Lie algebra of nonabelian group is $[T^a,T^b]=if^{abc}T^c$. For $SO(3)$ case, is the representation $T^a_{ij}=-i\epsilon^{aij}$ fundamental or adjoint? The fundamental representation is defined as identifying $T^a$ the original generator. The adjoint representation is defined as identifying $T^a$ as structure constant. For $SO(3)$ case, generators seem to be same as structure contant.


  1. Srednicki QFT
  • 3
    $\begingroup$ They coincide.. $\endgroup$
    – nervxxx
    Dec 7, 2013 at 17:34
  • 2
    $\begingroup$ @nervxxx Why not turn your comment into an answer? $\endgroup$ Dec 7, 2013 at 18:02

1 Answer 1


Defining a Lie algebra by the commutation relations $[T^a,T^b]=if^{abc} T^c$, the adjoint representation is defined by $(T_{adj}^a)^{bc}= if^{abc}$. Now it turns, that in the special case of $so(3)=su(2)$, you have $f^{abc} = \epsilon^{abc}$, where $\epsilon^{abc}$ is the totally antisymmetric tensor. So, your representation is the adjoint representation.


Due to OP comments, some precisions :

There is a subtelty. the Lie algebra defined by the commutation relations $[T^a,T^b]=i\epsilon^{abc} T^c$ is fundamentally $su(2)$, but it is also $so(3)$.

Why this? In fact, different Lie groups may correspond to the same Lie algebra. In the over way, if we take the "exponential" of a Lie algebra, we obtain only one simply connected Lie group. Starting from the Lie algebra $su(2)$, and exponentiate it, we obtain the simply connected Lie group $SU(2)$. However, from the Lie group $SU(2)$, we may consider other Lie groups which are the quotient of $SU(2)$ by a finite group, so these new groups are not simply connected, but they share with $SU(2)$ the same Lie algebra. For instance, one has $SO(3) = SU(2)/Z_2$. So two elements of $SU(2)$ correspond to the same element of $SO(3)$. But $SO(3)$ and $SU(2)$ share the same Lie algebra $su(2)$

We see that $SU(2)$ play a central role in all this, so it is natural to look at its fundamental representation, which is $2-$dimensional, the fundamental representation is simply given by the Pauli Matrices $(T_{f}^a)= \frac{1}{2}\sigma_a$. The adjoint representation for $SU(2)$ is 3-dimensional and is given by $(T_{adj}^a)^{bc}= if^{abc}$

Now, you may also think at the representations of $SO(3)$, and, yes, the fundamental representation is $3$-dimensional, and is the same as the adjoint representation of $SU(2)$.

It is in fact better to give to $SU(2)$ a central role, in the meaning, that the representations of $SO(3)$ are included in the representations of $SU(2)$ (the inverse is not true), which is more fundamental (because, as explained above, it is the simply connected group you obtain by exponentiating the Lie algebra).

Finally, a problem of vocabulary, maybe you have heard of will hear also about "spinorial" representations of $S0(3)$, but it is in fact simply the representations of $SO(3)$ which are not standard ("vectorial") representations of $SO(3)$, so these are representations of $SU(2)$ which are not standard representations of $SO(3)$. For instance, you may consider the fundamental representation of $SU(2)$ (which applies to a 2-spinor), as a "spinorial" representation of $SO(3)$

  • $\begingroup$ From $SO(3)$ piont of view, $\varepsilon$ is also generator, so it is also fundamental rep. ? $\endgroup$
    – thone
    Dec 8, 2013 at 3:46
  • $\begingroup$ @CraigThone : I updated the answer $\endgroup$
    – Trimok
    Dec 9, 2013 at 11:04
  • 2
    $\begingroup$ @CraigThone I talk about $SO(3)$, $SU(2)$, their shared Lie algebra and the relationship between their global topology in some detail here and here $\endgroup$ Dec 9, 2013 at 11:45
  • $\begingroup$ It is a little weird for me. $SO(3)$ and $SU(2)$ have the same algebra while this algebra has different representations for each. May I infer that the representation of the algebra contains the information of its group which is not known from the algebra itself? $\endgroup$
    – thone
    Dec 9, 2013 at 13:35
  • 1
    $\begingroup$ @CraigThone : Strictly speaking, I was talking about representation of Lie groups (which is the usual case in Physics), not about the representations of Lie algebras. In some sense, the representations of a Lie algebra, is the "same thing" that the representations of its simply connected exponential Lie group. But other Lie groups which share the same Lie algebra, have not all the representations, for instance $S0(3)$ has only one half of the representations (which correspond to integer spin), while $SU(2)$ has all representations (integer spin + half-integer spin) $\endgroup$
    – Trimok
    Dec 9, 2013 at 17:58

Your Answer

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

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