Twistor notation in space-time (Part 1)

Question

This is sort of a continuation of this and this previous discussions.

In the first of my links one sees the surjective isometry between real or complex $(1,3)$ signature Minkowski space and the real or complex (respectively) span of Pauli matrices to be given as, $V_\mu \mapsto V_{\alpha \dot{\alpha}} = V^\mu \sigma _{\mu \alpha \dot{\alpha}}$ using the standard Pauli matrices.

• In reference to the second of the above links one would need that another set of matrices (say A) to "reverse" the above map as $V^\mu = A^{\mu \dot{\alpha} \alpha} V_{\alpha \dot{\alpha}}$. What would be these matrices $A^\mu$?

(..in the expression similar to the above in Lubos's answer he has $V_\mu = \sigma _ \mu ^{\alpha \dot{\alpha}} V_{\alpha \dot{\alpha}}$..which seems to have the $\alpha$ and $\dot{\alpha}$ indices in unexpected (wrong?) places and doesn't seem to the inverse of $V_{\alpha \dot{\alpha}} = V^\mu \sigma _{\mu \alpha \dot{\alpha}}$..)

• In the second of my links the answer by Lubos says that the relationship between $V_\mu$ and $V_{\alpha \dot{\alpha}}$ is a reflection of the fact that $SL(2,C)$ is a double cover (hence locally isomorphic) of $SO^+(1,3)$.

  This is confusing to me - since as also pointed out in the same answer a $4$ vector can be thought of as a tensor product of the $2$ and $\bar{2}$ representations of $SL(2,C)$ (..the left and the right handed Weyl fermions..)

  Hence isn't the relationship between $V_\mu$ and $V_{\alpha \dot{\alpha}}$ and the above interpretation coming from the fact that the complexified $(1,3)$ signature M inkowski space supports a $(\frac{1}{2},\frac{1}{2})$ representation of $SL(2,C)\times SL(2,C)$?

  • In the light of the above isn't it necessary that one things of complex space-time vectors $V_\mu$ to do the interpretation of them as lying in tensor products of $2$ and $\bar{2}$ representations of $SL(2,C)$?

  Or is it there in some implicit form since Lubos doesn't seem to need the null twistor condition as required in the answer by Roy Simpson to map to real Minkowski space.

  • The above mapping between $V_\mu$ and $V_{\alpha \dot{\alpha}}$ seems to have in it built the most negative sign convention for the metric. ($det(V_{\alpha \dot{\alpha}}) = (V^0)^2 - \sum _{i =1} ^{i=3} (V^i)^2$) How does one change the mapping if one wants to work in the most positive sign convention?

  • Once one has mapped a space-time vector $V^\mu$ to $V_{\alpha \dot{\alpha}}$ when is it true that one can now find a left and right chiral Weyl spinors $\lambda_\alpha$ and $\bar{\lambda}_\dot{\alpha}$ such that $V_{\alpha \dot{\alpha}} = \lambda_\alpha \bar{\lambda}_\dot{\alpha}$ ?

  Is it only when $V$ is a null vector as is the case for modern application of scattering of fast gluons where one neglects their mass? (..but there I am confused to see why not both $\lambda$ and $\bar{\lambda}$ is used for each of the gluons but only one depending on identifications about incoming/outgoing (anti)article nature..)

I am using the convention of having both $\alpha$ and $\dot{\alpha}$ being down as I see in many recent string theory papers. I guess sometimes one wants to write the index of the conjugate representations ($\dot{\alpha}$) upstairs and $\alpha$ downstairs.

Answer 1

The matrices $A^\mu$ are clearly just the inverse matrices that multiply the "bispinor components" of a vector to get the usual vector component. So $A^{\mu\dot\alpha \alpha}$ is the inverse to $\sigma^\mu_{\alpha\dot\alpha}$ – you treat the $\alpha,\dot\alpha$ indices as the rows and columns – and this inverse may also be obtained by a simple raising of the vector indices via $\eta^{\mu\nu}$ and the spinor indices via $\epsilon_{\alpha\beta}$ etc.

So all this discussion was just to say that $$ A^{\mu\dot\alpha\alpha}\equiv \sigma^{\mu,\alpha\dot\alpha}$$ which is not shocking because it's the only "nicely covariant" object with one vector index and the two spinor indices.

$SL(2,C)$ is locally isomorphic to $SO(3,1)^+$ and vice versa. This means that the irreducible representations of these two groups (allowing any phase shifts for rotations by 360 degrees) may be obtained from tensor products of the fundamental representations of $SL(2,C)$. Because the fundamental representations are complex, there are two of them, ${\bf 2}$ and $\bar{\bf 2}$, and the irreducible representations are $$ {\bf 2}^{\otimes 2j_L,sym} \otimes \bar{\bf 2}^{\otimes 2j_R,sym} $$ So this is just a representation whose (complex) dimension is $(2j_L+1)(2j_R+1)$. If $j_L=j_R$, one may impose a reality condition on this representation (one that has to exchange $j_L$ and $j_R$ because the complex conjugation does interchange ${\bf 2}$ and $\bar{\bf 2}$) so $(2j_L+1)(2j_R+1)$ is also the real dimension.

Reality conditions on the representations

This answers another question of yours. If $j_L=j_R$, then it is not necessary to think about the complex representation because the representation may be made real: there is a natural reality condition (commuting with the action of the group) that may be imposed to turn the complex representation into a real one.

This representation specified by $(j_L,j_R)$ may also be thought of as a continuation of a similar representation of $SU(2)\times SU(2)$ which is locally isomorphic to $SO(4)$. However, $SO(4)$ clearly isn't locally isomorphic to $SO(3,1)$ – equivalently, $SU(2)\times SU(2)$ isn't locally isomorphic to $SL(2,C)$. However, the complexification of all these groups are the same – it is $SO(4,C)$ which is locally the same as $SL(2,C)\times SL(2,C)$ – which means that one may always obtain a representation of one of the groups by continuing the representations of another. Because the irreducible representations of $SU(2)\times SU(2)$ are obviously given by two independent angular momenta, $(j_L,j_R)$, one for each factor, the same is true for $SL(2,C)$ even though the group itself doesn't split into two factors (a direct product). The fact that two independent $j$'s have to be specified for $SL(2,C)$ has a different explanation – namely that its fundamental representation is complex so there are really two inequivalent fundamental reps.

Signature

If you want the determinant to switch the sign, you simply multiply $\sigma_{\mu\alpha\dot\alpha}$ by an $i$. That works simply because $i^2=-1$. If you thought that $i$ is a shocking number, it's not: some of the matrices $\sigma_\mu$ are inevitably pure imaginary regardless of your convention (note that the Pauli matrix $\sigma_y$ is pure imaginary, for example). Multiplying all of them by $i$ or $-i$ turns the real ones into pure imaginary ones and vice versa: that's equivalent to the change of the conventions for "mostly positive" or "mostly negative" metric tensors.

Factorization of vectors

One may sometimes factorize $V_{\alpha\dot\alpha} = \lambda_\alpha \bar\lambda_{\dot \alpha}$. However, it's clearly not the case for all vectors $V$. Just calculate $V_\mu V^\mu$ using this Ansatz. You will get it is $(\lambda)^2 (\bar\lambda)^2$ which identically vanishes because each factor is zero. Note that $\lambda^2=\epsilon_{\alpha\beta}\lambda^\alpha\lambda^\beta$ vanishes because it's an antisymmetric tensor contracted with a symmetric one. So only null vectors may be factorized in this way. But yes, all of them may. If $\bar\lambda$ is required to be complex conjugate to $\lambda$, then $V$ must be a real future-directed (or past-directed, depending on the signature convention discussed above) null vector for the decomposition to exist.

There is no "convention" involved if you only write expressions with all indices being manifest. One may always raise and lower spinor indices using $\epsilon$: one must only decide how the indices are ordered in this $\epsilon$ which affects the overall sign only.

I feel that the discussion above uses many simple identities and steps you're perhaps not familiar with, like the raising and lowering of spinor indices with the epsilon and things that follow from the translations such as $$ \sigma_{\mu,\alpha\dot\alpha} \sigma_{\nu}^{\alpha\dot\alpha} = \eta_{\mu\nu} $$ But there may be lots of these things. You have to rediscover them and check them yourself. They're inevitable by the symmetry and one can't list absolutely every aspect of these groups and representations in one Physics Stack Exchange answer.