# What's going on when we compute $d(\gamma(z)) = \frac{1}{|cz+d|^2}dz$, where $\gamma \in \operatorname{SL}_2(\mathbb Z)$

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?

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*}
• 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 – D_S Jan 24 at 21:45