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What's going on when we compute $d(gamma(z)) = frac{1}{|cz+d|^2}dz$, where $gamma in...
$begingroup$
Let $gamma = begin{pmatrix} a & b \ c & d end{pmatrix} in operatorname{SL}_2(mathbb Z)$. Consider the space $Omega^1(mathbb H)$ of smooth complex $1$-forms on $mathbb H$. These consist of all smooth sections of $mathbb H$ into the complexified tangent bundle of $mathbb H$, written formally as
$$omega = f(x,y)dx + g(x,y)dy$$
for $f, g$ smooth complex valued functions on $mathbb H$. In particular, we have the holomorphic differential forms, given by $h(x,y)dz$, where $dz = dx + i dy$, and $h$ holomorphic on $mathbb H$.
Since $gamma$ induces a diffeomorphism of $mathbb H$ to itself, it induces a diffeomorphism on the tangent bundle of $mathbb H$ to itself, and therefore an automorphism of $Omega^1(mathbb H)$.
We can compute the effect of $gamma$ on the form $dz$. I have seen the following done:
$$gamma.(dz) = d(gamma(z)) = d( frac{az+b}{cz+d}) = (frac{az+b}{cz+d})' = frac{1}{(cz+d)^2}dz tag{1}$$
I don't really understand what is going on here formally with differential forms. How do we formally justify what is happening in (1)? I did compute the action of $gamma$ on $dz$ and got the same answer in another way:
Another way:
Let's consider the map $d gamma$ on the tangent bundle. With our chart, we can do this by computing the Jacobian of $gamma$. Writing $gamma(x,y) = u + iv$, and using the fact that $gamma$ is holomorphic and $v(z) = v(x+iy) = frac{y}{|cz+d|^2}$, we have by the Cauchy-Riemann equations that
$$d gamma = begin{pmatrix} frac{partial u}{partial x} & frac{partial u}{partial y} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} = begin{pmatrix} frac{partial v}{partial y} & -frac{partial v}{partial x} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} $$
The dual map on the contangent bundle is then given by the tranpose:
$$begin{pmatrix} frac{partial v}{partial y} & frac{partial v}{partial x} \ -frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix}
tag{2} $$
with
$$frac{partial v}{partial x} = |cz+d|^{-3}(-4c^2x-4ycd)$$
$$frac{partial v}{partial y} = frac{|cz+d|^2-2y^2c^2}{|cz+d|^4} $$
Now by (2), $d gamma$ sends $dx$ to $frac{partial v}{partial y} dx - frac{partial v}{partial x}dy$, and it sends $dy$ to $frac{partial v}{partial x}dx + frac{partial v}{partial y} dy$. Therefore, $d gamma$ sends $dz = dx + i dy$ to
$$(frac{partial v}{partial y} dx - frac{partial v}{partial x}dy) + i(-frac{partial v}{partial x}dx + frac{partial v}{partial y} dy) = (frac{partial v}{partial y} + i frac{partial v}{partial x})dz $$
This last expression, $frac{partial v}{partial y} + i frac{partial v}{partial x}$, is equal to $frac{partial v}{partial x} + i frac{partial v}{partial x} = frac{partial}{partial x}(u+iv) = frac{partial}{partial x}(gamma(z)) = gamma'(z)$.
This shows that $gamma$ induces an automorphism on holomorphic $1$-forms which sends $dz$ to
$$dz mapsto gamma'(z)dz = frac{1}{(cz+d)^2}dz$$
exactly as in (1). But how can we justify this without going into the Cauchy-Riemann equations and Jacobian calculation?
complex-analysis differential-geometry differential-forms riemann-surfaces complex-manifolds
asked Jan 24 at 15:43
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
Let $gamma = begin{pmatrix} a & b \ c & d end{pmatrix} in operatorname{SL}_2(mathbb Z)$. Consider the space $Omega^1(mathbb H)$ of smooth complex $1$-forms on $mathbb H$. These consist of all smooth sections of $mathbb H$ into the complexified tangent bundle of $mathbb H$, written formally as
$$omega = f(x,y)dx + g(x,y)dy$$
for $f, g$ smooth complex valued functions on $mathbb H$. In particular, we have the holomorphic differential forms, given by $h(x,y)dz$, where $dz = dx + i dy$, and $h$ holomorphic on $mathbb H$.
Since $gamma$ induces a diffeomorphism of $mathbb H$ to itself, it induces a diffeomorphism on the tangent bundle of $mathbb H$ to itself, and therefore an automorphism of $Omega^1(mathbb H)$.
We can compute the effect of $gamma$ on the form $dz$. I have seen the following done:
$$gamma.(dz) = d(gamma(z)) = d( frac{az+b}{cz+d}) = (frac{az+b}{cz+d})' = frac{1}{(cz+d)^2}dz tag{1}$$
I don't really understand what is going on here formally with differential forms. How do we formally justify what is happening in (1)? I did compute the action of $gamma$ on $dz$ and got the same answer in another way:
Another way:
Let's consider the map $d gamma$ on the tangent bundle. With our chart, we can do this by computing the Jacobian of $gamma$. Writing $gamma(x,y) = u + iv$, and using the fact that $gamma$ is holomorphic and $v(z) = v(x+iy) = frac{y}{|cz+d|^2}$, we have by the Cauchy-Riemann equations that
$$d gamma = begin{pmatrix} frac{partial u}{partial x} & frac{partial u}{partial y} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} = begin{pmatrix} frac{partial v}{partial y} & -frac{partial v}{partial x} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} $$
The dual map on the contangent bundle is then given by the tranpose:
$$begin{pmatrix} frac{partial v}{partial y} & frac{partial v}{partial x} \ -frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix}
tag{2} $$
with
$$frac{partial v}{partial x} = |cz+d|^{-3}(-4c^2x-4ycd)$$
$$frac{partial v}{partial y} = frac{|cz+d|^2-2y^2c^2}{|cz+d|^4} $$
Now by (2), $d gamma$ sends $dx$ to $frac{partial v}{partial y} dx - frac{partial v}{partial x}dy$, and it sends $dy$ to $frac{partial v}{partial x}dx + frac{partial v}{partial y} dy$. Therefore, $d gamma$ sends $dz = dx + i dy$ to
$$(frac{partial v}{partial y} dx - frac{partial v}{partial x}dy) + i(-frac{partial v}{partial x}dx + frac{partial v}{partial y} dy) = (frac{partial v}{partial y} + i frac{partial v}{partial x})dz $$
This last expression, $frac{partial v}{partial y} + i frac{partial v}{partial x}$, is equal to $frac{partial v}{partial x} + i frac{partial v}{partial x} = frac{partial}{partial x}(u+iv) = frac{partial}{partial x}(gamma(z)) = gamma'(z)$.
This shows that $gamma$ induces an automorphism on holomorphic $1$-forms which sends $dz$ to
$$dz mapsto gamma'(z)dz = frac{1}{(cz+d)^2}dz$$
exactly as in (1). But how can we justify this without going into the Cauchy-Riemann equations and Jacobian calculation?
complex-analysis differential-geometry differential-forms riemann-surfaces complex-manifolds
asked Jan 24 at 15:43
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
Let $gamma = begin{pmatrix} a & b \ c & d end{pmatrix} in operatorname{SL}_2(mathbb Z)$. Consider the space $Omega^1(mathbb H)$ of smooth complex $1$-forms on $mathbb H$. These consist of all smooth sections of $mathbb H$ into the complexified tangent bundle of $mathbb H$, written formally as
$$omega = f(x,y)dx + g(x,y)dy$$
for $f, g$ smooth complex valued functions on $mathbb H$. In particular, we have the holomorphic differential forms, given by $h(x,y)dz$, where $dz = dx + i dy$, and $h$ holomorphic on $mathbb H$.
Since $gamma$ induces a diffeomorphism of $mathbb H$ to itself, it induces a diffeomorphism on the tangent bundle of $mathbb H$ to itself, and therefore an automorphism of $Omega^1(mathbb H)$.
We can compute the effect of $gamma$ on the form $dz$. I have seen the following done:
$$gamma.(dz) = d(gamma(z)) = d( frac{az+b}{cz+d}) = (frac{az+b}{cz+d})' = frac{1}{(cz+d)^2}dz tag{1}$$
I don't really understand what is going on here formally with differential forms. How do we formally justify what is happening in (1)? I did compute the action of $gamma$ on $dz$ and got the same answer in another way:
Another way:
Let's consider the map $d gamma$ on the tangent bundle. With our chart, we can do this by computing the Jacobian of $gamma$. Writing $gamma(x,y) = u + iv$, and using the fact that $gamma$ is holomorphic and $v(z) = v(x+iy) = frac{y}{|cz+d|^2}$, we have by the Cauchy-Riemann equations that
$$d gamma = begin{pmatrix} frac{partial u}{partial x} & frac{partial u}{partial y} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} = begin{pmatrix} frac{partial v}{partial y} & -frac{partial v}{partial x} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} $$
The dual map on the contangent bundle is then given by the tranpose:
$$begin{pmatrix} frac{partial v}{partial y} & frac{partial v}{partial x} \ -frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix}
tag{2} $$
with
$$frac{partial v}{partial x} = |cz+d|^{-3}(-4c^2x-4ycd)$$
$$frac{partial v}{partial y} = frac{|cz+d|^2-2y^2c^2}{|cz+d|^4} $$
Now by (2), $d gamma$ sends $dx$ to $frac{partial v}{partial y} dx - frac{partial v}{partial x}dy$, and it sends $dy$ to $frac{partial v}{partial x}dx + frac{partial v}{partial y} dy$. Therefore, $d gamma$ sends $dz = dx + i dy$ to
$$(frac{partial v}{partial y} dx - frac{partial v}{partial x}dy) + i(-frac{partial v}{partial x}dx + frac{partial v}{partial y} dy) = (frac{partial v}{partial y} + i frac{partial v}{partial x})dz $$
This last expression, $frac{partial v}{partial y} + i frac{partial v}{partial x}$, is equal to $frac{partial v}{partial x} + i frac{partial v}{partial x} = frac{partial}{partial x}(u+iv) = frac{partial}{partial x}(gamma(z)) = gamma'(z)$.
This shows that $gamma$ induces an automorphism on holomorphic $1$-forms which sends $dz$ to
$$dz mapsto gamma'(z)dz = frac{1}{(cz+d)^2}dz$$
exactly as in (1). But how can we justify this without going into the Cauchy-Riemann equations and Jacobian calculation?
complex-analysis differential-geometry differential-forms riemann-surfaces complex-manifolds
asked Jan 24 at 15:43
D_SD_S
13.9k61553
$endgroup$
Let $gamma = begin{pmatrix} a & b \ c & d end{pmatrix} in operatorname{SL}_2(mathbb Z)$. Consider the space $Omega^1(mathbb H)$ of smooth complex $1$-forms on $mathbb H$. These consist of all smooth sections of $mathbb H$ into the complexified tangent bundle of $mathbb H$, written formally as
$$omega = f(x,y)dx + g(x,y)dy$$
for $f, g$ smooth complex valued functions on $mathbb H$. In particular, we have the holomorphic differential forms, given by $h(x,y)dz$, where $dz = dx + i dy$, and $h$ holomorphic on $mathbb H$.
Since $gamma$ induces a diffeomorphism of $mathbb H$ to itself, it induces a diffeomorphism on the tangent bundle of $mathbb H$ to itself, and therefore an automorphism of $Omega^1(mathbb H)$.
We can compute the effect of $gamma$ on the form $dz$. I have seen the following done:
$$gamma.(dz) = d(gamma(z)) = d( frac{az+b}{cz+d}) = (frac{az+b}{cz+d})' = frac{1}{(cz+d)^2}dz tag{1}$$
I don't really understand what is going on here formally with differential forms. How do we formally justify what is happening in (1)? I did compute the action of $gamma$ on $dz$ and got the same answer in another way:
Another way:
Let's consider the map $d gamma$ on the tangent bundle. With our chart, we can do this by computing the Jacobian of $gamma$. Writing $gamma(x,y) = u + iv$, and using the fact that $gamma$ is holomorphic and $v(z) = v(x+iy) = frac{y}{|cz+d|^2}$, we have by the Cauchy-Riemann equations that
$$d gamma = begin{pmatrix} frac{partial u}{partial x} & frac{partial u}{partial y} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} = begin{pmatrix} frac{partial v}{partial y} & -frac{partial v}{partial x} \ frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix} $$
The dual map on the contangent bundle is then given by the tranpose:
$$begin{pmatrix} frac{partial v}{partial y} & frac{partial v}{partial x} \ -frac{partial v}{partial x} & frac{partial v}{partial y} end{pmatrix}
tag{2} $$
with
$$frac{partial v}{partial x} = |cz+d|^{-3}(-4c^2x-4ycd)$$
$$frac{partial v}{partial y} = frac{|cz+d|^2-2y^2c^2}{|cz+d|^4} $$
Now by (2), $d gamma$ sends $dx$ to $frac{partial v}{partial y} dx - frac{partial v}{partial x}dy$, and it sends $dy$ to $frac{partial v}{partial x}dx + frac{partial v}{partial y} dy$. Therefore, $d gamma$ sends $dz = dx + i dy$ to
$$(frac{partial v}{partial y} dx - frac{partial v}{partial x}dy) + i(-frac{partial v}{partial x}dx + frac{partial v}{partial y} dy) = (frac{partial v}{partial y} + i frac{partial v}{partial x})dz $$
This last expression, $frac{partial v}{partial y} + i frac{partial v}{partial x}$, is equal to $frac{partial v}{partial x} + i frac{partial v}{partial x} = frac{partial}{partial x}(u+iv) = frac{partial}{partial x}(gamma(z)) = gamma'(z)$.
This shows that $gamma$ induces an automorphism on holomorphic $1$-forms which sends $dz$ to
$$dz mapsto gamma'(z)dz = frac{1}{(cz+d)^2}dz$$
exactly as in (1). But how can we justify this without going into the Cauchy-Riemann equations and Jacobian calculation?
complex-analysis differential-geometry differential-forms riemann-surfaces complex-manifolds
complex-analysis differential-geometry differential-forms riemann-surfaces complex-manifolds
asked Jan 24 at 15:43
D_SD_S
13.9k61553
asked Jan 24 at 15:43
D_SD_S
13.9k61553
asked Jan 24 at 15:43
D_SD_S
13.9k61553
asked Jan 24 at 15:43
D_SD_S
13.9k61553
asked Jan 24 at 15:43
D_SD_S
13.9k61553
13.9k61553
add a comment |
add a comment |
1 Answer
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$begingroup$
Don't quite understand your question, are you asking how do we compute (1)? Since it's in $SL_2(mathbb{Z})$ we have $ad-bc=1$.
begin{align*}
d( frac{az+b}{cz+d}) &= frac{(az+b)'(cz+d)-(az+b)(cz+d)'}{(cz+d)^2}dz\
&= frac{a(cz+d)-(az+b)c}{(cz+d)^2}dz\
&= frac{ad-bc}{(cz+d)^2}dz\
&= frac{1}{(cz+d)^2}dz tag{1}
end{align*}
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
$endgroup$
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
add a comment |
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$begingroup$
Don't quite understand your question, are you asking how do we compute (1)? Since it's in $SL_2(mathbb{Z})$ we have $ad-bc=1$.
begin{align*}
d( frac{az+b}{cz+d}) &= frac{(az+b)'(cz+d)-(az+b)(cz+d)'}{(cz+d)^2}dz\
&= frac{a(cz+d)-(az+b)c}{(cz+d)^2}dz\
&= frac{ad-bc}{(cz+d)^2}dz\
&= frac{1}{(cz+d)^2}dz tag{1}
end{align*}
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
$endgroup$
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
add a comment |
$begingroup$
Don't quite understand your question, are you asking how do we compute (1)? Since it's in $SL_2(mathbb{Z})$ we have $ad-bc=1$.
begin{align*}
d( frac{az+b}{cz+d}) &= frac{(az+b)'(cz+d)-(az+b)(cz+d)'}{(cz+d)^2}dz\
&= frac{a(cz+d)-(az+b)c}{(cz+d)^2}dz\
&= frac{ad-bc}{(cz+d)^2}dz\
&= frac{1}{(cz+d)^2}dz tag{1}
end{align*}
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
$endgroup$
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
add a comment |
$begingroup$
Don't quite understand your question, are you asking how do we compute (1)? Since it's in $SL_2(mathbb{Z})$ we have $ad-bc=1$.
begin{align*}
d( frac{az+b}{cz+d}) &= frac{(az+b)'(cz+d)-(az+b)(cz+d)'}{(cz+d)^2}dz\
&= frac{a(cz+d)-(az+b)c}{(cz+d)^2}dz\
&= frac{ad-bc}{(cz+d)^2}dz\
&= frac{1}{(cz+d)^2}dz tag{1}
end{align*}
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
$endgroup$
Don't quite understand your question, are you asking how do we compute (1)? Since it's in $SL_2(mathbb{Z})$ we have $ad-bc=1$.
begin{align*}
d( frac{az+b}{cz+d}) &= frac{(az+b)'(cz+d)-(az+b)(cz+d)'}{(cz+d)^2}dz\
&= frac{a(cz+d)-(az+b)c}{(cz+d)^2}dz\
&= frac{ad-bc}{(cz+d)^2}dz\
&= frac{1}{(cz+d)^2}dz tag{1}
end{align*}
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
answered Jan 24 at 20:07
Xipan XiaoXipan Xiao
1,885611
1,885611
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
add a comment |
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
$begingroup$
I understand how (1) is computed with taking the derivative of $frac{az+b}{cz+d}$, my question is how to formally justify it with differential forms
$endgroup$
– D_S
Jan 24 at 21:45
add a comment |
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