I'm following a paper to solve this equation:

$y_{j}=y_{o}$ + A$\eta^{T}$ (Eq. 2)

My question is about the term $\eta^{T}$. In the paper says:

"With symbol $\eta$, we denoted a 1 × 6 contravariant matrix with each element being a random number from the interval (0, 1) and $\eta^{T}$ , figuring in Eq.(2), is its covariant form."

I'm not sure If I understood correctly what is $\eta^{T}$ .

Is it basically a matrix 6 x 1, with the elements being a random number from the interval (0, 1) ? Something like that:

$\eta^{T}$= $\left[ {\begin{array}{cc} 0.548 \\ 0.365 \\ 0.889 \\ 0.725 \\ 0.427 \\ 0.169 \\ \end{array} } \right]$

Just in case, this is the paper: http://articles.adsabs.harvard.edu/pdf/2017CoSka..47....7N

(In Pag. 6)


A is a 6x6 covariance matrix

$y_{j}$ is a 6x1 matrix

$y_{o}$ is a 6x1 matrix

Thanks in advance for any help


1 Answer 1


A covariant tensor is obtained by lowering the contravariant indexes of the tensor with the metric tensor. That is you must contract the indexes of the tensor with the metric like:

$g_{ij} \eta^j=\eta_i \,,$

where $\eta^j \rightarrow \eta$ and $\eta_i \rightarrow \eta^T$ In the paper this is not clear, it seems that the covariant tensor is just the matrix transpose of the contravariant one. It looks like just jargon. There is some metric in that 6 dimensional space?.

I'm not sure if this helps.

  • $\begingroup$ This what I was thinking, that they meant, the transpose. However, all the explanation is bit messy and it is not clear. Thanks a lot Gluoncito! $\endgroup$
    – Mileva
    Commented Jul 3, 2018 at 15:08

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