MHB The Dual Space and Differential Forms .... ....

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The discussion focuses on understanding the representation of a differential form as a linear combination of basis one-forms in the context of dual spaces. The key statement from Shifrin's text is that a differential form can be expressed as φ = a₁dx₁ + ... + aₙdxₙ. A participant demonstrates this by showing that for an arbitrary vector v in ℝⁿ, the linearity of φ and the basis forms leads to the conclusion that φ equals the linear combination of the basis forms. This clarification helps in grasping the relationship between vectors and differential forms in multivariable mathematics. The explanation provided is appreciated for its clarity and usefulness in understanding the concept.
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I am reading the book: Multivariable Mathematics by Theodore Shifrin ... and am focused on Chapter 8, Section 2, Differential Forms ...

I need some help in order to fully understand some statements of Shifrin at the start of Chapter 8, Section 2 on the dual space ...

The relevant text from Shifrin reads as follows:
View attachment 8791In the above text from Shifrin we read the following:

" ... ... Then $$\phi = a_1 dx_1 + \ ... \ ... \ + a_n dx_n$$ ... ... "
Can someone please demonstrate and explain how/why $$\phi = a_1 dx_1 + \ ... \ ... \ + a_n dx_n$$ ... ...
Help will be much appreciated ... ...

Peter
 

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Hi Peter,

To establish the result, let $v=\sum_{k=1}^{n}v^{k}e_{k}$ be a vector in $\mathbb{R}^{n}$. Then, using linearity of $\phi$ judiciously, $$\phi(v) = \sum_{k=1}^{n}v^{k}\phi\left(e_{k}\right)=\sum_{k=1}^{n}a_{k}v^{k}.$$ On the other hand, using linearity of the $dx^{k}$,
\begin{align*}
\left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(v\right) &= \left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
&= a_{1}dx^{1}\left(\sum_{k=1}^{n}v^{k}e_{k}\right) +\cdots + a_{n}dx^{n}\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
&=a_{1}v^{1}+\cdots +a_{n}v^{n}.
\end{align*}
Since the two are equal and $v$ was chosen arbitrarily, we have $\phi=a_{1}dx^{1}+ \cdots + a_{n}dx^{n}.$
 
GJA said:
Hi Peter,

To establish the result, let $v=\sum_{k=1}^{n}v^{k}e_{k}$ be a vector in $\mathbb{R}^{n}$. Then, using linearity of $\phi$ judiciously, $$\phi(v) = \sum_{k=1}^{n}v^{k}\phi\left(e_{k}\right)=\sum_{k=1}^{n}a_{k}v^{k}.$$ On the other hand, using linearity of the $dx^{k}$,
\begin{align*}
\left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(v\right) &= \left(a_{1}dx^{1}+\cdots + a_{n}dx^{n}\right)\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
&= a_{1}dx^{1}\left(\sum_{k=1}^{n}v^{k}e_{k}\right) +\cdots + a_{n}dx^{n}\left(\sum_{k=1}^{n}v^{k}e_{k}\right)\\
&=a_{1}v^{1}+\cdots +a_{n}v^{n}.
\end{align*}
Since the two are equal and $v$ was chosen arbitrarily, we have $\phi=a_{1}dx^{1}+ \cdots + a_{n}dx^{n}.$
Thanks GJA ... most helpful ...

Appreciate your help ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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