Conversation started Aug 31, 2020 at 16:22.
user434058
Aug 31, 2020 16:22
@Slereah Could you give me an example of such a $M$?
Slereah
Slereah
@FakeMod $\mathrm{diag}(1,1,1)$
user434058
@bolbteppa Yeah. That doubling the basis vectors works well as of now, but I am quite concerned by the lack of rigour in that method, that's why I was hoping for a mathematical method. Anyways, thanks :-)
bolbteppa
bolbteppa
user image
Technically what I said is rigoroous and this is the way the wiki does it
Covariance and contravariance of vectors
In multilinear algebra and tensor analysis, covariance and contravariance describe how the quantitative description of certain geometric or physical entities changes with a change of basis. In physics, a basis is sometimes thought of as a set of reference axes. A change of scale on the reference axes corresponds to a change of units in the problem. For instance, by changing scale from meters to centimeters (that is, dividing the scale of the reference axes by 100), the components of a measured velocity vector are multiplied by 100. Vectors exhibit this behavior of changing scale inversely to...
It's easier if you use a metric so projections can be defined and no crazy partial derivative basis is needed
user434058
@Slereah Now what would happen if I input $v,\omega$ in that particular $M$ such that $v= \begin{bmatrix} a\\b\end{bmatrix}$ and $\omega=\begin{bmatrix}c & d \end{bmatrix}$?
Slereah
Slereah
Well, if we treat $M$ as a map from vectors to vectors
user434058
Aug 31, 2020 16:29
@Slereah No, I am talking about $M:V\times V^*\to \mathbb R$.
Slereah
Slereah
Then $$Mv =\begin{pmatrix} 1 & 0\\0 & 1\end{pmatrix} \begin{pmatrix} a\\b\end{pmatrix} = \begin{pmatrix} a\\b\end{pmatrix}$$
user434058
@bolbteppa Hmmm... I see. If that's valid and good enough, then I don't see any reason behind chasing a more abstract definition. Thanks!
Slereah
Slereah
If you want to do the bilinear function directly, it's gonna be $$M(v, \omega) = M^\mu_\nu v^\nu \omega_\mu$$
user434058
@Slereah Is this a definition, or is there a reason why it is this way?
Slereah
Slereah
@FakeMod As it is a tensor, $M$ is linear
So you can decompose it via the base
$$M(v, \omega) = M(v^\mu e_\mu, \omega_\nu e^\nu) = v^\mu \omega_\nu M(e_\mu, e^\nu) $$
user434058
Aug 31, 2020 16:33
@Slereah Oh, alright. I get what you're saying, thanks.
Slereah
Slereah
Then you define $M(e_\mu, e^\nu)$ as the components of the matrix
 
Conversation ended Aug 31, 2020 at 16:33.