Metric Tensor Division: Is It Proper?

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SUMMARY

The discussion centers on the proper interpretation of the equation {{x}^{a}}{{g}_{ab}}={{x}_{b}} in the context of metric tensor division. It is established that this equation represents a summation over repeated indices, specifically in a 4-dimensional manifold. Consequently, the assertion that {{g}_{ab}} can be expressed as {{g}_{ab}}=\frac{{{x}_{b}}}{{{x}^{a}}} is incorrect, as the first equation is not a single term but a sum of terms. This distinction is crucial for understanding the properties of metric tensors in differential geometry.

PREREQUISITES
  • Understanding of tensor notation and operations
  • Familiarity with differential geometry concepts
  • Knowledge of 4-dimensional manifolds
  • Basic principles of index notation and summation conventions
NEXT STEPS
  • Study the properties of metric tensors in differential geometry
  • Learn about index notation and its implications in tensor calculus
  • Explore the concept of summation over repeated indices in tensor equations
  • Investigate the implications of tensor operations in 4-dimensional manifolds
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Mathematicians, physicists, and students of differential geometry who are working with tensor calculus and seeking to deepen their understanding of metric tensors and their properties.

redstone
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If you know that
[tex]{{x}^{a}}{{g}_{ab}}={{x}_{b}}[/tex]

is it proper to say that you also know
[tex]{{g}_{ab}}=\frac{{{x}_{b}}}{{{x}^{a}}}[/tex]
 
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redstone said:
If you know that
[tex]{{x}^{a}}{{g}_{ab}}={{x}_{b}}[/tex]

is it proper to say that you also know
[tex]{{g}_{ab}}=\frac{{{x}_{b}}}{{{x}^{a}}}[/tex]

No, because the expression [itex]{{x}^{a}}{{g}_{ab}}={{x}_{b}}[/itex] is not a single term, it's a sum of terms. Repeated indexes are summed over, so what your first equation really means is

[tex]\Sigma_a {{x}^{a}}{{g}_{ab}} = x^0 g_{0b} + x^1 g_{1b} + x^2 g_{2b} + x^3 g_{3b} = {{x}_{b}}[/tex]

(I've assumed that we're working in a 4-dimensional manifold.) There's no way to transform that into your second equation.
 
Ah, yes, of course. Thank you.
 

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