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5

1

$begingroup$

Suppose that we have a Riemannian metric $ds^2=Edu^2+2Fdudv+Gdv^2$ on a local coordinate neighborhood $(U;(u,v))$ prove that for the following vector fields:

$$e_1=frac1sqrtEfracpartialpartial u,quad e_2=frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)$$

The $1-$forms:
$$omega_1=sqrtEleft(du+fracFEdvright),quad omega_2=sqrtfracEG-F^2Edv$$
satisfies:
$$omega_i(e_k)=delta_ik$$

My work: Let $p(u,v)$ a differentiable function, I think that I must show fisrtly that $omega_1(e_1(p))=p$, i.e. $omega_1(e_1)=1$ the identity function.

$$omega_1left(tfrac1sqrtEtfracpartialpartial u(p)right)=tfrac1sqrtEtfracpartialpartial u(omega_1(p))$$
I do this beacuse an $1-form$ $alpha$ is such that $alpha(fX)=falpha(X)$. Then It is correct that
$omega_1(p)=sqrtEleft(pdu+fracFEpdvright)?$
and then apply the partial derivative, my problem is that I don't know ho operate the $1$-form. Anyone can guide me in how can I reach the result or a explicit form to operate with $omega_1$ and $omega_2$?

I'm using the book "Umehara, differential geometry of surfaces".

differential-geometry manifolds differential-forms

share|cite|improve this question

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

$endgroup$

add a comment | 

5

1

$begingroup$

Suppose that we have a Riemannian metric $ds^2=Edu^2+2Fdudv+Gdv^2$ on a local coordinate neighborhood $(U;(u,v))$ prove that for the following vector fields:

$$e_1=frac1sqrtEfracpartialpartial u,quad e_2=frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)$$

The $1-$forms:
$$omega_1=sqrtEleft(du+fracFEdvright),quad omega_2=sqrtfracEG-F^2Edv$$
satisfies:
$$omega_i(e_k)=delta_ik$$

My work: Let $p(u,v)$ a differentiable function, I think that I must show fisrtly that $omega_1(e_1(p))=p$, i.e. $omega_1(e_1)=1$ the identity function.

$$omega_1left(tfrac1sqrtEtfracpartialpartial u(p)right)=tfrac1sqrtEtfracpartialpartial u(omega_1(p))$$
I do this beacuse an $1-form$ $alpha$ is such that $alpha(fX)=falpha(X)$. Then It is correct that
$omega_1(p)=sqrtEleft(pdu+fracFEpdvright)?$
and then apply the partial derivative, my problem is that I don't know ho operate the $1$-form. Anyone can guide me in how can I reach the result or a explicit form to operate with $omega_1$ and $omega_2$?

I'm using the book "Umehara, differential geometry of surfaces".

differential-geometry manifolds differential-forms

share|cite|improve this question

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

$endgroup$

add a comment | 

$begingroup$

Suppose that we have a Riemannian metric $ds^2=Edu^2+2Fdudv+Gdv^2$ on a local coordinate neighborhood $(U;(u,v))$ prove that for the following vector fields:

$$e_1=frac1sqrtEfracpartialpartial u,quad e_2=frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)$$

The $1-$forms:
$$omega_1=sqrtEleft(du+fracFEdvright),quad omega_2=sqrtfracEG-F^2Edv$$
satisfies:
$$omega_i(e_k)=delta_ik$$

My work: Let $p(u,v)$ a differentiable function, I think that I must show fisrtly that $omega_1(e_1(p))=p$, i.e. $omega_1(e_1)=1$ the identity function.

$$omega_1left(tfrac1sqrtEtfracpartialpartial u(p)right)=tfrac1sqrtEtfracpartialpartial u(omega_1(p))$$
I do this beacuse an $1-form$ $alpha$ is such that $alpha(fX)=falpha(X)$. Then It is correct that
$omega_1(p)=sqrtEleft(pdu+fracFEpdvright)?$
and then apply the partial derivative, my problem is that I don't know ho operate the $1$-form. Anyone can guide me in how can I reach the result or a explicit form to operate with $omega_1$ and $omega_2$?

I'm using the book "Umehara, differential geometry of surfaces".

differential-geometry manifolds differential-forms

share|cite|improve this question

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

$endgroup$

share|cite|improve this question

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

share|cite|improve this question

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

asked May 6 at 1:02

Ragnar1204Ragnar1204

531417

2 Answers
2

active

oldest

votes

4

$begingroup$

I think you can do this by direct calculation. Fix $pin U$. If $X_pin TU_p$, then $X_p=apartial_u+bpartial_v$ for some $a,bin mathbb R$, and by definition of the $1-$forms $du$ and $dv$, we have $du(X_p)=a$ and $dv(X_p)=b$. Now, apply this to the tangent vectors $e_1$ and $e_2$:

$omega_1(e_1)=sqrtEleft(du+fracFEdvright)frac1sqrtEfracpartialpartial u=sqrt Efrac1sqrt E(partial_uu)=fracsqrt Esqrt E=1$

Similarly,

$omega_1(e_2)=sqrtEleft(du+fracFEdvright)left(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right )=-fracsqrt EsqrtEG-F^2left (fracFsqrt E-fracFsqrt EEright )=0.$

$omega_2(e_1)=sqrtfracEG-F^2Edvleft(frac1sqrtEfracpartialpartial u right)=sqrtfracEG-F^2E^2partial_u v=0$

and

$omega_2(e_2)=sqrtfracEG-F^2Edvleft(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right)=left(sqrtfracEG-F^2Eright)cdot left(fracsqrt EsqrtEG-F^2right)=1$

share|cite|improve this answer

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, what a surprising clear answer! thanks a lot, you've clarified this mess for me, really thanks! +1
  $endgroup$
  – Ragnar1204
  May 6 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Matematleta
  May 6 at 4:19
  

  
  
  
  

add a comment | 

3

$begingroup$

Note that $e_i$ is an orthonormal vector fields wrt $ds^2$. Hence we let $
omega_i=ds^2(e_i,cdot)$ which is dual frame.

share|cite|improve this answer

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363

$endgroup$

add a comment | 

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4

$begingroup$

I think you can do this by direct calculation. Fix $pin U$. If $X_pin TU_p$, then $X_p=apartial_u+bpartial_v$ for some $a,bin mathbb R$, and by definition of the $1-$forms $du$ and $dv$, we have $du(X_p)=a$ and $dv(X_p)=b$. Now, apply this to the tangent vectors $e_1$ and $e_2$:

$omega_1(e_1)=sqrtEleft(du+fracFEdvright)frac1sqrtEfracpartialpartial u=sqrt Efrac1sqrt E(partial_uu)=fracsqrt Esqrt E=1$

Similarly,

$omega_1(e_2)=sqrtEleft(du+fracFEdvright)left(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right )=-fracsqrt EsqrtEG-F^2left (fracFsqrt E-fracFsqrt EEright )=0.$

$omega_2(e_1)=sqrtfracEG-F^2Edvleft(frac1sqrtEfracpartialpartial u right)=sqrtfracEG-F^2E^2partial_u v=0$

and

$omega_2(e_2)=sqrtfracEG-F^2Edvleft(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right)=left(sqrtfracEG-F^2Eright)cdot left(fracsqrt EsqrtEG-F^2right)=1$

share|cite|improve this answer

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, what a surprising clear answer! thanks a lot, you've clarified this mess for me, really thanks! +1
  $endgroup$
  – Ragnar1204
  May 6 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Matematleta
  May 6 at 4:19
  

  
  
  
  

add a comment | 

4

$begingroup$

I think you can do this by direct calculation. Fix $pin U$. If $X_pin TU_p$, then $X_p=apartial_u+bpartial_v$ for some $a,bin mathbb R$, and by definition of the $1-$forms $du$ and $dv$, we have $du(X_p)=a$ and $dv(X_p)=b$. Now, apply this to the tangent vectors $e_1$ and $e_2$:

$omega_1(e_1)=sqrtEleft(du+fracFEdvright)frac1sqrtEfracpartialpartial u=sqrt Efrac1sqrt E(partial_uu)=fracsqrt Esqrt E=1$

Similarly,

$omega_1(e_2)=sqrtEleft(du+fracFEdvright)left(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right )=-fracsqrt EsqrtEG-F^2left (fracFsqrt E-fracFsqrt EEright )=0.$

$omega_2(e_1)=sqrtfracEG-F^2Edvleft(frac1sqrtEfracpartialpartial u right)=sqrtfracEG-F^2E^2partial_u v=0$

and

$omega_2(e_2)=sqrtfracEG-F^2Edvleft(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right)=left(sqrtfracEG-F^2Eright)cdot left(fracsqrt EsqrtEG-F^2right)=1$

share|cite|improve this answer

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, what a surprising clear answer! thanks a lot, you've clarified this mess for me, really thanks! +1
  $endgroup$
  – Ragnar1204
  May 6 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Matematleta
  May 6 at 4:19
  

  
  
  
  

add a comment | 

$begingroup$

I think you can do this by direct calculation. Fix $pin U$. If $X_pin TU_p$, then $X_p=apartial_u+bpartial_v$ for some $a,bin mathbb R$, and by definition of the $1-$forms $du$ and $dv$, we have $du(X_p)=a$ and $dv(X_p)=b$. Now, apply this to the tangent vectors $e_1$ and $e_2$:

$omega_1(e_1)=sqrtEleft(du+fracFEdvright)frac1sqrtEfracpartialpartial u=sqrt Efrac1sqrt E(partial_uu)=fracsqrt Esqrt E=1$

Similarly,

$omega_1(e_2)=sqrtEleft(du+fracFEdvright)left(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right )=-fracsqrt EsqrtEG-F^2left (fracFsqrt E-fracFsqrt EEright )=0.$

$omega_2(e_1)=sqrtfracEG-F^2Edvleft(frac1sqrtEfracpartialpartial u right)=sqrtfracEG-F^2E^2partial_u v=0$

and

$omega_2(e_2)=sqrtfracEG-F^2Edvleft(frac-1sqrtEG-F^2left(fracFsqrtEfracpartialpartial u-sqrtEfracpartialpartial vright)right)=left(sqrtfracEG-F^2Eright)cdot left(fracsqrt EsqrtEG-F^2right)=1$

share|cite|improve this answer

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

$endgroup$

share|cite|improve this answer

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

answered May 6 at 1:56

MatematletaMatematleta

12.5k21020

3

$begingroup$

Note that $e_i$ is an orthonormal vector fields wrt $ds^2$. Hence we let $
omega_i=ds^2(e_i,cdot)$ which is dual frame.

share|cite|improve this answer

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363

$endgroup$

add a comment | 

3

$begingroup$

Note that $e_i$ is an orthonormal vector fields wrt $ds^2$. Hence we let $
omega_i=ds^2(e_i,cdot)$ which is dual frame.

share|cite|improve this answer

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363

$endgroup$

add a comment | 

$begingroup$

Note that $e_i$ is an orthonormal vector fields wrt $ds^2$. Hence we let $
omega_i=ds^2(e_i,cdot)$ which is dual frame.

share|cite|improve this answer

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363

$endgroup$

share|cite|improve this answer

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363

answered May 6 at 1:43

HK LeeHK Lee

14.2k52363