In my Quantum Field Theory class we were discussing the finite dimensional representations of the Lorentz group. We discussed the vector representation which acts on 4 vectors, and then also the spinor representation that acts on spinors. My issue is that both of these representations are 4 dimensional.

I'm used to dealing with SO(3) where the representations are labeled by half integers and there is a unique mapping from each label to the dimension of the representation. Spin (1/2) gives 2 dimensional matrices, Spin (1) gives 3 dimensional matrices, and so on. But it seems we have lost that uniqueness in the Lorentz group. The vector representation and the spinor representation have the same dimension. Does this have to do with the non compactness of the lorentz group?


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    $\begingroup$ Related: physics.stackexchange.com/questions/47344/… $\endgroup$
    – user108787
    Commented Sep 28, 2016 at 16:12
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    $\begingroup$ I'm not sure if the existence of only a single representation for each dimension is tied to any specific property of a group... that's just sort of a lucky coincidence because $SO(3)$ is so simple, not because $SO(3, 1)$ is missing some property. $\endgroup$
    – knzhou
    Commented Sep 28, 2016 at 16:14
  • $\begingroup$ I'm not too keen on group theory, only what I've picked up from physics, which of course is not sufficiency. $\endgroup$ Commented Sep 28, 2016 at 16:14

2 Answers 2


Indeed, you have lost some sort of uniqueness of dimension, but not between the vector and the spinor representation: The vector representation of $\mathrm{SO}(1,3)$ is irreducible, while the four-dimensional Dirac-spinor representation is not - it is the sum of a left-chiral and a right-chiral Weyl representation.

In general, the finite-dimensional representations of the (connected component of the) Lorentz group are in bijection to the finite-dimensional representations of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. For the precise relation between $\mathrm{SO}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see this answer by Qmechanic. The representation theory of $\mathfrak{su}(2)$ is precisely that of spin as we know it, and therefore a finite-dimensional representation of the Lorentz group is labeled by two half-integers $(s_1,s_2)$. If one examines the way the $\mathfrak{su}(2)$ algebras actually related to the Lorentz algebra, one finds that the total spin of such a representation should be $s_1+s_2$.

The representation space associated to $(s_1,s_2)$ is just $\mathbb{C}^{2s_1 +1}\otimes\mathbb{C}^{2s_2+1}$, i.e. we tensor the spin-$s_i$ representations with each other. Of course, this shows you that the dimension of the space is no longer unique for a given representation, even if its irreducible.

However, this has nothing to do with the non-compactness of the Lorentz group, it's simply because it's a little more complicated than the "easy" $\mathrm{SO}(3)$. For instance, the compact $\mathrm{SU}(2)\times\mathrm{SU}(2)$ shares the same finite-dimensional representation theory.

  • $\begingroup$ Thanks so much!. In general are there certain properties that a group must have for the dimension of each representation to be unique? I know that is more of a math question, but I thought I would ask. $\endgroup$ Commented Sep 28, 2016 at 16:22

No, the loss of uniqueness does not have to do with the non-compactness of the Lorentz group. The fact that there is only one irreducible representation of any given dimension is special to the group $SU(2)$ (and by extension, to $SO(3)$). It is not true even for other compact semisimple Lie groups. For example, $SU(3)$ has two distinct irreducible representations of dimension 3, which also happen to be conjugates to each other. It also has four representations of dimension 15, two pairs of conjugate representations.

You hopefully will soon learn that the finite-dimensional irreducible representations of the Lorentz group can be labelled by two numbers, which happen to correspond (in a roundabout sense) to representations of $SU(2)$.

In general, an irreducible representation for a semisimple Lie algebra can be labelled by the values of its independent Casimirs. $SU(2)$ has one independent Casimir, so you only need one number to specify a representation. $SU(3)$ and $SO(1,3)$ have two Casimirs, so you need two numbers.

And, as ACuriousMind pointed out, the spinor representation you're talking about isn't actually irreducible. It is the direct sum of two irreducible representations, $(\tfrac{1}{2},0)$ and $(0,\tfrac{1}{2})$ in one common notation. The vector representation is irreducible, and is labelled $(\tfrac{1}{2},\tfrac{1}{2})$.


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