The Metric Tensor

• 22 June 2025

At every point of our coordinate space, we can use the tangent space to define the distance in any (infinitesimal) direction. In general the square of the distance will be a quadratic form in the (infinitesimal) coordinate displacements along the basis vectors of the tangent space.

Quadratic Form

We can represent the quadratic form as a symmetric matrix

For a general tangent space with basis $\{e_i\}$, and (infinitesmal) displacements along those basis vectors $\{dx^i\}$, then the square of the distance is:

$$ds^2=\sum_{ij}g_{ij}{dx}^i {dx}^j$$

where $g=\{g_{ij}\}$ is our matrix and $g_{ij}=e_i\cdot e_j$

and if $x=x^i e_i$ and $y=y^i e_i$, then we can define the dot product

$$x \cdot y = \sum_{ij}g_{i j} x^i y^j$$

which also lets us define the angle $\theta$ between $x$ and $y$ since we can take

$$x\cdot y=|x||y|\cos\theta$$

as a definition.

Law of Cosines

We can figure these out in our examples using the Law of cosines.

If we have our basis vectors $AB$, $AC$ with angle $alpha$ between them, to find the displacement of them together you put them head to tail (doesn't matter which is first), and the side $AD$ is the net displacement.

Consider the following diagram:

If $\angle BAC=\alpha$, then $\angle ACD=\pi-\alpha$.

And since $\cos(\pi-\alpha)=-\cos\alpha$ the law of cosines tells us

$$s^2=a^2+b^2+2ab\cos\alpha$$

Tangent Space for non-skew Sphere

We have calculated:

• $|e^1|=1$
• $|e^2|=\sin\theta$

which give us

$$g=\begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta \end{pmatrix}$$

Tangent Space for skew Sphere

We have calculated:

• $|e^1|=1$
• $|e^2|=\sin\theta$
• $|e^1||e^2|\cos\alpha = \frac{\gamma}{\theta} \sin\phi \sin\theta$

which give us

$$\large g=\def\arraystretch{1.5} \def\A{\frac{\gamma}{\theta} \sin\phi \sin\theta} \begin{pmatrix} 1 & \A \\ \A & \sin^2\theta \end{pmatrix}$$

Next Up

Next up will be "The Dual Basis"

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  The Skew Tangent Space