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does [math]\varepsilon^{\alpha_{1}...\alpha_{n}} T_{\alpha_{1}...\alpha_{n}}[/math] transform like a scalar? supposing that T is a rank n covariant tensor field.
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hur durr
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>>8821054
yes. It is a trivial question, but an important pedagogical question all the same.
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Does it transform like a scalar? no. The levi-civita symbols fail to transform like rank n contravariant tensors and so you cannot contract them to form invariants.
attempting to transform the levi civita symbol yields the following
[eqn]\varepsilon^{\alpha_{1}...\alpha_{1}}\prod_{i=1}^{n}\frac{\partial\bar{x}^{\beta_{i}}}{\partial x^{\alpha_{i}}} = \bar{\varepsilon}^{\beta_{1}...\beta_{1}}
|\partial \bar{x} |[/eqn]
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>>8821321
from this it is clear that the symbol is a tensor density of weight one, and that we can define a tensor from it thusly
[eqn]E^{\alpha_{1}...\alpha_{n}} := \frac{\varepsilon^{\alpha_{1}...\alpha_{n}}}{\sqrt{|g|}}[/eqn]
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>>8821331
as [math]E^{\alpha_{1}...\alpha_{n}} T_{\alpha_{1}...\alpha_{n}}[/math] now transforms like a scalar field, we are able to take the integral over D dimensional space time and add it to our action.
the term is
[eqn]\int d^{D}x \sqrt{|g|} E^{\alpha_{1}...\alpha_{n}} T_{\alpha_{1}...\alpha_{n}} = \int d^{D}x \ \varepsilon^{\alpha_{1}...\alpha_{n}}
T_{\alpha_{1}...\alpha_{n}} [/eqn]
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>>8821353
isn't it lovely? dependence on the metric has vanished, and we now have something that looks like it is invariant under all "topology preserving" maps.
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>>8821363
and now, if we consider a topological term like:
[eqn] \int d^{(3+1)}x \ \varepsilon^{\alpha \beta \gamma \delta} F_{\alpha \beta} F_{\gamma \delta} [/eqn]
which is rather similar to the standard maxwell action, just swap out the permutation symbol for a pair of metric tensors and carry a factor of [math]\sqrt{|g|}[/math] with the differential.
then we have a path to chern-simons theory.
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>>8821043
who is this man
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>>8822178
He's a branelet.
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>>8821396
Dude chill what the fuck stop dropping these redpills ffs
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This thread is my favorite D-meme
Thread posts: 12
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