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 $A, B \in sl_2(\mathbb{C})$.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.