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:

$$\begin{pmatrix} \frac{\partial v}{\partial y} & \frac{\partial v}{\partial x} \\ -\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} \tag{2} $$


$$\frac{\partial v}{\partial x} = |cz+d|^{-3}(-4c^2x-4ycd)$$

$$\frac{\partial v}{\partial y} = \frac{|cz+d|^2-2y^2c^2}{|cz+d|^4} $$

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?


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*}

  • $\begingroup$ 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

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