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Let $a > b > 0$. Consider the torus $\mathbb{T}^2$ obtained by rotating the circle of radius $b$ centered at $(a, 0, 0)$ (lying in the $xz$ plane) around the $z$ axis. It can be parametrized by local charts involving the restriction of the map $(θ, ϕ) → ((a + b \cos ϕ) \cos θ,(a + b \cos ϕ) \sin θ, b \sin ϕ)$ to squares $I × J$ with $|I|, |J| < 2π$. This is same as $z^2 = 4b^2 − (\sqrt{x^2 + y^2}−(a − b))^2$

Compute the Riemannian metric induced on $\mathbb{T^2}$ from $\mathbb{R}^3$ in the local coordinates $(θ, ϕ)$.

The only thing that I know is Riemannian metric on a manifold $M$ is an inner product on $T_pM$ such that for each chart $(U,x)$ on $M$ the functions $g_{ij}=\langle\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\rangle$ are differentiable on $U$. This definition is really abstract and it does make any sense to me to apply it here. Could somebody please tell me how to find that inner product here?

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  • $\begingroup$ Just substitute $x,y,z$ in metric $g=dx\otimes dx+dy\otimes dy+dz\otimes dz$ according your parameterization. $\endgroup$ – Canis Lupus Oct 1 '16 at 16:55
  • $\begingroup$ What is $dx\otimes dx$? $\endgroup$ – Extremal Oct 1 '16 at 16:56
  • $\begingroup$ I don't know your background. But it seems to be basics: do you know that given local coordinates $x_i$ every bilinear form (twice covariant tensor) can be written as $\sum_{i,j}g_{ij}\,dx_i\otimes dx_j$ (where $dx_i$ is the differentials of $x_i$)? Physical literature often omits everything besides $g_{ij}$. $\endgroup$ – Canis Lupus Oct 1 '16 at 17:18
  • $\begingroup$ Sorry I dont know. $\endgroup$ – Extremal Oct 1 '16 at 17:38
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So you have $x_1=\theta$ and $x_2=\phi.$ You have a parametrization $$f(\theta,\phi) = \big((a+b\cos\phi)\cos\theta,(a+b\cos\phi)\sin\theta,b\sin\phi\big).$$ The vector field $\dfrac{\partial}{\partial x_i}$ is nothing fancier than $\dfrac{\partial f}{\partial x_i}$, so compute the vectors $\dfrac{\partial f}{\partial\theta}$ and $\dfrac{\partial f}{\partial\phi}$ and calculate the dot products!

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  • $\begingroup$ Ok so I got $\langle\frac{\partial f}{\partial \theta},\frac{\partial f}{\partial \phi}\rangle=0$. Is that the Riemannian metric? $\endgroup$ – Extremal Oct 1 '16 at 16:52
  • $\begingroup$ You must compute all possibilities. $x_1 = \theta$ and $x_2 = \phi$. This is $g_{12} = g_{21}$, but you also need $g_{11}$ and $g_{22}$. $\endgroup$ – Alfred Yerger Oct 1 '16 at 16:56
  • $\begingroup$ Those are the off-diagonal terms. But you need the diagonal terms too!! :) $\endgroup$ – Ted Shifrin Oct 1 '16 at 16:56
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    $\begingroup$ Oh that means the metric is $(g_{ij})$. My teacher didn't tell us it. I think now it makes sense to me. Thanks. $\endgroup$ – Extremal Oct 1 '16 at 16:59

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