# Metric tensor times its inverse using Kronecker delta

From tensor calculus, we know that

$$\begin{equation} g^{\mu\nu}=\delta_{\lambda}^{\mu}\delta_{\phi}^{\nu}g^{\lambda\phi}.\tag{1} \end{equation}$$ Based on (1), is the following true? $$\begin{equation} g^{\mu\nu}g_{\lambda\phi}^{}=\delta_{\lambda}^{\mu}\delta_{\phi}^{\nu}.\tag{2} \end{equation}$$

• Clearly not. The right-hand side would always be 0 or 1, while the left-hand side can be anything. Both the $\lambda$ and $\phi$ indices have been contracted in your first equation, you cannot just move that tensor to the other side. May 31, 2021 at 13:35

The simplest possible concrete example demonstrates this to be wrong. Take $$[\eta]=\text{diag}(-1,+1,+1,+1)$$, then:

$$\eta^{00}\eta_{11}=-1\cdot +1=-1\tag{1},$$

on the other hand:

$$\delta^0_1\delta^0_1=0. \tag{2}$$

No. The first relation is a sum over $$\lambda$$ and $$\phi$$. You can't multiply the inverse metric $$g^{\lambda\phi}$$ on the RHS with anything to get the LHS.

No it wouldn't. First of all you can't have more than two indices repeating which happens when you multiplied with $$g_{\lambda \phi}$$, this is your biggest mistake. To confirm what you can get by this multiplication, multiply it by $$g_{\alpha \beta}$$

$$g^{\mu \nu} g_{\alpha \beta} = \delta^{\mu}_{\lambda} \delta^{\nu}_{\phi} g^{\lambda \phi} g_{\alpha \beta}$$

$$g^{\mu \nu} g_{\alpha \beta} = \delta^{\mu}_{\lambda} g^{\lambda \nu} g_{\alpha \beta}$$

$$g^{\mu \nu} g_{\alpha \beta} = g^{\mu \nu} g_{\alpha \beta}$$