as defined by $\phi$), however I was confused as to how to treat the full expression above.

The Jacobian matrices are $2 \times 3$ matrices, so the second has to be transposed to be a $3 \times 2$ matrix in order to give the required $2 \times 2$ metric $g_{\mu \nu}$. My question is, in the pullback equation above, how does the metric $g_{\alpha \beta}$ act on the Jacobian $\frac{\partial y^{\beta} }{\partial x^{\mu}}$, and how should I be writing this down? I can see that $y^{\beta}$ should be replaced with $y^{\alpha}$, but should any indices be lowered, and how should I interpret the metric tensor transposing the Jacobian matrix?


Edit - Ted Shriffin's comments are correct, the last component of the map should be $\cos \theta$ not $\cos \phi$, and all my matrices should be transposed.

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    $\begingroup$ Be careful. You have a typo in your definition of the mapping — and please don't use $\phi$ for both the mapping and one of the coordinates :) Actually, the mapping should be written as a vector (not a row), and the Jacobian will be $3\times 2$. $\endgroup$ Commented Aug 8, 2019 at 18:41
  • $\begingroup$ I agree having $\phi$ as the map and in the coords is a bit confusing but I was copying exactly from the Carroll textbook incase readers were familiar with it. Could you point out the typo? $\endgroup$
    – Eletie
    Commented Aug 8, 2019 at 18:52
  • $\begingroup$ As for the Jacobian $\frac{\partial y^{\alpha} }{\partial x^{\mu}}$, it is written explicitly (in the textbook) as a $2 \times 3$ matrix, so maybe I am confused as to what you mean? $\endgroup$
    – Eletie
    Commented Aug 8, 2019 at 18:59
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    $\begingroup$ The last coordinate needs to be $\cos\theta$. The mapping goes from $\Bbb R^2$ to $\Bbb R^3$, so the derivative matrix is $3\times 2$. This is enough sloppiness for me. I would find another text. $\endgroup$ Commented Aug 8, 2019 at 21:01
  • $\begingroup$ @Eletie Hi, I am studying your question now. Cam you please tell me what is $\frac{\partial y^{\beta} }{\partial x^{\mu}}$ explicitly? Moreover I want to know what is $y^{\beta}$ and $x^{\mu}$ in this case? $\endgroup$
    – user886636
    Commented Mar 6, 2021 at 7:26

1 Answer 1

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Seeing as I'm getting new comments on this old question I thought I'd post the answer myself. Thanks to Ted Shifrin's help in the comments and also pointing out the typo in the mapping: the map $\Phi:M \to N$ should of course be $$ \Phi(\stackrel{x^{\mu} \ \rm{coords}}{\theta \, , \, \phi})^T = (\stackrel{y^\alpha \ \rm{coords}}{\sin\theta \cos \phi , \sin \theta \sin \phi, \cos \theta})^T \ , $$ where the transpose makes clear these are vectors not rows. I've also made clear the $x^{\mu}$ and $y^{\alpha}$ coordinates on $S^2$ and $\mathbb{R}^3$ respectively. My mistake came from being unfamiliar with using matrix notation with tensors (it's been a while since I've done linear algebra). The final matrix of course must've been $2 \times 2$ because the indices $\alpha \beta$ were summed over whilst $\mu = \{0,1\}$. Similarly, all the expressions I had should have been transposed, but in the textbook they weren't written this way, probably to save space. e.g. the Jacobian was written as a $2 \times 3$ matrix but should have been $3 \times 2$, $$ \frac{\partial y^\alpha}{\partial x^{\mu}} = \begin{pmatrix} \frac{\partial y^1}{\partial x^1} & \frac{\partial y^1}{\partial x^2} \\ \frac{\partial y^2}{\partial x^1} & \frac{\partial y^2}{\partial x^2} \\ \frac{\partial y^3}{\partial x^1} & \frac{\partial y^3}{\partial x^2} \end{pmatrix} = \begin{pmatrix} \cos \theta \cos \phi & -\sin \theta \sin \phi \\ \cos \theta \sin \phi & \sin \theta \cos \phi \\ -\sin \theta & 0 \end{pmatrix} \ . $$

Then the induced metric can be written as $$ \begin{align} (\Phi^* g)_{\mu \nu} &= \frac{\partial y^{\alpha}}{\partial x^{\mu}} \frac{\partial y^{\beta}}{\partial x^{\nu}} g_{\alpha \beta} \\ &= \frac{\partial y^{\alpha}}{\partial x^{\mu}} \frac{\partial y^{\beta}}{\partial x^{\nu}} \delta_{\alpha \beta} \ , \end{align} $$ with the Kronecker delta from the flat Euclidean metric on $\mathbb{R}^3$. Here we don't just use matrix multiplication as I originally thought (this wouldn't make sense anyway). We can find the components of the induced metric by summing over the repeated index, $$ (\Phi^* g)_{\mu \nu} = \sum_{\alpha = 1}^3 \frac{\partial y^{\alpha}}{\partial x^{\mu}} \frac{\partial y^{\alpha}}{\partial x^{\nu}} . $$ E.g. the $\mu \nu = 0 0$ component is just $$ (\cos \theta \cos \phi)^2 + (\cos \theta \sin \phi)^2 + (- \sin \theta)^2 = 1 \ , $$ and repeat for the others. One then thankfully finds the induced metric to be $$ g_{\mu \nu}(x) = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2 \theta \\ \end{pmatrix} \ . $$

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  • $\begingroup$ This is very helpful. $\endgroup$
    – Elio Li
    Commented Jan 24 at 11:12

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