# Product of a specific $(0,2)$ and $(2,0)$ tensor (Minkowski Metric tensor) [closed]

How to calculate $$\eta_{\mu\nu}\eta^{\mu\nu}$$ Where $$\eta=\begin{bmatrix} -1 \\ &1 \\&&1\\&&&1\end{bmatrix}$$ All other entries are $$0$$.

• I don't understand where "But this is wrong since ..." comes from. Why is the equation following that not compatible with your result (which I believe is correct)? Feb 20, 2023 at 3:50
• Your last equation will become correct when you set $\gamma$ equal to $\mu.$ Then what is the trace of $\delta?$ Feb 20, 2023 at 4:53
• @joriki yeah, I was following a YT lecture and I had a misunderstanding of what the lecturer was saying, indeed the answer is 4 and not 2 which I assumed it to be, hence the question. I'll edit this post into a Q&A format now. Thank you for your comment. Feb 20, 2023 at 12:52
• @KurtG. Thank you for pointing out the error, also please refer to my above comment. Thank you. Feb 20, 2023 at 12:53

Writing $$\eta^{\mu\nu}$$ as $$\eta^{\mu\nu}=\eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}$$ Thus $$\eta_{\mu\nu}\eta^{\mu\nu}=\eta_{\mu\nu}\eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}$$ Rearranging $$\eta_{\mu\nu}\eta^{\mu\nu}=\eta^{\mu\alpha}\eta_{\alpha\beta}\eta^{\nu\beta}\eta_{\mu\nu} \\ =\eta^\mu_\beta \eta^\beta_\mu=4$$ Or equivalently since $$\eta^{\mu\nu}\eta_{\mu\chi}={\delta^\nu}_\chi\,$$ Where $$\delta$$ is Kronecker delta.
Substituting $$\chi=\nu$$ we get 4. Note that this only works because $$\eta$$ is symmetrical.
• In $\eta^{\mu\nu}\eta_{\mu\chi}=\delta^\chi_\nu$ you cannot have upper $\nu$ on the left and lower on the right. Ditto with $\chi$. Index notation is easy but not forgiving sloppyness. Feb 20, 2023 at 13:09
• Looks OK. Even better would be ${\delta^\nu}_\chi\,.$ You will soon deal with tensors where that matters. Feb 20, 2023 at 13:13