# If $v_{a \dot{b}}$ transforms like a four-vector, what does $v_{a}^{\dot{b}}$ describe?

The $( \frac{1}{2}, 0)$ representation of the Lorentz group acts on left-chiral spinors $\chi_a$, the $( 0,\frac{1}{2} )$ representation on right-chiral spinors $\chi^{\dot a}$.

The $( \frac{1}{2}, \frac{1}{2}) = ( \frac{1}{2}, 0) \otimes ( 0,\frac{1}{2} )$ representation acts therefore on objects with one dotted and one undotted index. My naive, first guess would be that the $( \frac{1}{2}, \frac{1}{2})$ representation acts on objects with one lower undotted and one upper dotted index: $v_{a}^{\dot{b}}$. An upper dotted index transforms like a right-chiral spinor, a lower undotted like a left-chiral spinor.

Quite surprising for me is that $v_{a \dot{b}}= v_\mu \sigma^\mu_{a \dot{b}}$ transforms like a four-vector and $v_{a}^{\dot{b}}$ transforms differently, because the transformation behaviour of a lower dotted index is different than that of an upper dotted index.

Why does $v_{a \dot{b}}$ transform like a four-vector and not the naive first guess $v_{a}^{\dot{b}}$? Is there any name for objects transforming like $v_{a}^{\dot{b}}$, just as left-chiral spinors, right-chiral spinors or four-vectors are defined by their transformation behaviour?

Firstly, it doesn't matter whether the indices are up or down here when you ask what representation an object lives in, because the raised and lowered versions are equivalent. Both $\chi_a$ and $\chi^a$ give objects in the $(\frac{1}{2},0)$ rep, and similarly for dotted. The relation between the two is the linear transformation given by $\chi^a=\epsilon^{ab}\chi_b$ (just a change of basis). The matrices that gives the transformations for upstairs and downstairs are then related by a similarity transform (conjugation with the epsilon tensor, or equivalently combined inverse and transpose).

Then an object in the $(\frac{1}{2},\frac{1}{2})$ rep can be described in a few different ways: $v_a^\dot{b}$, or equivalently $v_{a\dot{b}}$ are the obvious ones if you know $(\frac{1}{2},\frac{1}{2})=(\frac{1}{2},0)\otimes(0,\frac{1}{2})$ (again using some $\epsilon$ to move between different descriptions/bases). But there happens to be another natural, and more familiar, basis to choose. The change of basis matrix to get to this is the collection $\sigma^\mu_{a \dot{b}}$ of Pauli matrices. (You can think of this, if it helps, as a $4\times 4$ matrix, with $^\mu$ indices on the columns, and $_{a\dot{b}}$ indices on the rows). The appropriate conjugation by this takes a tensor product of $SU(2)$ matrices to a Lorentz transformation of a vector, or vice versa.

So the answer is that any of the index structures you suggest describe the same representation, just in a different basis. The choice you make is just a matter of convenience, and there is always some linear map that will move from any one to any other.

• Great answer, thank you! Two small questions: Firstly, with different basis you mean instead of using the usual Pauli matrices, $v_{a \dot{b} } = v_{\nu} \sigma^{ \nu}_{a \dot{b} } = v^0 \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + v^1 \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} +v^2 \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix} + v^3 \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}$ we use a different basis for hermitian $2 \times 2$ matrices? (How do we know this representation acts on hermitian matrices anyway?) – Tim Jan 21 '15 at 7:35
• And secondly, do you have an idea why we need to combine two objects in different bases in order to get sth Lorentz invariant, i.e. an undotted upper with an undotted lower index etc.? (I know that it works, as we can see explicitly by looking at the transformation behaviour, but now from the point of view that objects with upper and lower index are the same objects merely in a different basis, I think there must be a good reason.) – Tim Jan 21 '15 at 7:38
• By a new basis I mean the following: there are four components in $v_\mu$, labelled by $\mu=0,1,2,3$. There are also four components in $v_{a\dot{b}}$, labelled by $a=1,2$ and $\dot{b}=1,2$. These components are related by linear combinations: for example, $v_{1\dot{1}}=v_\mu \sigma^\mu_{1\dot{1}}=v_0 \sigma^0_{1\dot{1}}+v_1 \sigma^1_{1\dot{1}}+v_2 \sigma^2_{1\dot{1}}+v_3 \sigma^3_{1\dot{1}}=v_0+v_3$. It is this relation that is a change of basis; the collection of $\sigma$s gives a map from one basis of the representation to the other. – Holographer Jan 21 '15 at 11:49
• The reason that you combine objects in the specified ways to get a scalar is that they live in dual representations, which means that the transformations cancel. The fact that the dual is equivalent to the original representation is special for $SU(2)$, and is related to the fact that there is an invariant tensor $\epsilon_{ab}$ (see Schur's Lemma). – Holographer Jan 21 '15 at 12:05
• Thanks for your elaborations. Just one small thing: I do understand now what you mean by the "basis change" through the $\sigma$ matrices. Nevertheless, you write that $v_{a \dot{b} }$ and $v_{a }^\dot{b}$ are related by a basis change, too. Does this mean that we have for example: $v_{a \dot{b} } = v_{\nu} \sigma^{ \nu}_{a \dot{b} } = v^0 \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + v^1 \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} +v^2 \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix} + v^3 \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}$ – Tim Jan 24 '15 at 11:23

With the antisymmetric structure in Spinors, we need to define an ordering for the indices in the Einstein summation convention in order to omit indices without making sign errors all over the place. The standard (e.g. Wess & Bagger) convention is that for objects in the $(1/2, 0)$ we have a North-East to South-West rule, i.e. $$\phi \psi \equiv \phi^\alpha \psi_\alpha = - \psi_\alpha \phi^\alpha \neq \psi \phi$$ For the dotted indices of the $(0, 1/2)$ rep the order is the other way round $$\bar \chi \bar \xi = \bar \chi_{\dot \beta} \bar \xi^{\dot \beta}$$ A spinor bilinear that transforms like a vector is now construced by combining $$\phi \sigma^\mu \bar\xi \equiv \phi^\alpha \sigma^\mu_{\alpha \dot \beta} \bar \xi^{\dot\beta}$$ or by combining $$\bar \xi \bar \sigma^\mu \phi \equiv \bar \xi_{\dot \beta} \bar \sigma^{\mu\, \dot \beta \alpha} \phi_\alpha$$ Note that in the second case I made use of $\bar \sigma$, which has components $$\left( \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \sigma^i \right)$$ One could of course also use a non-strandard index ordering, but then one can not omit the spinor indices without confusing people $$V^\mu = \phi^\alpha {\sigma^\mu_\alpha}^{\dot \beta} \bar \xi_{\dot \beta} = - \phi^\alpha \sigma^\mu_{\alpha \dot\beta} \bar \xi^{\dot \beta} = - \phi \sigma^\mu \bar \xi = - V^\mu\vert_\text{standard}$$