Transformations between the fundamental domains of Eisenstein series I am self-learning elliptic functions and modular form but struggling with the very basics. 
I have programmed the Eisenstein series
$$E_6 = \sum_{m,n \in \mathbb{Z}} \frac{1}{(m +n \tau)^6} $$
and produced the following image (color denotes the argument)

which is similar to that on wiki.
Eisenstein series and automorphic representations by P. Fleig introduced the fundamental domain, as the figure below.

A similar figure finds in  L. Vepstak’s The Minkowski question Mark, GL(2,Z) and the Modular Group. 

My question is how to transform from the fundamental domain
$$
|Re(z)|<\frac{1}{2} \quad and \quad |z|>1|  \quad and \quad Im(z)\geqslant 0
$$
to other domains (appears as distorted triangles), or how to map a point in the fundamental domain to a corresponding point in another domain?
I guess that SL(2, $\mathbb{Z}$)-invariance should associate with the symmetries between the triangular domain
$$
G_{2k}\left( \frac{az+b}{cz+d} \right) = (cz+d)^{2k}G_{2k}(z)
$$
but no clues how this equation divides into triangles that correspond to the fundamental domain.
 A: First of all, I think you've chosen an excellent topic of self-study. It's very deep and beautiful, and there's still lots to be found.
I think the first thing you should ask yourself is what does it mean to be a fundamental domain?
You have a modular function $f(z)$. This means that for all $\gamma \in \textrm{SL}(2, \mathbb{Z})$, you have $f(\gamma z) = (cz + d)^k f(z)$ for some integer weight $k$. This is a transformation law. Let's better understand what each $\gamma$ does to $z$.
Two matrices of note in $\mathrm{SL}(2, \mathbb{Z})$ are
$$ T = \begin{pmatrix} 1&1\\0&1 \end{pmatrix}, \qquad S = \begin{pmatrix} 0&-1\\1&0\end{pmatrix}.$$
Unofficially, the $T$ stands for "translation" and $S$ is an "inverSion". ($I$ would probably mean the identity matrix).
The action of $T$ on $z$ is $Tz = z + 1$, and the action of $S$ on $z$ is $Sz = -1/z$. Under $T$, the strip $-\frac{1}{2} \leq \mathrm{Re} \; z \leq \frac{1}{2}$ is sent to the strip $\frac{1}{2} \leq \mathrm{Re} \; z \leq \frac{3}{2}$, and it's not hard to see that this tiles the plane in strips. Under the action of $S$, everything outside the unit circle is swapped with everything inside the unit circle.
A fundamental domain for the upper halfplane $\mathcal{H}$ under the action of $\mathrm{SL}(2, \mathbb{Z})$ is a set $D$ containing one representative of each orbit of $\mathcal{H}$ under $\mathrm{SL}(2, \mathbb{Z})$. Stated differently, for any $z \in \mathcal{H}$, there should exist some $\tau \in D$ and a $\gamma \in \mathrm{SL}(2, \mathbb{Z})$ such that $\gamma \tau = z$ (along with some uniqueness assumptions).
The standard picture of the fundamental domain for $\mathcal{H}$ under $\mathrm{SL}(2, \mathbb{Z})$ can be heuristically determined from the shifting property of $T$ (so that we can focus on just a unit strip) and $S$ (so that we can focus on just the outside of the unit circle). In fact,
$$ \mathrm{SL}(2, \mathbb{Z}) = \langle S, T \rangle,$$
and one can show that the standard fundamental really is a fundamental domain (with the proper definition) these generators.
Now, back to your question. Every point $z \in \mathcal{H}$ can be written uniquely as $\gamma \tau$ for a $\tau$ in the standard fundamental domain. Determining this $\gamma$ often occurs naturally in the proof that the standard fundamental domain really is a fundamental domain (and you would benefit from going through this proof). But it is also possible to handle it slightly differently: if $\textrm{Im} \; z \geq 1$, then as $f$ is invariant under translation, you can write $z = \tau + n$ for some integer $n$ and $\tau$ in the fundamental domain, and then $f(z) = f(\tau)$. If $\lvert z \rvert < 1$, then you can use $S$ to invert through the circle (and note that $f(Sz) = z^k f(z)$) and then shift appropriately, and then repeat if necessary. Describing particular steps completely is a bit tedious, but the process is not too hard.
This describes how to map points from the plane to the fundamental domain (or inversely, how to map points from the fundamental domain to the plane).
A: To answer your question, and to understand the picture of the "fundamental domains in the upper half plane" (your third picture), it's probably best if we forget about $E_6$ and $G_{2k}$ for the time being. They are not directly relevant to your question, although we will return to them.
Instead, I would recommend that we concentrate on what the picture is actually showing. 
The picture is showing the "fundamental domains" of the modular group. The modular group is a group of transformations on the (extended) upper half plane:
$$ \left\{ z \mapsto\frac{az + b}{cz+d} \ : \ a, b, c, d \in \mathbb Z, \ \  ad-bc=1\right\}.$$
In the picture, the upper half plane is divided into many regions (also known as "fundamental domains"). These regions obey a nice property: Given any two regions $R_1$, $R_2$, there exists a unique transformation $g$ in the group of modular transformations that maps $R_1$ onto $R_2$.

Shall we work through an explicit example? Let's consider the transformation
$$ g : z \mapsto - \frac 1 {z}$$
(i.e. $a = 0, b = 1, c = - 1, d = 0$). And let's consider the region
$$ R_1 = \left\{  |{\rm Re}(z) | \leq \tfrac 1 2, \ \ |z| \geq 1 \right\}.$$
Can we work out the image of $R_1$ under the action of $g$?
Well, $g$ is a Mobius transformation, and we know that Mobius transformations map (generalised) circles to (generalised) circles. The boundary of $R_1$ is made up of segments of generalised circles. So the boundary of $g(R_1)$ should be made up of segments of generalised circles too. To be more specific:


*

*The part of the boundary of $R_1$ that is a line segment containing the points $\frac 1 2 + \frac{i \sqrt 3}{2}$, $\frac 1 2 + i$ and $\infty$ is mapped to a circle segment containing the points $-\frac 1 2 + \frac{i\sqrt 3}{2}$, $- \frac 2 5 + \frac{4i}{5}$ and $0$.

*The part of the boundary of $R_1$ that is a line segment containing the points $-\frac 1 2 + \frac{i \sqrt 3}{2}$, $-\frac 1 2 + i$ and $\infty$ is mapped to a circle segment containing the points $\frac 1 2 + \frac{i\sqrt 3}{2}$, $ \frac 2 5 + \frac{4i}{5}$ and $0$. 

*The part of the boundary of $R_1$ that is a circle segment containing the points $- \frac 1 2 + \frac{i\sqrt 3}{2}$ and $i$, $-\frac 1 2 + \frac{i\sqrt 3}{2}$ is mapped to itself.
Thus $g$ maps our region $R_1$ to the region in your diagram with vertices at $0$, $- \frac 1 2 + \frac{i \sqrt 3}{2}$ and $\frac 1 2 + \frac{i \sqrt 3}{2}$. And for the remainder of this answer, I'll call this new region $R_2$.

So what does all this have to do with your Eisenstein series? Well, the Eisenstein series obey the relation
$$ G_{2k} \left( \frac{az+b}{cz+d}\right) = (cz+d)^{2k}G_{2k}(z).$$
You may like to think of this equation as a simple relationship  between the values of the function $G_{2k}$ in one region of your picture and the values in another region.
For example, we've seen that $g : z \mapsto - 1 / z$ maps the region $R_1$ to the region $R_2$. The equation
$$ G_{2k} \left( - \frac 1 z \right) = z^{2k} G_{2k}(z)$$
is then a nice relationship between the values that $G_{2k}(z)$ takes in the region $R_1$ and the values it takes in the region $R_2$: if you calculate the values of $G_{2k}$ in one region, you get the values of $G_{2k}$ in the other region for free.
