Let $(X,\omega)$ be a Riemann surface of genus $g$ with holomorphic 1-form $\omega$ (or equivalently a translation structure). Let $\Omega\mathcal{T}_g$ be the space of holomorphic 1-forms over genus $g$ surface. The famous $\mathrm{SL}_2\mathbb{R}$ action on $\Omega\mathcal{T}_g$ is defined by composing each coordinate chart of $(X,\omega)$ with matrices in $\mathrm{SL}_2\mathbb{R}$. It is claimed that by descending to $\mathbb{H}$ and $\mathcal{T}_g$ from their tangent/cotangent bundle, the $\mathrm{SL}_2\mathbb{R}$ action embeds the hyperbolic plane $\mathbb{H}$ isometrically into Teichmuller space $\mathcal{T}_g$ of genus $g$.
A few things I have learned so far:
- $\mathrm{PSL}_2\mathbb{R}$ is identified with $T^1\mathbb{H}$ by choosing $(i,i)\in T^1\mathbb{H}$ for identity matrix. For any $g\in \mathrm{PSL}_2\mathbb{R}$ identified with $(x,v)\in T^1\mathbb{H}$, $ga_t$ traces the geodesic in $\mathbb{H}$ passing through $x$ with tangent vector $v$. (Update: this part is explained in more detail in DMG's answer.)
- Let $a_t:=\begin{pmatrix}e^{t/2}&0\\ 0&e^{-t/2} \end{pmatrix}, t\in\mathbb{R}$. The action of $a_t$ streches the horizontal direction and shrinks the vertical direction of $(X,\omega)$. By the Teichmuller theorem, $a_t\cdot (X,\omega)$ traces the geodesic in $\mathcal{T_g}$ passing through $X$ with marking $id_X$ as $t$ varies.
So if the embedding is given by the identification of $T^1\mathbb{H}$ with $\mathrm{PSL}_2\mathbb{R}\cdot (X_0,\omega_0)$ above, the curve $t\mapsto a_tg$ in $T^1\mathbb{H}$ is sent to the geodesic $t\mapsto a_t\cdot (X,\omega)$ in $\mathcal{T}_g$ given that $(X,\omega)=g\cdot (X_0,\omega_0)$. However, the former curve $t\mapsto a_tg$ is NOT the geodesic but a scaling of the complex number $x$ to which $g$ is identified with. The fact that a non-geodesic is sent to a geodesic contradicts the claim that the embedding of $\mathbb{H}$ is isometric.
Where does my argument go wrong? Or do we define the identification of $\mathbb{H}$ with $\mathrm{SO(2)}\backslash \mathrm{SL}_2\mathbb{R}\cdot (X_0,\omega_0)$ differently?
The same post on MO https://mathoverflow.net/questions/303017/teichmueller-disk-and-the-mathrmsl-2-mathbbr-action?noredirect=1#comment755009_303017
I would appreciate any help very much. Thank you in advance.
The identification of $PSL_2(\mathbb{R})$ and $T^1 \mathbb{H}$ stems from the fact that for any $(x, v) \in T^1 \mathbb{H}$, there is a unique Moebius transformation $M \in PSL_2(\mathbb{R})$ sending the imaginary axis in $\mathbb{H}$ to the geodesic $A_{M}$ passing through $x$ with direction $v$ at $x$. To put it shortly: $PSL_2(\mathbb{R})$ acts simply transitively on $T^1 \mathbb{H}$. This gives bijection $PSL_2(\mathbb{R})(i,i) \leftrightarrow T^1\mathbb{H}$. The notation $(i,i)$ means the point $i \in \mathbb{H}$ in the first factor, and the unit vector based at $i$ and tangent to the imaginary axis in the second factor.
Under this identification, the action of the geodesic flow $g_t$ on $T^1 \mathbb{H}$ corresponds to right multiplication in $PSL_2(\mathbb{R})$ by the matrix $a_t:=\begin{pmatrix} e^{t/2} & 0 \\ 0 & e^{-t/2} \end{pmatrix}$, i.e. if $M \in PSL_2(\mathbb{R})$ is represented by $(x,v)$ in $T^1 \mathbb{H}$ under the above bijection, then
$$g_t \cdot (x,v)=Ma_t.$$
Here is why: $a_t$ corresponds to a hyperbolic transformation with translation axis the imaginary axis, and with translation distance exactly $t$. Under the bijection, we are doing $(Ma_t) \cdot (i,i) = M \cdot (a_t \cdot (i,i))$, i.e. we first translate along the imaginary axis a distance $t$ (we flow along the imaginary axis), and then we send the imaginary axis to the geodesic determined by $(x,v)$ by the isometry $M$, which we named $A_M$. Since isometries preserve distances, the image point is the translate of $(x,v)$ along the geodesic $A_M$ a distance $t$, i.e. the action of the geodesic flow.
As a final remark, note that the choice of the family of matrices $a_t$ depends on the choice of base point (in our case, the basepoint is $(i,i)$).
Edit:
Now we discuss the action of $SL_2(\mathbb{R})$ on $\Omega \cal{T}_{g,n}$.
Fix $(X_0,\omega_0)$, where $X_0$ is a conformal structure on $S_{g,n}$ and $\omega_0$ is a holomorphic quadratic differential on $X_0$. $\omega_0$ gives normal coordinates of the quadratic differential $\omega_0$ on $X_0$ (outside of zeroes of $\omega_0$). The element $B \in SL_2(\mathbb{R})$ can be interpreted as an affine map in these coordinates. $B \cdot (X_0, \omega_0)=(X_0,\omega_B)$. The data $\mathcal{X}_B=[(X_0,\omega_B),id]$ gives a point in Teichmueller space with marking the identity, where we understand $\omega_B$ gives a gives a new conformal structure on $X_0$ in by composing the coordinate patch of normal coordinates with the affine map $B$ and completing the complex structure at the zeroes of $\omega_B$ by removable singularity theorem. This yields a quasi-conformal map $f : \mathcal{X}_0 \to \mathcal{X}_B$, and a terminal quadratic differential $\omega_B$. If $B \in SO(2)$, then the marking $f : \mathcal{X}_0 \to \mathcal{X}_B$ is conformal, so we get the same point in Teichmueller space. Thus, the action descends to a faithful action of $SL_2(\mathbb{R})/SO(2)$ on $ \cal{T}_{g,n}$. We claim that this descended action of $SL_2(\mathbb{R})/SO(2)$ on $\cal{T}_{g,n}$ yields an isometric injection $SL_2(\mathbb{R})/SO(2)(X_0,\omega_0) \to \cal{T}_{g,n}$. It is easy to check that any matrix $A \in SL_2(\mathbb{R})$ with $A \notin SO(2)$ can be written as $A=U D_K V$ where $U,V \in SO(2)$ and $D_K=\begin{pmatrix} \sqrt{K} & 0 \\ 0 & \frac{1}{\sqrt{K}} \end{pmatrix}$ and $K>1$. The Teichmuller distance between $\mathcal{X}_0$ and $\mathcal{X}_{D_K}$ is given by the $\log K$ where $K$ is the infimum of the quasi-conformal dilatation, which is realized by the affine map described in the construction above. More generally, to compute the distance between $\mathcal{X}_A$ and $\mathcal{X}_B$, we can take the base point at $(X_0, \omega_B)$, and decompose $AB^{-1} = U_1 D_K U_2$. Finally, we can show that the embedding $SL_2(\mathbb{R})/SO(2)(X_0,\omega_0) \to \cal{T}_{g,n}$ is isometric, by using the bijection with $\mathbb{H}$, and noting that if $\rho$ denotes the hyperbolic distance on $\mathbb{H}$, then $\rho(A(i),B(i))=\rho(AB^{-1}(i),i)=\rho(D_K(i),i)=\log(K)$, which is the same as the Teichmueller distance computed above.