# Homework Help: A question about covariant representation of a vector

Tags:
1. Jun 22, 2015

### guv

1. The problem statement, all variables and given/known data
Hi I am reviewing the following document on tensor:
https://www.grc.nasa.gov/www/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf

2. Relevant equations

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

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

3. 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.

2. Jun 24, 2015

### Fredrik

Staff Emeritus
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 $V_j{}^* e^{(j)}{}^*$.

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.

3. Jul 2, 2015

### 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.