The Mystery of the Minus Sign in Geometry, Topology, and Physics

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SUMMARY

The discussion centers on the minus sign in equations (10.85) and (10.108) of "Geometry, Topology and Physics" by Mikio Nakahara. The participant highlights the relationship between the Levi-Civita (pseudo-)tensor and the metric tensor, specifically the Minkowski metric represented as ##\eta=\mathrm{diag}(1,-1,-1,-1)##. The participant asserts that the covariant components of the Levi-Civita tensor yield a negative sign due to the determinant of the metric tensor, confirming the established conventions in the high-energy physics (HEP) community.

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  • Understanding of Levi-Civita (pseudo-)tensor notation
  • Familiarity with Minkowski metric tensor, specifically ##\eta=\mathrm{diag}(1,-1,-1,-1)##
  • Knowledge of tensor calculus in the context of geometry and physics
  • Basic concepts of high-energy physics (HEP) conventions
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  • Explore the implications of the Minkowski metric in relativity
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nenyan
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I am reading GEOMETRY, TOPOLOGY AND PHYSICS written by MIKIO NAKAHARA (second edition). I have a problem on page 400.
I wonder why the sign is minus in Eq. (10.85).

And the same problem appears on page 405 in Eq.(10.108). I think it should be minus one half.
 

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Have you considered that with the standard definition (at least in the HEP community) ##\epsilon^{\mu \nu \rho \sigma}=\text{sign}(\mu,\nu,\rho,\sigma)## the covariant components of the Levi-Civita (pseudo-)tensor reads
$$\epsilon_{\alpha \beta \gamma \delta}=\eta_{\alpha \mu} \eta_{\beta \nu} \eta_{\gamma \rho} \eta_{\delta \sigma} \epsilon^{\mu \nu \rho \sigma}=\det \eta \epsilon^{\alpha \beta \gamma \delta}=-\epsilon^{\alpha \beta \gamma \delta},$$
because ##\eta=\mathrm{diag}(1,-1,-1,-1)## (or in the east-coast convention ##\mathrm{diag}(-1,1,1,1)##).
 
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vanhees71 said:
Have you considered that with the standard definition (at least in the HEP community) ##\epsilon^{\mu \nu \rho \sigma}=\text{sign}(\mu,\nu,\rho,\sigma)## the covariant components of the Levi-Civita (pseudo-)tensor reads
$$\epsilon_{\alpha \beta \gamma \delta}=\eta_{\alpha \mu} \eta_{\beta \nu} \eta_{\gamma \rho} \eta_{\delta \sigma} \epsilon^{\mu \nu \rho \sigma}=\det \eta \epsilon^{\alpha \beta \gamma \delta}=-\epsilon^{\alpha \beta \gamma \delta},$$
because ##\eta=\mathrm{diag}(1,-1,-1,-1)## (or in the east-coast convention ##\mathrm{diag}(-1,1,1,1)##).
Thank you! I see.
 

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