Are $v,u$ orthogonal on a surface if $v^{\alpha}u^{\beta}g_{\alpha\beta}=0$?

Question

I already know that for euclidean vectors, $u\cdot v=u^{\alpha}v^{\beta}\delta_{\alpha\beta}$, and that if this is zero then the two vectors are orthogonal. If we let the metric tensor describe a surface that is not globally flat however, like a sphere, then does $v^{\alpha}u^{\beta}g_{\alpha\beta}=0$ confirm their orthogonality in the surface?

Answer 1

If you mean that you have two vectors $u$ and $v$ tangent to a curved surface, like a sphere, embedded in regular 3D Euclidean space then, yes, if your product is zero, they are orthogonal within the surface. In this case $u^{\alpha}v^{\beta}\delta_{\alpha\beta}$ is, in fact, the dot product of two 3D Euclidean vectors, even if it doesn't look like it, and will give zero if the vectors are normal. The whole point to the machinery of differential geometry of surfaces embedded in 3D Euclidean space is to recast 3D calculations in such a way as to need only two coordinates, i.e. the local coordinates covering the surface. Now, if you mean something beyond this, abstract surfaces and what have you, you're going beyond what I know.