# Proving Something is a Rank-4 Tensor

Whilst going over my undergraduate notes on General Relativity, I came across the Quotient Rule for tensors: Briefly, if $$\mathbf{X \, A} = \mathbf{B}$$ with $$\mathbf{B}$$ being a non-zero tensor and $$\mathbf{A}$$ any arbitrary tensor, then $$\mathbf{X}$$ is a tensor as well. For now, we can assume all tensors considered are Cartesian. I am trying to apply this rule to the following example.

Let's say the components of $$\mathbf{B}$$ are given by $$B_{ij} = X_{ijkl} A_{kl}.$$ I want to verify that $$\mathbf{X}$$ is a tensor as well. I start off by writing out the transformation of $$\mathbf{B}$$ under a "rotation" to new, primed coordinates: $$B_{ij}' = X_{ijkl}' A_{kl}'.$$ Since $$\mathbf{A}$$ and $$\mathbf{B}$$ are rank-2 tensors, their components transform like so: $$B_{ij}' = a_{i p } a_{j q} B_{pq} \\ A_{kl}' = a_{kx } a_{ly} A_{xy}.$$ I substitute these into the previous equation to get $$X_{ijkl}' a_{kx}a_{ly} A_{xy} = a_{ip}a_{jq}B_{pq},$$ and I further substitute in $$\mathbf{B}$$, so I have $$X_{ijkl}' a_{kx}a_{ly} A_{xy} = a_{ip}a_{jq}X_{pqkl}A_{kl}.$$ Here, I am unsure how to proceed. I believe the problem is well-defined, and I know that in order to show $$\mathbf{X}$$ is a tensor, I need to show it transforms something like $$X_{ijkl}' = (\mathrm{four \, terms})\, X_{pqkl}$$ with 4 free indices. How do I get these extra terms? Any help is greatly appreciated.

• If you're wanting to use valid tensor notation then your very first equation is invalid since you have your $k$ and $l$ index are down twice. Commented Apr 7, 2021 at 20:00
• Would this be more suited for math.stackexchange.com? Also in GR one would use raised and lowered indices with the Einstein summation convention. Commented Apr 7, 2021 at 20:00
• @Charlie For now, we can assume all tensors considered are Cartesian. So all indices can be lower. Commented Apr 7, 2021 at 20:14
• @G.Smith I wasn't aware that this meant the index convention could be broken like this Commented Apr 8, 2021 at 0:06
• @Charlie In Cartesian components you raise and lower indices with the Kronecker delta so it doesn’t really make any difference from a computational viewpoint because the upper and lower components have the same value. My classical mechanics textbook used all-lower indices to discuss, for example, the inertia tensor and how it relates the angular momentum to the angular velocity. Commented Apr 8, 2021 at 0:17

So, when you applied the equation $$B_{ij}=X_{ijkl}A_{kl}$$ to get $$X_{ijkl}' a_{kx}a_{ly} A_{xy} = a_{ip}a_{jq}X_{pqkl}A_{kl}$$, note that in the expression $$X_{ijkl}A_{kl}$$, $$k$$ and $$l$$ are essentially dummy indices, and can be renamed to whatever we want. In particular, the next step is much clearer if we rename them to $$x$$ and $$y$$!
Then we instead get $$X_{ijkl}' a_{kx}a_{ly} A_{xy} = a_{ip}a_{jq}X_{pqxy}A_{xy}$$ Now, we use the fact that this equation is assumed to hold for an arbitrary tensor $$A$$, and so holds componentwise for all $$x,y$$. (Explicitly, you can imagine getting each equation for each particular $$x,y$$ by considering $$A$$ to be a basis tensor, i.e. such that it is $$1$$ in its $$xy$$ component and $$0$$ otherwise.) So we have, for every $$x,y$$, $$X_{ijkl}' a_{kx}a_{ly} = a_{ip}a_{jq}X_{pqxy}$$ Now we use the fact that these are Cartesian tensors, and so $$a_{ij}$$ is the inverse of $$a_{ji}$$, i.e. $$\delta_{ik}=a_{ij}a^{-1}_{jk}=a_{ij}a_{kj}$$. (This is much clearer with upper and lower indices and in a coordinate basis, so you can essentially use only one symbol, and not need to worry about transposes!) So, we have \begin{align} X_{ijkl}' a_{kx}a_{ly}a_{mx}a_{ny} &= a_{ip}a_{jq}X_{pqxy}a_{mx}a_{ny} \\ X_{ijkl}' \delta_{km}\delta_{ln} &= a_{ip}a_{jq}a_{mx}a_{ny}X_{pqxy}\\ X_{ijmn}' &= a_{ip}a_{jq}a_{mx}a_{ny}X_{pqxy}\end{align}
and we're done. (The swap in the order of $$X$$ and $$a$$ on the rhs from step 1 to 2 is just commutativity of ordinary multiplication for a nice layout, not any meaningful tensor operation.)
You might find this PDF helpful, which slips in the inversion at a slightly different step! (For them, $$X$$ is $$A$$, and $$A$$ is $$\xi$$.)