# Is this a correct definition of a Riemannian metric tensor?

I am a total beginner in Riemannian geometry but I'm trying to teach myself the basics. So the following could contain many horrible mistakes.

Suppose I describe the x,y-plane by curvilinear coordinates $$q_1$$ and $$q_2$$:

$$q_1: x^2-2-y = 0$$

$$q_2: -x^2+2-y =0$$

Then the normalized tangent basis at each point along $$q_1$$ and $$q_2$$ is

$$\left\{ \frac{1}{1+4x^2} \begin{pmatrix} 1 \\ 2x \end{pmatrix}, \frac{1}{1+4x^2} \begin{pmatrix} 1 \\ -2x \end{pmatrix} \right\}$$

No I simply use the classical dot product on the tangent basis for the metric tensor:

$$g_{12} = \left< \frac{1}{1+4x^2} \begin{pmatrix} 1 \\ 2x \end{pmatrix},\frac{1}{1+4x^2} \begin{pmatrix} 1 \\ -2x \end{pmatrix} \right> = \frac{1-4x^2}{(1+4x^2)^2} = g_{21}$$

And, equivalently $$g_{11} = g_{22} = 1$$.

Is that a correct definition of a Riemannian metric tensor?

• Your $q_1$ and $q_2$ are merely two curves in the plane, not coordinate grids. Do you mean without the $= 0$? But that looks strange as well. Is this an example from an exercise, or just something you came up with? – Arthur Feb 14 at 11:44
• I came up with it. I would kill for exercises with solutions but I cannot find anything online. – AlphaOmega Feb 14 at 11:58
• Fair enough. But your coordinate system still needs some work before it becomes, you know, and actual coordinate system. Why not stick to the known coordinate systems, like polar, or elliptical? – Arthur Feb 14 at 12:00
• Ok yeah that's a good point. So the elliptic coordinates are $x = a\cosh\mu \cos\nu$, $y = a\sinh\mu \sin \nu$. Now I have no idea where to go from here. I've seen the metric tensor defined as $g_{ij}=g_{ij}\left( \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right)$. As far as I was able to understand $\frac{\partial}{\partial x^i}$ is the $ith$ tangent vector, right? How would I compute that? – AlphaOmega Feb 14 at 14:23