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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
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 ... ...


Section 13.1 reads as follows:



Browder - Defn of a differential form .png
Browder - 2 - Start of Section 13.1 ... ... PART 2  ... .png




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
Jan 16, 2013
255
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.
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
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 ...

Your post was very helpful indeed ...

Thanks again ...

Peter