MATHEMATICS

M. A. NAIMARK

ON IRREDUCIBLE LINEAR REPRESENTATIONS OF THE FULL LORENTZ GROUP

(Presented by Academician A. N. Kolmogorov, 1 X 1956)

In a note by the author (¹), an exact formulation and solution were given of the problem of describing, up to equivalence, all completely irreducible representations of the proper Lorentz group. In the present note an analogous formulation and solution are given for the full Lorentz group; here, without any qualifications, the notation and results of note (¹) are used.

A non-rigorous derivation of the formulas for representations of the full Lorentz group in infinitesimal form was given earlier in the paper (²) by I. M. Gelfand and A. M. Yaglom*; the formulas presented in the present note for representations of the full Lorentz group in integral form apparently appear here for the first time.

1. Formulation of the problem. Let (\mathfrak G) denote the full Lorentz group, and (\mathfrak G_+) the proper Lorentz group. Then (\mathfrak G=\mathfrak G_+\cup s\mathfrak G), where (s) is reflection with respect to the first three coordinate axes

[
s:\ x'_1=-x_1,\quad x'_2=-x_2,\quad x'_3=-x_3,\quad x'_4=x_4.
]

Here (s) commutes with all rotations of three-dimensional space and, moreover, satisfies the conditions: (s^2=1;\ sb(t)s=b(-t)), where

[
b(t)=
\left|
\begin{array}{cccc}
1&0&0&0\
0&1&0&0\
0&0&\operatorname{ch}t&\operatorname{sh}t\
0&0&\operatorname{sh}t&\operatorname{ch}t
\end{array}
\right|.
]

It follows that a representation (g\to T_g) of the group (\mathfrak G) is completely determined by its restriction (g\to T_g) to the group (\mathfrak G_+) and by the operator (S=T_s), satisfying the conditions: (\alpha)\ S^2=1,\ \beta)\ ST_{b(t)}S=T_{b(-t)},\ \gamma)\ ST_g=T_gS) for all rotations (g) of three-dimensional space.

Passing from representations of the group (\mathfrak G_+) to representations of the group (\mathfrak A), we can write these conditions in the form

[
S^2=1,\quad ST_\varepsilon S=T_{\varepsilon^{-1}}\quad \text{for all matrices } \varepsilon=
\left|\begin{array}{cc}
e^{-t}&0\
0&e^t
\end{array}\right|,\quad \operatorname{Im} t=0
\tag{1a}
]

and

[
ST_u=T_uS \quad \text{for all } u\in\mathfrak U.
\tag{1b}
]

Finally, these conditions can also be written in the form (ST_gS=T_{g^\wedge}), where the notation is introduced: (g^\wedge=g^{*-1}).

* We note that for the case of unitary representations of the full Lorentz group the derivation (²) can be made rigorous, so that the non-rigorous character of this derivation essentially pertains to the case of nonunitary representations.

We shall call the group ring (\mathfrak C) of the group (\mathfrak G) the totality of all formal sums (c=x+sy), where (x, y) belong to the group ring (X) of the group (\mathfrak A); here the operations in (\mathfrak C) are defined by the formulas

[
\lambda c=\lambda x+s(\lambda y),\qquad
c_1+c_2=(x_1+x_2)+s(y_1+y_2),
]

[
c_1c_2=(x_1x_2+y_1^{\wedge}y_2)+s(y_1x_2+x_1^{\wedge}y_2)
]

for (c=x+sy,\ c_1=x_1+sy_1,\ c_2=x_2+sy_2), where (x^{\wedge}(g)=x(g^{\wedge})). If a representation (g\to T_g) of the group (\mathfrak G) is given, then, putting (T_c=T_x+ST_y) for (c=x+sy), we obtain, as is easy to verify, a representation of the ring (\mathfrak C).

A representation (g\to T_g) of the group (\mathfrak G) is called completely irreducible if the operators (T_c,\ c\in\mathfrak C), form a completely irreducible set. The problem consists in describing, up to equivalence, all completely irreducible representations of the group (\mathfrak G); here equivalence is understood in the sense of the definition given in (1).

1. Construction of the representations. We shall indicate a complete set of completely irreducible representations of the group (\mathfrak G) that are inequivalent to one another. These representations are constructed as follows:

(\alpha)) Representations (D_{0,\rho}^{+},\ D_{0,\rho}^{-}). Let, in general, (S_{m,\rho}) denote the completely irreducible representation (a\to T_a) of the group (\mathfrak A), defined by the numbers ((m,\rho)) (infinite-dimensional for (\rho^2\ne-(|m|+2n)^2,\ n=1,2,\ldots), and spinorial for (\rho^2=-(|m|+2n)^2,\ n=1,2,\ldots)), and let (R_{m,\rho}) be the space of this representation.

Consider the representation (S_{0,\rho}) of the group (\mathfrak A), and define in (R_{0,\rho}) the operator (S) by one of the two formulas

[
Sf(u)=f(su),\qquad Sf(u)=-f(su),
\tag{2}
]

where*

[
s=
\begin{pmatrix}
0 & 1\
-1 & 0
\end{pmatrix}.
]

It is not hard to verify that each of these operators (S) satisfies conditions (1a), (1b) and, consequently, these operators together with the representation (S_{0,\rho}) of the group (\mathfrak A) define two representations of the group (\mathfrak G), which we shall denote by (D_{0,\rho}^{+},\ D_{0,\rho}^{-}), respectively.

If ({\xi_p^k}) is the canonical basis in (R_{0,\rho}), then (S\xi_p^k=(-1)^k\xi_p^k) in the case of the representation (D_{0,\rho}^{+}), and (S\xi_p^k=(-1)^{k+1}\xi_p^k) in the case of the representation (D_{0,\rho}^{-}).

(\beta)) Representations (D_{m,0}^{+},\ D_{m,0}^{-},\ m>0). Consider the representations (S_{m,0}) and (S_{-m,0}), (m>0), of the group (\mathfrak A). These representations are unitary and unitarily equivalent. Let (W) be an isometric operator taking (S_{m,0}) into (S_{-m,0}), and let (A_s) be the operator defined by the formula (A_sf(u)=\varkappa f(su)) for (f\in R_{m,0}), where (\varkappa=(-1)^{m/2}) for even (m) and (\varkappa=(-1)^{(m+1)/2}) for odd (m). Then (A_s) isometrically maps (R_{m,0}) onto (R_{-m,0}). We now define the operator (S) in (R_{m,0}) by one of the two formulas (S=A_sW,\ S=-A_sW). Again one can verify that in either of these two cases the operator (S), together with the representation (S_{m,0}) of the group (\mathfrak A), defines a representation of the group (\mathfrak G). The two representations thus obtained will be denoted, respectively, by (D_{m,0}^{+},\ D_{m,0}^{-}).

If ({\xi_p^k}) is the canonical basis in (R_{m,0}), then (S\xi_p^k=(-1)^{[k]}\xi_p^k) in the case of the representation (D_{m,0}^{+}), and (S\xi_p^k=(-1)^{[k]+1}\xi_p^k) in the case of the representation (D_{m,0}^{-}), where ([k]) denotes the integer part of the number (k).

* We denote by the same letter (s) the matrix
[
\begin{pmatrix}
0 & 1\
-1 & 0
\end{pmatrix}
]
and the reflection with respect to the first three axes; this, however, will not lead to any misunderstanding in what follows.

γ) Representations (D_{m,\rho}), (m>0,\ \rho\ne 0). Consider the representations (S_{m,\rho}), (S_{-m,\rho}) of the group (\mathfrak A). Let (\overline R_{m,\rho}) be the unitary space which is the direct sum of the unitary spaces (R_{m,\rho}), (R_{-m,\rho}); define in (\overline R_{m,\rho}) the operators (T_a) and (S) by setting

[
T_a{f_1,f_2}={T'_a f_1,T''_a f_2},\quad
S{f_1(u),f_2(u)}=\varkappa{f_1(su),(-1)^m f_2(su)},
]

where (T'a), (T''_a) are the operators of the representations (S), and (\varkappa) is defined in the same way as in case β).}), (S_{-m,\rho

It is easy to verify that the operators (T_a) and (S) define a representation of the group (\mathfrak G); we denote this representation by (D_{m,\rho}). If ({\xi_p^k}), ({\eta_p^k}) are the canonical bases of the representations (S_{m,\rho}), (S_{-m,\rho}), then the vectors (\overline \xi_p^k={\xi_p^k,0}), (\overline \eta_p^k={0,\eta_p^k}) form a basis in (\overline R_{m,\rho}), and the action of the operator (S) on these vectors is given by the formulas:

[
S\overline \xi_p^k=(-1)^{[k]}\overline \eta_p^k,\quad
S\overline \eta_p^k=(-1)^{[k]}\overline \xi_p^k.
]

Theorem 1. The representations (D_{0,\rho}^{+}), (D_{0,\rho}^{-}), (D_{m,0}^{+}), (D_{m,0}^{-}), (D_{m,\rho}) ((m>0,\ \rho\ne 0)), corresponding to all possible (m>0) and (\rho), are completely irreducible and pairwise inequivalent.

Let us note that the operator (S) in each of these representations is unitary; consequently, these representations are unitary if and only if the corresponding (S_{m,\rho}) are unitary. It is also evident that the representations (D_{0,\rho}^{+}), (D_{0,\rho}^{-}), (D_{m,0}^{+}), (D_{-m,0}^{-}), (D_{m,\rho}) ((m>0,\ \rho\ne 0)) are finite-dimensional if and only if the corresponding representations (S_{m,\rho}) are finite-dimensional.

3. Main result.

Theorem 2. Every completely irreducible representation of the full Lorentz group (\mathfrak G) is equivalent to one of the representations

[
D_{0,\rho}^{+},\ D_{0,\rho}^{-},\ D_{m,0}^{+},\ D_{m,0}^{-},\ D_{m,\rho}\ (m>0,\ \rho\ne 0).
]

4. Idea of the proof. Put (\mathfrak C_j^k={c=x+sy:x,y\in X_j^k}), (X_{j+}^{k}={x:x\in X_j^k,\ x^{\wedge}=x}), (X_{j-}^{k}={x:x\in X_j^k,\ x^{\wedge}=-x}); then (\mathfrak C_j^k), (X_{j+}^{k}) are subrings of the rings (\mathfrak C) and (X_j^k), respectively.

Let (g\to T_g) be a completely irreducible representation of the group (\mathfrak G); let (\mathfrak M_j^k) be some subspace different from ((0)) and corresponding to this representation. Then (\mathfrak M_j^k) is invariant with respect to the operators (T_c), (c\in\mathfrak C_j^k). Put (A_c\xi=T_c\xi) for (\xi\in\mathfrak M_j^k). The correspondence (c\to A_c) is a representation of the ring (\mathfrak C_j^k).

I. If the representation (g\to T_g) of the group (\mathfrak G) is completely irreducible, then the corresponding representation (c\to A_c) of the ring (\mathfrak C_j^k) is also completely irreducible.

II. Two completely irreducible representations (g\to T'_g), (g\to T''_g) of the group (\mathfrak G), for which (\mathfrak M_j^{\prime k}\ne(0)), (\mathfrak M_j^{\prime\prime k}\ne(0)) for fixed (j,k), are equivalent if and only if the corresponding representations (c\to A'_c), (c\to A''_c) of the ring (\mathfrak C) are equivalent.

Now put (T_s=S). Since (S^2=1), we have (\mathfrak M_j^k=\mathfrak P_j^k\oplus\mathfrak D_j^k), where (\mathfrak P_j^k={\xi:\xi\in\mathfrak M_j^k,\ S\xi=-\xi}), (\mathfrak D_j^k={\xi:\xi\in\mathfrak M_j^k,\ S\xi=\xi}); (\mathfrak P_j^k), (\mathfrak D_j^k) are invariant with respect to the operators (T_x), (x\in X_{j+}^{k}), and (T_x\mathfrak P_j^k\subset\mathfrak D_j^k), (T_x\mathfrak D_j^k\subset\mathfrak P_j^k) for (x\in X_{j-}^{k}). Setting (\Lambda_x\xi=T_x\xi) for (\xi\in\mathfrak P_j^k), (x\in X_{j+}^{k}), we obtain a representation (x\to\Lambda_x) of the ring (X_{j+}^{k}). Taking into account that (S=-1) on (\mathfrak P_j^k), we have:

III. If the original representation (g\to T_g) is completely irreducible, then the representation (x\to\Lambda_x) of the ring (X_{j+}^{k}) is also completely irreducible.

Since the ring (X_j^k) is commutative, we conclude from this that either (\mathfrak P_j^k=(0)), or (\mathfrak P_j^k) is one-dimensional. Similarly, either (\mathfrak D_j^k=(0)), or (\mathfrak D_j^k) is one-dimensional; consequently, if (\mathfrak M_j^k\ne(0)), then (\mathfrak M_j^k) is either one-dimensional or two-dimensional.

Let us consider these cases separately.

1) (\mathfrak M_j^k) is one-dimensional. The representation (c\to A_c) is then one-dimensional and therefore is defined by some multiplicative linear functional (\lambda(c)) on (\mathfrak C_j^k), and hence on (X_j^k). To this latter functional there corresponds a certain representation (S_{m,\rho}) of the group (\mathfrak A). Since either (\mathfrak P_j^k=(0)), or (\mathfrak D_j^k=(0)), we have (\lambda(x)=0) on (X_{j-}^k); consequently, (\lambda(x^*)=\lambda(x)). Hence we conclude that either (m=0), or (\rho=0); moreover, either (S=1) on (\mathfrak M_j^k), or (S=-1) on (\mathfrak M_j^k). But then the functional (\lambda(c)) for the given representation coincides with the functional (\lambda(c)) for one of the representations (D_{0,\rho}^{+}), (D_{0,\rho}^{-}), (D_{m,0}^{+}), (D_{m,0}^{-}), so that, by II, the given representation is equivalent to one of these representations.

2) (\mathfrak M_j^k) is two-dimensional; consequently, (\mathfrak P_j^k), (\mathfrak D_j^k) are one-dimensional. By virtue of the complete irreducibility of the representation (c\to A_c) of the ring (\mathfrak C_j^k), there exist a function (y_0\in X_{j-}^k) and vectors (\xi_1\in\mathfrak P_j^k), (\xi_2\in\mathfrak D_j^k) such that
(A_{y_0}\xi_1=\xi_2), (A_{y_0}\xi_2=\xi_1). Moreover, from the commutativity of the ring (X_j^k) it follows easily that
(A_x\xi_1=\lambda'(x)\xi_1), (A_x\xi_2=\lambda'(x)\xi_2) for (x\in X_{j+}^k), and
(A_x\xi_1=\lambda'(xy_0)\xi_2), (A_x\xi_2=\lambda'(xy_0)\xi_1) for (x\in X_j^k), where (\lambda'(x)) is a certain multiplicative linear functional in (X_{j+}^k). At the same time
(\lambda'(x_1y_0)\lambda'(x_2y_0)=\lambda'(x_1x_2)), where (x_1,x_2\in X_{j-}^k).

Putting (\lambda(x)=\lambda(x'+x'')=\lambda'(x')+\lambda'(x''y_0)) for (x'\in X_{j+}^k), (x''\in X_{j-}^k), and taking into account the last relation for (\lambda'), we easily find that (\lambda) is a multiplicative linear functional in (X_j^k), and consequently defines a certain representation (S_{m,\rho}) of the group (\mathfrak A). From the complete irreducibility of the representation (c\to A_c) and the formulas

[
S\xi=-\xi \text{ on } \mathfrak P_j^k,\qquad S\xi=\xi \text{ on } \mathfrak D_j^k
\tag{3}
]

we conclude that (m\ne0), (\rho\ne0), so that one may assume (m>0).

Moreover, from formulas (3) and the definition of the functional (\lambda(x)) it follows that, for
(c=(x_1+y_1)+s(x_2+y_2)), (x_1,x_2\in X_{j+}^k), (y_1,y_2\in X_j^k), the matrix of the operator (A_c) in the basis (\xi_1,\xi_2) has the form

[
\left|
\begin{matrix}
\lambda(x_1-x_2) & \mu(y_1+y_2)\
\mu(y_1-y_2) & \lambda(x_1+x_2)
\end{matrix}
\right|.
]

But a direct computation shows that, with a suitable choice of basis, the matrix of the operator (A_c) for the representation (D_{m,\rho}) has the same form. On the basis of Proposition II we therefore conclude that the given representation is equivalent to the representation (D_{m,\rho}).

Moscow Institute of Physics and Technology

Received
28 VII 1956

References

({}^{1}) M. A. Naimark, DAN, 97, No. 6, 969 (1954). ({}^{2}) I. M. Gelfand, A. M. Yaglom, ZhETF, 18, No. 8, 703 (1948).