We know the irreps of $\mathfrak{su}(2)$ are indexed by (half) integer j with representation space $\mathcal{V}_j$ which has dimensionality $dim(\mathcal{V}_j)=2j+1$.

If we take the tensor product of two of these irreducible presentations say $\rho_{j_1}$ and $\rho_{j_2}$ then we know we can decompose this into a direct sum of irreducible reps. For example $j_1=j_2=1/2$, then representation space decomposes as $\mathcal{V}_{\frac{1}{2}} \otimes \mathcal{V}_{\frac{1}{2}}= \mathcal{V}_0 \oplus \mathcal{V}_1 $.

Now lets consider irreps of the Lorentz group $\mathfrak{so(3,1)}$. One can show that the Lie algebra decomposes $\mathfrak{su(2)} \oplus \mathfrak{su(2)} \simeq \mathfrak{so(3,1)} $ and so irreps of the Lorentz group are indexed by two half integer numbers $(j_1,j_2)$. We can form tensor product reps too analogous to the $\mathfrak{su(2)}$ example above, for example $\mathcal{V}_{(\frac{1}{2},\frac{1}{2})} \otimes \mathcal{V}_{(\frac{1}{2},\frac{1}{2})}= \mathcal{V}_{(1,1)} \oplus \mathcal{V}_{(1,0)} \oplus\mathcal{V}_{(0,1)}\oplus\mathcal{V}_{(0,0)} $. So far so good.

Now the problem is that in literature I have read, including my uni lecture notes and an specific example 'Quantum Field Theory and the Standard Model' by Matthew D. Schwartz p163, they begin to decompose reps of the lorentz lie algebra into reps of SU(2) lie algebra. A specific example will clarify what I mean.

Take the irrep indexed by $(\frac{1}{2},\frac{1}{2})$. Matthew says we can decompose this into $1 \oplus 0$ but this doesn't make sense to me. Matthew then uses this to show that a 4-vector rep of the lorentz group can represent a particle of spin 0 or 1, then using an appropriate lagrangian (Proca lagrangian) we can make the spin 0 component non propagating and we end up with a theory describing a massive spin 1 particle.

Since $(\frac{1}{2},\frac{1}{2})$ really means $\frac{1}{2} \oplus \frac{1}{2}$ then how do we end up with $1 \oplus 0$? Generally if we take any rep of the lorentz group $(j_1,j_2)$ and had representations $\rho_{j_1}$ and $\rho_{j_2}$ mapping to representation spaces $\mathcal{V_{j_1}}$ and $\mathcal{V_{j_2}}$ respectively then the resulting representation of the lie alegbra would be $\mathcal{V_{j_1}} \oplus \mathcal{V_{j_2}}$? I am not sure what is happening at this stage.

Any help would be very appreciated.

  • $\begingroup$ If I understand your question, you are missing one point. Textbooks talk about how a $(\frac{1}{2}, \frac{1}{2})$ rep is seen as a $1 \oplus 0$ when a $SO(3)$ rotation is made. $\endgroup$ Commented Feb 5, 2021 at 14:51
  • 2
    $\begingroup$ " Since $(\frac{1}{2},\frac{1}{2})$ really means $\frac{1}{2} \oplus \frac{1}{2}$"... is very-very wrong. It means $\frac{1}{2} \otimes \frac{1}{2}$. Look at the group element in the Cartesian product. $\endgroup$ Commented Feb 5, 2021 at 15:15
  • $\begingroup$ You probably misunderstand the coproduct in the algebra. $\endgroup$ Commented Feb 5, 2021 at 15:21
  • $\begingroup$ Do you appreciate $(j_1,j_2)$ has dimension $(2j_1+1)(2j_2+1)$? $\endgroup$ Commented Feb 5, 2021 at 16:02

1 Answer 1


First one minor detail, the isomorphism of Lie algebras you mention demands a complexification. The Lorentz algebra $\mathfrak{so}(1,3)$ by default is a real Lie algebra. Observe that when you prove the aforementioned isomorphism you need to take linear combinations of the generators multiplying by $i$. In order to do so you must complexify, so the correct statement would be $\mathfrak{so}_\mathbb{C}(1,3)\simeq \mathfrak{su}_\mathbb{C}(2)\oplus \mathfrak{su}_\mathbb{C}(2)$. For more details on this check "Quantum Field Theory for Mathematicians" by Robin Ticciati which includes a nice discussion.

Now a more severe mistake is that, as explained in comments, you say that $\left(\frac{1}{2},\frac{1}{2}\right)$ is $\frac{1}{2}\oplus \frac{1}{2}$. That is not true at all. The true construction is this: as soon as you complexify the Lorentz algebra to $\mathfrak{so}_\mathbb{C}(1,3)$ you can construct the elements $$\mathscr{A}_i=\dfrac{1}{2}(\mathscr{J}_i+i\mathscr{K}_i),\quad \mathscr{B}_i=\dfrac{1}{2}(\mathscr{J}_i-i\mathscr{K}_i),\tag{1}$$

where $\mathscr{J}_i$ are the angular momentum operators and $\mathscr{K}_i$ the boost generators. You can easily check that the Lorentz algebra implies $\mathscr{A}_i$ and $\mathscr{B}_j$ commute with each other and independently obey the angular momentum algebra. To construct these you then pick two representations of the angular momentum algebra, say the irreps with spins $A$ and $B$. The first acts on $\mathbb{C}^{2A+1}$ and has generators $J_i^{(A)}$ and the second on $\mathbb{C}^{2B+1}$ and has generators $J_i^{(B)}$. Then you construct the space $\mathbb{C}^{2A+1}\otimes \mathbb{C}^{2B+1}$ and you embbed these generators as $$\mathscr{A}_i = J_i^{(A)}\otimes 1_B,\quad \mathscr{B}_i = 1_A\otimes J_i^{(B)}.\tag{2}$$

It is again very easy to show that $\mathscr{A}_i$ commutes with the $\mathscr{B}_j$ and that individually they satisfy the same algebra that $J_i^{(A)}$ and $J_i^{(B)}$ satisfies. You then use (1) in the reverse direction to obtain the action of the Lorentz generators. Since you have one representation space $\mathbb{C}^{2A+1}\otimes \mathbb{C}^{2B+1}$ and operators defined there obeying the Lorentz algebra you have a representation of the Lorentz algebra. As you can see this is nothing like a direct sum of two representations of $\mathfrak{su}_\mathbb{C}(2)$ which is what $\frac{1}{2}\oplus \frac{1}{2}$ means.

Now to your question, what people really mean when they say that $(\frac{1}{2},\frac{1}{2})$ decomposes as $0\oplus 1$ is that the rotation subgroup of the Lorentz group is reducibly represented and its representation decomposes as $0\oplus 1$.

Mathematically, take ${\rm SO}(1,3)$. Since rotations are Lorentz transformations we have ${\rm SO}(3)\subset {\rm SO}(1,3)$. More rigorously there is a subgroup of ${\rm SO}(1,3)$ isomorphic to ${\rm SO}(3)$. Now if $V$ is a representation space of ${\rm SO}(1,3)$ and $\rho: {\rm SO}(1,3)\to {\rm GL}(V)$ is a representation, if you consider $\rho(R)$ just for $R\in {\rm SO}(3)$ then by restriction to a subgroup you have a representation of the rotation group $\rho|_{\rm SO}(3):{\rm SO}(3)\to {\rm GL}(V)$. This representation can be either reducible or irreducible, and in particular you can decompose it into the irreps that you already know.

That is what is going on. In the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group, the associated ${\rm SO}(3)$ representation is $0\oplus 1$.

This is important because to study because it is tied to the spins of the particles that the field is capable of representing. For an in-depth discussion see Chapter 5 of Weinberg's "The Quantum Theory of Fields, Volume 1".


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