
MHB Master
#1
February 10th, 2019,
00:21
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:
In the above text from Shifrin we read the following:
" ... ... Then $ \displaystyle \phi = a_1 dx_1 + \ ... \ ... \ + a_n dx_n$ ... ... "
Can someone please demonstrate and explain how/why $ \displaystyle \phi = a_1 dx_1 + \ ... \ ... \ + a_n dx_n$ ... ...
Help will be much appreciated ... ...
Peter

February 10th, 2019 00:21
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#2
February 10th, 2019,
00:37
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 $\phi=a_{1}dx^{1}+ \cdots + a_{n}dx^{n}.$

MHB Master
#3
February 10th, 2019,
01:22
Thread Author
Originally Posted by
GJA
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