The representation $(\frac{1}{2},\frac{1}{2})$ of the Lorentz group correspond to a four- vector or a spin-one object. Right? Does it imply that any four-vector is identical to a spin-one object or any scalar is identical to a spin-0 object? This can't be correct, right? Because although $A^\mu$ is a four vector and a spin-one object at the same time (which is photon), there is no concept of spin associated with $p^\mu$ or $J^\mu$. I'm confused by terminologies of representation.

Edit- How can I show that $A^\mu$ represent a spin-1 object?

  • 1
    $\begingroup$ Technically, the representation $\left(\frac 12, \frac 12\right)$ is either for the group $\mbox{SL}(2,\mathbb C)$ or for the group $\mbox{SU}(2,\mathbb C) \times \mbox{SU}(2,\mathbb C)$, i.e. it's not a representation of the Lorentz group into a finite dimensional vector space. $\endgroup$
    – DanielC
    Dec 2, 2017 at 16:06

2 Answers 2


I) Firstly, we are talking about the direct or Cartesian product $SU(2)\times SU(2)$ of groups, not the tensor product$^1$ $SU(2)\otimes SU(2)$ of groups.

II) Secondly, $SU(2)\times SU(2)$ is not isomorphic to the Lorentz group $SO(3,1)$ but rather to a compact cousin

$$[SU(2)\times SU(2)]/\mathbb{Z}_2~\cong~ SO(4).$$

In particular, a $(\frac{1}{2},\frac{1}{2})$ irrep under $su(2)\oplus su(2)$ corresponds to a 4-dimensional fundamental vector representation under $o(4)$.

III) Thirdly, OP might be thinking of the complexified Lorentz group $SO(3,1;\mathbb{C})$, which has double cover $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$,

$$[SL(2,\mathbb{C})\times SL(2,\mathbb{C})]/\mathbb{Z}_2~\cong~ SO(3,1;\mathbb{C}).$$

cf. this Phys.SE post. In particular, a $(\frac{1}{2},\frac{1}{2})$ irrep under $sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$ corresponds to a 4-dimensional fundamental vector representation under $o(3,1;\mathbb{C})$.


$^1$ Note that there exist various Abelian and non-Abelian tensor product constructions for groups. E.g. for the Abelian group $(\mathbb{R}^n,+)$, the tensor product is $\mathbb{R}^n\otimes\mathbb{R}^m\cong \mathbb{R}^{nm}$, while the Cartesian product is $\mathbb{R}^n\times\mathbb{R}^m\cong \mathbb{R}^{n+m}$.

  • $\begingroup$ -@Qmechanic♦ -Is there a difference between cartesian product and tensor product? I thought they are same and also same as direct product. $\endgroup$
    – SRS
    Feb 16, 2014 at 4:47
  • $\begingroup$ How does the dirac spinor transform in terms of $SO(3,1)$? I thought $SL(2, \mathbb{C})$ is the double cover of $SO^+(3,1)$ (the connected component of the lorentz group), and so the left and ring hand weyl spinor correspond to the fundamental rep of $sl(2,\mathbb{C})$. But according to your post, $(1/2,1/2)$ irrep corresponds to the 4-dim vector rep. $\endgroup$
    – user7757
    Feb 16, 2014 at 15:10
  • $\begingroup$ @ramanujan_dirac The left handed weyl spinor irrep is $(1/2, 0)$ while the right-handed weyl spinor irrep is $(0,1/2)$. The irrep $(1/2, 1/2)$ is distinct from these and is equivalent to the vector rep. $\endgroup$ Feb 19, 2014 at 5:56

The problem here is with the identification of the $(A,B)$ values of a representation with spin. $A$ and $B$ do not correspond to spin (they are not even Hermitian!), they just happen to obey $SU(2)$ Lie algebras, and as such they add up in the same way that spins do. When we say that $A_\mu,J_\mu,p_\mu,...$ are all in the $(\frac{1}{2},\frac{1}{2}) $ representation of the Lorentz group we mean that they transform as a four-vector, that's all. People may get lazy and say they are spin 1 objects, but what they really mean is $(A,B)$ spin 1 objects.


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