Worked example on a covariant vector transformation

[Dyatlov]

Hello.
I would like to check my understanding of how you transform the covariant coordinates of a vector between two bases.
I worked a simple example in the attached word document.
Let me know what you think.

 

[fresh_42]

Hello.
I would like to check my understanding of how you transform the covariant coordinates of a vector between two bases.
I worked a simple example in the attached word document.
Let me know what you think.

You should know, that to download a word document from unknown origin is a big hurdle not many of us want to risk. Definitely not me, as I additionally consider it as bad behavior to force me to take action rather than trying to concentrate on help. But this is my personal attitude towards users who don't take the effort of posting their questions adequately while simultaneously demanding efforts from others. I think you should know this and consider to type in the entire question instead. We have a LaTex library in place that helps a lot to even type in complicated formulas.

 

[jedishrfu]

Here's a PDF version:

 

[jedishrfu]

Here's the content of the PDF (limited to 10 image files per post so had to use latex codes for some stuff) via copy and paste (equations are images)

----------------------// example.docx as pdf //----------------------------------------------

The formula for the covariant vector transformation from the $R^2$ coordinate system to the $B$ coordinate system is:

For our example, the vector

, where

and

are the basis for

.

Our new coordinate system

; spanned by

To find $V_1$ in the $B$ basis in terms of $V_i$ in $R^2$:

 

[zwierz]

First, it is not a good idea to use superscripts for components of a covector. Simple way to restore these formulas is as follows. Vector's coordinates form the column: $x=(x^1,\ldots, x^m)^T$ and covector's coordinates form the row $\xi=(\xi_1,\ldots, \xi_m)$. The pairing $\xi x=\xi_ix^i$ is a scalar.

If vector's coordinates transform with the help of a matrix $C$ by the rule $x'=Cx$ and the covector's coordinates transform as follows $\xi'=\xi D$
then formula $\xi'x'=\xi x$ implies $D=C^{-1}$