Timeline for Correct order for tensor operations, when expressed in matrix form

7 events

when toggle format	what		by	license	comment
Jan 11 at 12:03	vote	accept	Tiago
Jan 11 at 10:49	comment	added	basics		I performed the product you need in the edit at the end of the answer. Now, you should easily realize that the "lq" component of the 2-nd order tensor $\eta \hspace{-4pt} \eta \cdot \mathbb{A} \cdot \eta \hspace{-4pt} \eta $ is $ \eta_{li} A^{ik} \eta_{kq}$, and thus this should solve your doubts.
Jan 11 at 10:47	history	edited	basics	CC BY-SA 4.0	added 970 characters in body
Jan 11 at 10:38	comment	added	basics		Try to compute $\mathbb{B} \cdot \mathbb{A}$ with the notation I used above. And then try to compute $\eta \hspace{-4pt} \eta \cdot \mathbb{A} \cdot \eta \hspace{-4pt} \eta$ and compare it with $\eta \hspace{-4pt} \eta\cdot \eta \hspace{-4pt} \eta \cdot \mathbb{A} $ or any other combination you like
Jan 11 at 9:22	comment	added	Tiago		This does not explain why the right order is $(A_{\alpha\beta}) = (\eta_{\alpha\mu})(A^{\mu\nu})(\eta_{\nu\beta})$.
Jan 10 at 20:08	comment	added	basics		In the answer above, the vectors $\mathbf{b}^j$ are vectors of the reciprocal base of the base $\mathbf{b}_i$, defined as $\mathbf{b}^j \cdot \mathbf{b}_i = \delta_i^j$, so that it's easier to evaluate the dot product.
Jan 10 at 20:06	history	answered	basics	CC BY-SA 4.0