0
$\begingroup$

What is meant by the vector representation of $\mathrm{SO}(6)$?

I have only encountered the term vector representation in the context of the Lorentz group $\mathrm{SO}(1,3)$, where it refers to the $(1/2,1/2)$ representation, derived from the isomorphism $\mathrm{so}(1,3) \cong \mathrm{su}(2) \oplus \mathrm{su}(2)$ of the Lie algebra.

$\endgroup$
1
  • $\begingroup$ The isomorphism you mention does not exist. The $\mbox{so}(1,3)$ needs to be complexified and the same for the $\mbox{su}(2)$. $\endgroup$
    – DanielC
    Commented Feb 13, 2020 at 14:19

2 Answers 2

5
$\begingroup$

Unless there is any further context, it's the fundamental 6-dimensional representation: if $$ M^T M=1_6\,,\quad \det M=1\,,\qquad M \;\text{is}\; 6\times 6 $$ so $M$ is an $SO(6)$ matrix, then the representation space is $\mathbb R^6$ and $M$ acts as $$ \mathbb R^6\ni v \to Mv. $$

$\endgroup$
3
  • $\begingroup$ Thanks! Yes, it seems to be the case from the context. Is this terminology standard? It seems a bit strange. In another place in the same paper, they talk about the fundamental representation, which seems to refer to the same thing. It seems to me it would be better to use the term defining representation. $\endgroup$ Commented Feb 13, 2020 at 11:27
  • 1
    $\begingroup$ Yes, for $so(\dots)$ things. "Fundamental representation" can be ambiguous in the presence of exceptional isomorphisms for low rank algebras, notably in this very case of $su(4)\cong so(6)$ (as Lie algebras). $\endgroup$
    – user21299
    Commented Feb 13, 2020 at 11:31
  • $\begingroup$ Alright, thanks! $\endgroup$ Commented Feb 13, 2020 at 11:32
2
$\begingroup$

As already stated in another answer, the vector representation of $SO(6)$ is the fundamental representation. Since you mentioned the $(1/2,1/2)$ representation of $so(1,3)$ I would like to add some examples of non-vector representations.

A simple example of a representation different from the vector representation is the adjoint representation acting on the vector space of $6\times 6$ matrices. Here, $g\in SO(6)$ acts as $A\in M_6(\mathbb{R})\mapsto g^{-1}A g$. Also a spinor does not transform in the vector representation of $SO(1,3)$, but in a projective one, e.g. in $(1/2,0)$ and $(0,1/2)$.

You might check this related question discussing the terminology of the expression "vector representation".

$\endgroup$
1
  • $\begingroup$ Thanks for the example and for the link! $\endgroup$ Commented Feb 13, 2020 at 16:08

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.