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How to calculate the pullback of a $k$-form explicitly
I'm having trouble doing actual computations of the pullback of a $k$-form. I know that a given differentiable map $alpha: mathbb{R}^{m} rightarrow mathbb{R}^{n}$ induces a map $alpha^{*}: Omega^{k}(mathbb{R}^{n}) rightarrow Omega^{k}(mathbb{R}^{m})$ between $k$-forms. I can recite how this map is defined and understand why it is well defined, but when I'm given a particular $alpha$ and a particular $omega in Omega^{k}(mathbb{R}^{n})$, I cannot compute $alpha^{*}omega$.
For example I found an exercise (Analysis on Manifolds, by Munkres) where $omega = xy , dx + 2z , dy - y , dzin Omega^{k}(mathbb{R}^{3})$ and $alpha: mathbb{R}^{2} rightarrow mathbb{R}^{3}$ is defined as $alpha(u, v) = (uv, u^{2}, 3u + v)$, but I got lost wile expanding the definition of $alpha^{*} omega$. How can I calculate this?
Note: This exercise is not a homework, so feel free to illustrate the process with any $alpha$ and $omega$ you wish.
differential-geometry differential-forms
edited Nov 22 '13 at 18:25
gofvonx
1,042828
asked Nov 22 '13 at 0:31
Tony Burbano
23643
add a comment |
I'm having trouble doing actual computations of the pullback of a $k$-form. I know that a given differentiable map $alpha: mathbb{R}^{m} rightarrow mathbb{R}^{n}$ induces a map $alpha^{*}: Omega^{k}(mathbb{R}^{n}) rightarrow Omega^{k}(mathbb{R}^{m})$ between $k$-forms. I can recite how this map is defined and understand why it is well defined, but when I'm given a particular $alpha$ and a particular $omega in Omega^{k}(mathbb{R}^{n})$, I cannot compute $alpha^{*}omega$.
For example I found an exercise (Analysis on Manifolds, by Munkres) where $omega = xy , dx + 2z , dy - y , dzin Omega^{k}(mathbb{R}^{3})$ and $alpha: mathbb{R}^{2} rightarrow mathbb{R}^{3}$ is defined as $alpha(u, v) = (uv, u^{2}, 3u + v)$, but I got lost wile expanding the definition of $alpha^{*} omega$. How can I calculate this?
Note: This exercise is not a homework, so feel free to illustrate the process with any $alpha$ and $omega$ you wish.
differential-geometry differential-forms
edited Nov 22 '13 at 18:25
gofvonx
1,042828
asked Nov 22 '13 at 0:31
Tony Burbano
23643
Can you post your calculation so we can see where you get lost?
– Bruno Joyal
Nov 22 '13 at 0:57
add a comment |
I'm having trouble doing actual computations of the pullback of a $k$-form. I know that a given differentiable map $alpha: mathbb{R}^{m} rightarrow mathbb{R}^{n}$ induces a map $alpha^{*}: Omega^{k}(mathbb{R}^{n}) rightarrow Omega^{k}(mathbb{R}^{m})$ between $k$-forms. I can recite how this map is defined and understand why it is well defined, but when I'm given a particular $alpha$ and a particular $omega in Omega^{k}(mathbb{R}^{n})$, I cannot compute $alpha^{*}omega$.
For example I found an exercise (Analysis on Manifolds, by Munkres) where $omega = xy , dx + 2z , dy - y , dzin Omega^{k}(mathbb{R}^{3})$ and $alpha: mathbb{R}^{2} rightarrow mathbb{R}^{3}$ is defined as $alpha(u, v) = (uv, u^{2}, 3u + v)$, but I got lost wile expanding the definition of $alpha^{*} omega$. How can I calculate this?
Note: This exercise is not a homework, so feel free to illustrate the process with any $alpha$ and $omega$ you wish.
differential-geometry differential-forms
edited Nov 22 '13 at 18:25
gofvonx
1,042828
asked Nov 22 '13 at 0:31
Tony Burbano
23643
I'm having trouble doing actual computations of the pullback of a $k$-form. I know that a given differentiable map $alpha: mathbb{R}^{m} rightarrow mathbb{R}^{n}$ induces a map $alpha^{*}: Omega^{k}(mathbb{R}^{n}) rightarrow Omega^{k}(mathbb{R}^{m})$ between $k$-forms. I can recite how this map is defined and understand why it is well defined, but when I'm given a particular $alpha$ and a particular $omega in Omega^{k}(mathbb{R}^{n})$, I cannot compute $alpha^{*}omega$.
For example I found an exercise (Analysis on Manifolds, by Munkres) where $omega = xy , dx + 2z , dy - y , dzin Omega^{k}(mathbb{R}^{3})$ and $alpha: mathbb{R}^{2} rightarrow mathbb{R}^{3}$ is defined as $alpha(u, v) = (uv, u^{2}, 3u + v)$, but I got lost wile expanding the definition of $alpha^{*} omega$. How can I calculate this?
Note: This exercise is not a homework, so feel free to illustrate the process with any $alpha$ and $omega$ you wish.
differential-geometry differential-forms
differential-geometry differential-forms
edited Nov 22 '13 at 18:25
gofvonx
1,042828
asked Nov 22 '13 at 0:31
Tony Burbano
23643
edited Nov 22 '13 at 18:25
gofvonx
1,042828
asked Nov 22 '13 at 0:31
Tony Burbano
23643
edited Nov 22 '13 at 18:25
gofvonx
1,042828
edited Nov 22 '13 at 18:25
gofvonx
1,042828
edited Nov 22 '13 at 18:25
gofvonx
1,042828
1,042828
asked Nov 22 '13 at 0:31
Tony Burbano
23643
asked Nov 22 '13 at 0:31
Tony Burbano
23643
asked Nov 22 '13 at 0:31
Tony Burbano
23643
23643
Can you post your calculation so we can see where you get lost?
– Bruno Joyal
Nov 22 '13 at 0:57
add a comment |
Can you post your calculation so we can see where you get lost?
– Bruno Joyal
Nov 22 '13 at 0:57
Can you post your calculation so we can see where you get lost?
– Bruno Joyal
Nov 22 '13 at 0:57
Can you post your calculation so we can see where you get lost?
– Bruno Joyal
Nov 22 '13 at 0:57
add a comment |
3 Answers
3
active
oldest
votes
Instead of thinking of $alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,qquad y=u^2,qquad z =3u+v.
$$
Then
$$
dx ;=; frac{partial x}{partial u}du+frac{partial x}{partial v}dv ;=; v,du+u,dv
$$
and similarly
$$
dy ;=; 2u,duqquadtext{and}qquad dz;=;3,du+dv.
$$
Therefore,
$$
begin{align*}
xy,dx + 2z,dy - y,dz ;&=; (uv)(u^2)(v,du+u,dv)+2(3u+v)(2u,du)-(u^2)(3,du+dv)\[1ex]
&=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
end{align*}
$$
We conclude that
$$
alpha^*(xy,dx + 2z,dy - y,dz) ;=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
$$
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
add a comment |
In some of my research I encountered $2$-forms that were given by $$omega = dx wedge dp + dy wedge dq$$
and a map $$i : (u,v) mapsto (u,v,f_u,-f_v)$$
for a general smooth map $f : (u,v) mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*omega$. So,
begin{align}
i^*omega &= i^*(dx wedge dp + dy wedge dq) \
&= d(x circ i)wedge d(p circ i) + d(y circ i)wedge d(q circ i).
end{align}
Now, calculating each terms gives
begin{align}
d(x circ i) &= d(u) = du, \
d(y circ i) &= d(v) = dv, \
d(p circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \
d(q circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv.
end{align}
Then, the pullback is given by
begin{align}
i^*omega &= du wedge (f_{uu}du + f_{uv}dv) - dv wedge (f_{vu}du + f_{vv}dv) \
&= du wedge (f_{uu}du) + du wedge (f_{uv}dv) - dv wedge (f_{vu}du) - dv wedge (f_{vv}dv).
end{align}
Now here, since the wedge product is $C^infty(M)$-bilinear rather than just $Bbb R$-bilinear, and $du wedge dv = -dvwedge du$. Using this property we have
$$i^*omega =2f_{uv} du wedge dv.$$
I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $alpha$ and $omega$!
answered Jul 24 '17 at 13:42
Kevin
5,410822
add a comment |
An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too verbose...
If $omegainOmega^1(N)$ and $alpha:Mrightarrow N$. The aim of the pullback is to define a form $alpha^*omegainOmega^1(M)$ from a form $omegainOmega^1(N)$.
A 1-form $omega$ evaluated at $n=(x,y,z)in N$ is $$omega[n]=omega_x(n)dx+omega_y(n)dy+omega_z(n)dz$$ where
- $(omega_x(n),omega_y(n),omega_z(n))inmathbb{R}^3$
$dx, dy, dz$ are also 1-forms, the dual basis of $frac{partial}{partial x},frac{partial}{partial y},frac{partial}{partial z}$.
$omega[n]in T_nN^*$ eats one vector $bin T_nN$ (a tangent vector at point $n$):
begin{align*}
omega[n](b^xfrac{partial}{partial x}+b^yfrac{partial}{partial y}+b^zfrac{partial}{partial z}) &=omega_x(n)b^x+omega_y(n)b^y+omega_z(n)b^zinmathbb{R}
end{align*}
Now the objective is to define $alpha^*omegainOmega^1(M)$ from $omegainOmega^1(N)$. Given a point $m=(u,v)in M$ and a vector $ain T_m M$, a natural idea is to use the point $n=alpha(m)in N$ and to pushforward a vector $ain T_mM$ to get a vector $alpha_*(a)in T_{alpha(m)}N$. Actually this is how $alpha^*omega$ is defined:
$$(alpha^*omega)[m](a)=omega[alpha(m)](alpha_*(a))$$
Now we must compute $alpha_*(a)in T_{alpha(m)}N$, see at the end for details, we get:
$$
alpha_*(a)=dalpha^x[alpha(m)](a)frac{partial}{partial x}+dalpha^y[alpha(m)](a)frac{partial}{partial y}+dalpha^z[alpha(m))](a)frac{partial}{partial z}
$$
Thus:
begin{align*}
(alpha^*omega)[m](a)&=overbrace{omega_x[alpha(m)]dalpha^x[alpha(m)](a)}^{text{term }dx(frac{partial}{partial x})=1}+
overbrace{omega_y[alpha(m)]dalpha^y[alpha(m)](a)}^{text{term }dy(frac{partial}{partial y})=1}+
overbrace{omega_z[alpha(m)]dalpha^z[alpha(m)](a)}^{text{term }dz(frac{partial}{partial z})=1} in mathbb{R}
end{align*}
We can drop the argument, vector $a=a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v}$, it remains:
$$
(alpha^*omega)[m]=omega_x[alpha(m)]dalpha^x[alpha(m)]+
omega_y[alpha(m)]dalpha^y[alpha(m)]+
omega_z[alpha(m)]dalpha^z[alpha(m)]in T_mM^*
$$
In the example: $(omega_x,omega_y,omega_z)=(xy,2z,-y)$ and $alpha: (u,v)mapsto (uv,u^2,3u+v)$, therefore:
- $(omega_x[alpha(m)],omega_y[alpha(m)],omega_z[alpha(m)])=(u^3v,6u+2v,-u^2)$
- $dalpha^x[alpha(m)]=vdu+udv$
$dalpha^y[alpha(m)]=2udu$- $dalpha^z[alpha(m)]=3du+dv$
By substitution we get the expected result:
$$
(alpha^*omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)
$$
$$
alpha^*omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dvinOmega^1(M)
$$
It remains to explain how to compute $alpha_star(a)in T_{alpha(m)}N$ from a tangent vector $ain T_mM$
One can interpret a vector $ain T_mM$ as a first order differential operator acting on functions $f:Mtomathbb{R}$.
More explicitly we have:
$$
a[f]=(a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v})f=df[m](a) inmathbb{R}
$$
The pushforward $alpha_*$ transform a vector $ain T_mM$ into a vector $alpha_star(a)in T_{alpha(m)}N$, thus it must act on functions $g:Ntomathbb{R}$. A natural definition is:
$$
alpha_star(a)[g]=d(gcircalpha)[m](a)inmathbb{R}
$$
Expanding this formula we get:
begin{align*}
d(gcircalpha)[m](a)&=left((frac{partial g}{partial x}frac{partial alpha^x}{partial u}+frac{partial g}{partial y}frac{partial alpha^y}{partial u}+frac{partial g}{partial z}frac{partial alpha^z}{partial u})du+(frac{partial g}{partial x}frac{partial alpha^x}{partial v}+frac{partial g}{partial y}frac{partial alpha^y}{partial v}+frac{partial g}{partial z}frac{partial alpha^z}{partial v})dvright)(a) \
&= left((frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)frac{partial g}{partial z}right)(a) \
&= (frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)(a)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)(a)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)(a)frac{partial g}{partial z} \
&= left( dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z} right) g
end{align*}
we get the expected result:
$$
alpha_star(a)=dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z}in T_{alpha(m)}N
$$
CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $alpha$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
add a comment |
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3 Answers
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3 Answers
3
active
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votes
Instead of thinking of $alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,qquad y=u^2,qquad z =3u+v.
$$
Then
$$
dx ;=; frac{partial x}{partial u}du+frac{partial x}{partial v}dv ;=; v,du+u,dv
$$
and similarly
$$
dy ;=; 2u,duqquadtext{and}qquad dz;=;3,du+dv.
$$
Therefore,
$$
begin{align*}
xy,dx + 2z,dy - y,dz ;&=; (uv)(u^2)(v,du+u,dv)+2(3u+v)(2u,du)-(u^2)(3,du+dv)\[1ex]
&=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
end{align*}
$$
We conclude that
$$
alpha^*(xy,dx + 2z,dy - y,dz) ;=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
$$
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
add a comment |
Instead of thinking of $alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,qquad y=u^2,qquad z =3u+v.
$$
Then
$$
dx ;=; frac{partial x}{partial u}du+frac{partial x}{partial v}dv ;=; v,du+u,dv
$$
and similarly
$$
dy ;=; 2u,duqquadtext{and}qquad dz;=;3,du+dv.
$$
Therefore,
$$
begin{align*}
xy,dx + 2z,dy - y,dz ;&=; (uv)(u^2)(v,du+u,dv)+2(3u+v)(2u,du)-(u^2)(3,du+dv)\[1ex]
&=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
end{align*}
$$
We conclude that
$$
alpha^*(xy,dx + 2z,dy - y,dz) ;=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
$$
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
add a comment |
Instead of thinking of $alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,qquad y=u^2,qquad z =3u+v.
$$
Then
$$
dx ;=; frac{partial x}{partial u}du+frac{partial x}{partial v}dv ;=; v,du+u,dv
$$
and similarly
$$
dy ;=; 2u,duqquadtext{and}qquad dz;=;3,du+dv.
$$
Therefore,
$$
begin{align*}
xy,dx + 2z,dy - y,dz ;&=; (uv)(u^2)(v,du+u,dv)+2(3u+v)(2u,du)-(u^2)(3,du+dv)\[1ex]
&=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
end{align*}
$$
We conclude that
$$
alpha^*(xy,dx + 2z,dy - y,dz) ;=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
$$
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
Instead of thinking of $alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,qquad y=u^2,qquad z =3u+v.
$$
Then
$$
dx ;=; frac{partial x}{partial u}du+frac{partial x}{partial v}dv ;=; v,du+u,dv
$$
and similarly
$$
dy ;=; 2u,duqquadtext{and}qquad dz;=;3,du+dv.
$$
Therefore,
$$
begin{align*}
xy,dx + 2z,dy - y,dz ;&=; (uv)(u^2)(v,du+u,dv)+2(3u+v)(2u,du)-(u^2)(3,du+dv)\[1ex]
&=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
end{align*}
$$
We conclude that
$$
alpha^*(xy,dx + 2z,dy - y,dz) ;=; (u^3v^2+9u^2+4uv),du,+,(u^4v-u^2),dv.
$$
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
edited May 5 '14 at 15:15
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
answered Nov 22 '13 at 1:39
Jim Belk
37.4k285150
37.4k285150
add a comment |
add a comment |
In some of my research I encountered $2$-forms that were given by $$omega = dx wedge dp + dy wedge dq$$
and a map $$i : (u,v) mapsto (u,v,f_u,-f_v)$$
for a general smooth map $f : (u,v) mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*omega$. So,
begin{align}
i^*omega &= i^*(dx wedge dp + dy wedge dq) \
&= d(x circ i)wedge d(p circ i) + d(y circ i)wedge d(q circ i).
end{align}
Now, calculating each terms gives
begin{align}
d(x circ i) &= d(u) = du, \
d(y circ i) &= d(v) = dv, \
d(p circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \
d(q circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv.
end{align}
Then, the pullback is given by
begin{align}
i^*omega &= du wedge (f_{uu}du + f_{uv}dv) - dv wedge (f_{vu}du + f_{vv}dv) \
&= du wedge (f_{uu}du) + du wedge (f_{uv}dv) - dv wedge (f_{vu}du) - dv wedge (f_{vv}dv).
end{align}
Now here, since the wedge product is $C^infty(M)$-bilinear rather than just $Bbb R$-bilinear, and $du wedge dv = -dvwedge du$. Using this property we have
$$i^*omega =2f_{uv} du wedge dv.$$
I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $alpha$ and $omega$!
answered Jul 24 '17 at 13:42
Kevin
5,410822
add a comment |
In some of my research I encountered $2$-forms that were given by $$omega = dx wedge dp + dy wedge dq$$
and a map $$i : (u,v) mapsto (u,v,f_u,-f_v)$$
for a general smooth map $f : (u,v) mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*omega$. So,
begin{align}
i^*omega &= i^*(dx wedge dp + dy wedge dq) \
&= d(x circ i)wedge d(p circ i) + d(y circ i)wedge d(q circ i).
end{align}
Now, calculating each terms gives
begin{align}
d(x circ i) &= d(u) = du, \
d(y circ i) &= d(v) = dv, \
d(p circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \
d(q circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv.
end{align}
Then, the pullback is given by
begin{align}
i^*omega &= du wedge (f_{uu}du + f_{uv}dv) - dv wedge (f_{vu}du + f_{vv}dv) \
&= du wedge (f_{uu}du) + du wedge (f_{uv}dv) - dv wedge (f_{vu}du) - dv wedge (f_{vv}dv).
end{align}
Now here, since the wedge product is $C^infty(M)$-bilinear rather than just $Bbb R$-bilinear, and $du wedge dv = -dvwedge du$. Using this property we have
$$i^*omega =2f_{uv} du wedge dv.$$
I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $alpha$ and $omega$!
answered Jul 24 '17 at 13:42
Kevin
5,410822
add a comment |
In some of my research I encountered $2$-forms that were given by $$omega = dx wedge dp + dy wedge dq$$
and a map $$i : (u,v) mapsto (u,v,f_u,-f_v)$$
for a general smooth map $f : (u,v) mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*omega$. So,
begin{align}
i^*omega &= i^*(dx wedge dp + dy wedge dq) \
&= d(x circ i)wedge d(p circ i) + d(y circ i)wedge d(q circ i).
end{align}
Now, calculating each terms gives
begin{align}
d(x circ i) &= d(u) = du, \
d(y circ i) &= d(v) = dv, \
d(p circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \
d(q circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv.
end{align}
Then, the pullback is given by
begin{align}
i^*omega &= du wedge (f_{uu}du + f_{uv}dv) - dv wedge (f_{vu}du + f_{vv}dv) \
&= du wedge (f_{uu}du) + du wedge (f_{uv}dv) - dv wedge (f_{vu}du) - dv wedge (f_{vv}dv).
end{align}
Now here, since the wedge product is $C^infty(M)$-bilinear rather than just $Bbb R$-bilinear, and $du wedge dv = -dvwedge du$. Using this property we have
$$i^*omega =2f_{uv} du wedge dv.$$
I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $alpha$ and $omega$!
answered Jul 24 '17 at 13:42
Kevin
5,410822
In some of my research I encountered $2$-forms that were given by $$omega = dx wedge dp + dy wedge dq$$
and a map $$i : (u,v) mapsto (u,v,f_u,-f_v)$$
for a general smooth map $f : (u,v) mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*omega$. So,
begin{align}
i^*omega &= i^*(dx wedge dp + dy wedge dq) \
&= d(x circ i)wedge d(p circ i) + d(y circ i)wedge d(q circ i).
end{align}
Now, calculating each terms gives
begin{align}
d(x circ i) &= d(u) = du, \
d(y circ i) &= d(v) = dv, \
d(p circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \
d(q circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv.
end{align}
Then, the pullback is given by
begin{align}
i^*omega &= du wedge (f_{uu}du + f_{uv}dv) - dv wedge (f_{vu}du + f_{vv}dv) \
&= du wedge (f_{uu}du) + du wedge (f_{uv}dv) - dv wedge (f_{vu}du) - dv wedge (f_{vv}dv).
end{align}
Now here, since the wedge product is $C^infty(M)$-bilinear rather than just $Bbb R$-bilinear, and $du wedge dv = -dvwedge du$. Using this property we have
$$i^*omega =2f_{uv} du wedge dv.$$
I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $alpha$ and $omega$!
answered Jul 24 '17 at 13:42
Kevin
5,410822
answered Jul 24 '17 at 13:42
Kevin
5,410822
answered Jul 24 '17 at 13:42
Kevin
5,410822
answered Jul 24 '17 at 13:42
Kevin
5,410822
5,410822
add a comment |
add a comment |
An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too verbose...
If $omegainOmega^1(N)$ and $alpha:Mrightarrow N$. The aim of the pullback is to define a form $alpha^*omegainOmega^1(M)$ from a form $omegainOmega^1(N)$.
A 1-form $omega$ evaluated at $n=(x,y,z)in N$ is $$omega[n]=omega_x(n)dx+omega_y(n)dy+omega_z(n)dz$$ where
- $(omega_x(n),omega_y(n),omega_z(n))inmathbb{R}^3$
$dx, dy, dz$ are also 1-forms, the dual basis of $frac{partial}{partial x},frac{partial}{partial y},frac{partial}{partial z}$.
$omega[n]in T_nN^*$ eats one vector $bin T_nN$ (a tangent vector at point $n$):
begin{align*}
omega[n](b^xfrac{partial}{partial x}+b^yfrac{partial}{partial y}+b^zfrac{partial}{partial z}) &=omega_x(n)b^x+omega_y(n)b^y+omega_z(n)b^zinmathbb{R}
end{align*}
Now the objective is to define $alpha^*omegainOmega^1(M)$ from $omegainOmega^1(N)$. Given a point $m=(u,v)in M$ and a vector $ain T_m M$, a natural idea is to use the point $n=alpha(m)in N$ and to pushforward a vector $ain T_mM$ to get a vector $alpha_*(a)in T_{alpha(m)}N$. Actually this is how $alpha^*omega$ is defined:
$$(alpha^*omega)[m](a)=omega[alpha(m)](alpha_*(a))$$
Now we must compute $alpha_*(a)in T_{alpha(m)}N$, see at the end for details, we get:
$$
alpha_*(a)=dalpha^x[alpha(m)](a)frac{partial}{partial x}+dalpha^y[alpha(m)](a)frac{partial}{partial y}+dalpha^z[alpha(m))](a)frac{partial}{partial z}
$$
Thus:
begin{align*}
(alpha^*omega)[m](a)&=overbrace{omega_x[alpha(m)]dalpha^x[alpha(m)](a)}^{text{term }dx(frac{partial}{partial x})=1}+
overbrace{omega_y[alpha(m)]dalpha^y[alpha(m)](a)}^{text{term }dy(frac{partial}{partial y})=1}+
overbrace{omega_z[alpha(m)]dalpha^z[alpha(m)](a)}^{text{term }dz(frac{partial}{partial z})=1} in mathbb{R}
end{align*}
We can drop the argument, vector $a=a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v}$, it remains:
$$
(alpha^*omega)[m]=omega_x[alpha(m)]dalpha^x[alpha(m)]+
omega_y[alpha(m)]dalpha^y[alpha(m)]+
omega_z[alpha(m)]dalpha^z[alpha(m)]in T_mM^*
$$
In the example: $(omega_x,omega_y,omega_z)=(xy,2z,-y)$ and $alpha: (u,v)mapsto (uv,u^2,3u+v)$, therefore:
- $(omega_x[alpha(m)],omega_y[alpha(m)],omega_z[alpha(m)])=(u^3v,6u+2v,-u^2)$
- $dalpha^x[alpha(m)]=vdu+udv$
$dalpha^y[alpha(m)]=2udu$- $dalpha^z[alpha(m)]=3du+dv$
By substitution we get the expected result:
$$
(alpha^*omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)
$$
$$
alpha^*omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dvinOmega^1(M)
$$
It remains to explain how to compute $alpha_star(a)in T_{alpha(m)}N$ from a tangent vector $ain T_mM$
One can interpret a vector $ain T_mM$ as a first order differential operator acting on functions $f:Mtomathbb{R}$.
More explicitly we have:
$$
a[f]=(a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v})f=df[m](a) inmathbb{R}
$$
The pushforward $alpha_*$ transform a vector $ain T_mM$ into a vector $alpha_star(a)in T_{alpha(m)}N$, thus it must act on functions $g:Ntomathbb{R}$. A natural definition is:
$$
alpha_star(a)[g]=d(gcircalpha)[m](a)inmathbb{R}
$$
Expanding this formula we get:
begin{align*}
d(gcircalpha)[m](a)&=left((frac{partial g}{partial x}frac{partial alpha^x}{partial u}+frac{partial g}{partial y}frac{partial alpha^y}{partial u}+frac{partial g}{partial z}frac{partial alpha^z}{partial u})du+(frac{partial g}{partial x}frac{partial alpha^x}{partial v}+frac{partial g}{partial y}frac{partial alpha^y}{partial v}+frac{partial g}{partial z}frac{partial alpha^z}{partial v})dvright)(a) \
&= left((frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)frac{partial g}{partial z}right)(a) \
&= (frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)(a)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)(a)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)(a)frac{partial g}{partial z} \
&= left( dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z} right) g
end{align*}
we get the expected result:
$$
alpha_star(a)=dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z}in T_{alpha(m)}N
$$
CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $alpha$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
add a comment |
An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too verbose...
If $omegainOmega^1(N)$ and $alpha:Mrightarrow N$. The aim of the pullback is to define a form $alpha^*omegainOmega^1(M)$ from a form $omegainOmega^1(N)$.
A 1-form $omega$ evaluated at $n=(x,y,z)in N$ is $$omega[n]=omega_x(n)dx+omega_y(n)dy+omega_z(n)dz$$ where
- $(omega_x(n),omega_y(n),omega_z(n))inmathbb{R}^3$
$dx, dy, dz$ are also 1-forms, the dual basis of $frac{partial}{partial x},frac{partial}{partial y},frac{partial}{partial z}$.
$omega[n]in T_nN^*$ eats one vector $bin T_nN$ (a tangent vector at point $n$):
begin{align*}
omega[n](b^xfrac{partial}{partial x}+b^yfrac{partial}{partial y}+b^zfrac{partial}{partial z}) &=omega_x(n)b^x+omega_y(n)b^y+omega_z(n)b^zinmathbb{R}
end{align*}
Now the objective is to define $alpha^*omegainOmega^1(M)$ from $omegainOmega^1(N)$. Given a point $m=(u,v)in M$ and a vector $ain T_m M$, a natural idea is to use the point $n=alpha(m)in N$ and to pushforward a vector $ain T_mM$ to get a vector $alpha_*(a)in T_{alpha(m)}N$. Actually this is how $alpha^*omega$ is defined:
$$(alpha^*omega)[m](a)=omega[alpha(m)](alpha_*(a))$$
Now we must compute $alpha_*(a)in T_{alpha(m)}N$, see at the end for details, we get:
$$
alpha_*(a)=dalpha^x[alpha(m)](a)frac{partial}{partial x}+dalpha^y[alpha(m)](a)frac{partial}{partial y}+dalpha^z[alpha(m))](a)frac{partial}{partial z}
$$
Thus:
begin{align*}
(alpha^*omega)[m](a)&=overbrace{omega_x[alpha(m)]dalpha^x[alpha(m)](a)}^{text{term }dx(frac{partial}{partial x})=1}+
overbrace{omega_y[alpha(m)]dalpha^y[alpha(m)](a)}^{text{term }dy(frac{partial}{partial y})=1}+
overbrace{omega_z[alpha(m)]dalpha^z[alpha(m)](a)}^{text{term }dz(frac{partial}{partial z})=1} in mathbb{R}
end{align*}
We can drop the argument, vector $a=a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v}$, it remains:
$$
(alpha^*omega)[m]=omega_x[alpha(m)]dalpha^x[alpha(m)]+
omega_y[alpha(m)]dalpha^y[alpha(m)]+
omega_z[alpha(m)]dalpha^z[alpha(m)]in T_mM^*
$$
In the example: $(omega_x,omega_y,omega_z)=(xy,2z,-y)$ and $alpha: (u,v)mapsto (uv,u^2,3u+v)$, therefore:
- $(omega_x[alpha(m)],omega_y[alpha(m)],omega_z[alpha(m)])=(u^3v,6u+2v,-u^2)$
- $dalpha^x[alpha(m)]=vdu+udv$
$dalpha^y[alpha(m)]=2udu$- $dalpha^z[alpha(m)]=3du+dv$
By substitution we get the expected result:
$$
(alpha^*omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)
$$
$$
alpha^*omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dvinOmega^1(M)
$$
It remains to explain how to compute $alpha_star(a)in T_{alpha(m)}N$ from a tangent vector $ain T_mM$
One can interpret a vector $ain T_mM$ as a first order differential operator acting on functions $f:Mtomathbb{R}$.
More explicitly we have:
$$
a[f]=(a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v})f=df[m](a) inmathbb{R}
$$
The pushforward $alpha_*$ transform a vector $ain T_mM$ into a vector $alpha_star(a)in T_{alpha(m)}N$, thus it must act on functions $g:Ntomathbb{R}$. A natural definition is:
$$
alpha_star(a)[g]=d(gcircalpha)[m](a)inmathbb{R}
$$
Expanding this formula we get:
begin{align*}
d(gcircalpha)[m](a)&=left((frac{partial g}{partial x}frac{partial alpha^x}{partial u}+frac{partial g}{partial y}frac{partial alpha^y}{partial u}+frac{partial g}{partial z}frac{partial alpha^z}{partial u})du+(frac{partial g}{partial x}frac{partial alpha^x}{partial v}+frac{partial g}{partial y}frac{partial alpha^y}{partial v}+frac{partial g}{partial z}frac{partial alpha^z}{partial v})dvright)(a) \
&= left((frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)frac{partial g}{partial z}right)(a) \
&= (frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)(a)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)(a)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)(a)frac{partial g}{partial z} \
&= left( dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z} right) g
end{align*}
we get the expected result:
$$
alpha_star(a)=dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z}in T_{alpha(m)}N
$$
CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $alpha$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
add a comment |
An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too verbose...
If $omegainOmega^1(N)$ and $alpha:Mrightarrow N$. The aim of the pullback is to define a form $alpha^*omegainOmega^1(M)$ from a form $omegainOmega^1(N)$.
A 1-form $omega$ evaluated at $n=(x,y,z)in N$ is $$omega[n]=omega_x(n)dx+omega_y(n)dy+omega_z(n)dz$$ where
- $(omega_x(n),omega_y(n),omega_z(n))inmathbb{R}^3$
$dx, dy, dz$ are also 1-forms, the dual basis of $frac{partial}{partial x},frac{partial}{partial y},frac{partial}{partial z}$.
$omega[n]in T_nN^*$ eats one vector $bin T_nN$ (a tangent vector at point $n$):
begin{align*}
omega[n](b^xfrac{partial}{partial x}+b^yfrac{partial}{partial y}+b^zfrac{partial}{partial z}) &=omega_x(n)b^x+omega_y(n)b^y+omega_z(n)b^zinmathbb{R}
end{align*}
Now the objective is to define $alpha^*omegainOmega^1(M)$ from $omegainOmega^1(N)$. Given a point $m=(u,v)in M$ and a vector $ain T_m M$, a natural idea is to use the point $n=alpha(m)in N$ and to pushforward a vector $ain T_mM$ to get a vector $alpha_*(a)in T_{alpha(m)}N$. Actually this is how $alpha^*omega$ is defined:
$$(alpha^*omega)[m](a)=omega[alpha(m)](alpha_*(a))$$
Now we must compute $alpha_*(a)in T_{alpha(m)}N$, see at the end for details, we get:
$$
alpha_*(a)=dalpha^x[alpha(m)](a)frac{partial}{partial x}+dalpha^y[alpha(m)](a)frac{partial}{partial y}+dalpha^z[alpha(m))](a)frac{partial}{partial z}
$$
Thus:
begin{align*}
(alpha^*omega)[m](a)&=overbrace{omega_x[alpha(m)]dalpha^x[alpha(m)](a)}^{text{term }dx(frac{partial}{partial x})=1}+
overbrace{omega_y[alpha(m)]dalpha^y[alpha(m)](a)}^{text{term }dy(frac{partial}{partial y})=1}+
overbrace{omega_z[alpha(m)]dalpha^z[alpha(m)](a)}^{text{term }dz(frac{partial}{partial z})=1} in mathbb{R}
end{align*}
We can drop the argument, vector $a=a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v}$, it remains:
$$
(alpha^*omega)[m]=omega_x[alpha(m)]dalpha^x[alpha(m)]+
omega_y[alpha(m)]dalpha^y[alpha(m)]+
omega_z[alpha(m)]dalpha^z[alpha(m)]in T_mM^*
$$
In the example: $(omega_x,omega_y,omega_z)=(xy,2z,-y)$ and $alpha: (u,v)mapsto (uv,u^2,3u+v)$, therefore:
- $(omega_x[alpha(m)],omega_y[alpha(m)],omega_z[alpha(m)])=(u^3v,6u+2v,-u^2)$
- $dalpha^x[alpha(m)]=vdu+udv$
$dalpha^y[alpha(m)]=2udu$- $dalpha^z[alpha(m)]=3du+dv$
By substitution we get the expected result:
$$
(alpha^*omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)
$$
$$
alpha^*omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dvinOmega^1(M)
$$
It remains to explain how to compute $alpha_star(a)in T_{alpha(m)}N$ from a tangent vector $ain T_mM$
One can interpret a vector $ain T_mM$ as a first order differential operator acting on functions $f:Mtomathbb{R}$.
More explicitly we have:
$$
a[f]=(a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v})f=df[m](a) inmathbb{R}
$$
The pushforward $alpha_*$ transform a vector $ain T_mM$ into a vector $alpha_star(a)in T_{alpha(m)}N$, thus it must act on functions $g:Ntomathbb{R}$. A natural definition is:
$$
alpha_star(a)[g]=d(gcircalpha)[m](a)inmathbb{R}
$$
Expanding this formula we get:
begin{align*}
d(gcircalpha)[m](a)&=left((frac{partial g}{partial x}frac{partial alpha^x}{partial u}+frac{partial g}{partial y}frac{partial alpha^y}{partial u}+frac{partial g}{partial z}frac{partial alpha^z}{partial u})du+(frac{partial g}{partial x}frac{partial alpha^x}{partial v}+frac{partial g}{partial y}frac{partial alpha^y}{partial v}+frac{partial g}{partial z}frac{partial alpha^z}{partial v})dvright)(a) \
&= left((frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)frac{partial g}{partial z}right)(a) \
&= (frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)(a)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)(a)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)(a)frac{partial g}{partial z} \
&= left( dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z} right) g
end{align*}
we get the expected result:
$$
alpha_star(a)=dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z}in T_{alpha(m)}N
$$
CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $alpha$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too verbose...
If $omegainOmega^1(N)$ and $alpha:Mrightarrow N$. The aim of the pullback is to define a form $alpha^*omegainOmega^1(M)$ from a form $omegainOmega^1(N)$.
A 1-form $omega$ evaluated at $n=(x,y,z)in N$ is $$omega[n]=omega_x(n)dx+omega_y(n)dy+omega_z(n)dz$$ where
- $(omega_x(n),omega_y(n),omega_z(n))inmathbb{R}^3$
$dx, dy, dz$ are also 1-forms, the dual basis of $frac{partial}{partial x},frac{partial}{partial y},frac{partial}{partial z}$.
$omega[n]in T_nN^*$ eats one vector $bin T_nN$ (a tangent vector at point $n$):
begin{align*}
omega[n](b^xfrac{partial}{partial x}+b^yfrac{partial}{partial y}+b^zfrac{partial}{partial z}) &=omega_x(n)b^x+omega_y(n)b^y+omega_z(n)b^zinmathbb{R}
end{align*}
Now the objective is to define $alpha^*omegainOmega^1(M)$ from $omegainOmega^1(N)$. Given a point $m=(u,v)in M$ and a vector $ain T_m M$, a natural idea is to use the point $n=alpha(m)in N$ and to pushforward a vector $ain T_mM$ to get a vector $alpha_*(a)in T_{alpha(m)}N$. Actually this is how $alpha^*omega$ is defined:
$$(alpha^*omega)[m](a)=omega[alpha(m)](alpha_*(a))$$
Now we must compute $alpha_*(a)in T_{alpha(m)}N$, see at the end for details, we get:
$$
alpha_*(a)=dalpha^x[alpha(m)](a)frac{partial}{partial x}+dalpha^y[alpha(m)](a)frac{partial}{partial y}+dalpha^z[alpha(m))](a)frac{partial}{partial z}
$$
Thus:
begin{align*}
(alpha^*omega)[m](a)&=overbrace{omega_x[alpha(m)]dalpha^x[alpha(m)](a)}^{text{term }dx(frac{partial}{partial x})=1}+
overbrace{omega_y[alpha(m)]dalpha^y[alpha(m)](a)}^{text{term }dy(frac{partial}{partial y})=1}+
overbrace{omega_z[alpha(m)]dalpha^z[alpha(m)](a)}^{text{term }dz(frac{partial}{partial z})=1} in mathbb{R}
end{align*}
We can drop the argument, vector $a=a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v}$, it remains:
$$
(alpha^*omega)[m]=omega_x[alpha(m)]dalpha^x[alpha(m)]+
omega_y[alpha(m)]dalpha^y[alpha(m)]+
omega_z[alpha(m)]dalpha^z[alpha(m)]in T_mM^*
$$
In the example: $(omega_x,omega_y,omega_z)=(xy,2z,-y)$ and $alpha: (u,v)mapsto (uv,u^2,3u+v)$, therefore:
- $(omega_x[alpha(m)],omega_y[alpha(m)],omega_z[alpha(m)])=(u^3v,6u+2v,-u^2)$
- $dalpha^x[alpha(m)]=vdu+udv$
$dalpha^y[alpha(m)]=2udu$- $dalpha^z[alpha(m)]=3du+dv$
By substitution we get the expected result:
$$
(alpha^*omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)
$$
$$
alpha^*omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dvinOmega^1(M)
$$
It remains to explain how to compute $alpha_star(a)in T_{alpha(m)}N$ from a tangent vector $ain T_mM$
One can interpret a vector $ain T_mM$ as a first order differential operator acting on functions $f:Mtomathbb{R}$.
More explicitly we have:
$$
a[f]=(a^ufrac{partial}{partial u}+a^vfrac{partial}{partial v})f=df[m](a) inmathbb{R}
$$
The pushforward $alpha_*$ transform a vector $ain T_mM$ into a vector $alpha_star(a)in T_{alpha(m)}N$, thus it must act on functions $g:Ntomathbb{R}$. A natural definition is:
$$
alpha_star(a)[g]=d(gcircalpha)[m](a)inmathbb{R}
$$
Expanding this formula we get:
begin{align*}
d(gcircalpha)[m](a)&=left((frac{partial g}{partial x}frac{partial alpha^x}{partial u}+frac{partial g}{partial y}frac{partial alpha^y}{partial u}+frac{partial g}{partial z}frac{partial alpha^z}{partial u})du+(frac{partial g}{partial x}frac{partial alpha^x}{partial v}+frac{partial g}{partial y}frac{partial alpha^y}{partial v}+frac{partial g}{partial z}frac{partial alpha^z}{partial v})dvright)(a) \
&= left((frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)frac{partial g}{partial z}right)(a) \
&= (frac{partial alpha^x}{partial u}du+frac{partial alpha^x}{partial v}dv)(a)frac{partial g}{partial x}+(frac{partial alpha^y}{partial u}du+frac{partial alpha^y}{partial v}dv)(a)frac{partial g}{partial y}+(frac{partial alpha^z}{partial u}du+frac{partial alpha^z}{partial v}dv)(a)frac{partial g}{partial z} \
&= left( dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z} right) g
end{align*}
we get the expected result:
$$
alpha_star(a)=dalpha^x[m](a)frac{partial}{partial x}+dalpha^y[m](a)frac{partial}{partial y}+dalpha^z[m](a)frac{partial}{partial z}in T_{alpha(m)}N
$$
CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $alpha$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
edited Nov 20 '18 at 19:04
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
answered Nov 20 '18 at 15:20
Picaud Vincent
1,21838
1,21838
add a comment |
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Can you post your calculation so we can see where you get lost?
– Bruno Joyal
Nov 22 '13 at 0:57