I'm having a hard time understanding some stuff about the first fundamental form. Since I think I have decently understood what the first fundamental form means (inner product restricted to the tangent space of a surface) I think what confuses me is the notation.. Specifically I'm having a hard time understanding how to calculate the angle of two curves on a surface

Let's take this problem as an example

Let $g_{11} = 1, g_{12}=0, g_{22}=4+u^2$ be the coefficients of the first fundamental form of some surface $Σ$ calculate the angle between the curves that lie on the surface given by the equations:

$u+v=0$ and $u-v=0$

So typically a general type of solution for this would be to calculate the tangent vectors of each of their curves at their point of intersection and use their inner product to find their angle. Let's try to do this here:

Let $c_1(t)=(t,t)$ and $c_2(t)=(t,-t)$ be a parametrization of the two curves. These two curves have tangent vectors $x_1 = (1,1) $ and $x_2 = (1,-1)$.

Now let $θ$ be their angle, it can be found as:

$cosθ = \frac{\left \langle x_1,x_2 \right \rangle}{\left \| x_1 \right \| \left \| x_2 \right \|}$

Of course since we are talking about angles on a surface we have to use the first fundamental form. In my textbook this inner product is written as: $\left \langle a,b \right \rangle = g_{ij} a^i b^j$

Can someone please use the example to help me understand this notation?

Thanks in advance!


1 Answer 1


This is Einstein summation notation. You are summing over indices which appear twice, once lowered and once raised. So

$$\langle a, b\rangle = g_{ij}a^ib^j = \sum_{i,j = 1}^ng_{ij}a^ib^j.$$

If $a$, $b$ are the column vectors with components $a^i$ and $b^j$ respectively, then we can also view this sum as computing $a^TGb$ where $G$ is the $n\times n$ matrix with $(i, j)^{\text{th}}$ entry $g_{ij}$. This should be reminiscent of the situation in $\mathbb{R}^n$ where every inner product is of this form for some positive-definite symmetric $n\times n$ matrix. The difference here is that the matrix $G$ is a matrix of functions (indicating that the inner product varies from tangent space to tangent space).

In this case, we see that

$$\langle x_1, x_2\rangle = \begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix} 1 & 0\\\ 0 & 4 + u^2\end{bmatrix}\begin{bmatrix} 1\\ -1\end{bmatrix} = 1 - (4 + u)^2 = -3 - u^2.$$

You can compute similarly to find $\|x_1\|$ and $\|x_2\|$ which you can then use to compute $\theta$. However, you only need to do this at the point where the two curves intersect, namely $u = v = 0$.

  • $\begingroup$ Great thank you so much! It's clear now, I've never seen this notation before glad to know what it's called. Thanks again $\endgroup$
    – Eva K.
    Jan 27, 2021 at 22:24

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