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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.

JerryCastilla
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  • 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 '20 at 16:48
  • @TorstenSchoeneberg its a problem froma book. I edited my question for reference – JerryCastilla Jul 15 '20 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 '20 at 22:40

1 Answers1

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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.

Stephen Montgomery-Smith
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