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I am studying the Dirac gamma matrices and have encountered the Kronecker delta $\delta_{ij}$ That I am accustomed to. However, I have also come across a different form, $\delta_{\mu}^{\nu} $.

This particular example comes from deriving a very simple expression: $$\gamma_\mu\gamma^\mu=\delta_\mu^\mu$$ How is this delta different from the one with both indexes lowered? Is there any additional meaning besides emphasizing the position of the indexes on the LHS?

Qmechanic
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Nick Heumann
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  • Do you understand covariant and contravariant indices and how to raise and lower them with the metric tensor? – Ghoster Dec 23 '22 at 01:12
  • Yes, I understand how to do it in any given matrix, say $\gamma_\mu=\eta_{\mu\nu}\gamma^\nu$. But I don't understand that meaning in the delta matrix. Isnt the delta matrix just a matrix with 1s in the diagonal and 0s elsewhere? – Nick Heumann Dec 23 '22 at 01:24
  • Yes. When you raise an index on $\eta_{\mu\nu}$ you get $\eta^\mu_\nu$, but this is just $\delta^\mu_\nu$ because of how the inverse metric is defined: $\eta_{\mu\lambda}\eta^{\lambda\nu}\equiv\delta_\mu^\nu$. – Ghoster Dec 23 '22 at 01:37
  • Note that $\delta_{ij}$ is a tensor under rotations but $\delta_{\mu\nu}$ is not a tensor under Lorentz transformations; it isn’t even defined as far as I know. – Ghoster Dec 23 '22 at 01:42
  • Possible duplicates: https://physics.stackexchange.com/q/119126/2451 and links therein. – Qmechanic Dec 23 '22 at 04:25

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