Differential Geometry-Wedge product












3












$begingroup$


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







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  • 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












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














3












3








3


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

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oldest

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5












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















5












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














5












5








5





$begingroup$

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*}






share|cite|improve this answer











$endgroup$



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*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 26 '15 at 3:06

























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


















  • $begingroup$
    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


















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A Topological Invariant for $pi_3(U(n))$

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3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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