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How many independant components does the Riemann-Tensor have in n- dimensional space?

- Thread starter Thaakisfox
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- #1

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How many independant components does the Riemann-Tensor have in n- dimensional space?

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this is related to the number of components related to riemanian tensor and the space that contains itHow many independant components does the Riemann-Tensor have in n- dimensional space?

as a simple example if u have a mixed tensor of 5 components 3 contarvariant and 2 covariant in 4 dimensional space so u have 4^5 compnents i.e 1024

u see how many equations are contracted to single one that's why einstein begin his general relativity by studying tensors with his friend Grassmann

my name is mina, i study QFT

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for more information see Shaum vector analysis chap8How many independant components does the Riemann-Tensor have in n- dimensional space?

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Chris Hillman

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Just thought I'd stress that you are probably asking about the number ofHow many independant components does the Riemann-Tensor have in n- dimensional space?

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Chris Hillman

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Hi, Mina, I think you are confusing Marcel Grossmann with Hermann Grassmann here! The latter was a fellow graduate student with Einstein at ETH (but Grossmann studied math not physics); the former was the completely different mathematician who introduced what is now called Grassmann or exterior algebra, later adopted by Cartan to give exterior calculus, aka the study of differential forms.this is related to the number of components related to riemanian tensor and the space that contains it

as a simple example if u have a mixed tensor of 5 components 3 contarvariant and 2 covariant in 4 dimensional space so u have 4^5 compnents i.e 1024

u see how many equations are contracted to single one that's why einstein begin his general relativity by studying tensors with his friend Grassmann

my name is mina, i study QFT

Also, the example you gave overlooks the possibility of algebraic symmetries which will in general reduce the number of algebraically independent components. For example the Riemann tensor (more or less by definition) satisfies [itex]R_{abcd} = -R_{bacd}[/itex].

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