Write $omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.

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At each point $p ∈ mathbb R^3$, define a bilinear function $omega_p$
on $T_p(mathbb R^3)$ by

$$omega_p(textbfa,textbfb)=omega_p(beginbmatrix
a^1 \
a^2 \
a^3 \ endbmatrix, beginbmatrix
b^1 \
b^2 \
b^3 \ endbmatrix)= p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix $$ , for tangent vectors $textbfa,textbfb ∈ T_p(mathbb R^3)$, where $p^3$ is the third component of $p = (p^1,
p^2, p^3)$. Since $omega_p$ is an alternating bilinear function on
$T_p(mathbb R^3)$, $omega$ is a $2-$form on $mathbb R^3$. Write
$omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.

I know that $omega=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3,$ $textbfa=a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,textbfb=b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3.$

$omega(textbfa,textbfb)=omega(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)$ $$=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3 (a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=a_12a^1b^2+a_13a^1b^3+a_23a^2b^3.$$

How does this answer related to $$p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix. $$ Please help me.

differential-geometry proof-verification tensors exterior-algebra

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asked Jul 15 at 3:09

N. Maneesh

2,4271924

add a comment | 

up vote
1
down vote

favorite

At each point $p ∈ mathbb R^3$, define a bilinear function $omega_p$
on $T_p(mathbb R^3)$ by

$$omega_p(textbfa,textbfb)=omega_p(beginbmatrix
a^1 \
a^2 \
a^3 \ endbmatrix, beginbmatrix
b^1 \
b^2 \
b^3 \ endbmatrix)= p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix $$ , for tangent vectors $textbfa,textbfb ∈ T_p(mathbb R^3)$, where $p^3$ is the third component of $p = (p^1,
p^2, p^3)$. Since $omega_p$ is an alternating bilinear function on
$T_p(mathbb R^3)$, $omega$ is a $2-$form on $mathbb R^3$. Write
$omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.

I know that $omega=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3,$ $textbfa=a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,textbfb=b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3.$

$omega(textbfa,textbfb)=omega(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)$ $$=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3 (a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=a_12a^1b^2+a_13a^1b^3+a_23a^2b^3.$$

How does this answer related to $$p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix. $$ Please help me.

differential-geometry proof-verification tensors exterior-algebra

share|cite|improve this question

asked Jul 15 at 3:09

N. Maneesh

2,4271924

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

At each point $p ∈ mathbb R^3$, define a bilinear function $omega_p$
on $T_p(mathbb R^3)$ by

$$omega_p(textbfa,textbfb)=omega_p(beginbmatrix
a^1 \
a^2 \
a^3 \ endbmatrix, beginbmatrix
b^1 \
b^2 \
b^3 \ endbmatrix)= p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix $$ , for tangent vectors $textbfa,textbfb ∈ T_p(mathbb R^3)$, where $p^3$ is the third component of $p = (p^1,
p^2, p^3)$. Since $omega_p$ is an alternating bilinear function on
$T_p(mathbb R^3)$, $omega$ is a $2-$form on $mathbb R^3$. Write
$omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.

I know that $omega=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3,$ $textbfa=a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,textbfb=b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3.$

$omega(textbfa,textbfb)=omega(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)$ $$=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3 (a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=a_12a^1b^2+a_13a^1b^3+a_23a^2b^3.$$

How does this answer related to $$p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix. $$ Please help me.

differential-geometry proof-verification tensors exterior-algebra

share|cite|improve this question

asked Jul 15 at 3:09

N. Maneesh

2,4271924

At each point $p ∈ mathbb R^3$, define a bilinear function $omega_p$
on $T_p(mathbb R^3)$ by

$$omega_p(textbfa,textbfb)=omega_p(beginbmatrix
a^1 \
a^2 \
a^3 \ endbmatrix, beginbmatrix
b^1 \
b^2 \
b^3 \ endbmatrix)= p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix $$ , for tangent vectors $textbfa,textbfb ∈ T_p(mathbb R^3)$, where $p^3$ is the third component of $p = (p^1,
p^2, p^3)$. Since $omega_p$ is an alternating bilinear function on
$T_p(mathbb R^3)$, $omega$ is a $2-$form on $mathbb R^3$. Write
$omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.

I know that $omega=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3,$ $textbfa=a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,textbfb=b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3.$

$omega(textbfa,textbfb)=omega(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)$ $$=a_12dx^1 ∧dx^2+a_13dx^1 ∧dx^3+a_23dx^2 ∧dx^3 (a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=a_12a^1b^2+a_13a^1b^3+a_23a^2b^3.$$

How does this answer related to $$p^3det beginbmatrix
a^1&a^2 \
b^1 &b^2 \ endbmatrix. $$ Please help me.

differential-geometry proof-verification tensors exterior-algebra

share|cite|improve this question

asked Jul 15 at 3:09

N. Maneesh

2,4271924

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 3:09

N. Maneesh

2,4271924

asked Jul 15 at 3:09

N. Maneesh

2,4271924

asked Jul 15 at 3:09

N. Maneesh

2,4271924

2,4271924

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

The general formula for $omega_p=a_12dx_1wedge dx_2+a_13dx_1wedge dx_3+a_23dx_2wedge dx_3$ is $$omega_p(a,b)=a_12det beginpmatrixa^1&a^2\b^1&b^2endpmatrix+a_13detbeginpmatrixa^1&a^3\b^1&b^3endpmatrix+a_23detbeginpmatrixa^2&a^3\b^2&b^3endpmatrix.$$

Choose a special $omega_p=p^3dx^1wedge dx^2$,

$omega_p(a,b)=omega_p(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=p^3det beginpmatrixa^1&a^2\b^1&b^2endpmatrix$

share|cite|improve this answer

edited Jul 15 at 3:32

answered Jul 15 at 3:20

W. mu

54319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
  – N. Maneesh
  Jul 15 at 3:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. I got it.
  – N. Maneesh
  Jul 15 at 4:47
  

  
  
  
  

add a comment | 

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The general formula for $omega_p=a_12dx_1wedge dx_2+a_13dx_1wedge dx_3+a_23dx_2wedge dx_3$ is $$omega_p(a,b)=a_12det beginpmatrixa^1&a^2\b^1&b^2endpmatrix+a_13detbeginpmatrixa^1&a^3\b^1&b^3endpmatrix+a_23detbeginpmatrixa^2&a^3\b^2&b^3endpmatrix.$$

Choose a special $omega_p=p^3dx^1wedge dx^2$,

$omega_p(a,b)=omega_p(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=p^3det beginpmatrixa^1&a^2\b^1&b^2endpmatrix$

share|cite|improve this answer

edited Jul 15 at 3:32

answered Jul 15 at 3:20

W. mu

54319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
  – N. Maneesh
  Jul 15 at 3:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. I got it.
  – N. Maneesh
  Jul 15 at 4:47
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

The general formula for $omega_p=a_12dx_1wedge dx_2+a_13dx_1wedge dx_3+a_23dx_2wedge dx_3$ is $$omega_p(a,b)=a_12det beginpmatrixa^1&a^2\b^1&b^2endpmatrix+a_13detbeginpmatrixa^1&a^3\b^1&b^3endpmatrix+a_23detbeginpmatrixa^2&a^3\b^2&b^3endpmatrix.$$

Choose a special $omega_p=p^3dx^1wedge dx^2$,

$omega_p(a,b)=omega_p(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=p^3det beginpmatrixa^1&a^2\b^1&b^2endpmatrix$

share|cite|improve this answer

edited Jul 15 at 3:32

answered Jul 15 at 3:20

W. mu

54319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
  – N. Maneesh
  Jul 15 at 3:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. I got it.
  – N. Maneesh
  Jul 15 at 4:47
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The general formula for $omega_p=a_12dx_1wedge dx_2+a_13dx_1wedge dx_3+a_23dx_2wedge dx_3$ is $$omega_p(a,b)=a_12det beginpmatrixa^1&a^2\b^1&b^2endpmatrix+a_13detbeginpmatrixa^1&a^3\b^1&b^3endpmatrix+a_23detbeginpmatrixa^2&a^3\b^2&b^3endpmatrix.$$

Choose a special $omega_p=p^3dx^1wedge dx^2$,

$omega_p(a,b)=omega_p(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=p^3det beginpmatrixa^1&a^2\b^1&b^2endpmatrix$

share|cite|improve this answer

edited Jul 15 at 3:32

answered Jul 15 at 3:20

W. mu

54319

The general formula for $omega_p=a_12dx_1wedge dx_2+a_13dx_1wedge dx_3+a_23dx_2wedge dx_3$ is $$omega_p(a,b)=a_12det beginpmatrixa^1&a^2\b^1&b^2endpmatrix+a_13detbeginpmatrixa^1&a^3\b^1&b^3endpmatrix+a_23detbeginpmatrixa^2&a^3\b^2&b^3endpmatrix.$$

Choose a special $omega_p=p^3dx^1wedge dx^2$,

$omega_p(a,b)=omega_p(a^1fracpartialpartial x^1+a^2fracpartialpartial x^2+a^3fracpartialpartial x^3,b^1fracpartialpartial x^1+b^2fracpartialpartial x^2+b^3fracpartialpartial x^3)=p^3det beginpmatrixa^1&a^2\b^1&b^2endpmatrix$

share|cite|improve this answer

edited Jul 15 at 3:32

answered Jul 15 at 3:20

W. mu

54319

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 3:32

edited Jul 15 at 3:32

edited Jul 15 at 3:32

answered Jul 15 at 3:20

W. mu

54319

answered Jul 15 at 3:20

W. mu

54319

answered Jul 15 at 3:20

W. mu

54319

54319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
  – N. Maneesh
  Jul 15 at 3:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. I got it.
  – N. Maneesh
  Jul 15 at 4:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
  – N. Maneesh
  Jul 15 at 3:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. I got it.
  – N. Maneesh
  Jul 15 at 4:47
  

  
  
  
  

How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
– N. Maneesh
Jul 15 at 3:24

How do I find this? for getting $omega_p=p^3dx^1 wedge dx^2$. I need to get $a_12=p^3$ , $a_13=a_23=0.$ How do I get that condition?
– N. Maneesh
Jul 15 at 3:24

Thank you very much. I got it.
– N. Maneesh
Jul 15 at 4:47

Thank you very much. I got it.
– N. Maneesh
Jul 15 at 4:47

add a comment | 

 

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