To construct the irreducible representations of the homogeneous Lorentz group, starting with the $J^{\mu\nu}$ generators, one defines $\mathbf{J}=\{J^1,J^2,J^3\}$ as $J^1 = J^{23}$ and so on by cyclic permutations and $\mathbf{K} = \{K^1, K^2, K^3\}$ as $K^i = J^{0i}$. Then by defining

$$ \mathbf{A}=\frac{1}{2}(\mathbf{J}+i\mathbf{K}), \mathbf{B}=\frac{1}{2}(\mathbf{J}-i\mathbf{K})\tag1$$

one gets two independent $SU(2)$ algebras, since


$$[B_i, B_j] = i\epsilon_{ijk} B_k,\tag3$$

$$[A_i, B_j]=0.\tag4$$

Thus $\mathbf{A}$ can be represented by the usual angular momentum matrices $\mathbf{J}^{(A)}_{a,a^\prime}$, where $A$ is the maximum value (integer or half integer) and $a,a^\prime = -A,...,+A$. Similarly for $\mathbf{B}$.

Now the $\mathbf{A}$ and $\mathbf{B}$ that satisfy the above commutaion relations are given by,

$$\mathbf{A} = \mathbf{J}^{(A)} \otimes \mathbf{1}_{2b+1}\tag5$$


$$\mathbf{B} = \mathbf{1}_{2a+1} \otimes \mathbf{J}^{(B)}.\tag6$$

$\mathbf{1}$ stands for the unit matrix of the appropriate dimensions and $\otimes$ is the Kronecker product.

But the commutation relations can also be satisfied by choosing $\mathbf{A}$ and $\mathbf{B}$ as follows,

$$\mathbf{A} = \mathbf{J}^{(A)} \oplus \mathbf{0}_{2b+1}\tag7$$


$$\mathbf{B} = \mathbf{0}_{2a+1} \oplus \mathbf{J}^{(B)}.\tag8$$

$\mathbf{0}$ stands for the zero matrix of the appropriate dimensions and $\oplus$ is the direct sum.

Usual choice (5) and (6) gives all irreducible representations. But what does choice (7) and (8) give?


2 Answers 2


Frankly, I don't know; they are block matrices that yield legitimate fully reduced representations of the Lorentz group, obviously unsuited to fields with different vector and spinor indices. You might take this as a note to focus your question.

The conventional set (I), $$\mathbf{A} = \mathbf{J}^{(A)} \otimes \mathbf{1}_{(B)}\tag5$$ $$\mathbf{B} = \mathbf{1}_{(A)} \otimes \mathbf{J}^{(B)},\tag6$$ and the set (II), $$\mathbf{A} = \mathbf{J}^{(A)} \oplus \mathbf{0}_{ (B)}\tag7$$ $$\mathbf{B} = \mathbf{0}_{ (A)} \oplus \mathbf{J}^{(B)},\tag8$$ both realize the same Lorentz Lie algebra, as you noticed. In fact, mathematicians prefer the set (II) for abstract generators of the Lie algebra, for obvious reasons: it displays the subalgebras (and subgroups) manifestly, by dint of the blocking of the respective matrices!

You have taken (A) and (B) to be irreducible representations of su(2), so the set (I) acts on the tensor product space of two such reps, i.e. $(2a+1)\times(2b+1)$-dimensional vectors. To avoid dangerous nebulous generality, take a=1/2 and b=2.

So the set (I) acts on 10-vectors, which reduce to 4⊕6 under the action of the rotation subgroup. But the already reduced set (II) consists of block matrices, acting on reduced 7-dimensional vectors, 2⊕5.

Perhaps, if you wanted to compare (I) and (II), you might take as a lark (II) to act on irreps (A') and (B'), instead, of spin 3/2 and 4/2, respectively, what you'd get after Clebsching rotations in (I) by a similarity transformation!

  • The set (II) contains the reduced rep of the rotation subalgebra su(2), $$ \mathbf{A} +\mathbf{B}= \mathbf{J}^{(A)} \oplus \mathbf{J}^{(B)}, $$ a far cry from the reducible, but not reduced rep of set (I), $$ \mathbf{J}^{(A)} \otimes \mathbf{1}_{(B)} + \mathbf{1}_{(A)} \otimes \mathbf{J}^{(B)} , $$ the conventional coproduct of tensored representations ("addition of spins"). (As a rep of the full Lorentz group, of course, (I) is irreducible: the boosts snag.)

You may wish to observe corresponding group elements in this reduced representation, e.g., $$ e^{i\theta J^j+i\beta K^j}= e^{(i\theta-\beta)A^j+(i\theta +\beta)B^j } \\ = e^{(i\theta-\beta)J_{(A)}^j}\oplus e^{(i\theta +\beta)J^j_{(B)} } , $$ which looks counterintuitive until you consider the block matrices involved.

NB And yet... (geeky)

However, the world does not run on irreducible representations alone. The cornerstone in physics is the reduced (1/2,0)⊕(0,1/2) Dirac spinor, which is in the form (II), unlike the conventional vector representation which is in the reduced (Clebsched) version of (I). Check this.

Specifically, for both A and B being the fundamental doublet representation, (I) and (II) coincidentally have the same dimension, since 2+2=2×2 : In both cases, you are dealing with 4×4 matrices! WP has both the II construction, the π one, $\vec J= \tfrac{1}{2}\operatorname {diag}(\vec \sigma, \vec \sigma) $, $\vec K= \tfrac{i}{2}\operatorname {diag}(\vec \sigma, -\vec \sigma) $; but also the Clebsched (I) 4-vector, where the singlet part of the reduction for $\vec J$ has been moved to the (11) entry by the Clebsch matrices, and the $\vec K$ has been moved to the first row and first column, vacating the 3×3 block where $\vec J$ lives.

The takeaway is that your (I) handles 4-vectors, while (II) handles Dirac (bi-)spinors.

  • $\begingroup$ So if i understand this correctly, every $(m,n)$ representation of set(II) is equivalent to $(m,0)\oplus(0,n)$ representation of set(I) ? $\endgroup$
    – Alec Demus
    Commented Feb 14, 2022 at 11:08
  • $\begingroup$ No! You saw they have different dimensionalities, in general! In the NB case you added two (I)s to get your (II), and you focused on the J sub algebra… $\endgroup$ Commented Feb 14, 2022 at 11:59
  • $\begingroup$ but they do have same dimensions! dimension of $(m,n)$ representation in second construction is (2m+1)+(2n+1) which is same as the dimension of $(m,0)\oplus(0,n)$ in first construction. $\endgroup$
    – Alec Demus
    Commented Feb 14, 2022 at 12:39
  • $\begingroup$ i think there is a mismatch of terminology. maybe i should add subscripts $(m,n)_I$ and $(m,n)_{II}$ to distinguish both constructions. then $dim[(m,n)_I]=(2m+1)\times(2n+1)$ and $dim[(m,n)_{II}]=dim[(m,0)_I\oplus(0,n)_I]=(2m+1)+(2n+1)$ $\endgroup$
    – Alec Demus
    Commented Feb 14, 2022 at 12:56
  • $\begingroup$ No, you misunderstood the initial explanation of the answer! For arbitrary m, n, (m,n) is irreducible and cannot be represented in the second construction which is always reduced, (𝑚,0)⊕(0,𝑛)! That's the key point! The two constructions are not equivalent, in general! $\endgroup$ Commented Feb 14, 2022 at 12:56

I had the same trouble for a while, and I've never seen a physics book explain the reasoning. But it is a general theorem that all the irreducible representations of a direct sum of Lie algebras are given as tensor products V x U (x here means tensor product, not Cartesian product!), where V is an irreducible representation of the first Lie algebra, and U an irreducible representation of the second. It's quite easy to find this result googling around.


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.