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In cartesian coordinates, the unit vectors $\{u_x, u_y, u_z\}$ are universal. That is, $u_x(x, y, z)$ is constant and so on for the rest of them. Because of that, the dot product $\langle v | w \rangle$ is defined everywhere in $\mathbb{R}^n$.

But in spherical, $u_\phi(r, \phi, \theta)$ is not constant. It varies. So the dot product $\langle v | w \rangle$ is only defined when $u$ and $v$ share an origin.

But still, I kind of miss that old dot product. Maybe I can get a bilinear form that is defined everywhere in spherical coordinates, by doing something different. What do I need to do to get my universal bilinear form in spherical coordinates?

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  • $\begingroup$ Even In Euclidean space, you should be taking the dot product of tangent vectors at different points. What shoukd this mean? $\endgroup$ – Ted Shifrin Apr 24 at 17:14
  • $\begingroup$ @TedShifrin I think I wrote my question incorrectly. I mean, at point P <v, w> is going to have one formula and at point Q != P <v, w> is going to have a differen onet. Contrast the case in cartesian coordinates where <v, w> is the same formula at all points. $\endgroup$ – ErotemeObelus Apr 24 at 17:19
  • $\begingroup$ Aha. Ultimately, the fact that the sphere has nonzero curvature prevents your having a coordinate system in which the inner product $g_{ij}$ is given by a constant matrix. That can happen only on a flat (locally Euclidean) space. $\endgroup$ – Ted Shifrin Apr 25 at 5:52
  • $\begingroup$ Can you talk more about this $g_{ij}$ thingie? $\endgroup$ – ErotemeObelus Apr 25 at 21:48

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