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When are two vectors orthogonal in curved spacetime?

From wikipedia: "In 2-D or higher-dimensional Euclidean space, two vectors are orthogonal if and only if their dot product is zero, i.e. they make an angle of 90°, or π/2 radians. Hence orthogonality of vectors is an extension of the concept of perpendicular vectors into higher-dimensional spaces." [http://en.wikipedia.org/wiki/Orthogonality]

How does this generalize for non-Euclidean space?

Is the following true?

Two vectors, $V_1$ and $V_2$, are orthogonal if and only if: $$G_{MN} V_1^M V_2^N = 0$$ Where $G_{MN}$ is the metric of the space.

P.S.: I used Einstein notation: Summation over $M$,$N$ implied and $M$,$N$ $\in$ $\left\{1, 2, ..., d\right\}$.

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    $\begingroup$ No, two vectors at two different points, are part of different vector spaces (the tangent space at that point), and hence cannot be compared. You cannot define orthogonality of vectors at two different points. At the same point of ofcourse what you say is true. $\endgroup$ – user7757 Jun 16 '14 at 17:10
  • $\begingroup$ @ramanujan_dirac that should be an answer $\endgroup$ – David Z Jun 16 '14 at 17:58
  • $\begingroup$ A coordinate free notion (i.e. not based on "tangent vectors", but on spacetime intervals) of "angle" (and hence of "orthogonality") would be for instance for any three events $A$, $B$, $C$ to evaluate $$\angle [ A B C ] := \text{ArcSin} \left[ \frac{1}{2} \sqrt{ 2 + 2 \frac{s^2[ A C ]}{s^2[ A B ]} + 2 \frac{s^2[ A C ]}{s^2[ B C ]} - \frac{s^2[ B C ]}{s^2[ A B ]} - \frac{s^2[ A B ]}{s^2[ B C ]} - \frac{s^2[ A C ]}{s^2[ A B ]} \frac{s^2[ A C ]}{s^2[ B C ]} } \right];$$ with cases involving light-like intervals $s[ A B ] = 0$ and/or $s[ B C ] = 0$ to be discussed separately. $\endgroup$ – user12262 Jun 17 '14 at 4:49
  • $\begingroup$ Related: the question "What is the notion of a spatial angle in general relativity?" (to which I have submitted a more complete answer already). $\endgroup$ – user12262 Jun 18 '14 at 0:53
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(Comment posted as an answer)

No, two vectors at two different points, are part of different vector spaces (the tangent space at that point), and hence cannot be compared. You cannot define orthogonality of vectors at two different points. At the same point of ofcourse what you say is true.

For more information read about Parallel Transport.

Suggestions and corrections are welcome, if you feel this is too short to be an answer (which I felt and hence posted as a comment), I will try to be more pedagogic.

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Vectors in a curved space are really tangent vectors. For tangent avectors at the same point and thus in the same tangent space orthogonality is defined by $$G_{MN} V_1^M V_2^N = 0$$ as in you expected.

As the other answer says, inner products and orthogonality are not defined for vectors at different point since they are in different tangent spaces. If there is a connection specified on the space space (in a metric space you can always take the Levi-Civita connection) you can parallel transport one vector to the other, but for a curved space the result of that is path dependent, so the inner product is still not uniquely determined.

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