In QFT fields are classified according to representations of the Lorentz group $O(1,3)$.

Now, most books when getting into this say that in order to understand the representations of $O(1,3)$ we need to study the representations of the Lie algebra $\mathfrak{so}(1,3)$.

This in turn can be done by complexifying $\mathfrak{so}(1,3)$, if I understand well because any representation of a Lie algebra has a unique holomorphic extension to a representation of the complexified Lie algebra (by the way the reason for this was the topic of a quite old question of mine here).

So one actually has to study representations of $\mathbb{C}\otimes \mathfrak{so}(1,3)$. It turns out we have the isomorphism $$\mathbb{C}\otimes \mathfrak{so}(1,3)\simeq \mathbb{C}\otimes \mathfrak{su}(2)\oplus \mathbb{C}\otimes \mathfrak{su}(2)$$

So that the complexified $\mathfrak{so}(1,3)$ is a direct sum of two copies of the angular momentum algebra whose representation theory is well-known from QM.

In summary representations of $\mathfrak{so}(1,3)$ are $D_{A,B}$ labelled by $A,B\in\frac{1}{2}\mathbb{Z}_{>}$ and act on $\mathbb{C}^{(2A+1)(2B+1)}$.

But this is the representation theory of the algebra, not of the Lorentz group!

This is usually remedied in QFT texts by saying: well, exponentiate. In other words, for every $X\in \mathfrak{so}(1,3)$ we have $\exp \theta X \in SO(1,3)$ and we define the representation $\mathscr{D}_{A,B}$ by

$$\mathscr{D}_{A,B}(\exp \theta X)=\exp \theta D_{A,B}(X) .$$

So far so good. But this is not a priori a representation of $SO(1,3)$. Rather it is a representation of $\exp(\mathfrak{so}(1,3))$ since in general the exponential is not surjective.

In this case however we have $\exp(\mathfrak{so}(1,3))=SO_e^+(1,3)$ the proper ortochronous Lorentz group and after googling a little this seems to be a specific property of the groups $SO(p,q)$, not a specific version of a general Lie group result. In fact, it actually seems rather a non-trivial property of $SO(p,q)$!

The questions:

  1. Are we allowed in QFT to derive the representations from the Lie algebra just because of the happy coincidence that for $O(1,3)$ we get the whole $SO_e^+(1,3)$ exponentiating the Lie algebra?

    Or there is some reason that makes $\exp(\mathfrak{so}(1,3))$ the really important group to be studied for QFT, the identification with $SO_e^+(1,3)$ being secondary?

  2. If what in fact matters is $SO_e^+(1,3)$, why this subgroup is what should be studied to classify fields instead of the whole $O(1,3)$?

  3. I've read somewhere a long time ago, which I unfortunatelly don't remember where, that this has to do with the exponential relating to the universal cover of $O(1,3)$ somehow, but I don't see how this fits here. Is this relation the reason why we are allowed to get the representations out of the Lorentz algebra?


4 Answers 4


The full Lorentz group is $$ O(1,3) = SO^+(1,3)\rtimes D_4\,, $$ where $D_4$ is generated by $P$ and $T$, the discrete symmetries of parity and time reversal. Its representations can be labeled by representations of $SO^+(1,3)$, hence of $\mathfrak{so}(1,3)$, and of $D_4$. In physics' language this means that a particle is defined by the spin quantum numbers that you explained in the question, plus the $P$ and $T$ quantum numbers.

The representations of $D_4 \cong \mathbb{Z}_2\times \mathbb{Z}_2$ are simply a pair $(\pm1,\pm1)$ that specifies for both factors of $\mathbb{Z}_2$ whether the representation is the trivial or the alternating.

So the answer is that we do classify particles with irreducible representations of the full $O(1,3)$, it's just that sometimes when we do not care too much about $P$ and $T$ or when they both act trivially, we omit specifying them.

As an example: when you hear that pions are "pseudoscalar," that means that we are specifying a representation of $O(1,3)$.

  • $\begingroup$ So let's see if I got it. When one classifies fields according to representations of $SO_e^+(1,3)$ one is just not specifying the behavior under $P$ and $T$ which can be specified separately? And in turn what allows the study of representations of $SO_e^+(1,3)$ to be reduced to the study of representations of $\mathfrak{so}(1,3)$ is the very nice situation that $SO_e^+(1,3)=\exp(\mathfrak{so}(1,3))$? $\endgroup$
    – Gold
    Apr 21, 2019 at 1:34
  • $\begingroup$ I'd say yes to the first question. As for the second, you phrase it as if it is a lucky accident, but it's always true that $\exp(\mathfrak{g})$ is the connected component of $G$ containing the identity, and usually (or probably always, I can't think of counterexamples), the full $G$ is the connected component $\rtimes$ some discrete group. $\endgroup$
    – MannyC
    Apr 21, 2019 at 4:09
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    $\begingroup$ @MannyC it's quite trivial to show that last part - fix any element $x$ from $G$ and act with the $G_e$ (connected component of identity) on it from the left. Because the action is continuous, you'll never leave the connected component $G_x$. So $G_e x \subset G_x$. By the same logic $G_x x^{-1} \subset G_e$, which means that $G_x \subset G_e x$. Therefore $G_e x = G_x$: all connected components are diffeomorphic and the group structure on $G_x$ is related to the group structure on $G_e$ through conjugation by any chosen element $x$ from the connected component. $\endgroup$ Apr 21, 2019 at 6:55
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    $\begingroup$ @MannyC I think that this is not really true in general. See for instance: physics.stackexchange.com/q/160939, math.stackexchange.com/q/1573944 and mathoverflow.net/q/23478. Maybe I'm missing somethng really basic, but what I got from these posts is that this isn't true in general, but is (non-trivially) true for the Lorentz group. $\endgroup$
    – Gold
    Apr 21, 2019 at 22:22
  • $\begingroup$ @MannyC I think in the end I got the general idea. I've organized it and posted one answer with a summary of what I understood after reading the answers and thinking about it a little. Corrections or comments are welcome. Thanks for the help! $\endgroup$
    – Gold
    Apr 25, 2019 at 0:47

I want to add one comment: there is no good reason for a QFT Hilbert space to even transform under a representation of the full $O(1,3)$ group!

Say you have a QFT Hilbert space. There exist momentum $\hat P_\mu$, angular momentum $\hat L_i$, and boost operators $\hat K_i$ satisfying the regular algebra when acting on this Hilbert space.

\begin{align} &[\hat L_i, \hat L_j] = i \varepsilon_{ijk} \hat L_k && [\hat K_i, \hat K_j] = -i \varepsilon_{ijk}\hat L_k && [\hat L_i, \hat K_j] = i\varepsilon_{ijk}\hat K_k \\ &[\hat L_i, \hat P_j] = i\varepsilon_{ijk} \hat P_k && [\hat K_i, \hat P_i] =i \hat P_t && [\hat K_i, \hat P_t] = i\hat P_i. \nonumber \end{align}

Particles are eigenstates of the momentum operators $\hat{P}_\mu$ with eigenvalues $p_\mu$. They also contain extra data (spin, charge, etc.) which I will just label by $\lambda$. So we can write particle states as $$ |p,\lambda\rangle. $$

Now, by definition, a parity operator $\hat{\mathsf{P}}$ would be one that satisfies \begin{equation*} \hat{\mathsf{P}}^{-1} \hat L_i \hat{\mathsf{P}} = \hat L_i \hspace{0.5cm} \hat{\mathsf{P}}^{-1} \hat K_i \hat{\mathsf{P}} = \hat K_i \hspace{0.5cm} \hat{\mathsf{P}}^{-1} \hat P_0 \hat{\mathsf{P}} = \hat P_0 \hspace{0.5cm} \hat{\mathsf{P}}^{-1} \hat P_i \hat{\mathsf{P}} = -\hat P_i. \end{equation*}

This means that the state $$ \hat{\sf{P}}|p,\lambda\rangle $$ would be a state with its value of $p_0$ left invariant, $\vec{p}$ negated, and its angular momentum left invariant due to the simple algebra: \begin{align*} \hat P_i \hat{\mathsf{P}} |p, \lambda\rangle & = \hat{\mathsf{P}} \hat{\mathsf{P}}^{-1} \hat P_i \hat{\mathsf{P}} |p, \lambda\rangle = - \hat{\mathsf{P}} \hat P_i|p, \lambda\rangle \\ \hat L_i \hat{\mathsf{P}} |p, \lambda\rangle & = \hat{\mathsf{P}} \hat{\mathsf{P}}^{-1} \hat L_i |p, \lambda\rangle = \hat{\mathsf{P}} \hat L_i|p, \lambda\rangle \\ \hat P_0 \hat{\mathsf{P}} |p, \lambda\rangle & = \hat{\mathsf{P}} \hat{\mathsf{P}}^{-1} \hat P_0 \hat{\mathsf{P}} |p, \lambda\rangle= \hat{\mathsf{P}} \hat P_0|p, \lambda\rangle. \end{align*}

Finally, note that the helicity of a massless particle is given by \begin{equation*} h = \frac{\vec p \cdot \vec L}{p_0}. \end{equation*} Therefore, if $|p,\lambda\rangle$ has some helicity $h$, then $\hat{\mathsf{P}} |p,\lambda\rangle$ has some helicity $-h$. Note that, as per Wigner's classification, helicities label irreducible massless representations of the connected component of the Poincare group.

Okay, here's the kicker: say you have a QFT Hilbert space consisting of ONLY right handed Weyl fermions with $h = -1/2$. We have then proven that if this Hilbert space carries a parity operator $\hat{\mathsf{P}}$ then there must also be left handed Weyl fermion states with $h = 1/2$. However, it is obviously possible to construct a Hilbert space with only right handed fermions.

So what have we learned? We have learned that there's no reason for a QFT to transform under the whole $O(3,1)$ group! They don't necessarily have $\hat{\mathsf{C}}$, $\hat{\mathsf{P}}$, or $\hat{\mathsf{T}}$ operators.

However, even though individual $\hat{\mathsf{C}}$, $\hat{\mathsf{P}}$, or $\hat{\mathsf{T}}$ operators usually don't exist, there does always exist a combined $CPT$ operator. That's the CPT theorem. See more in the answer here: What are the assumptions that $C$, $P$, and $T$ must satisfy?

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    $\begingroup$ +1. This is reminiscent of the same argument from classical physics – there's in fact no physical reason that the full $O(3, 1)$ should be a symmetry even in the completely classical case. If we limit ourselves to special orthochronous transformations, we'll still get the whole relativistic machinery, and there's no consistency condition requiring that $P$ or $T$ should mean anything. As an example, consider a self-dual 2-form field. Its image under $P$ is anti-self-dual, meaning that $P$ can not by definition be realized as a symmetry of a self-dual field. $\endgroup$ Apr 21, 2019 at 7:05

$\def\SO{\rm SO^+_e(1,3)} \def\so{\mathfrak{so}(1,3)} \def\SL{{\rm SL}(2,\Bbb C)}$ As to your final questions. You're mixing two different topics. The first is the relationship between O(1,3) and $\SO$. This has been elucidated by @MannyC. The second is your statement that by exponentiating the Lie algebra you get $\SO$. (Surely you couldn't get the full O(3), which isn't connected. $\SO$ is its connected component of the unity. O(1,3) has four connected subsets, but only $\SO$ is a group.)

But this isn't true: by exponentiating $\so$ you get the covering group, which is simply connected - $\rm SO^+_e(1,3)$ isn't, as well as SO(3). This covering group is $\SL$ (complex $2\times2$ matrices with det=1). The relationship between $\SL$ and $\SO$ is a $2\to1$ homomorphism: if $M\in\SL$ then also $\,-M\in\SL$ and both are mapped into the same element of $\SO$.

A parallel situation prevails for representtions. Irreps of $\SL$ - or of $\so$ - fall into two classes

  • faithful reps meaning that there is a $1\to1$ correspondence between elements of $\SL$ and their representatives
  • unfaithful $2\to1$ reps, where both $M$ and $\,-M$ are mapped over the same matrix.

Only the latter class gives true irreps of $\SO$. The former gives the so-called "two-value" reps which aren't true reps of $\SO$.

However both classes are of physical interest. Not only because the former consists of the half-integer-spin irreps, but - more fundamentally - because in QM the true representative of a physical state isn't an element of the Hilbert space, but a unitary ray, i.e. a vector up to a phase factor. Then $M$ and $\,-M$ (and their faithful representatives) are equally acceptable as their actions on a state are indistinguishable.

Shortly said, in QM we're interested not in true representations but in projective ones.

  • $\begingroup$ but we know that the exponential is surjective onto $SO_e^+(1,3)$. So what you mean is that it is not injective? So that in fact it is 2 to 1? And in general the exponential of $\mathfrak{g}$ will be a bijection onto the universal cover? A reference to this last result - if true - would be highly appreciated. $\endgroup$
    – Gold
    Apr 21, 2019 at 22:27
  • $\begingroup$ @user1620696 So what you mean is that it is not injective? So that in fact it is 2 to 1? Yes. And in general the exponential of $\mathfrak g$ will be a bijection onto the universal cover? I believed so, but in "en.wikipedia.org/wiki/…" I've found the following statement: the exponential map of $SL(2, \Bbb R)$ is not surjective. $\endgroup$
    – Elio Fabri
    Apr 22, 2019 at 10:03
  • $\begingroup$ I think I got the point. I've added one answer with what I understood. Comments and corrections are welcome! Thanks for the help, you answer was really helpful. $\endgroup$
    – Gold
    Apr 25, 2019 at 0:49

After reading all the very good answers carefully I think I understood what is going on and decide to post a summary of it.

First, let $G$ be an arbitrary Lie group with Lie algebra $\mathfrak{g}$. Let further $\mathscr{D} : G\to GL(V)$ be a representation of $G$. Then, $\mathscr{D}$ gives rise to a representation of the Lie algebra $\mathfrak{g}$ by differentiation at the origin.

In fact, the exponential map $\exp : \mathfrak{g}\to G$ is surjective onto a neighbhorhood $U\subset G$ of the origin. Thus we are able to write for any $g\in U$ \begin{equation}g=\exp \lambda X,\quad X\in \mathfrak{g}.\end{equation}

This allows us to define $D : \mathfrak{g}\to \operatorname{End}(V)$ by \begin{equation}D(X)v=\dfrac{d}{d\lambda}\bigg|_{\lambda=0} \mathscr{D}(\exp \lambda X)v\end{equation}

One would question if all representations of $\mathfrak{g}$ arise in this way. In fact there is a result saying that this is true only when $G$ is simply connected. So the first point of importance is the following:

If $G$ is a Lie group with Lie algebra $\mathfrak{g}$ the representations of $G$ descend to representations of $\mathfrak{g}$. On the other hand, in general not all representations of $\mathfrak{g}$ arise in this form. In the special case that $G$ is simply connected, then it is true and all representations of $\mathfrak{g}$ arise as derivatives of those of $G$.

When $G$ is not simply connected, then its universal cover $\tilde{G}$ is, and they both share the same Lie algebra. So in that case, we might say that the representations of $\mathfrak{g}$ which do not arise from derivatives of representations of $G$ instead arise from derivatives of representations of the universal cover.

Second, let now a representation $D : \mathfrak{g}\to \operatorname{End}(V)$ be given and suppose that it is one of these representations that arise from a representation $\mathscr{D}:G\to GL(V)$ of $G$. The question is: can $\mathscr{D}$ be determined from $D$?

The answer is that we know how to do it in a neigbhorhood of the identity. Again, $\exp : \mathfrak{g}\to G$ is surjective onto a neigbhorhood of the identity $U = \exp(\mathfrak{g})$. In that case, for $g\in U$ there's $\lambda$ and $X$ such that $g = \exp (\lambda X)$.

We thus define $\mathscr{D}$ by $$\mathscr{D}(\exp \lambda X)=\exp \lambda D(X).$$

This recovers $\mathscr{D}$ out of $D$ inside the open set $U$ only. So the second important result is the following one:

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $\mathscr{D}:G\to GL(V)$ be a representation of $G$ that descends to $D : \mathfrak{g}\to \operatorname{End}(V)$. Then we can recover $\mathscr{D}$ out of $D$ in the neighborhood $U = \exp(\mathfrak{g})$ of the identity reconstructed by the exponential.

Third, we ask when this is enough to understand $\mathscr{D}$ completely. The answer lies in the following theorem: Let $G$ be a connected topological group, then any neighborhood of the identity generates $G$.

To prove this fact recall that in a connected topological space the only sets which are open and closed at the same time are $\emptyset$ and the whole space, thus define $S$ the set generated by a neighborhood of the identity and show $S$ is both open and closed. In other words: in a connected topological group any group element is a finite product of elements in a neighborhood of the identity.

So now suppose $G$ is connected. The theorem applies in particular to the exponential neigbhorhood $U = \exp(\mathfrak{g})$. In that case, knowing the representation as an exponential inside $U$ is enough to determine it everywhere since a general group element is just a product of such exponentials! If $G$ is not connected, this applies to the connected component of $G$ containing the identity. In fact, this leads to the third point of importance:

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $\mathscr{D}: G\to GL(V)$ be a representation of $G$ that descends to $D : \mathfrak{g}\to \operatorname{End}(V)$. Since $U = \exp(\mathfrak{g})$ is a neighborhood of the identity, it generates the connected component of $G$ containing the identity $G_e$. Thus, we recover $\mathscr{D}$ inside $G_e$ by noticing that for a general $g\in G_e$ there are $g_1,\dots,g_k\in U$ with $ g= g_1\cdots g_k$ and that $g_i = \exp \lambda_i X_i$. Hence we get \begin{equation}\mathscr{D}(g)=\exp \lambda_1 D(X_1)\cdots \exp \lambda_k D(X_k).\end{equation}

This is the general story that enables us to look at the Lorentz group. So let $O(1,3)$ be given, we wish to understand its representations. Now by the above methods, we are able to get the representations of $O(1,3)$ in the connected component with the identity by exponentiating its Lie algebra $\mathfrak{so}(1,3)$ representations. Since $O(1,3) = SO_e^+(1,3)\rtimes D_4$ where $D_4$ is generated by parity and time reversal, understanding the representations of $SO_e^+(1,3)$ is already enough.

Finally, in general the elements of $SO_e^+(1,3)$ would be given by finite products in the image of the exponential so that the representation of a general $g\in SO_e^+(1,3)$ would be a finite product of exponentials and this would be fine. It so happens, however, that the neighborhood $U = \exp(\mathfrak{so}(1,3))$ recreated by the exponential is in fact $SO_e^+(1,3)$, the consensus in the literature being that this is already a non-trivial result.

Being more precise: for $SO(1,3)$ the exponential is surjective onto the connected component with the identity. So in fact, this simplifies matters and any $g\in SO_e^+(1,3)$ takes just a single exponential to be represented. This is not an essential point though, because even if it weren't the case, knowledge of the representation on the exponential neighborhood would still give the understanding of the representation as a whole, it is just one very nice state of affairs which makes matters much simpler.


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