Graduate Decomposing Rank-2 Tensors in Dirac's "General Theory of Relativity

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Dirac's "General Theory of Relativity" states that a general rank-2 tensor can be decomposed into a sum of outer products, specifically expressed as T^{\mu\nu} = A^\mu B^\nu + A'^\mu B'^\nu + ... This concept is reiterated in the context of the covariant derivative, where Dirac emphasizes that a tensor T_{\mu\nu} can similarly be represented. The discussion explores whether this decomposition is obvious and seeks clarification on the notation used. A participant explains that by defining vectors A^\nu and B_\nu, the decomposition aligns with Dirac's definition of tensors. The conversation highlights the straightforward nature of this tensor decomposition while addressing LaTeX formatting issues.
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Dirac says that a general rank-2 tensor ## T^{\mu\nu} ## can be decomposed as ## A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + \cdots\, ##. Is this obvious?
Dirac's book "General Theory of Relativity" says on p. 2 that a general rank-2 tensor can be written as a sum of outer products: $$ T^{\mu\nu} = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + \cdots $$ Importantly, he repeats this on p. 18, in developing the covariant derivative, where he mentions that a tensor ## T_{\mu\nu} ## is "expressible as a sum of terms like ## A_\mu B_\nu ##".

Is this obvious? Can someone show or explain this?
 
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Kostik said:
TL;DR Summary: Dirac says that a general rank-2 tensor can be decomposed: ##T^\mu\nu = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + ...##. Is this obvious?

Dirac's book "General Theory of Relativity" says on p. 2 that a general rank-2 tensor can be written as a sum of outer products:

$$T^\mu\nu = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + ...$$

Importantly, he repeats this on p. 18, in developing the covariant derivative, where he mentions that a tensor ##T_\mu\nu$ is "expressible as a sum of terms like $A_\muB_\nu##".

Is this obvious? Can someone show or explain this?
By definition, a tensor of rank two can be written as
$$
T = T^{\mu\nu} e_\mu \otimes e_\nu
$$
We can introduce the vectors ##A^\nu = T^{\mu\nu} e_\mu## and ##B_\nu = e_\nu## (note that here ##\nu## is being used as a counter rather than a component index) and therefore
$$
T = A^\nu \otimes B_\nu
$$
 
I'm not familiar with your notation, I wonder if Dirac's decomposition can be explained using only his definition of tensors.
 
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Kostik said:
I added the missing braces, but the LaTex still doesn't seem to be working in the original post.
It's a known issue when you make the first post to use LaTeX (OP or reply) on a page. The parser doesn't get loaded until you refresh the page. Your LaTeX looks fine to me, and will look fine to you once you've hit refresh.
 
Ibix said:
It's a known issue when you make the first post to use LaTeX (OP or reply) on a page. The parser doesn't get loaded until you refresh the page. Your LaTeX looks fine to me, and will look fine to you once you've hit refresh.
Aha, yes, I see it now.
 
Oh, I think it's actually fairly straightforward. Write (showing the summation explicitly): $$T^{\mu\nu}=\sum_{\lambda,\kappa}T^{\lambda\kappa}{\delta_\lambda}^\mu{\delta_\kappa}^\nu \,\,\,\,\text{(no Einstein summation)}$$ Then ##A^\mu = T^{\lambda\kappa}{\delta_\lambda}^\mu## (not summed over ##\lambda##) and ##B^\nu={\delta_\kappa}^\nu##. (Regard ##\lambda## and ##\kappa## as fixed.) Since everything in sight is a tensor, the ##A^\mu## and ##B^\nu## are obviously vectors (no need to worry about constructing a non-vector).
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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