# Why is the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group realized as the vector space of Hermitian $2\times 2$ matrices?

Why can we write an arbitrary object $v_{a \dot{b} }$ our transformations in this basis act on as

$$v_{a \dot{b} } = v_{\nu} \sigma^{ \nu}_{a \dot{b} } = v^0 \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + v^1 \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} +v^2 \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix} + v^3 \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}$$

Formulated differently: How do we know that the vector space for the $(\frac{1}{2},\frac{1}{2})= (\frac{1}{2},0) \otimes (0,\frac{1}{2})$ representation of the Lorentz group is the space of hermitian $2\times2$ matrices? The vector space for the $(\frac{1}{2},0)$ representation is $\mathbb{C}^2$ and I guess the same is true for the $(0,\frac{1}{2})$ representation, but I can't put it together to end up with hermitian matrices.

EDIT: I found in the book Symmetry and the Standard Model: Mathematics and Particle Physics by Matthew Robinson the following explanation.

Recall that just as any real matrix can be written as the sum of a symmetric matrix and an antisymmetric matrix, any complex matrix can be written as the sum of a Hermitian matrix and an anti-Hermitian matrix. However, the two indices on our matrix $v^{a \dot b}$ transform under representations of $SU(2)$. Notice that in the generators of these copies of $SU(2)$, both sets of generators $N^-$ and $N^+$ are Hermitian (cf. (3.229)). So, we’ll limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times 2$ matrix.

If anyone could help me understand this line of thought my problem would be solved.

Why does this allow us to " limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times 2$ matrix"?

I understand that our representation here acts on complex $2\times2$ matrices. But I don't understand why we can restrict to hermitian matrices.

• I don't understand your question. Finite vector spaces are isomorphic if (and only if) they are of equal dimension. – Prof. Legolasov Jan 22 '15 at 10:34
• @Hindsight Thanks for your comment. My problem is understanding why the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz transformations act on hermitian $2\times2$ matrices. The second part of my question is only my try towards an answer. Because we have $(\frac{1}{2},\frac{1}{2})= (\frac{1}{2},0) \otimes (0,\frac{1}{2})$ I thought, maybe the corresponding vector space is $\mathbb{C}^2 \otimes \mathbb{C}^2$ and this justifies somehow that the objects living in this representation are hermitian $2\times2$ matrices. – Tim Jan 22 '15 at 10:42
• Tim, your comment is correct, I think. The Dirac spinor representation acts upon $\mathbb{C}^2 \otimes \mathbb{C}^2 = \mathbb{C}^4$. The Hermitian 2x2 matrices are $\mathbb{R}^4$ as a vector space, and hence can be embedded into the Dirac representation, but I don't think they actually are the full representation space for the complex representation. Maybe it is implied that this is the real form of the usual representation. – ACuriousMind Jan 22 '15 at 14:29
• As far as I know Dirac spinors transform according to the $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$ representation of the Lorentz group. Is then $\mathbb{C}^2 \otimes \mathbb{C}^2 = \mathbb{C}^4$ or $\mathbb{C}^2 \oplus \mathbb{C}^2 = \mathbb{C}^4$ correct? – Tim Jan 23 '15 at 9:13
• – glS Jan 24 '15 at 13:11

## 2 Answers

A point from the comments that should be emphasised is that any finite-dimensional vector space is isomorphic to any other of the same dimension.

What defines a representation of a group is the action of the group elements on the vector space. Then it is often convenient to choose some particular manifestation of the representation space in which the action looks nice, or familiar.

In the case of the proper Lorentz group, to classify the representations we use the fact that its complexification is isomorphic (up to a $\mathbb{Z}_2$) to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$, so an element can be given by a pair $(A,B)$ of $SL(2,\mathbb{C})$ matrices. Then the $(\frac12,\frac12)$ representation can be conveniently described as acting on the space of $2\times2$ matrices as $$(A,B):M\mapsto A M B^\dagger.$$ This is natural if you think of $M$ as a tensor product of vectors in the $(\frac12,0)$ and $(0,\frac12)$ representations, like $M=u v^\dagger$ (or a sum of such terms).

To see the relation to the vector representation, write the matrices as $$M=\begin{pmatrix}t+z & x-i y\\ x+iy & t-z \end{pmatrix}$$ and notice that the determinant is $t^2-x^2-y^2-z^2$, and further that this is preserved under the action of the group. Using the $(t,x,y,z)$ basis instead of these matrices would give the usual Lorentz transformation of vectors. Going from the complexification to the real section of $SO(3,1)_+$ restricts you to a single $SL(2,\mathbb{C})$, which corresponds to $A=B$ the way it's written here, so Hermiticity (or reality of $(t,x,y,z)$) is also preserved.

• Thanks a lot for your answer. In fact, I was about to ask you in the other question if you could maybe have a look at this question :) Unfortunately, I still don't understand why $M$ must be hermitian and nothing else. I understand that it must be $2\times2$, because we use the two-dimensional representation for both copies of $SL(2,\mathbb{C})$ and the transformation behaviour you describe, but hermiticity is still not obvious for me. – Tim Jan 24 '15 at 13:18
• Hermiticity comes from a reality condition: the space of Hermitian matrices is the real vector space preserved by the action of the group. The fact that $SO(3,1)$ corresponds to $A=B$ follows because the two $\mathfrak{sl}(2)$ algebras are related by complex conjugation (as they come from $J+i K$ and $J-i K$). – Holographer Jan 24 '15 at 14:28
• Okay, but then I do not understand the reality condition you mention. Why does it follow from $(\frac{1}{2},0)^\star= (0,\frac{1}{2})$ that $(\frac12,\frac12)$ acts on a real vector space? – Tim Jan 24 '15 at 15:23
• I edited the question with a quote from a book, which, I think, uses the same line of thoughts as you here. Maybe this helps to understand what my problem is – Tim Jan 25 '15 at 11:17
• You can have the group act on all the matrices if you like: an 8 -(real)-dimensional space. But the set of Hermitian matrices (or the antiHermitian ones) is invariant, so this representation is reducible, as the direct sum of two representations of 4 real dimensions. – Holographer Jan 25 '15 at 19:27

Holographer has already given a correct answer. Let us here just try to emphasize the main points.

1. As a check of dimensions, note that the lhs. and rhs. of the isomorphism of representations $$\tag{1} (\frac{1}{2},\frac{1}{2})~\cong~ {\rm Mat}_{2\times 2}(\mathbb{C})$$ both have 4 complex dimensions, cf. e.g. this Phys.SE post. It would not make sense to replace the rhs. of (1) with the vector space $u(2)$ of Hermitian $2\times 2$ matrices, because that has only 4 real dimensions.

2. The reason the vector space $u(2)$ of Hermitian $2\times 2$ matrices shows up is because the vector representation of the Lorentz group is the Minkowski space $M(1,3;\mathbb{R})\cong u(2)$, which in turn is isomorphic to the vector space $u(2)$ of Hermitian $2\times 2$ matrices. The complexified Minkowski space $M(1,3;\mathbb{C})\cong {\rm Mat}_{2\times 2}(\mathbb{C})$ is then isomorphic to the vector space ${\rm Mat}_{2\times 2}(\mathbb{C})$ of complex $2\times 2$ matrices. Further details and justification are e.g. given in this Phys.SE post.