# Induced metric on $S^2$ from pullback of metric on $\mathbb{R}^3$

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.

• 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 $3\times 2$. Commented Aug 8, 2019 at 18:41
• 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? Commented Aug 8, 2019 at 18:52
• 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? Commented Aug 8, 2019 at 18:59
• 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. Commented Aug 8, 2019 at 21:01
• @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?
– user886636
Commented Mar 6, 2021 at 7:26

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 $$g_{\mu \nu}(x) = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2 \theta \\ \end{pmatrix} \ .$$