I'm having some difficulty understanding the representation theory of the Lorentz group. While it's a fundamentally mathematical question, mathematicians and physicists use very different language for representation theory. I think a particle physicist will be more likely than a mathematician to be able to explain this in a way I can follow.(adsbygoogle = window.adsbygoogle || []).push({});

Suppose we have some Lie algebra L(G) over the real numbers, already equipped with a set of irreducible representations. We can complexify L(G) by extending scalar multiplication to the complex numbers. We can then reinterpret this complexified Lie algebra, L(G)^{C}, as arealalgebra whose dimension is twice as large as the original L(G). A bit more concretely, if L(G) has basis {T_{i}}, then the real interpretation of the complexified L(G) (which we'll call Re{L(G)^{C}}) has basis {X_{i},Y_{i}}, where X_{i}= T_{i}, Y_{i}= iT_{i}. We now want to figure how to generate the irreps of Re{L(G)^{C}} in terms of those of the original L(G). The reason why we're interested in doing such a convoluted-seeming construction is that the Lorentz group's Lie algebra (either the whole thing or just the component connected to the identity, I can't remember which) is precisely Re{L(SU(2))^{C}}.

If [itex]c_{ijk}[/itex] are the structure constants of L(G), then by definition: [itex][T_i,T_j]=c_{ijk}T_k[/itex]. Hence, Re{L(G)^{C}}) has the commutation relations:

[itex][X_i,X_j]=c_{ijk}X_k[/itex]

[itex][X_i,Y_j]=c_{ijk}Y_k[/itex]

[itex][Y_i,Y_j]=-c_{ijk}X_k[/itex]

Thus, if L(G) has an irrep [itex]{d(T_i)}[/itex], we can write down two distinct representations of Re{L(G)^{C}}) that respect these commutation relations:

[itex]d(X_i) = d(T_i), d(Y_i) = i d(T_i)[/itex] (type 1)

and

[itex]d(X_i) = d(T_i), d(Y_i) = -i d(T_i)[/itex] (type 2)

This is where I get confused: I'm then told that if we want to get the most general irrep for Re{L(G)^{C}}) we need to take the tensor product of irreps of L(G): one of type 1 and the other of type 2.

[itex]d(X_i) = d^{(1)}(T_i) \otimes I + I \otimes d^{(2)}(T_i)[/itex]

[itex]d(Y_i) = i d^{(1)}(T_i) \otimes I - i I \otimes d^{(2)}(T_i)[/itex]

where I is the identity matrix of the representation space and [itex]d^{(1)},d^{(2)}[/itex] are in general allowed to be two different irreps of L(G). I don't understand how this construction gives us irreps of Re{L(G)^{C}}). I can see how the special cases work: if [itex]d^{(1)}[/itex] or [itex]d^{(2)}[/itex] is the trivial irrep then we obviously recover the type 2 and type 1 (respectively) representations from earlier. But why in general does this give us an irrep of the complexified algebra? Tensor product representations generated from irreps are not, in general, irreducible—I don't understand why this is an exception.

This is clearly a very important result since it allows us to conclude that [itex]SU(2) \otimes SU(2)[/itex] provides us with the representations of the Lorentz group: (1/2,1/2) is the 4-vector (fundamental) rep; (1/2,0) and (0,1/2) are the left- and right- handed Weyl spinor reps, and their direct sum is the Dirac spinor rep.; etc. I just don't see how the L(G) irreps yield Re{L(G)^{C}}) irreps in the construction given above. Can someone explain that part?

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# Irreducible representations of the Lorentz group

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