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?

Question

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.

Answer 1

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.

Answer 2

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.