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!
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 $$ E\vec x = 0. $$