# Commutativity and symmetric property in tensor manipulation

I have been trying to express $\eta^{\mu\nu}$ in terms of $\eta_{\mu\nu}$ and I have stumble upon the following relation:

$\eta^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}$

I can see how the indices $\alpha$ and $\beta$ contract to form the contravariant tensor $\eta^{\mu\nu}$, but I do not understand why the ordering of the tensors $\eta^{\mu\alpha}$, $\eta^{\nu\beta}$ and $\eta_{\alpha\beta}$ does not matter. If I am correct, only the inner upper and lower indices can contract, so that the ordering $\eta^{\mu\alpha}\eta_{\alpha\beta}\eta^{\nu\beta}$ is more appropriate. So, $\eta^{\mu\alpha}\eta_{\alpha\beta}\eta^{\nu\beta} = \eta^{\mu}_{\beta}\eta^{\nu\beta}$, so that inner upper and lower index $\alpha$ contracts. Then, $\eta^{\mu}_{\beta}\eta^{\nu\beta} = \eta^{\mu}_{\beta}\eta^{\beta\nu}$, because the metric tensor is symmetric. Next, $\eta^{\mu}_{\beta}\eta^{\beta\nu} = \eta^{\mu\nu}$ and we end up with the desired form.

Do you think ordering of the tensors in a tensor product matters? I say this, because matrix multiplication is non-commutative. Therefore, shouldn't tensor multiplication also be non-commutative?

Also, I could simplify the RHS into LHS only because the metric tensor is symmetric? What if the tensor were not symmetric? How could we then have managed to simplify?

Well they are real numbers... if you like you can express to another physicist your relation just as well like so:

Suppose $\eta_{\mu \nu}$ is a metric in $N$ dimensions with components who vary contravariantly and $\bar{\eta}_{\mu \nu}$ the corresponding covariant components. Then $\eta_{\mu\nu} = \sum_{\alpha=1}^N\sum_{\beta=1}^N\eta_{\mu\alpha}\eta_{\nu\beta}\bar{\eta}_{\alpha\beta}$

the summand is just a product of real numbers, and products of real numbers can be arranged however you please: $abc=cab=bca=cba=\cdots$.

You'll see this kind of phrasing in books just before they introduce the Einstein summation convention! "whose components vary contravariantly" is replaced with a superscript and "whose components vary covariantly" is replaced with a subscript and the sums are dropped, so you can write the whole thing more quickly. But you're always talking about sums and your components are actual numbers.

In matrix multiplication $AB\neq BA$ can occur. This equation has no indices. It is still the case that $\sum_j A_{ij}B_{jk}=\sum_j B_{jk}A_{ij}$, because these are just real numbers.

You really should take a second look at the definition of the Einstein tensor notation. There's absolutely no magic to it (excluding covariance/contravariance!)

• Thanks for the kind reply. I was also wondering why $\eta^{\mu\alpha}\eta^{\mu\beta} = 1$. Is it because the metric tensor is symmetric, so that we can rewrite the product as $\eta^{\alpha\mu}\eta^{\mu\beta}$, and then we sum over the index $\mu$ to form the product of 1? – nightmarish Mar 12 '15 at 6:03
• watch your upper and lower indices! It's also not true it equals $1$. See property 18 on mathworld.wolfram.com/MetricTensor.html – user12029 Mar 12 '15 at 6:37