A question about covariant representation of a vector

[guv]

Homework Statement

Hi I am reviewing the following document on tensor:
https://www.grc.nasa.gov/www/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf

Homework Equations

In the middle of page 27, the author says:
Now, using the covariant representation, the expression $$\vec V=\vec V^*$$
then becomes

$$\vec V = V_j \vec e^{(j)}= V_j^* \vec e^{(j)*} = \vec V^*$$

The Attempt at a Solution

How does this work? Earlier in the document, the Vector is always represented as
$$\vec V = V_j \vec e_{(j)} = V^j \vec e^{(j)} $$

I don't see how suddenly the coordinates or components of covariant basis combing with the contracovariant basis can represent the same vector? It's been made clear earlier,

$$V_j = V^i g_{ij}$$ and $$g_{ij}$$ is not 1 in general.

Thank you for clarification.

 

[Fredrik]

Homework Statement

Hi I am reviewing the following document on tensor:
https://www.grc.nasa.gov/www/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf

Homework Equations

In the middle of page 27, the author says:
Now, using the covariant representation, the expression $$\vec V=\vec V^*$$
then becomes

$$\vec V = V_j \vec e^{(j)}= V_j^* \vec e^{(j)*} = \vec V^*$$

First of all, I should mention that this author's conventions are pretty unusual. You won't find them in math books.

K and K* denotes two arbitrary coordinate systems. There are two bases for $\mathbb R^3$ associated with each coordinate system. So there's a total of four bases. Any element of $\mathbb R^3$ can be expressed as a linear combination of the three elements of any of these four bases.

$\{\vec e^{(i)}\}$ is the basis such that each $\vec e^{(i)}$ is in the direction of increasing $i$ coordinate. He calls this the contravariant basis associated with K. $\{\vec e_{(i)}\}$ is a basis such that each $\vec e_{(i)}$ is in a direction perpendicular to the plane in which the other two coordinates are constant. He calls this the covariant basis associated with K. The other two bases, $\{\vec e^{(i)}{}^*\}$ and $\{\vec e_{(i)}{}^*\}$ are defined similarly with respect to K*.

The reason why $\vec V=V_j e^{(j)}=V_j{}^* e^{(j)}{}^*$ is just that $\{\vec e^{(i)}\}$ and $\{\vec e^{(i)}{}^*\}$ are bases, and the $V_j$ and the $V_j{}^*$ are defined as the numbers such that these equalities hold. The reason why these things are equal to $\vec V^*$ is just that this symbol is a (pointless) notation for the linear combination $V_j{}^* e^{(j)}{}^*$.

The Attempt at a Solution

How does this work? Earlier in the document, the Vector is always represented as
$$\vec V = V_j \vec e_{(j)} = V^j \vec e^{(j)} $$

It looks like he was just using a different notation for the components earlier. What he wrote as $V^j$ earlier is written as $V_j$ now.

I wouldn't recommend that you use this document to learn about tensors unless you have to. I like chapter 3 in Schutz's GR book. And the chapter on tensors in Sergei Treil's linear algebra book looks good too. Their conventions are more standard: They start with a vector space V and define V* as the vector space of linear functions from V into $\mathbb R$. V* is called the dual space of V. Given an ordered basis $(e_1,\dots,e_n)$ for V, one can define an ordered basis $(e^1,\dots,e^n)$ for V* by $e^i(e_j)=\delta^i_j$ for all i,j. The latter ordered basis is said to be the dual of the former. Note that these two bases are bases for two different vector spaces.

 

[guv]

Thanks for the clarification. I suspected that the author was a bit careless with that part of the discussion. The notations are actually okay for me coming from a physics background. I'll take a look at Schutz and Sergei Treil's books.