1
$\begingroup$

Thinking about $SO(3)$. Any rotation matrix $R$ can be written $$ R = e^{\theta \hat{n}\cdot J} $$ where $J$ is a vector the three skew-symmetric generators of rotation $J_x$, $J_y$, and $J_z$. In this case $R$ representation a rotation of angle $\theta$ about axis $\hat{n}$. These three $J$ matrices can be written as \begin{align*} J_x =& \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\\ J_y =& \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\\ J_z =& \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ \end{align*} We have $$ [J_i, J_j] = \epsilon_{ijk} J_k $$ Thinking of the space spanned by the $J_i$ matrices as the Lie algebra $\mathfrak{so}(3)$ of the Lie group $SO(3)$ we can understand that $\epsilon_{ijk}$ are the structure coefficients of the Lie algebra.

However, in this case we ALSO have the property that $$ (J_i)_{kj} = -\epsilon_{ijk} $$ So the matrix elements of the action of the $J_i$ matrices on $\mathbb{R}^3$ are also given by $\epsilon_{ijk}$. We have $$ [J_i, J_j] = -(J_i)_{jk} J_k $$ Is this a coincidence? Or is there some deeper Lie or representation theory reason why this is expected?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ this is called the adjoint representation. it exists for any simple lie algebra $\endgroup$
    – AccidentalFourierTransform
    Commented Jun 28 at 16:06
  • $\begingroup$ Have you learned about the Jacobi identity? $\endgroup$
    – Cosmas Zachos
    Commented Jun 28 at 16:07
  • $\begingroup$ @AccidentalFourierTransform do you mean "simple" Lie algebra in the technical sense like here: en.wikipedia.org/wiki/Simple_Lie_algebra $\endgroup$
    – Jagerber48
    Commented Jun 28 at 16:11
  • $\begingroup$ @CosmasZachos I know what the Jacobi identity is, I know that Lie algebras satisfy it. That's all. $\endgroup$
    – Jagerber48
    Commented Jun 28 at 16:16
  • 1
    $\begingroup$ Well, you then know that the structure constants satisfy the commutation relations for the Lie algebra, and you are done: not a coincidence. $\endgroup$
    – Cosmas Zachos
    Commented Jun 28 at 16:20

1 Answer 1

Reset to default
4
$\begingroup$

This is not a coincidence. The three matrices you have chosen as generators for your rotations live in the adjoint representation of $\mathfrak{so}(3)$. This explicit matrix elements of the adjoint representation are always given by the structure constants themselves (https://en.wikipedia.org/wiki/Adjoint_representation)

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ Ok, so in particular, if we had another representation of $\mathfrak{so}(3)$ like $\sigma_{x,y,z}$, the Pauli matrices, we wouldn't be surprised to find $[\sigma_i, \sigma_j] \propto (J_i)_{jk}\sigma_k$. That is, commutation relations of any representation of a Lie algebra are always going to be given by matrix coefficients of elements of THE adjoint representation? $\endgroup$
    – Jagerber48
    Commented Jun 28 at 16:10
  • 1
    $\begingroup$ That is correct, the structure constants always give the matrix elements of the adjoint representation. $\endgroup$
    – CStarAlgebra
    Commented Jun 28 at 16:18
  • 1
    $\begingroup$ I feel like this is a dumb question but asking anyways: I didn't know that "the three matrices I have chosen as generators for [my] rotations live in the adjoint representation of $\mathfrak{so}(3)$". How can I tell that my matrices are in the adjoint representation? $\endgroup$
    – Jagerber48
    Commented Jun 28 at 16:35
  • 1
    $\begingroup$ That is a non trivial question (see mathoverflow.net/questions/416420/…), I would say that for $\mathfrak{so}(3)$ one would know that the $\mathbf{3}$ rep is the defining one and it just so happens to be the adjoint here. $\endgroup$
    – CStarAlgebra
    Commented Jun 28 at 17:17
  • 1
    $\begingroup$ Perhaps the fact that $(J_i)_{jk} \propto f_{ijk}$ is the evidence that $J_i$ generate the adjoint representation. And the thing I should not be surprised by is the fact that there is some representation which satisfies this (namely the adjoint representation exists). $\endgroup$
    – Jagerber48
    Commented Jun 28 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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