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1

$begingroup$

I am reading a proof of a basic result in differential geometry, which I am having trouble following.

The statement is the following: Suppose $X$ is a vector field on $M$, a $C^infty$ manifold, and $(U; x_1, ..., x_n)$ be a coordinate chart. Then $X|_U$ can be uniquely expressed as
$$
X|_U = sum_j=1^n a_j fracpartialpartial x_j
$$
where $a_j in C^infty(M)$.

Proof presented goes as follows: Define $a_j = X x_j.$ Let $ f in C^infty(U)$ and $p in U$. Then we may write
$$
f = f(p) + sum_i fracpartial fpartial x_i (p) (x_i - x_i(p)) + h,
$$
where $h$ is a linear combination of products $h_1h_2$ with $h_1(p) = h_2(p) = 0$.
Then since $X$ is a derivation
$$
X(h_1 h_2)(p) = h_1(p) X h_2(p) + h_2(p) Xh_1(p) =0.
$$
From which point the statement is easy enough to deduce.

My question is: We only know $f$ is in $C^infty(U)$. How does one deduce about $h$ only from this information? I am having difficulty following this because $f$ may not have a converging Taylor series in which case I don't see how one can deduce that $h$ is a linear combination of products $h_1h_2$ with $h_1(p) = h_2(p) = 0$ (I can see this if we assume $f$ has a Taylor series expansion everywhere... but not sure how to see this here).

Thank you.

calculus multivariable-calculus differential-geometry vector-spaces taylor-expansion

share|cite|improve this question

edited Mar 29 at 15:03

Takeshi Gouda

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

$endgroup$

add a comment | 

1

$begingroup$

I am reading a proof of a basic result in differential geometry, which I am having trouble following.

The statement is the following: Suppose $X$ is a vector field on $M$, a $C^infty$ manifold, and $(U; x_1, ..., x_n)$ be a coordinate chart. Then $X|_U$ can be uniquely expressed as
$$
X|_U = sum_j=1^n a_j fracpartialpartial x_j
$$
where $a_j in C^infty(M)$.

Proof presented goes as follows: Define $a_j = X x_j.$ Let $ f in C^infty(U)$ and $p in U$. Then we may write
$$
f = f(p) + sum_i fracpartial fpartial x_i (p) (x_i - x_i(p)) + h,
$$
where $h$ is a linear combination of products $h_1h_2$ with $h_1(p) = h_2(p) = 0$.
Then since $X$ is a derivation
$$
X(h_1 h_2)(p) = h_1(p) X h_2(p) + h_2(p) Xh_1(p) =0.
$$
From which point the statement is easy enough to deduce.

My question is: We only know $f$ is in $C^infty(U)$. How does one deduce about $h$ only from this information? I am having difficulty following this because $f$ may not have a converging Taylor series in which case I don't see how one can deduce that $h$ is a linear combination of products $h_1h_2$ with $h_1(p) = h_2(p) = 0$ (I can see this if we assume $f$ has a Taylor series expansion everywhere... but not sure how to see this here).

Thank you.

calculus multivariable-calculus differential-geometry vector-spaces taylor-expansion

share|cite|improve this question

edited Mar 29 at 15:03

Takeshi Gouda

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

$endgroup$

add a comment | 

$begingroup$

I am reading a proof of a basic result in differential geometry, which I am having trouble following.

The statement is the following: Suppose $X$ is a vector field on $M$, a $C^infty$ manifold, and $(U; x_1, ..., x_n)$ be a coordinate chart. Then $X|_U$ can be uniquely expressed as
$$
X|_U = sum_j=1^n a_j fracpartialpartial x_j
$$
where $a_j in C^infty(M)$.

Proof presented goes as follows: Define $a_j = X x_j.$ Let $ f in C^infty(U)$ and $p in U$. Then we may write
$$
f = f(p) + sum_i fracpartial fpartial x_i (p) (x_i - x_i(p)) + h,
$$
where $h$ is a linear combination of products $h_1h_2$ with $h_1(p) = h_2(p) = 0$.
Then since $X$ is a derivation
$$
X(h_1 h_2)(p) = h_1(p) X h_2(p) + h_2(p) Xh_1(p) =0.
$$
From which point the statement is easy enough to deduce.

My question is: We only know $f$ is in $C^infty(U)$. How does one deduce about $h$ only from this information? I am having difficulty following this because $f$ may not have a converging Taylor series in which case I don't see how one can deduce that $h$ is a linear combination of products $h_1h_2$ with $h_1(p) = h_2(p) = 0$ (I can see this if we assume $f$ has a Taylor series expansion everywhere... but not sure how to see this here).

Thank you.

calculus multivariable-calculus differential-geometry vector-spaces taylor-expansion

share|cite|improve this question

edited Mar 29 at 15:03

Takeshi Gouda

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

$endgroup$

share|cite|improve this question

edited Mar 29 at 15:03

Takeshi Gouda

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

share|cite|improve this question

edited Mar 29 at 15:03

Takeshi Gouda

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

asked Mar 29 at 13:54

Takeshi GoudaTakeshi Gouda

264

1 Answer
1

active

oldest

votes

0

$begingroup$

The standard proof is to write
$$f(x)=f(p) + sum g_i(x)(x_i-x_i(p))$$
where $g_i$ is smooth and $g_i(p) = partial f/partial x_i(p)$. Working in $Bbb R^n$ with $p=0$, just write
$$f(x) = f(0) + int_0^1 frac ddt f(tx),dt$$
and use the chain rule.

share|cite|improve this answer

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This makes so much more sense! Thank you!
  $endgroup$
  – Takeshi Gouda
  Mar 30 at 12:39
  

  
  
  
  

add a comment | 

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0

$begingroup$

The standard proof is to write
$$f(x)=f(p) + sum g_i(x)(x_i-x_i(p))$$
where $g_i$ is smooth and $g_i(p) = partial f/partial x_i(p)$. Working in $Bbb R^n$ with $p=0$, just write
$$f(x) = f(0) + int_0^1 frac ddt f(tx),dt$$
and use the chain rule.

share|cite|improve this answer

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This makes so much more sense! Thank you!
  $endgroup$
  – Takeshi Gouda
  Mar 30 at 12:39
  

  
  
  
  

add a comment | 

0

$begingroup$

The standard proof is to write
$$f(x)=f(p) + sum g_i(x)(x_i-x_i(p))$$
where $g_i$ is smooth and $g_i(p) = partial f/partial x_i(p)$. Working in $Bbb R^n$ with $p=0$, just write
$$f(x) = f(0) + int_0^1 frac ddt f(tx),dt$$
and use the chain rule.

share|cite|improve this answer

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This makes so much more sense! Thank you!
  $endgroup$
  – Takeshi Gouda
  Mar 30 at 12:39
  

  
  
  
  

add a comment | 

$begingroup$

The standard proof is to write
$$f(x)=f(p) + sum g_i(x)(x_i-x_i(p))$$
where $g_i$ is smooth and $g_i(p) = partial f/partial x_i(p)$. Working in $Bbb R^n$ with $p=0$, just write
$$f(x) = f(0) + int_0^1 frac ddt f(tx),dt$$
and use the chain rule.

share|cite|improve this answer

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692

$endgroup$

share|cite|improve this answer

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692

answered Mar 29 at 16:13

Ted ShifrinTed Shifrin

64.8k44692