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Author Topic: Rotation Group and (0, 2) Tensor  (Read 28255 times)
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เกียรติศักดิ์
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« Reply #15 on: February 18, 2007, 02:20:17 PM »

Oh maybe I haven't taken enough care when reading English. If a group is transitive, then its representation is transitive too right? And "the representation" in that sentence refers to the representation of an adjoint group (whatever it is). redfaced
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« Reply #16 on: February 18, 2007, 06:28:02 PM »

...
I think the ability to consider additional operation like contraction does not arise from \Lambda^{\text{T}}g\Lambda=g and the fact that we are talking about  \mathrm{SO} (3, 1) , but because our present representation comprises not ordinary matrices, but ones that portray (0, 2) tensors. So all the properties we can use to reduce them are those of such tensors; contraction can always be done. The problem is what property the contraction does make use of.

...
Am I correct? Smiley

Consider orthogonal transformations where \Lambda^{\text{T}}\Lambda = \Lambda\Lambda^{\text{T}}=\mathbf{1} or \Lambda_{ij}\Lambda_{ik}=\Lambda_{ji}\Lambda_{ki}=\delta_{jk} (summation convention over repeated indices is assumed here).
Now X^{\prime}_{ii}=\Lambda_{ij}\Lambda_{ik}X_{jk}=\delta_{jk}X_{jk}=X_{ii} using the above condition. Therefore the trace is invariant because of the above condition.
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« Reply #17 on: February 18, 2007, 08:28:40 PM »

Oh I see. redfaced

Thank you very much. smitten smitten smitten I had a chance to really think about many things during the discussion.
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« Reply #18 on: July 19, 2007, 08:00:39 PM »

Sorry, I'm afraid I can't. Smiley Smiley Smiley
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