# Understanding the tensor notation in special/general theory of relativity

So I at the moment I am struggling with the tensor notation in special/general theory of relativity.

So we had this in university:

$$\sum_{\nu = 0}^{3} \eta_{\mu \nu } \Lambda ^{\nu}_{\;\beta} = (\eta \Lambda)_{\mu\beta}$$

$$\Rightarrow \sum_{\mu = 0}^{3} \Lambda ^{\mu}_{\;\alpha}(\eta \Lambda)_{\mu\beta} = (\Lambda^{T}\eta \Lambda)_{\alpha\beta}$$

So my question is: how do I know that I can write the summation term

$$\sum_{\mu = 0}^{3} \Lambda ^{\mu}_{\;\alpha}(\eta \Lambda)_{\mu\beta}$$

as a matrix product with the transposed of lambda ?

And here I have the same problem:

$$\sum_{\mu = 0}^{3}\sum_{\nu = 0}^{3}\eta_{\alpha \nu }\eta_{\beta \mu }F^{\mu\nu} = F_{\beta\alpha}$$

So I know that I can write this like this:

$$\sum_{\mu = 0}^{3}\sum_{\nu = 0}^{3}\eta_{\alpha \nu }\eta_{\beta \mu }F^{\mu\nu} = (\eta F\eta^{T})_{\beta\alpha}=F_{\beta\alpha}$$

but I don't know why.

After continuing searching through math.stack, I found this: Matrix notation in tensor transformations. So I would see it like this:

for the first one: $$\sum_{\nu = 0}^{3} \eta_{\mu \nu } \Lambda ^{\nu}_{\;\beta} = (\eta \Lambda)_{\mu\beta}$$ $$\Rightarrow \sum_{\mu = 0}^{3} \Lambda ^{\mu}_{\;\alpha}(\eta \Lambda)_{\mu\beta} =\sum_{\nu = 0}^{3} \sum_{\mu = 0}^{3} \Lambda ^{\mu}_{\;\alpha}\eta_{\mu \nu } \Lambda ^{\nu}_{\;\beta}= \sum_{\nu = 0}^{3} \sum_{\mu = 0}^{3} (\Lambda^{T})_{\alpha}^{\;\;\mu}\;\eta_{\mu \nu } \;\Lambda ^{\nu}_{\;\beta} =(\Lambda^{T}\eta \Lambda)_{\alpha\beta}$$

and for the second one:

$$\sum_{\mu = 0}^{3}\sum_{\nu = 0}^{3}\eta_{\alpha \nu }\eta_{\beta \mu }F^{\mu\nu} =\sum_{\mu = 0}^{3}\sum_{\nu = 0}^{3}\eta_{\beta \mu }F^{\mu\nu}(\eta^{T})_{\nu\alpha }= (\eta F\eta^{T})_{\beta\alpha}=F_{\beta\alpha}$$

So am I seeing this right ?

So my question is: how do I know that I can write the summation term $$\sum_{\mu = 0}^{3} \Lambda ^{\mu}_{\;\alpha}(\eta \Lambda)_{\mu\beta}$$ as a matrix product with the transposed of lambda ?

You can write $$\sum_{\mu=0}^3 \Lambda^\mu_{\;\alpha}(\eta\Lambda)_{\mu\beta} = \sum_{\mu=0}^3 (\Lambda^\top)^{\;\mu}_{\alpha}(\eta\Lambda)_{\mu\beta} = (\Lambda^\top \eta \Lambda)_{\alpha\beta},$$just keep in mind the definition of matrix products.

And here I have the same problem: $$\sum_{\mu = 0}^{3}\sum_{\nu = 0}^{3}\eta_{\alpha \nu }\eta_{\beta \mu }F^{\mu\nu} = F_{\beta\alpha}$$ So I know that I can write this like this: $$\sum_{\mu = 0}^{3}\sum_{\nu = 0}^{3}\eta_{\alpha \nu }\eta_{\beta \mu }F^{\mu\nu} = (\nu F\nu^{T})_{\beta\alpha}=F_{\beta\alpha}$$ but I don't know why.

I guess you meant $\eta F \eta^\top$ there instead. Anyway, take baby steps: \begin{align} \sum_{\mu=0}^3 \sum_{\nu=0}^3 \eta_{\alpha \nu}\eta_{\beta\mu}F^{\mu \nu} &= \sum_{\nu=0}^3 \eta_{\alpha \nu}\left(\sum_{\mu=0}^3 \eta_{\beta\mu}F^{\mu \nu}\right) \\ &= \sum_{\nu =0}^3\eta_{\alpha \nu}(\eta F)^{\;\nu}_{\beta} \\ &= \sum_{\nu = 0}^3 (\eta F)^{\;\nu}_{\beta} (\eta^\top)_{\nu \alpha} \\ &=(\eta F \eta^\top)_{\beta \alpha}.\end{align}

As a last remark, if $F$ is a $(2,0)$-tensor, that double sum can be seen as the definition of $F_{\beta \alpha}$. More precisely, from $F$ and the metric tensor $\eta$, we define a $(0,2)$-tensor $\widetilde{F}$ by $\widetilde{F}(x,y) = F(x_\flat, y_\flat)$, where $\flat$ is the musical isomorphism from the vector space to its dual, induced by $\eta$ (that is, $x_\flat(y) = \eta(x,y)$ for all $y$). The components $\widetilde{F}_{\beta\alpha}$ are given by the double sum, but since we regard $\widetilde{F}$ and $F$ as "the same tensor", we write $F_{\beta\alpha}$ in favour of $\widetilde{F}_{\beta\alpha}$.