# Proof that Kronecker's Delta is invariant under Lorentz transformation

There's an exercise in my book that says

"Prove that the Kronecker $$\delta$$ is invariant under Lorentz transformations".

The solution says that from the property $$\Lambda^Tg\Lambda=g$$ of Lorentz transformations follows that $$\delta^{\mu}_{\nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\beta}_{\nu}\delta^{\alpha}_{\beta}$$but I don't understand why.

From the first relation if we multiply for $$g$$ at right, since the metric tensor $$g$$ is the inverse of itself, we obtain $$\Lambda^Tg\Lambda g=I$$ which means that $$(\Lambda^T)^{\mu}_{\alpha}g_{\alpha\beta}\Lambda^{\beta}_{\gamma}g_{\gamma\nu}=\delta ^{\mu}_{\nu}$$ or $$\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\gamma}g_{\alpha\beta}g_{\gamma\nu}=\delta ^{\mu}_{\nu}$$ but I dont's see why $$\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\gamma}g_{\alpha\beta}g_{\gamma\nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\beta}_{\nu}\delta^{\alpha}_{\beta}$$

• Which book? Which page? Sep 19, 2021 at 16:59
• "Problem book in quantum field theory" by Voja Radovanovic', page 68. Sep 19, 2021 at 17:02

Just use the fact that $$g^\alpha_\beta = \delta^\alpha_\beta$$. So in components: $$\begin{equation} g_{\mu \nu} =\Lambda^\alpha_\mu \Lambda^\beta_\nu g_{\alpha \beta} \Longleftrightarrow \delta^\mu_\nu = \Lambda_\alpha^\mu \Lambda_\nu^\beta \delta_\beta^\alpha \end{equation}$$