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In tensor calculus, I recently came across the formula for the angle between two vectors (non null) in Riemannian Space, which is as follows:

$ cos \theta = \frac{g_{ij}A^iB^j}{\sqrt {g_{ij}A^iA^j}\sqrt {g_{ij}B^iB^j}}$; and the distance formula $|A|^2=g_{ij} A^iA^j $. I came across a problem related to this topic, where it says:

If $ X^i =\frac{1}{\sqrt {g_{pq}Y^pY^q}}Y^i $ (where $X^i$ and $Y^i$ are vector components and $g_{ij}$ is the fundamental tensor), show that $X^i$ is a unit vector.

My question is, whether , the dummy indices in the denominators imply this:

$ X^i =\frac{1}{\sqrt {\sum_p \sum_q {g_{pq}Y^pY^q}}}Y^i $

Or,

this: $ X^i =\sum_p \sum_q {\frac{1}{\sqrt {g_{pq}Y^pY^q}}Y^i} $

If the first one is implied, then $|X|^2=g_{ij}X^iX^j= \frac{g_{ij}Y^iY^j}{\sqrt {g_{pq}Y^pY^q}\sqrt {g_{pq}Y^pY^q}}=\frac{g_{ij}Y^iY^j}{ {g_{pq}Y^pY^q}}= \frac{\sum_i \sum_j g_{ij}Y^iY^j }{\sum_p \sum_q g_{pq}Y^pY^q}=1 $.

If my interpretation is wrong, then I don't know how to proceed. Kindly clear my doubts.

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It's the first interpretation that is correct: $$X^i =\frac{1}{\sqrt {\sum_p \sum_q {g_{pq}Y^pY^q}}}Y^i$$

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  • $\begingroup$ So, the same is implied for that angle formula? (Summing over separately on Numerator and Denominator) $\endgroup$ – Subhasis Biswas Sep 19 '18 at 8:31
  • $\begingroup$ Yes, $$ \cos \theta = \frac{g_{ij}A^iB^j}{\sqrt {g_{ij}A^iA^j}\sqrt {g_{ij}B^iB^j}} = \frac{\sum_{ij} g_{ij}A^iB^j}{\sqrt {\sum_{ij} g_{ij}A^iA^j}\sqrt {\sum_{ij} g_{ij}B^iB^j}} $$ $\endgroup$ – md2perpe Sep 19 '18 at 11:18
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The right interpretation is your first one: $$ X^i =\frac{1}{\sqrt {\sum_p \sum_q {g_{pq}Y^pY^q}}}Y^i $$

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