What is meant by saying that $\theta$ is orthogonal? I am reading the materials discussed in lecture $5$ from the lecture notes on quantum groups about Belavin-Drinfeld classification theorem written by Pavel Etingof and Oliver Schiffmann.
In the first half of this lecture the authors proved that any Lie bialgebra structure on a complex simple Lie algebra $\mathfrak g$ is a quasitriangular structure and hence in order to classify all the Lie bialgebra structures on a complex simple Lie algebra $\mathfrak g$ it is enough to classify all the quasitriangular structures on $\mathfrak g$ which amounts to classify all the $r$-matrices whose symmetrization $r + r^{21} \in \mathbb C \Omega,$ where $\Omega$ is a Casimir tensor in $S^2 \mathfrak g$ corresponding to the Killing form $K$ on $\mathfrak {g}.$
So we need to search for $r$-matrices $r$ such that $r + r^{21} = \varepsilon \Omega$ for some $\varepsilon \in \mathbb C.$ By normalizing if necessary we can assume that $\varepsilon = 1$ so that $r$ is an $r$-matrix satisfying $r + r^{21} = \Omega.$ Then the authors showed that this is equivalent to finding out a linear map $f : \mathfrak g \longrightarrow \mathfrak g$ satisfying
$\tag {1} f + f^{\ast} = 1$ and $$\begin{equation}\tag{2} (f - 1) [f (x), f (y)] = f \left ( [(f - 1) (x), (f - 1) (y)] \right ). \end{equation}$$ After having obtained this equivalence the authors  argued that if $f - 1$ is invertible then the map $\theta : \mathfrak g \longrightarrow \mathfrak g$ given by $x \mapsto \frac {f (x)} {f(x) - x}$ is an orthogonal automorphism of $\mathfrak g.$
But Belavin and Drinfeld already showed that such an $f$ (and hence $f - 1$) cannot be invertible and hence this procedure of finding an orthogonal automorphism of $\mathfrak g$ doesn't really work. Hence the following adjustments need to be made.
First note that $\text {Ker}\ (f) \subseteq \text {Im}\ (f - 1)$ and $\text {Ker}\ (f - 1) \subseteq \text {Im}\ (f)$ for any linear operator $f.$ So we will consider $\theta$ as a map $$\theta : \frac {\text {Im}\ (f - 1)} {\text {Ker}\ (f)} \longrightarrow \frac {\text {Im}\ (f)} {\text {Ker}\ (f - 1)}.$$
Luckily, the above claim still holds in a modified form $:$

Lemma $:$ Let $f : \mathfrak g \longrightarrow \mathfrak g$ be linear map satisfying $(1).$ Then we have $\text {Ker}\ (f) = \text {Im}\ (f - 1)^{\perp},$ $\text {Ker}\ (f - 1) = \text {Im}\ (f)^{\perp}$ and the map $\theta$ is orthogonal. Furthermore, $f$ satisfies $(2)$ if and only if $\text {Im}\ (f)$ and $\text {Im}\ (f - 1)$ are Lie subalgebras of $\mathfrak {g},$ and $\theta$ is a Lie algebra isomorphism.

The first part of the proof has been skipped and left as an exercise. In the first part I don't understand what is meant by saying $\theta$ is orthogonal. I have few questions here $:$

*

*How is the modified $\theta$ defined?


*In order to talk about orthogonality of $\theta$ we need to have some bilinear forms on the quotients induced by the Killing form so as to define the adjoint $\theta^{\ast}$ of $\theta.$ What are they?


*In order to show that $\theta$ is a Lie algebra isomorphism we need to have brackets on the quotients induced by the bracket of $\mathfrak {g}$  giving rise to the structure of Lie algebras on the quotients. What are even they?
These questions are not addressed at all in the notes which need to be clarified first. Could anyone please shed some light on it?
Thanks for your time.
 A: Let me answer the questions I posted one by one.


*

*How is $\theta$ defined?


We define $\theta$ in the obvious way as follows $:$
$$\theta \left ((f - 1) (x) + \text {Ker}\ (f) \right ) = f(x) + \text {Ker}\ (f - 1).$$
For well-definedness of  $\theta$ if $x' \in \mathfrak g$ is such that $f (f - 1) (x -  x') = 0$ then $(f - 1) f (x - x') = 0$ (since $f$ and $f - 1$ commute) and hence $f (x) - f(x') \in \text {Ker}\ (f - 1),$ which is what we need to show.

What is the bilinear form defined on the domain and the range of $\theta$ induced by the Killing form?

The bilinear form induced by the Killing form defined on the domain of $\theta$ is given by $$K_D \left ((f - 1) (x) + \text {Ker} (f), (f - 1) (y) + \text {Ker}\ (f) \right ) = K \left ((f - 1) (x), (f - 1) (y) \right ).$$
In order to show that it is well-defined it is enough to show that it is well-defined with respect to the first component. For that let us take $x' \in \mathfrak g$ such that $f (f - 1) (x - x') = 0.$ Then $$\begin{align*} K \left ((f - 1) (x - x'), (f - 1) (y) \right ) & = K \left ((f - 1) (x - x'), -f^{\ast} (y) \right )\ \ (\because f + f^{\ast} = 1) \\ & = - K \left (f (f - 1) (x - x'),  y \right ) \\ & = 0\ \ (\because f(f - 1) (x - x') = 0) \end{align*}$$
So we are through.
Similarly, we can define a bilinear form on the range as follows $:$
$$K_R \left (f(x) + \text {Ker}\ (f - 1), f(y) + \text {Ker}\ (f - 1) \right ) = K \left (f(x), f(y) \right ).$$
Moreover these forms are symmetric and non-degenerate. Symmetricity of both the forms are inherited from the symmetricity of $K.$ Let me now show that $K_D$ is non-degenerate. The non-degeneracy of $K_R$ is similar.
Suppose for all $y \in \mathfrak g$ we have $K ((f - 1) (x), (f - 1) (y)) = 0$ then $K (f(f - 1) (x), y) = 0$ for all $y \in \mathfrak g.$ This shows by the non-degeneracy of $K$ that $f(f - 1) (x) = 0$ i.e. $(f - 1) (x) \in \text {Ker}\ (f),$ which is what we want to show.
From here it's fairly easy to show that $\theta^{\ast} = \theta^{-1}.$

What are brackets on the domain and the codomain of $\theta\ $?

If $f$ satisfies $(2)$ then the bracket on the domain of $\theta$ is given by $$\begin{align*} \left [(f - 1) (x) + \text {Ker}\ (f), (f - 1) (y) + \text {Ker}\ (f) \right ] & = \left [(f - 1) (x), (f - 1) (y) \right ] + \text {Ker}\ (f). \end{align*}$$ To see that this bracket is well-defined it is enough to show that the bracket is well-defined with respect to the first component. For that let $x, x' \in \mathfrak g$ such that $f(f - 1) (x - x') = 0.$ Now if $f$ satisfies $(2)$ we have $$\tag{3} \left [(f - 1) (y), (f - 1) (z) \right ]  = (f - 1) (w)$$ where $$w = [f(y), z] + [y, f(z)] - [y,z].$$ Then by invariance of $K$ we have for all $z \in \mathfrak g$ $$\begin{align*} K \left (f \left (\left [(f - 1) (x - x'), (f - 1) (y) \right ] \right ), z \right ) & = - K \left (f(f - 1) (x - x'), w \right )\ \ (\because f + f^{\ast} = 1\ \text {and by}\ (3)) \\ & = 0 \end{align*}$$ So by non-degeneracy of $K$ it follows that $$\begin{align*} f \left (\left [(f - 1) (x - x'), (f - 1) (y) \right ] \right ) & = 0. \end{align*}$$ Which is exactly what we need to show for well-definedness of the bracket.
In a similar way, we can define the bracket on the codomain of $\theta$ as $$\begin{align*} \left [f(x) + \text {Ker}\ (f - 1), f(y) + \text {Ker}\ (f - 1) \right ] & = [f(x), f(y)] + \text {Ker}\ (f - 1). \end{align*}$$
Now it is easy to show that $\theta$ is a Lie algebra map with respect these brackets.
