On Wikipedia and often in physics books it is stated, that the Lorenz group's finite dimensional non-unitary representations are completely reducible. We heavily use this in physics, however I do not know of a direct proof or reference for this. Could anyone help in providing such? It is clear to me that the Lorentz algebra $so(3,1)$ is semisimple, thus completely reducible. However, I don't see how we go to the group level, cause the exponential map is not surjective, since $SO(3,1)$ is non-compact and disconnected.

In particular, this is of interest because fields transform in irreps of $SO(3,1)$, so complete reducibility would be useful to have.

  • 1
    $\begingroup$ See I.M. Gel'fand, R.A. Minlos, and Z.Ya. Shapiro. Representations of the Rotation and Lorentz Groups and their Applications. Pergamon Press, New York. 1963. For finite dimensional irreps, one works with the complexification which is isomorphic to $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$. $\endgroup$ Commented Jan 31, 2022 at 22:13
  • $\begingroup$ that is clear, but that is on the Lie algebra level. How to get to the Lie group from there? The exponential is not surjective. $\endgroup$
    – ProphetX
    Commented Jan 31, 2022 at 22:18
  • $\begingroup$ there are lots of details in this text by Gel’fand. $\endgroup$ Commented Jan 31, 2022 at 22:25
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    $\begingroup$ Concerning surjectivity of exponential map, see physics.stackexchange.com/q/160939/2451 $\endgroup$
    – Qmechanic
    Commented Feb 1, 2022 at 0:56
  • $\begingroup$ Thanks! So indeed, the exponential map is only surjective for the proper ortochronous Lorentz group. $\endgroup$
    – ProphetX
    Commented Feb 1, 2022 at 11:29

1 Answer 1

  1. To every Lie algebra $\mathfrak{g}$ there is an associated simply-connected Lie group $G_s$. By the general correspondence of Lie groups and algebras, every representation of $G_s$ induces one of $\mathfrak{g}$ and vice versa, regardless of surjectivity of the exponential map. Hence a representation of $G_s$ is completely reducible if and only if the corresponding representation of $\mathfrak{g}$ is.

  2. For any connected Lie group $G_c$ with Lie algebra $\mathfrak{g}$, this is still true: If we have a representation $V_r$ of $G_c$ that has a subrepresentation $W_r\subset V_r$ of $G_c$ and $\mathfrak{g}$ is semi-simple, then we get $V_r = W_r\oplus W'_r$ for some other representation $W'_r$ of $\mathfrak{g}$. But $W'_r$ must be a subrepresentation of $G_c$ also, since you can always use the exponential map locally to get the representation of $G_c$ from the representation of $\mathfrak{g}$ - very similar to the general proof of the algebra-group correspondence, surjectivity is not needed here either, only connectedness so that you can "shift" the exponential map along paths in the group.

  3. For a disconnected Lie group $G_d$ with Lie algebra $\mathfrak{g}$, we have that all connected components are diffeomorphic to the identity component $G_0$ and there is an exact sequence $$ 1 \to G_0 \to G \to \pi_0(G)=G/G_0\to 1. \tag{1}$$ If this sequence splits, we have $G = \pi_0(G)\rtimes G_0$ where $\rtimes$ is a semi-direct product. If $\pi_0(G)$ is additionally Abelian, Mackey's theory of induced representations tells us that every representation of $G$ comes from representations of the factors $\pi_0(G)$ and $G_0$ and every pair of irreducible representations of the factors defines an irreducible representation of $G$. Since Maschke's theorem tells us that finite group have only completely reducible finite-dimensional representations, this means that a Lie group for which (1) splits and for which $\pi_0(G)$ is Abelian has completely reducible representations if $\mathfrak{g}$ is semi-simple.

Applying this to the Lorentz group, we have $\pi_0(G) = \mathbb{Z}_2\times\mathbb{Z}_2$ as the group of space parity and time reversal. It is Abelian, and one can show that the Lorentz group is indeed the semi-direct product of this and its connected component by explicitly writing out its matrix structure. Therefore, all finite-dimensional representations of the Lorentz group are completely reducible, and its irreducible representations can be given by an irreducible representation of the proper orthochronous Lorentz group $(V,\rho)$ + an irreducible representation of time reversal (which under the tensor product of the induced representation becomes just some operator on $V$ that squares to the identity) + an irreducible representation of space parity (likewise).

  • $\begingroup$ Thanks for the insightful answer! I have two short questions if possible: could you quote/reference the theorems you used in point 3? In particular, I am a bit confused as on how to obtain $\rho_c$ on $G/G_0$. My problem is that I can only show that $\rho_0$ is completely reducible, hence I don't know neither $\rho$ nor $\rho_c$ in the equation you gave. Do I have to use the fact that $G/G_0$ is isomorphic to a discrete group and then use Maschke to conclude $G/G_0$ reps are completely reducible, hence restrictions $\rho_c$ of $\rho$ are completely reducible too? $\endgroup$
    – ProphetX
    Commented Feb 1, 2022 at 11:28
  • $\begingroup$ EDIT: Moreover, I think I should be more precise in the sense that I allow projective representations, since Lie's theorem you quoted gives the reps of the universal cover/simply connected Lie group, which are in one-to-one correspondence with the projective representations of the connected component to the identity, i.e. proper ortochronous Lorentz group, whose reps are denoted by $\rho_0$ in your notation. $\endgroup$
    – ProphetX
    Commented Feb 1, 2022 at 11:30
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    $\begingroup$ @ProphetX 1. Yes, I'm using Maschke at the end there, that's why I stated the claim only for group with finitely many connected components. I've made that more explicit 2. Allowing projective representations simply means you can skip my point 2, so that doesn't really change anything. $\endgroup$
    – ACuriousMind
    Commented Feb 1, 2022 at 11:39
  • $\begingroup$ My intuition tells me that the issue in my example above is that $g_i$ do not form a subgroup, and we can get the counterexample because of that. I (intuitively, but am not sure) think that the argument can be fixed by requiring the set $\{ g_i \}$ to be a subgroup, however I do not see how could I choose this set explicitly for the Lorentz group. And how does complete reducibility of $\rho_c,\rho_0$ imply complete reducibility of $\rho$? $\endgroup$
    – ProphetX
    Commented Feb 1, 2022 at 12:37
  • $\begingroup$ I am not sure that the third point is true in general. Let $G=SO(2) \times \mathbb{Z}_2$. Consider the $2$-dimensional representation $\rho$ given by $\rho(g,h)(v)=gv$ for $v \in \mathbb{R}^2$. In this case $G_0=SO(2)$, such that $h^2\neq$ identity. Now, $\rho_c(g_1)=\rho(g_1)$ acts by $h$ on $\mathbb{R}^2$. But in $G/G_0$ we have $g_1^2$, so we would need $\rho_c(g_1)$ to be the identity, which is not the case by choice of $h$. $\endgroup$
    – ProphetX
    Commented Feb 1, 2022 at 13:26

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