Irreducible representations of the Lorentz group

[VantagePoint72]

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.

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 a real algebra 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 $c_{ijk}$ are the structure constants of L(G), then by definition: $[T_i,T_j]=c_{ijk}T_k$. Hence, Re{L(G)^C}) has the commutation relations:
$[X_i,X_j]=c_{ijk}X_k$
$[X_i,Y_j]=c_{ijk}Y_k$
$[Y_i,Y_j]=-c_{ijk}X_k$
Thus, if L(G) has an irrep ${d(T_i)}$, we can write down two distinct representations of Re{L(G)^C}) that respect these commutation relations:
$d(X_i) = d(T_i), d(Y_i) = i d(T_i)$ (type 1)
and
$d(X_i) = d(T_i), d(Y_i) = -i d(T_i)$ (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.
$d(X_i) = d^{(1)}(T_i) \otimes I + I \otimes d^{(2)}(T_i)$
$d(Y_i) = i d^{(1)}(T_i) \otimes I - i I \otimes d^{(2)}(T_i)$
where I is the identity matrix of the representation space and $d^{(1)},d^{(2)}$ 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 $d^{(1)}$ or $d^{(2)}$ 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 $SU(2) \otimes SU(2)$ 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?

 

[dextercioby]

First of all, read here: https://www.physicsforums.com/showthread.php?t=659011

Second, I don't understand why you chase the local approach, the global one based on SL(2,C) is standard for every theoretical physicist and is pretty well covered in textbooks on field theory.

The local approach is only sketched in the physics books, what you really need is a mathematician's treatment on this matter.

 

[Bill_K]

Tensor product representations generated from irreps are not, in general, irreducible—I don't understand why this is an exception.

A tensor product of two representations of the same group is in general reducible, but that's quite different from what we have here. The Lorentz group is a direct product of two subgroups, G = G₁ x G₂. Each element g of G can be written uniquely as a product of elements of G₁ and G₂: g = g₁g₂.

Given any representation of G, you can construct a representation of G₁ or G₂ from it by restriction. Conversely from representations of G₁ and G₂, one constructs the representation of G. The theorem is that if the representations of G₁ and G₂ are irreducible, so is the representation of G.

Proof by contradiction. If the representation of G is reducible, it means that a subset of its basis elements transform only among themselves under all group transformations. In particular, they would do so under transformations belonging only to G₁. Then the representation of G₁ would be reducible also.

 

[VantagePoint72]

Thanks, Bill. Is what I've written above enough to show that, in the general case, the group corresponding to the Lie algebra Re{L(G)^C} is the direct product of two copies of the group corresponding to L(G)? I don't see how to prove that from what I've given—but I do see that if it's true, it answers my question.

 

[Bill_K]

From the commutation relations you've quoted,
[X_i,X_j] = c_ijk X_k
[X_i,Y_j] = c_ijk Y_k
[Y_i,Y_j] = −c_ijk X_k
form the combinations J_i = X_i + iY_i and K_i = X_i - iY_i. Then
[J_i, J_i] = 2 c_ijk J_k
[K_i, K_i] = 2 c_ijk K_k
[J_i, K_i] = 0

which shows the group is a direct product of G₁ generated by the J's and G₂ generated by the K's.

 

[VantagePoint72]

Fantastic, thank you.

 

[samalkhaiat]

From the commutation relations you've quoted,
[X_i,X_j] = c_ijk X_k
[X_i,Y_j] = c_ijk Y_k
[Y_i,Y_j] = −c_ijk X_k
form the combinations J_i = X_i + iY_i and K_i = X_i - iY_i. Then
[J_i, J_i] = 2 c_ijk J_k
[K_i, K_i] = 2 c_ijk K_k
[J_i, K_i] = 0

which shows the group is a direct product of G₁ generated by the J's and G₂ generated by the K's.

Be careful here, in general if $G_{ 1 }$ and $G_{ 2 }$ are two COMPACT groups then $G_{ 1 } \times G_{ 2 } \equiv G$ is also compact. So, if $G_{ 1 } = SU( 2 )$ and $G_{ 2 } = SU( 2 )$, then their direct product can not be the NON-COMPACT Lorentz group $SO( 1 , 3 )$. Stated differently: the real algebra $\mathcal{ so } ( 1 , 3 ; \mathbb{ R } )$ does not contain two perpendicular copies of the real algebra $\mathcal{ su } ( 2 ; \mathbb{ C } )$. Depending on the signature of the metric in $\mathbb{ R }^{ 4 }$, one can show the following:
$$\mathcal{ so } ( 4 ; \mathbb{ R } ) \cong \mathcal{ su } ( 2 ; \mathbb{ C } ) \oplus \mathcal{ su } ( 2 ; \mathbb{ C } ) ,$$
$$\mathcal{ so } ( 1 , 3 ; \mathbb{ C } ) \cong \mathcal{ sl } ( 2 , \mathbb{ C } ) \oplus \mathcal{ sl } ( 2 , \mathbb{ C } ) ,$$
and,
$$\mathcal{ o } ( 2 , 2 ; \mathbb{ R } ) \cong \mathcal{ sl } ( 2 , \mathbb{ R } ) \oplus \mathcal{ sl } ( 2 , \mathbb{ R } ) .$$

Sam

 

[dextercioby]

From the commutation relations you've quoted,
[X_i,X_j] = c_ijk X_k
[X_i,Y_j] = c_ijk Y_k
[Y_i,Y_j] = −c_ijk X_k
form the combinations J_i = X_i + iY_i and K_i = X_i - iY_i. Then
[J_i, J_i] = 2 c_ijk J_k
[K_i, K_i] = 2 c_ijk K_k
[J_i, K_i] = 0

which shows the group is a direct product of G₁ generated by the J's and G₂ generated by the K's.

From $\frak{so}(4,\mathbb{R}) \simeq \frak{su}(2)\oplus\frak{su}(2)$ the global isomorphism

$$\mbox{SO}(4,\mathbb{R}) \simeq \mbox{SU}(2)\otimes\mbox{SU}{2}$$


will not follow, because the group in the LHS is not simply connected, while the one in the RHS is.

Actually by forming complex linear combinations of the generators of the initial algebra (call it $\mathcal{l}_{1}$) you're transforming it into its complexification, $\mathcal{l}_{1}^{C}$. Isomorphic LIe algebras lead by exponentiation to isomorphic groups iff the groups are simply connected and obviously both groups either compact or non-compact.

 

[dextercioby]

[...]
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 a real algebra 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}.[...]

Precisely this quoted part is addressed on page 228 in Willard Miller's text on symmetry groups and their applications.

 

[VantagePoint72]

Thank you, I'll try to track down a copy of that book.