I'm looking through Introduction to Supersymmetry by Muller-Kirsten and Wiedemann, along with any other resource I can find. I'm specifically trying to understand the concepts and notations for different representations of $SL(2, \mathbb{C})$, namely the dotted and undotted notation for Weyl spinors. I'm aware there are many conventions. What I was hoping for was one that was clear and internally consistent. I've seen this source recommended a lot, and it seems to explain more details than most.


Given some algebraic object, be it a group or an algebra or whatever, if we would like to study how it can act on things, we often map the elements to the operators/matrices acting on some vector space. Step one is to specify the vector space (typically real or complex in the Physics case). Step two is to specify a map from the group to the matrices acting on the chosen vector space. Because we are interested in capturing the behavior of the algebraic object we are "representing", we also require that the map from the algebraic object to the matrices is consistent with the algebraic operation (group multiplication, lie bracket, or whatever).

We would like to consider the linear representations of $SL(2, \mathbb{C})$. Since this group is defined to the the 2x2 matrices with complex entries, with determinant 1, we already have the obvious ("fundamental") representation. That is, our vector space is $\mathbb{C}^2$, and the representation maps the elements of the group to themselves. So we just have the 2x2 complex matrices (of determinant 1) acting on the space of complex 2-entry column vectors, by normal matrix multiplication.

Since our group consists of complex matrices, we could also pick the vector space $\mathbb{C}^2$, but instead map each group element to its complex conjugate (just complex conjugate each entry). I think this is called the "conjugate fundamental" representation.

The book (eqn 1.82 on page 38), denotes these two reps as follows: \begin{align} \psi_A' &= M_A^{\ B}\psi_B \\ \bar{\psi}_{\dot{A}} &= (M^*)_{\dot{A}}^{\ \dot{B}}\bar{\psi}^{\dot{B}}. \end{align}

Although the index placement is unexplained, I believe this can only mean that the down-left index on the matrices indicates the row, and the up-right index on the matrices indicates the column. And that the downstairs index on the spinor indicates the component of a column vector. It seems to me this is the most common convention. This is the opposite from what seems to be standard convention in GR and geometry, where one would have expected an up index on the column, but ok, as long as we proceed consistently, it should be fine.

Given a representation acting on some vector space, we can define its "dual" representation. This is where we take our representing matrices and the vector space they act on, and map the representing matrices to their inverse transpose, and have them act on the dual space.

So the book summarises this as (eqn 1.82 page 38):

\begin{align} {\psi_A}' &= M_A^{\ B}\psi_B \\ {\psi^A}' &= ((M^{-1})^T)^A_{\ B} \psi^B \\ {\bar{\psi}'}_{\dot{A}} &= (M^*)_{\dot{B}}^{\ \dot{A}}\bar{\psi}^{\dot{B}} \\ {\bar{\psi}'}^{\dot{A}} &= (((M^*)^{-1})^T)^{\dot{A}}_{\ \dot{B}}\bar{\psi}^{\dot{B}}. \end{align}

Or, since I think it is generally less confusing to mess around with index structures on the matrices, I prefer as written in THIS answer:

\begin{align} {\psi_A}' &= M_A^{\ B}\psi_B \\ {\psi^A}' &= (M^{-1})_B^{\ A} \psi^B \\ {\bar{\psi}'}_{\dot{A}} &= (M^*)_{\dot{B}}^{\ \dot{A}}\bar{\psi}^{\dot{B}} \\ {\bar{\psi}'}^{\dot{A}} &= ((M^*)^{-1})_{\dot{B}}^{\ \dot{A}}\bar{\psi}^{\dot{B}}. \end{align}

The indices go up and down using the "metric", the antisymmetric epsilon symbol.

The above is all a lengthy preamble. I summarised more than I probably needed to, because I thought it could be helpful to someone who is also struggling to understand this, and because it perhaps clarifies to any potential answerer what I do and don't know already.

My issue is that when people write Dirac spinors, they consist of 4 component columns. But they are usually built from $\psi_A$ and $\bar{\psi}^{\dot{A}}$ objects. From above, we are thinking of the second object as a row. Is this not a problem?

This issue is avoided in the book on page 42. They use the fact that given a vector space, its dual vector space is in particular a vector space. And that the dual vector space of the dual vector space is the original vector space. They say that we should actually think of the $\bar{\psi}^{\dot{A}}$ objects as the columns, and the $\bar{\psi}_{\dot{A}}$ objects as the rows.

I'm not sure there is an outright inconsistency, but why have we introduced the objects $\bar{\psi}_A$ and $\bar{\psi}_{\dot{A}}$ in such a way that it is natural to think of them as columns, only to then leave one of them as being considered a column, and invert things to consider the other a row? Why couldn't we have just used the $\psi_A$ and $\bar{\psi}_{\dot{A}}$ objects to form the Dirac spinor?

I have read some other posts on the topic of dotted and undotted indices here, but I didn't find any that answered this question.

EDITED: to fix a typo in the indices


1 Answer 1


First of all, I don't know the book to which you refer to. By the way in both block of formulas the third row does not follow correctly the index rules, since the index $\dot{B}$ appears twice up instead of one up and one down (or vice-versa). But this is just a detail.

Usually Einstein's double index summation convention applies which does not know of row or columns. The multiplication of vectors with matrices can often written conveniently as matrix products with vectors or with their transpose or even both of them, but this is done just for convenience, I would not derive a strict rule from that. In particular sometimes it can be quite tricky to translate formulas written in Einstein's index summation convention in matrix equations thereby correctly transposing etc.

When it comes do Dirac's bispinors they are composed of undotted and dotted spinor like

$$\Psi = \left(\begin{array}{c} \psi_\alpha \\ \bar{\chi}^\dot{\alpha}\end{array}\right)$$

also for convenience. This particular convenience which is called Weyl-representation. $\Psi$ would transform like $\Psi' = D\Psi$

Of course it could be written like

$$\Xi = \left(\begin{array}{c} \psi_\alpha \\ \ \bar{\chi}_\dot{\alpha}\end{array}\right)$$

which would transform like $\Xi' = T\Xi$. We would still call it a Dirac spinor because one could find a matrix S which does the following:

$$D= S T S^{-1}$$

because the spinors $\bar{\chi}_\dot{\alpha}$ and $\bar{\chi}^\dot{\alpha}$ transform according to two representations which are equivalent. This equivalence is well explained in Hyperon's post physics.stackexchange.com/q/743730/ but also in the post you refer to.

(just as side remark $S$ would have block diagonal form as it is a reducible Lorentz group representation where the first block would consist of the identity matrix and the second block would be the actual transformation between two equivalent representations of the dotted spinor.)

Dirac spinors in Weyl representation transform infinitesimally like (according to standard convention indices of Dirac spinors are usually suppressed):

$$\Psi \rightarrow (I -\frac{i}{4}\omega_{\lambda\nu}\Sigma^{\lambda\nu})\Psi$$

which if the Weyl representation is used can be conveniently decomposed in 2 infinitesimal Lorentz transformations for its 2 constituents:

$$\psi_\alpha \rightarrow (I +\frac{1}{2}\omega_{\lambda\nu}\sigma^{\lambda\nu})_\alpha^\beta \psi_\beta $$


$$\bar{\chi}^\dot{\alpha} \rightarrow (I +\frac{1}{2}\omega_{\lambda\nu}\bar{\sigma}^{\lambda\nu})^\dot{\alpha}_\dot{\beta} \bar{\chi}^\dot{\beta}$$

if we use the decomposition of the $\Sigma$-matrix (provided by the $\gamma$-matrix decomposition)

$$\Sigma^{\lambda\nu}=2i\left(\begin{array}{cc} \sigma^{\lambda\nu} & 0 \\ 0 & \bar{\sigma}^{\lambda\nu} \\ \end{array}\right)$$

where $\sigma^{\lambda\nu}=\frac{1}{4}(\sigma^\lambda\bar{\sigma}^\nu - \sigma^\nu\bar{\sigma}^\lambda)$ and $\bar{\sigma}^{\lambda\nu}=\frac{1}{4}(\bar{\sigma}^\lambda\sigma^\nu - \bar{\sigma}^\nu\sigma^\lambda)$ themselves can be conveniently decomposed into the "extended" Pauli-matrices $\sigma^\lambda =(1,\mathbf{\vec{\sigma}})$ and $\bar{\sigma}^\nu=(1,-\mathbf{\vec{\sigma}})$.

This is one of the main reasons why this convention is chosen. It is full-fledged explained in the Srednicki's book on QFT in chapter 35. Srednicki's book is accessible online, so you can check it there. Choosing another convention would make the so-called van der Waerden formalism even more complicated, so I think, that's the reason why it is widely used.

  • $\begingroup$ Thanks. I'll think about what you have said, and check out that book. You also have a similar index issue in your $\bar{\chi}^{\dot{\alpha}}$ equation. $\endgroup$
    – Gleeson
    Commented Feb 11 at 21:17
  • $\begingroup$ you are right, it is has been a typo. Corrected now. $\endgroup$ Commented Feb 11 at 22:12

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