Mixing
Differential Geometry-Wedge product
$begingroup$
How can we prove the following relation for differentiating the wedge product of a p-form $alpha_p$ and a q-form $beta_q$$$d(alpha_pwedgebeta _q)=dalpha_pwedgebeta_q+(-1)^{p}alpha_pwedge dbeta_q$$
differential-geometry
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
$endgroup$
add a comment |
$begingroup$
How can we prove the following relation for differentiating the wedge product of a p-form $alpha_p$ and a q-form $beta_q$$$d(alpha_pwedgebeta _q)=dalpha_pwedgebeta_q+(-1)^{p}alpha_pwedge dbeta_q$$
differential-geometry
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
$endgroup$
3
$begingroup$
There are many equivalent ways of organizing the definitions that go into that, and the answer to your question depends on which one of them has been used.
$endgroup$
– Harald Hanche-Olsen
Feb 25 '15 at 13:15
$begingroup$
I want to prove the above relation in simple way by using the definition of exterior derivative and wedge product
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 13:27
add a comment |
$begingroup$
How can we prove the following relation for differentiating the wedge product of a p-form $alpha_p$ and a q-form $beta_q$$$d(alpha_pwedgebeta _q)=dalpha_pwedgebeta_q+(-1)^{p}alpha_pwedge dbeta_q$$
differential-geometry
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
$endgroup$
How can we prove the following relation for differentiating the wedge product of a p-form $alpha_p$ and a q-form $beta_q$$$d(alpha_pwedgebeta _q)=dalpha_pwedgebeta_q+(-1)^{p}alpha_pwedge dbeta_q$$
differential-geometry
differential-geometry
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
asked Feb 25 '15 at 13:10
Roshan ShresthaRoshan Shrestha
28329
28329
3
$begingroup$
There are many equivalent ways of organizing the definitions that go into that, and the answer to your question depends on which one of them has been used.
$endgroup$
– Harald Hanche-Olsen
Feb 25 '15 at 13:15
$begingroup$
I want to prove the above relation in simple way by using the definition of exterior derivative and wedge product
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 13:27
add a comment |
3
$begingroup$
There are many equivalent ways of organizing the definitions that go into that, and the answer to your question depends on which one of them has been used.
$endgroup$
– Harald Hanche-Olsen
Feb 25 '15 at 13:15
$begingroup$
I want to prove the above relation in simple way by using the definition of exterior derivative and wedge product
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 13:27
3
3
$begingroup$
There are many equivalent ways of organizing the definitions that go into that, and the answer to your question depends on which one of them has been used.
$endgroup$
– Harald Hanche-Olsen
Feb 25 '15 at 13:15
$begingroup$
There are many equivalent ways of organizing the definitions that go into that, and the answer to your question depends on which one of them has been used.
$endgroup$
– Harald Hanche-Olsen
Feb 25 '15 at 13:15
$begingroup$
I want to prove the above relation in simple way by using the definition of exterior derivative and wedge product
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 13:27
$begingroup$
I want to prove the above relation in simple way by using the definition of exterior derivative and wedge product
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 13:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It suffices to take $alpha = f, dx^{i_1}wedgedotswedge dx^{i_p}$ and $beta=g,dx^{j_1}wedge dotswedge dx^{j_q}$. Then $alphawedgebeta = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $dalphawedgebeta$ and $alphawedge dbeta$ appearing in your formula.
EDIT: Doing the exercise for you:
begin{align*}
d(alphawedgebeta) &= d(fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (g,df + f,dg) wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (dfwedge dx^{i_1}wedgedotswedge dx^{i_p})wedge (g,dx^{j_1}wedge dotswedge dx^{j_q}) + \
&qquad (f,dg)wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalpha wedgebeta + (-1)^p (f,dx^{i_1}wedgedotswedge dx^{i_p})wedge (dgwedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalphawedgebeta + (-1)^p alphawedge dbeta.
end{align*}
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
1
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
It suffices to take $alpha = f, dx^{i_1}wedgedotswedge dx^{i_p}$ and $beta=g,dx^{j_1}wedge dotswedge dx^{j_q}$. Then $alphawedgebeta = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $dalphawedgebeta$ and $alphawedge dbeta$ appearing in your formula.
EDIT: Doing the exercise for you:
begin{align*}
d(alphawedgebeta) &= d(fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (g,df + f,dg) wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (dfwedge dx^{i_1}wedgedotswedge dx^{i_p})wedge (g,dx^{j_1}wedge dotswedge dx^{j_q}) + \
&qquad (f,dg)wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalpha wedgebeta + (-1)^p (f,dx^{i_1}wedgedotswedge dx^{i_p})wedge (dgwedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalphawedgebeta + (-1)^p alphawedge dbeta.
end{align*}
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
1
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
|
show 2 more comments
$begingroup$
It suffices to take $alpha = f, dx^{i_1}wedgedotswedge dx^{i_p}$ and $beta=g,dx^{j_1}wedge dotswedge dx^{j_q}$. Then $alphawedgebeta = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $dalphawedgebeta$ and $alphawedge dbeta$ appearing in your formula.
EDIT: Doing the exercise for you:
begin{align*}
d(alphawedgebeta) &= d(fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (g,df + f,dg) wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (dfwedge dx^{i_1}wedgedotswedge dx^{i_p})wedge (g,dx^{j_1}wedge dotswedge dx^{j_q}) + \
&qquad (f,dg)wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalpha wedgebeta + (-1)^p (f,dx^{i_1}wedgedotswedge dx^{i_p})wedge (dgwedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalphawedgebeta + (-1)^p alphawedge dbeta.
end{align*}
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
1
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
|
show 2 more comments
$begingroup$
It suffices to take $alpha = f, dx^{i_1}wedgedotswedge dx^{i_p}$ and $beta=g,dx^{j_1}wedge dotswedge dx^{j_q}$. Then $alphawedgebeta = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $dalphawedgebeta$ and $alphawedge dbeta$ appearing in your formula.
EDIT: Doing the exercise for you:
begin{align*}
d(alphawedgebeta) &= d(fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (g,df + f,dg) wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (dfwedge dx^{i_1}wedgedotswedge dx^{i_p})wedge (g,dx^{j_1}wedge dotswedge dx^{j_q}) + \
&qquad (f,dg)wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalpha wedgebeta + (-1)^p (f,dx^{i_1}wedgedotswedge dx^{i_p})wedge (dgwedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalphawedgebeta + (-1)^p alphawedge dbeta.
end{align*}
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
It suffices to take $alpha = f, dx^{i_1}wedgedotswedge dx^{i_p}$ and $beta=g,dx^{j_1}wedge dotswedge dx^{j_q}$. Then $alphawedgebeta = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$. Write out the product rule, and then keep track of the switches you must do to get $dalphawedgebeta$ and $alphawedge dbeta$ appearing in your formula.
EDIT: Doing the exercise for you:
begin{align*}
d(alphawedgebeta) &= d(fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (g,df + f,dg) wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= (dfwedge dx^{i_1}wedgedotswedge dx^{i_p})wedge (g,dx^{j_1}wedge dotswedge dx^{j_q}) + \
&qquad (f,dg)wedge (dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalpha wedgebeta + (-1)^p (f,dx^{i_1}wedgedotswedge dx^{i_p})wedge (dgwedge dx^{j_1}wedge dotswedge dx^{j_q}) \
&= dalphawedgebeta + (-1)^p alphawedge dbeta.
end{align*}
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
edited Feb 26 '15 at 3:06
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
answered Feb 25 '15 at 15:31
Ted ShifrinTed Shifrin
63.6k44591
63.6k44591
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
1
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
|
show 2 more comments
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
1
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Here is what I did like you said-$$alpha_p = f, dx^{i_1}wedgedotswedge dx^{i_p}$$ $$beta_q=g,dx^{j_1}wedge dotswedge dx^{j_q}$$ Then,$$alpha_pwedgebeta_q = fg, dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to apply product rule and the exterior derivative
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:19
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Also, we also know the formula for the wedge product $$alpha_pwedgebeta_q=(-1)^{pq}beta_qwedgealpha_p$$ Now, how to proceed from here if I want to prove the above relation through this
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 23:26
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Now,$$alpha_pwedgebeta_q=(fdg+gdf)dx^{i_1}wedgedotswedge dx^{i_p}wedge dx^{j_1}wedge dotswedge dx^{j_q}$$ Now, how to proceed from here.
$endgroup$
– Roshan Shrestha
Feb 26 '15 at 1:13
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
$begingroup$
Match up the two terms to the two terms in the formula.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 1:50
1
1
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
$begingroup$
Switching $dg$ with the $p$-form $dx^{i_1}wedgedotswedge dx^{i_p}$.
$endgroup$
– Ted Shifrin
Feb 26 '15 at 3:20
|
show 2 more comments
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$begingroup$
There are many equivalent ways of organizing the definitions that go into that, and the answer to your question depends on which one of them has been used.
$endgroup$
– Harald Hanche-Olsen
Feb 25 '15 at 13:15
$begingroup$
I want to prove the above relation in simple way by using the definition of exterior derivative and wedge product
$endgroup$
– Roshan Shrestha
Feb 25 '15 at 13:27