# Differential Forms ... Another question ... Browder, Section 13.1 ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am reading Chapter 13: Differential Forms ... ... and am currently focused on Section 13.1 Tensor Fields ...

I need some help in order to fully understand some statements by Browder in Section 13.1 ... ...  In the above text we read the following:

" ... ... We observe also that $$\displaystyle \text{dx}_P^j (h^1, \cdot \cdot \cdot , h^n) = h^j$$ ... ... "

My question is what is the nature of the $$\displaystyle h^i$$ ... given that $$\displaystyle \text{dx}_P^j$$ is a constant mapping from an open subset $$\displaystyle U$$ of $$\displaystyle \mathbb{R}^n$$ it seems that the $$\displaystyle h^i$$ are real numbers ...

... BUT ... then it seems strange that $$\displaystyle \text{dx}_P^j (h^1, \cdot \cdot \cdot , h^n) = h^j$$ ... that is ... how can the function evaluate to a real number when it is a constant mapping into $$\displaystyle \mathbb{R}^{ n \ast }$$ ... ... it should evaluate to a linear functional, surely ....

... indeed ... should Browder have written ...

$$\displaystyle \text{dx}_P^j (h^1, \cdot \cdot \cdot , h^n) = \tilde{e}^j$$

Can someone please clarify the above ...

Peter

#### GJA

##### Well-known member
MHB Math Scholar
Hi Peter ,

My question is what is the nature of the $$\displaystyle h^i$$ ... given that $$\displaystyle \text{dx}_P^j$$ is a constant mapping from an open subset $$\displaystyle U$$ of $$\displaystyle \mathbb{R}^n$$ it seems that the $$\displaystyle h^i$$ are real numbers ...
The nature of the $h$'s is that they are the components of a vector expressed in the basis $e_{1},\ldots, e_{n}$. The $h$'s are real numbers.

... BUT ... then it seems strange that $$\displaystyle \text{dx}_P^j (h^1, \cdot \cdot \cdot , h^n) = h^j$$ ... that is ... how can the function evaluate to a real number when it is a constant mapping into $$\displaystyle \mathbb{R}^{ n \ast }$$ ... ... it should evaluate to a linear functional, surely ....
There are two variables at work here: (i) The vector being input into $dx^{j}_{p}$ (i.e., $h=(h_{1},\ldots, h_{n})$) and (ii) The point $p\in \mathbb{R}^{n}.$ When Browder says that the mapping from $U$ into $\mathbb{R}^{n\ast}$ is constant, he means that $dx^{j}_{p}$ as a function of $p$ is constant; i.e., the 1-form $dx^{j}_{p}$ is the same functional at every point $p\in U$.

... indeed ... should Browder have written ...

$$\displaystyle \text{dx}_P^j (h^1, \cdot \cdot \cdot , h^n) = \tilde{e}^j$$
Based on your previous questions, I think you wanted Browder to write $$dx^{j}_{p}(h_{1},\ldots, h_{n})=\tilde{e}^{j}(h_{1},\ldots, h_{n})$$ for all $p\in U$. Note that writing things this way does accentuate the fact that $dx^{j}_{p}$ is a constant with respect to $p$ because the $p$ variable does not appear on the right hand side.

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#### Peter

##### Well-known member
MHB Site Helper
Hi Peter ,

The nature of the $h$'s is that they are the components of a vector expressed in the basis $e_{1},\ldots, e_{n}$. The $h$'s are real numbers.

There are two variables at work here: (i) The vector being input into $dx^{j}_{p}$ (i.e., $h=(h_{1},\ldots, h_{n})$) and (ii) The point $p\in \mathbb{R}^{n}.$ When Browder says that the mapping from $U$ into $\mathbb{R}^{n\ast}$ is constant, he means that $dx^{j}_{p}$ as a function of $p$ is constant; i.e., the 1-form $dx^{j}_{p}$ is the same functional at every point $p\in U$.

Based on your previous questions, I think you wanted Browder to write $$dx^{j}_{p}(h_{1},\ldots, h_{n})=\tilde{e}^{j}(h_{1},\ldots, h_{n})$$ for all $p\in U$. Note that writing things this way does accentuate the fact that $dx^{j}_{p}$ is a constant with respect to $p$ because the $p$ variable does not appear on the right hand side.

Thanks GJA ...