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  Constructing SUSY algebra via index structure

+ 2 like - 0 dislike

Often in literature the SUSY algebra is simply given, but various books, for example Bailin and Love, goes through the trouble of showing how the SUSY commutation relations are the only possible ones that you can write down. This is the content of my question.

SUSY adds two spinor generators to the Poincare algebra, our complete list of generators is given by the independent components of $$ M^{\mu\nu}, P^\mu, Q_\alpha, \bar{Q}_{\dot\beta} $$ Where the first two are the usual Lorentz and translation generators, the last two are the added spinor generators.

Among other thing, Bailin and Love makes statements such as:

$[P^\mu,Q_\alpha]$ must yield a spinor, the only possibility is $c \sigma^\mu _{\alpha\dot\beta}\bar Q^\dot\beta$.

Where $c$ is a constant, and $\sigma^\mu=(I,\sigma^i)$. They then go on to show that the Jacobi identity implies $c=0$.

The quoted statement is not obvious to me, in particular I cannot fully understand why the expression written above is the only possible commutation relation.

Let me make this a little more specific:

  1. From Bailin and Love, as well as other pedagogical literature, it appears that the structure constant must be some combination of $$c,\sigma^\mu_{\alpha \dot\beta},\sigma^{\mu\nu}_{\alpha\beta}$$ Where the last symbol is the usual SL(2) generators constructed from the Paulis. I want to know why these are the only things one is allowed to write. In particular, can one not construct other matrices that have the same index structure? Is it that these are the most general things with these index structures, or is it that they're fixed by symmetry?
  2. Restricting myself to constructing the structure constants from the things I wrote above, I was able to verify that as claimed, the index structure of the commutators of interest fix unique structure constants (up to a scaling factor). However I verified this by exhaustion, is there a systematic way to find these combinations?
  3. It turned out that all of the commutators yielded linear combination of one type of generator: for example in the quoted statement the result is a sum of right handed spinor generators. Is the result that commutators yield linear combinations of a single type of generator a general one? Or is it only a side effect of the uniqueness of the form of the structure constant fixed by index structure, as mentioned in point 2?
This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user bechira
asked Aug 10, 2014 in Theoretical Physics by bechira (80 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

It is maybe simpler to consider all the generators as representations of $SL(2,C)$, so, using spinor indices, you will have : $M^{\alpha \dot \alpha \beta \dot \beta}, P^{\beta \dot \beta}, Q_\alpha, \bar Q^\dot\beta$

Indices are raised and lowered with the Levi-Civita symbols $\epsilon_{\alpha \beta}, \epsilon^{\alpha \beta},\epsilon_{\dot \alpha \dot \beta},\epsilon^{\dot \alpha \dot \beta}$

Now, what is $[P^{\beta \dot \beta}, Q_\alpha]$ ?

We see that there is no generator with the form $G^{\beta \dot \beta}_\alpha$.

Levi-Civita symbols are not useful too, because they have $2$ lower or upper indices of same kind, so we cannot write something like $[P^{\beta \dot \beta}, Q_\alpha] = \epsilon_{\alpha \beta}Q^\dot\beta$ (there would be an obvious problem with the $_\beta$ indice).

So the only solution is a contraction on indices $\alpha$ and $\beta$, that is :

$[P^{\beta \dot \beta}, Q_\alpha] = \delta_{\alpha} ^{\beta} \bar Q^\dot\beta$

With $P^\mu = \sigma^\mu_{\beta \dot \beta}P^{\beta \dot \beta}$, (which means simply that the $(\frac{1}{2}, \frac{1}{2})$ representation of $SL(2,C)$ is equivalent to the fundamental representation of $SO(3,1)$ ) we get finally :

$[P^\mu, Q_\alpha] = \sigma^\mu_{\beta \dot \beta}\delta_{\alpha} ^{\beta} \bar Q^\dot\beta = \sigma^\mu_{\alpha \dot \beta} \bar Q^\dot\beta$

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user Trimok
answered Aug 11, 2014 by Trimok (955 points) [ no revision ]
Thanks a bunch. I am still slightly confused about my first question: in this case,are the Levi-Civita symbol and the identity the only tensors that transform like tensors with 2 spinor indices?

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user bechira
The Levi-Civita symbol is the only quantity which transforms as a representation with $2$ lower or upper spinor indices of the same kind : $\epsilon_{\alpha \beta}, \epsilon^{\alpha \beta},\epsilon_{\dot \alpha \dot \beta},\epsilon^{\dot \alpha \dot \beta}$. "Identity" is a quantity $\delta^a_b$ or $\delta^\dot a_\dot b$, so you have one upper and one lower indices of the same kind. The momentum transforms as $ P^{\beta \dot \beta}$ or $P_{\beta \dot \beta}$ (if you lower the indices), so you have $2$ lower or upper indices of different kind.

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user Trimok
This is probably a silly question, but can you point me somewhere that shows your first sentence, that the Levi-Civita symbol is the only tensor we can construct which transforms as shown?

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user bechira
Levi-Civita symbols are the spinor metrics (used to raise and lower indices). Independently of the metrics, the only possible operations are contractions on identical indices (one lower and one upper of the same kind), and the generators. There is no other possibility.

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user Trimok
+ 2 like - 0 dislike

This answer is mostly rephrasing Trimok's correct answer in other words.

  1. The super-Poincare group is supposed to be an extension of the Poincare group, which contains the Lorentz group and translations. We will complexify the Lorentz group. The Lie group $G:=SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (isomorhic to the double cover of) the complexified Lorentz group $SO(1,3;\mathbb{C})$, cf. e.g. this Phys.SE post. This fact gives rise to the undotted and dotted irreps. An irrep $(s,\dot{s})$ of $G$ is characterized by two non-negative half-integers $s,\dot{s}\in \frac{1}{2}\mathbb{N}_0$.

  2. The investigated commutator $[P_{\beta\dot{\beta}}, Q_{\alpha}]$ belongs to a tensor product representation of $G$, $$ (\frac{1}{2},\frac{1}{2}) \otimes (\frac{1}{2},0) ~\cong~(\frac{1}{2},0)^{\otimes 2}\otimes (0,\frac{1}{2})$$ $$\tag{1}~\cong~[(0,0)\oplus(1,0)]\otimes (0,\frac{1}{2}) ~\cong~(0,\frac{1}{2})\oplus (1,\frac{1}{2}),$$ which we, in turn, have decomposed in irreps.

  3. Of the 14 super-Poincare generators $t_a$, only the dublet $\bar{Q}_{\dot{\gamma}}$ transforms in one of the two irreps on the rhs. of eq. (1), namely first irrep $(0,\frac{1}{2})$. If we would like the super-Poincare algebra to close on the 14 generators $t_a$ without introducing new generators; in particular, if we would like the investigated commutator $$\tag{2}[P_{\beta\dot{\beta}}, Q_{\alpha}]~\in~ {\rm span}(t_a),$$ then the first irrep $(0,\frac{1}{2})$ on the rhs. of eq. (1) must be proportional to $\bar{Q}_{\dot{\gamma}}$, and the second irrep $(1,\frac{1}{2})$ must be annihilated.

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user Qmechanic
answered Aug 11, 2014 by Qmechanic (3,120 points) [ no revision ]

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