How can I obtain a normal unitary field to higher dimensional surfaces?

Question

More specifically, say I've got a surface in the form $X(u,v) = (x(u,v), y(u,v),z(u,v),t(u,v))$. The tangent fields are obviously $X_u$ and $X_v$, but I don't know how to give meaning to $X_u \times X_v$ in order to obtain a normal unitary field of $X$.

Answer 1

Indeed, the expression $X_u\times X_v$ is meaningless in $4$-dimensional Euclidean space.

Let us go back to the $3$-dimensional case and understand what is done there. Let $S\subset\mathbb{R}^3$ be a regular surface, and let $p\in S$. Then $T_pS\subset\mathbb{R}^3$ is a $2$-dimensional linear subspace, and so, its orthogonal complement is of dimension $1$. This means that there are exactly two different unit normal vectors to $S$ at $p$, corresponding to two different orientations. Very conveniently, these unit normal vectors are given by $$n=\pm\frac{X_u\times X_v}{|X_u\times X_v|}.$$

Now, if the surface $S$ is embedded in $\mathbb{R}^4$, the orthogonal complement $(T_pS)^\perp$ is of dimension $2$ and consequently, there is a circle of unit normal vectors to $S$ at $p$. Unlike the $3$-dimensional case, here there is no "magic formula" that produces a normal vector. However, you can find the orthogonal complement $(T_pS)^\perp$ by solving a system of two linear equations in four variables, and then choose a unit vector in it. In fact, you can even choose two unit normal vectors that are orthogonal to one another.