Understanding 4-Vector Representations in the Lorentz Group

[Silviu]

Hello! I am reading some notes on Lorentz group and at a point it is said that the irreducible representations (IR) of the proper orthochronous Lorentz group are labeled by 2 numbers (as it has rank 2). They describe the 4-vector representation $D^{(\frac{1}{2},\frac{1}{2})}$ and initially I thought this is an IR (also being a fundamental representation). However, further on they say that $D^{(\frac{1}{2},\frac{1}{2})} = D^{(\frac{1}{2}, 0)} \oplus D^{(0,\frac{1}{2})}$, which implies that $D^{(\frac{1}{2},\frac{1}{2})}$ is not an IR. So I am confused, is it or is it not IR? The way I was thinking about it, is that the 4 dimensional vector representation (i.e. under Lorentz group) is an IR while the 4 dim spinor representation (i.e. under $SL(2,C)$) is not IR. But wouldn't you need different notations for them? Then, $D^{(0,\frac{3}{2})}$ and $D^{(\frac{3}{2},0)}$ are also 4 dimensional, so what should I do with them? Are they IR, too? Can someone clarify this for me? Thank you!

 

[lpetrich]

For the first part, I will show that if a representation of a product group is constructed from an irreducible representation for each of the groups in it, than that representation is also irreducible.

For irreducibility: for a representation D(a) of a group, if a matrix X satisfies D(a).X = X.D(a), then X must be a multiple of the identity matrix.

For a group G that is a product G1 * G2, its elements are a = (a1,a2) where a1 ranges over all of G1 and a2 does so for G2. A representation D(a) = D1(a1)*D2(a2) where D1 is a rep of G1 and D2 of G2. In component form,

D(a)(i1j1,i2j2) = D1(a1)(i1,j1) * D2(a2)(i2,j2)

To test reducibility, we must find the possible matrices X(i1j1,i2j2) that satisfy D(a)(i1j1,i2j2) * X(j1k1,j2k2) = X(i1j1,i2j2) * D(a)(j1k1,j2k2) over all values of i1,i2,k1,k2 and summed over dummy indices j1,j2. Using the above decomposition of D(a) gives us

D1(a1)(i1,j1) * D2(a2)(i2,j2) * X(j1j2,k1k2) = X(i1i2,j1j2) * D1(a1)(j1,k1) * D2(a2)(j2,k2)

Setting a2 to the identity of G2 gives us D1(a1)(i1,j1) * X(j1i2,k1k2) = X(i1i2,j1k2) * D1(a1)(j1,k1) and since D1 is irreducible,

X(i1i2,j1j2) = δ(i1,j1) * X(i2,j2)

This in turn reduces the equation to D2(a2)(i2,j2) * X(j2,k2) = X(i2,j2) * D2(j2,k2) and that in turn gives us

X(i1i2,j1j2) = δ(i1,j1) * δ(i2,j2) * X

A multiple of the identity matrix. Thus, if component-group reps D1 and D2 are irreducible, then the product-group irrep D = D1 * D2 is also irreducible.

Thus, for the Lorentz group, every representation with form (j1,j2) is irreducible, since each part is irreducible.

 

[lpetrich]

In general, a product representation of two irreducible representations will be reducible, but if one of the irreps is the unit representation D(a) = 1, then the product will also be an irrep. Thus, (j1,j2) = (j1,0) * (0,j2) where 0 is the unit representation.

The vector rep of the Lorentz group is (1/2,1/2), and is thus irreducible.

The spinor rep is (1/2,0) + (0,1/2) and is thus reducible to two smaller spinors, (1/2,0) and (0,1/2), both irreducible. Each one of them is a rep of SL(2,C), but each one is the complex conjugate of the other.

The reps (3/2,0) and (0,3/2) are both irreducible.

Though all of (1/2,1/2), (1/2,0) + (0,1/2), (3/2,0), and (0,3/2) have size 4, they are all inequivalent.