# Every Lie algebra endomorphism of $\mathfrak{so}(3)$ is given by the anticommutator with a symmetric matrix.

I want to prove that for every Lie algebra endomorphism $$T$$ on $$\mathfrak g=\mathfrak{so}(3)$$, there exists a symmetric $$3\times 3$$ matrix $$B$$ such that $$T( x)=Bx+xB$$ for all $$x \in \mathfrak g$$. I cannot figure this out.

Edit: This is a problem in the book Quantum Mechanics for Mathematicians by Takhtajan

Edit The endomorphism is assumed to be symmetric.

• Since $\mathfrak g$ is simple and the assertion is trivial for $T=0$, we can w.l.o.g. assume $T$ to be an automorphism. If everything else fails, the automorphism group can be written down explicitly, but I admit I don't see how to go on from there. Is there any context to this question? – Torsten Schoeneberg Jul 15 at 16:48
• @TorstenSchoeneberg its a problem froma book. I edited my question for reference – JerryCastilla Jul 15 at 22:30
• Thanks. I feel like one should use somehow that $x=-x^{tr}$ for $x \in \mathfrak g$ as well as $B^{tr}=B$; and possibly, $Aut(\mathfrak g) = SO(3)$ meaning that $A^{-1}=A^{tr}$ for $A \in Aut(\mathfrak g)$, but I cannot make it work so far. The strange thing is that even for $T=id$ one needs a non-trivial matrix like $B= \frac12 I_3$. – Torsten Schoeneberg Jul 15 at 22:40

We can identify $$x\in \mathfrak{so}(3)$$ with $$\boldsymbol v \in \mathbb R^3$$: $$\begin{bmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{bmatrix} \mapsto \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$$ with the cross product as the Lie bracket.

I remember hearing that every non-zero endomorphism of $$\mathbb R^3$$ that preserves the cross product is of the form $$\boldsymbol v \mapsto R \boldsymbol v .$$ where $$R \in SO(3)$$. This corresponds to $$x \mapsto R^T x R.$$

But $$x \mapsto xB + Bx$$ corresponds to $$\boldsymbol v \mapsto (\text{tr}(B)I - B)\boldsymbol v.$$

Now we use that the endomorphism is symmetric. It seems to me that the Killing form on $$\mathbb R^3$$ with the cross product must be the standard inner product. So it follows that $$R$$ is symmetric.

That leaves two possibilities: $$R = I$$, when $$B = \frac12 I$$ works, or $$R = I - 2 \boldsymbol n \otimes \boldsymbol n$$ with $$\|\boldsymbol n\| = 1$$, in which case $$B = -\frac12 I +2 \boldsymbol n \otimes \boldsymbol n$$ works.

Added later: Why is a non-zero endomorphism on $$\mathbb R^3$$ that preserves the cross product necessarily an element of $$SO(3)$$? So suppose the endormorphism is $$\boldsymbol v \mapsto R \boldsymbol v$$. Let the three columns of $$R$$ be $$\boldsymbol a$$, $$\boldsymbol b$$, and $$\boldsymbol c$$. Then we have $$\boldsymbol a \times \boldsymbol b = \boldsymbol c, \quad \boldsymbol b \times \boldsymbol c = \boldsymbol a, \quad \boldsymbol c \times \boldsymbol a = \boldsymbol b .$$ Now if $$\boldsymbol a$$ and $$\boldsymbol b$$ are linearly dependent, then $$\boldsymbol c = \boldsymbol 0$$, from which it follows that $$\boldsymbol a = \boldsymbol b = \boldsymbol 0$$, which contradicts that the endomorphism is non-zero.

Now consider: $$\boldsymbol a \times (\boldsymbol a \times \boldsymbol b) = \boldsymbol a \times \boldsymbol c = - \boldsymbol b,$$ and $$\boldsymbol a \times (\boldsymbol a \times \boldsymbol b) = (\boldsymbol a \cdot \boldsymbol b) \boldsymbol a - \|\boldsymbol a\|^2 \boldsymbol b .$$ Then we see that $$\boldsymbol a \cdot \boldsymbol b = 0$$ and $$\|\boldsymbol a\| = 1$$. Similarly for any other pair of them. Thus $$\boldsymbol a$$, $$\boldsymbol b$$ and $$\boldsymbol c$$ are orthogonal unit vectors. Furthermore, they form a right handed pair. So $$R \in SO(3)$$.

Note: If $$B$$ were positive definite, then $$\text{tr}(B)I - B$$ could be a moment of inertia matrix created from the second moment tensor $$B = \int_{\mathbb R^3} \rho(\boldsymbol r) \boldsymbol r \otimes \boldsymbol r \, d \boldsymbol r$$ (here $$\rho$$ is the density function), and thus the map $$x \mapsto xB + Bx$$ is really a map from angular acceleration to angular momentum.

• I looked it up, the question in the book makes the hypothesis of the endomorphism to be symmetric. I would think w.r.t the killing form. – Victor Gustavo May Jul 17 at 18:12
• Then $R$ is symmetric, and it should be much easier. – Stephen Montgomery-Smith Jul 17 at 18:18
• I think you need to edit your problem. – Stephen Montgomery-Smith Jul 17 at 18:26
• Yes, I already edited the question. Thanks for pointing that out. – JerryCastilla Jul 18 at 5:03
• Re "Added even later": $(\mathfrak {so}_3, [,]) \simeq (\mathbb (R^3, \times))$ is simple, so every non-zero endomorphism is an automorphism, and it's fairly well-known that the automorphism group of $\mathfrak{so}_3$ is $SO(3)$ acting via matrix conjugation, as you basically (note $R^{-1}=R^T$ for $R \in SO(3)$) say yourself earlier in this (good, upvoted) answer. – Torsten Schoeneberg Jul 18 at 5:06