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Given a regular surface $\Sigma\in\mathbb{R}^4$ by a parametrisation of four coordinates $(x^1(x,y), x^2(x,y), x^3(x,y), x^4(x,y))$ how one can find two normal vectors to $\Sigma$?

For me, it's clear how to find a normal in 3D (by cross product). But for 4D my only idea is to find an orthogonal plane to the tangent one to $\Sigma$.

Please, help me!

Denis D. Bavrin
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    Your only idea is correct. Finding the orthogonal complement is straight-forward. The alternative is to guess two vectors not in the tangent plane and use Gram-Schmidt. I personally prefer the former. – Ted Shifrin Jul 19 '23 at 15:06
  • Does the section "Finding A Perpendicular Vector" in this answer answer your question? – robjohn Jul 21 '23 at 05:14

1 Answers1

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Here is the standard algorithm for such a problem:

Write down a basis for your tangent space. In this case, at the point parametrized by $(x,y)$, the basis is given by the column space of the matrix: $$ A = \begin{pmatrix} \partial x_1/\partial x & \partial x_1/\partial y \\ \partial x_2/\partial x & \partial x_2/\partial y \\ \partial x_3/\partial x & \partial x_3/\partial y \\ \partial x_4/\partial x & \partial x_4/\partial y \\ \end{pmatrix} $$ The orthogonal complement is the null space of the matrix $A^T$.

To compute the null space of a matrix, convert $A^T$ to reduced row echelon form $E$ (row operations will not change the null space). In this case, you will have two non-zero rows in the row echelon form.

It is then easy to solve for the null space - you have to solve the equation $$ E\vec x = 0. $$

Amritanshu Prasad
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