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I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ...

I need help in order to fully understand Theorem 2.2 and the remarks after the theorem ...Theorem 2.2 and the remarks after the theorem read as follows:View attachment 8642

View attachment 8643My questions on the above text from Tu are as follows:

At the end of the proof of the above theorem by Tu we read the following:

" ... ... This proves that \(\displaystyle D = D_v\) for \(\displaystyle v = \langle D x^1 / ... / ... / D x^n \rangle\) ... ..."The above implies \(\displaystyle v^i = D x^i\) ... ... but what does \(\displaystyle D x^i\) mean and how exactly is it equal to \(\displaystyle v^i\) ... ... ?

" ... ... Under the vector space isomorphism \(\displaystyle T_p ( \mathbb{R}^n ) \simeq D_p ( \mathbb{R}^n )\), the standard basis \(\displaystyle e_1 \ ... \ ... \ e_n\) for \(\displaystyle T_p ( \mathbb{R}^n )\) corresponds to the set \(\displaystyle \partial / \partial x^1 \mid_p \ ... \ ... \ \partial / \partial x^1 \mid_p\) of partial derivatives. From now on, we will make this identification and write a tangent vector \(\displaystyle v = \langle v^1 \ ... \ ... \ v^n \rangle = \sum v^i e_i\) as

\(\displaystyle v = \sum v^i \frac{ \partial }{ \partial x^i } \mid_p\) ... ... "I must say I was surprised and a little perplexed at a basis consisting of unevaluated partial derivatives ... but on reflection ... \(\displaystyle D_p ( \mathbb{R}^n )\) is a space of derivations, that is linear maps satisfying the Leibniz rule ... so having a basis consisting of linear maps \(\displaystyle \frac{ \partial }{ \partial x^i } \mid_p\) is not too surprising ... but how to think of a tangent vector as

\(\displaystyle v = \sum v^i \frac{ \partial }{ \partial x^i } \mid_p\) ... ...is a bit of a mystery ... certainly it is hard to imagine the above quantity as an arrow at point \(\displaystyle p\) ... also why we are dong this is a bit of a mystery too ...

Can someone please clarify ... ...

Hope someone can help to clarify the above issues ... ...

Peter

==========================================================================================It may help MHB readers of the above post to have access to the start of Tu's section on derivations ... so I am providing access to the same ... as follows:

View attachment 8644

Hope the above text helps with context and meaning ...

Peter

I need help in order to fully understand Theorem 2.2 and the remarks after the theorem ...Theorem 2.2 and the remarks after the theorem read as follows:View attachment 8642

View attachment 8643My questions on the above text from Tu are as follows:

**Question 1**

At the end of the proof of the above theorem by Tu we read the following:

" ... ... This proves that \(\displaystyle D = D_v\) for \(\displaystyle v = \langle D x^1 / ... / ... / D x^n \rangle\) ... ..."The above implies \(\displaystyle v^i = D x^i\) ... ... but what does \(\displaystyle D x^i\) mean and how exactly is it equal to \(\displaystyle v^i\) ... ... ?

__In some remarks after the end of the proof of the above theorem by Tu we read the following:__**Question 2**" ... ... Under the vector space isomorphism \(\displaystyle T_p ( \mathbb{R}^n ) \simeq D_p ( \mathbb{R}^n )\), the standard basis \(\displaystyle e_1 \ ... \ ... \ e_n\) for \(\displaystyle T_p ( \mathbb{R}^n )\) corresponds to the set \(\displaystyle \partial / \partial x^1 \mid_p \ ... \ ... \ \partial / \partial x^1 \mid_p\) of partial derivatives. From now on, we will make this identification and write a tangent vector \(\displaystyle v = \langle v^1 \ ... \ ... \ v^n \rangle = \sum v^i e_i\) as

\(\displaystyle v = \sum v^i \frac{ \partial }{ \partial x^i } \mid_p\) ... ... "I must say I was surprised and a little perplexed at a basis consisting of unevaluated partial derivatives ... but on reflection ... \(\displaystyle D_p ( \mathbb{R}^n )\) is a space of derivations, that is linear maps satisfying the Leibniz rule ... so having a basis consisting of linear maps \(\displaystyle \frac{ \partial }{ \partial x^i } \mid_p\) is not too surprising ... but how to think of a tangent vector as

\(\displaystyle v = \sum v^i \frac{ \partial }{ \partial x^i } \mid_p\) ... ...is a bit of a mystery ... certainly it is hard to imagine the above quantity as an arrow at point \(\displaystyle p\) ... also why we are dong this is a bit of a mystery too ...

Can someone please clarify ... ...

Hope someone can help to clarify the above issues ... ...

Peter

==========================================================================================It may help MHB readers of the above post to have access to the start of Tu's section on derivations ... so I am providing access to the same ... as follows:

View attachment 8644

Hope the above text helps with context and meaning ...

Peter