From wikipedia and some other sources, I've read that if $G_1$ and $G_2$ are two irreps of some group, then $G_1\otimes G_2$ can be a representation of both the group $G$ and the new group $G \times G$.

In the former case, the new representation is not in general, irreducible. A common example would be angular momentum addition, and the Clebsch-Gordan coefficients. According to wikipedia however, when viewed as a representation of $G\times G$, the tensor product of representations is irreducible.

Let us now consider the group $SO(3,1)$. We know that in terms of algebra,

$$so(3,1)=su(2)\oplus su(2)$$

I'm being a little sloppy with the notation, ignoring complexification etc.

Similarly $SO(3,1) \sim SU(2) \times SU(2)$ ( again, forgive the general sloppyness of notation ).

Now, according to the book physics from symmetry, it is mentioned,

$$(\frac{1}{2},\frac{1}{2})=(\frac{1}{2},0)\otimes (0,\frac{1}{2}).$$

So, the 'vector' representation of the lorentz group is a tensor product of the left and right handed 'weyl' representations of the same. This seems similar to the Clebsch-Gordan and addition of angular momentum to me. Moreover, when tensor product of two representations of the same group is used to represent the group itself, then the representation, is in general reducible.

However, from a different perspective, we can say that $(\frac{1}{2},\frac{1}{2})$ is basically the tensor product of two $2d$ representations of $SU(2)$, and thus is a representation of $SU(2)\times SU(2)$. From this perspective however, it must be irreducible, according to the first two paragraphs.

So, is the representation irreducible or not? I am having trouble understanding where my reasoning is going wrong.

Like if $G_1,G_2$ represent irreps of $SU(2)$, then $G_1\otimes G_2$ represents an irrep of $SU(2)\times SU(2)$. At the same time however, if $G_1,G_2$ represent irreps of $SO(3,1)$, then $G_1\otimes G_2$ is another representation of $SO(3,1)$ which in general is not an irrep.

Since $SU(2)\times SU(2) \sim SO(3,1)$, I seem to have reached a contradiction or something. Please tell me where my reasoning is flawed and what is the correct way to think about this.

  • $\begingroup$ Which wiki pages? $\endgroup$
    – Qmechanic
    Oct 14, 2023 at 8:25

2 Answers 2


This is a case where you can't be "ignoring complexification". The representation is reducible over the complex field but irreducible over the reals.

  • $\begingroup$ Can you please elaborate on this answer, as I'm only just starting to learn representation theory, if you don't mind. Also, from your answer can I say that, instead of $SO(3,1)$ if I had taken $SO(4)$, then it would have been irreducible ? $\endgroup$ Oct 13, 2023 at 22:18
  • $\begingroup$ The generators of $su(2)\oplus su(2)$ are complex combinations of the boosts and rotations that are generators of $so(3,1)$ and that's what makes reducibility possible over the complex, but obstruct reducibility over the reals as such combos are not allowed over the reals. $\endgroup$ Oct 13, 2023 at 22:58
  1. With the shorthand notation $$H~:=~SU(2)~\subseteq~SL(2,\mathbb{C})~=: ~G,$$

    • then $G$ is the double cover of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$,

    • and $G_L\times G_R$ is the double cover of the complexified proper Lorentz group $SO(1,3;\mathbb{C})$,

    cf. e.g. this related Phys.SE post.

  2. $G$ is isomorphic to the diagonal imbedding $$G~\cong~\{(g,g) \mid g\in G\} ~\subseteq~ G_L\times G_R,$$ and hence a subgroup of $G_L\times G_R$.

  3. OP asks about the $(\frac{1}{2},\frac{1}{2})$ representation $V_L\otimes V_R$, where $V_L$ and $V_R$ denote the left-handed and the right-handed Weyl-spinor representation, respectively, cf. e.g. this related Phys.SE post.

    Note that $V_L$ and $V_R$ are inequivalent representations of $G$.

  4. Any representation of a group induces a restricted representation on a subgroup.

  5. $V_L\otimes V_R$ is an irreducible representation wrt. the groups $$ G_L\times G_R, \quad G, \quad H_L\times H_R, $$ but it is an reducible representation via Clebsch-Gordan decomposition $$\frac{1}{2}\otimes\frac{1}{2} ~\cong~ 0\oplus 1$$ wrt. the subgroup $H$.

  6. This last fact is related to that the fundamental representation of $H$ is equivalent to the complex conjugate representation, cf. e.g. my Phys.SE answer here. The same does not hold for $G$.


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