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Let $L = sl_2(\mathbb{C})$. We call $L = H_0 \oplus H_1 \oplus H_2$ an $orthogonal \ decomposition$ if all components $H_i$ of decomposition are Cartan subalgebras and pairwise orthogonal with respect to the Killing form.

Now, suppose that $H_0$ is a Cartan subalgebra consisting of the diagonal matrices and $H_1$ is another Cartan subalgebra orthogonal to $H_0$ via the Killing Form. One can check easily that $H_1$ has a basis of the form: $H_1 = \Big < \begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix} \Big >_\mathbb{C}$ for some $a \neq 0$. Thus, $L = H_0 \oplus \Big < \begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix} \Big >_\mathbb{C}\oplus \Big < \begin{pmatrix} 0 & 1 \\ b & 0 \end{pmatrix} \Big >_\mathbb{C}$ for some $a, b \neq 0$.

Can we prove that either $a$ or $b$ (from above) must be $1$ in order to have the orthogonal decomposition? Or they don't have to be. If the answer is negative, can we still show that for each $a \neq 0$, there exists an automorphism $\phi \in Aut(L)$ such that $\phi(H_0) = H_0$ and $\phi(\begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix}) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$?

Note that the Killing form can be given by $K(A, B) = 4Tr(AB)$ for all $A, B \in sl_2(\mathbb{C})$.

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The orthogonal complement of $H_0$ in $\mathfrak sl_2$ with respect to the killing form is

$$\{\begin{pmatrix} 0 & a \\ b & 0 \end{pmatrix} : a, b \in \mathbb{C} \}$$

and as you mentioned if you want to decompose the two dimensional space $H_0^{\perp}$ into a direct sum of Cartan subalgebras $H_1 \oplus H_2$, then you're forced to have $H_1 = \langle \begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix} \rangle, H_2 = \langle \begin{pmatrix} 0 & 1 \\ b & 0 \end{pmatrix} \rangle$ for distinct, nonzero $a, b$.

If you want $H_1, H_2$ to be orthogonal, then you just need the trace $\begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ b & 0 \end{pmatrix} = \begin{pmatrix}b & 0 \\ 0 & a \end{pmatrix}$ to be zero. So neither $a$ nor $b$ has to be one; you just need $a = -b$.

For your question about the automorphism, I think the answer is no. I believe every Lie algebra automorphism of $\mathfrak{sl}_2$ is the differential of an automorphism of algebraic groups of $\textrm{SL}_2$. Every automorphism $\phi$ of $\textrm{SL}_2$ is know to be inner, say $x \mapsto gxg^{-1}$ for some $g \in \textrm{SL}_2$. The differential $d\phi$ of such an automorphism is given by the same formula: $X \mapsto gXg^{-1}$.

In order for $d\phi$ to fix $H_0$, $\phi$ has to fix the standard torus in $\textrm{SL}_2$, which means $g$ needs to be in that torus: so $g = \begin{pmatrix} c \\ & \frac{1}{c} \end{pmatrix}$. From here you can see that what you want is impossible.

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  • $\begingroup$ I am not sure why they would say that. Maybe I'm wrong somewhere? $\endgroup$ – D_S Sep 17 '17 at 3:24
  • $\begingroup$ @D_S: No, you are right. $\endgroup$ – Moishe Kohan Sep 17 '17 at 3:28
  • $\begingroup$ By the way, anyone knows if the statement "$sl_2(\mathbb(C))$ possesses an orthogonal decomposition that is unique up to conjugacy" is true? The book has this statement as a theorem. But from what we have discussed, I believe the statement is not true because @D_S is absolutely right. $\endgroup$ – NongAm Sep 17 '17 at 3:44
  • $\begingroup$ @NongAm Are you sure they require the summands to be Cartan subalgebras and not just abelian subalgebras? The first seems like an odd requirement. $\endgroup$ – Tobias Kildetoft Sep 17 '17 at 7:10
  • $\begingroup$ I recommend you take a screenshot of the claim and post a new question $\endgroup$ – D_S Sep 17 '17 at 19:11

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