I come from a maths background and am struggling with some of the more physical texts on SUSY. In particular they claim that the fermionic generators $Q_A^i$ carry a representation of the Lorentz group. What does this mean? I have never heard the word 'carry' applied to representations in a mathematical framework.

I would appreciate it if someone could

  1. give me a general mathematical definition of this term
  2. explain specifically why it is used in this context (see edit below)

Edit: most books I have read note that $$[Q_A, J_{ab}]= (b_{ab})_A^BQ_B$$ and use the super-Jacobi identity to conclude that the structure constant matrices $b_{ab}$ form a representation for the Lorentz algebra.

They use this to immediately conclude that $Q_A$ "carry a representation" of the Lorentz group. What is the logic here?

  • $\begingroup$ Presumably just that they have a spinor index, i.e. are acted on by $SL(2:\mathbb{C})$, which is a double cover of the proper orthochr. Lorentz group $\endgroup$
    – twistor59
    Sep 5 '13 at 12:23
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    $\begingroup$ Hmm... but wouldn't that just be trivial? If you give me $\mathbb{C}^n$ I can definitely define a linear action of $SL(2;\mathbb{C})$ on it. The physics books make this sounds like something deep though. $\endgroup$ Sep 5 '13 at 12:30
  • $\begingroup$ @EdwardHughes Yeah it's pretty trivial once you are already convinced that group theory is the right language to use and $SL(2;C)$ is the appropriate group etc. The nontrivial thing is that the fermionic generators have to carry a specific (half-spin) representation for the SUSY algebra to be consistent... $\endgroup$
    – Michael
    Sep 5 '13 at 12:46
  • $\begingroup$ But books usually go through a massive rigmarole about the structure constants to 'prove' this trivial fact. What's the point, when it essentially says nothing? $\endgroup$ Sep 5 '13 at 12:52
  • $\begingroup$ I get the feeling there's some specific way in which the Lorentz group is meant to act which you derive from the commutator $[J_{ab}, Q_A]$ (in $N=1$ SUSY say). But nowhere explains this. $\endgroup$ Sep 5 '13 at 12:54

People have essentially explained the details, but let me make an attempt to formulate it in a language more familiar to a mathematician. I will ignore subtleties that enter for more general Lie superalgebras.

Let $\mathfrak g$ be a Lie superalgebra with the $\mathbb Z_2$ grading $\mathfrak g = \mathfrak g_e\oplus\mathfrak g_o$, where the two factors are the even ("bosonic") and odd ("fermionic") part respectively. The even part $\mathfrak g_e$ form a closed Lie algebra and acts on the odd part $\mathfrak g_o$ by the adjoint action

\begin{equation} ad_g:\mathfrak g_o\rightarrow\mathfrak g_o, \qquad q\rightarrow ad_g(q)=[g,q], \end{equation} where $g\in\mathfrak g_e$ and $[.,.]$ is the commutator of the Lie superalgebra. Now, $\mathfrak g_o$ is a vector space and thus form a representation space of the even part $\mathfrak g_e$ (under the adjoint action). You can now decompose $\mathfrak g_o$ into irreducible representations of $\mathfrak g_e$. Thus you can construct a basis of $\mathfrak g_o$ that transforms under a representation of $\mathfrak g_e$ under the adjoint action, or in other words their commutators just correspond to some representation of $\mathfrak g_e$.

In the case you are talking about, $\mathfrak g_e$ is just the Poincare algebra and $\mathfrak g_o$ transforms under certain spinor-representation of it (under the adjoint action/commutator).

Edit: I think I earlier misunderstood the questions regarding the role of super-Jacobi identities. Let me, following joshphysics' suggestion, elaborate on this using the slightly more mathematical (basis independent) language. For a more basis-dependent approach, i recommend joshphysics' answer below.

As I explained above, the adjoint action $\text{ad}:\mathfrak g_e\rightarrow\mathfrak{gl}(\mathfrak g_o)$, or in other words $\text{ad}_x:\mathfrak g_o\rightarrow\mathfrak g_o$ (where $x\in\mathfrak g_e$), is actually a $\text{dim}(\mathfrak g_o)$ dimensional representation of the Lie algebra $\mathfrak g_e$ on the vector space $\mathfrak g_o$. This means that it's a Lie algebra homomorphism

$$ \left[ \text{ad}_x,\text{ad}_y \right](z) = \text{ad}_{[x,y]}(z),\qquad x,y\in\mathfrak g_e, z\in \mathfrak g_o,$$

where I use the notation

$$ \left[ \text{ad}_x,\text{ad}_y \right] = \text{ad}_x\circ \text{ad}_y - \text{ad}_y\circ \text{ad}_x.$$

One can very easily show that the adjoint action satisfy the above identity and is thus a representation, by making use of the Jacobi identites. Thus the Jacobi identities make sure the adjoint action is a Lie algebra homomorphism. If you chose a basis, you can easily see that this is equivalent to what joshphysics states in his answer. In particular, the coefficients $s_{\alpha,i}^\beta$ (in the notation of joshphysics), correspond to a representation of the even part $\mathfrak g_e$. Although I don't think it has to be the adjoint representation in general (that's not the case for the super-Poincare algebra for example).

  • $\begingroup$ Ah right - so the stuff about using the super-Jacobi identity is pretty pointless then? I guess they are trying to say what you've said above in a manifest way? If you could clarify that these thoughts are right I'll certainly accept! Many thanks. $\endgroup$ Sep 5 '13 at 17:53
  • $\begingroup$ What I explain above is the very general structure of Lie superalgebras, that explains what it means that the odd part transforms under the even part. I think what physics book do is trying to use this structure to explicitly construct a particular Lie superalgebra. $\endgroup$
    – Heidar
    Sep 5 '13 at 18:06
  • $\begingroup$ They start with the Poincare algebra and by hand add the odd part to get the $\mathbb Z_2$ grading. The only unknowns are the structure constants of the odd-odd and even-odd parts. Here they can use the fact the even-odd is related to representations of the Poincare algebra (by the above reasoning) and further use super jacobi identities to constrain the structure constants. Then they find the structure constants of the super Poincare algebra in a particular basis. $\endgroup$
    – Heidar
    Sep 5 '13 at 18:07
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    $\begingroup$ Right I get it completely now. The important thing is ask whether $\mathfrak{g}_o$ forms a representation for $\mathfrak{g}_e$ under the adjoint action. This is a direct consequence of the super-Jacobi identity. So the answer to my original question is that $\{Q_A\}$ naturally transform under a representation of the Lorentz group by considering the adjoint action. Obviously they can be made to in other ways mathematically, but this way comes immediately out of the appropriate super-Jacobi identity. $\endgroup$ Sep 5 '13 at 19:51
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    $\begingroup$ +1: Just read your response more carefully, and I see that what I said is essentially the same. I think including something like Edward Hughes's phrase "$\{Q_A\}$ naturally transform under a representation of the Lorentz group by considering the adjoint action" and "This is a direct consequence of the super-Jacobi identity" might be nice additions to you answer for future users who read it since he asks how the Jacobi identities are relevant. $\endgroup$ Sep 5 '13 at 22:50

I think this is probably equivalent to Heidar's answer, but I'll include it anyway for those who are less mathy. We consider a Lie superalgebra with a basis $\{B_i, F_\alpha\}$ satisfying the following structure relations: \begin{align} [B_i, B_j] &= ic_{ij}^{\phantom{ij}k}B_k, \qquad [F_\alpha, B_i] = s_{\alpha i}^{\phantom{\alpha i}\beta} F_\beta, \qquad \{F_\alpha, F_\beta\} = \gamma_{\alpha\beta}^{\phantom{\alpha\beta}i}B_i. \end{align} The $B_i$'s are called bosonic generators and the $F_i$'s are called Fermionic generators. Now we ask ourselves: can the structure constants be chosen arbitrarily? Well no; part fo the definition of a Lie superalgebra is that the brackets $[\cdot, \cdot]$ and $\{\cdot, \cdot\}$ are antisymmetric and symmetric respectively. Moreover, the super-Jacobi identities must be satisfied as part of the definition. Antisymmetry of the bracket $[\cdot, \cdot]$, for example, says that that $c_{ij}^{\phantom{ij}k}=-c_{ji}^{\phantom{ij}k}$. One can then show that enforcing the super-Jacobi identities requires that the matrices $S_i$ defined as \begin{align} (S_i)_\alpha^{\phantom\alpha\beta} = s_{\alpha i}^{\phantom{\alpha i}\beta} \end{align} form an adjoint representation of the bosonic Lie subalgebra given by the first structure relation above.

You could now ask, ok well all of this is well and good, but why do we care about algebras that are defined in this way (like requiring super-Jacobi identities)? Well, the answer to that is given by a famous theorem due to Haag, Lopuszanski, and Sohnius.

  • $\begingroup$ +1 This is indeed equivalent to what I was trying to say, from a slightly different perspective. $\endgroup$
    – Heidar
    Sep 5 '13 at 22:29
  • $\begingroup$ @EdwardHughes You might find this supersymmetry intro (by Sohnius himself) helpful inspirehep.net/record/222335 I think it's awesome. $\endgroup$ Sep 5 '13 at 22:54
  • $\begingroup$ I just updated my answer and noticed a small detail here. The matrix $(S_i)_\alpha^\beta$ has to be a representation of the bosonic Lie subalgebra, but it does not necessarily have to be the adjoint rep. right? That's for example not the case for the super-Poincare algebra. $\endgroup$
    – Heidar
    Sep 16 '13 at 21:02
  • $\begingroup$ @Heidar Yes that's right. One can, for example, have fermionic generators that are "Weyl Spinors" in the sense that the corresponding representation of the bosonic subalgebra is the Weyl-Spinor representation of the Lorentz algebra. $\endgroup$ Sep 18 '13 at 6:38

In general saying that some objects $A_{i}$ carry a (linear) representation $R$ of a group $G$ just means that you're considering the action of $G$ on the set of $A$s corresponding to the representation $R$ of $G$ on $\displaystyle \mathrm{span}(\{A_i\})$.

Physicists often use indices to quickly identify linear representations (e.g. 1 "vector index" = fundamental representation), in particular for the Lorentz group.

The case of fermionic indices is slightly more involved, since they are not really representations of the Lorentz group and as twistor59 said you need to consider the double cover. I think that this is not the main point of your question, and you can find some detail on Wikipedia.

Of course, this looks trivial to a mathematician since you can consider any set of n objects to carry an n-dimensional representation of any group. The point is that you choose what groups In field theory, relativistic invariance is implemented by making every field carry some representation of the Lorentz group, and building scalar objects (Lagrangians, amplitudes etc) with them.

Carry a certain representation $R$ then means that the Lorentz group acts as defined by $R$ on your objects (field and other operators). As you suggested in a comment it tells you what the commutator with the operators representing the Lorentz group generators are.

  • $\begingroup$ Re your last paragraph - why does the commutator with the Lorentz generators determine the representation (and vice versa)? I'll happily accept if you can give me a mathematical derivation. Apologies if I'm missing something obvious, because it feels like I am. I'll probably work it out myself later when I'm not at work! $\endgroup$ Sep 5 '13 at 13:14

Comments to the question (v3):

The fermionic SUSY generators belong to a super vector space $V$. That a vector space $V$ carries a representation of a Lie algebra $L$, e.g. the Lorentz Lie algebra, simply means that it is a Lie algebra representation of the Lie algebra $L$.

There is a similar terminology with the Lie algebra $L$ replaced with a Lie group $G$.

Warning: Note that in the literature one often finds authors talking about a Lie group $G$ when they really mean the corresponding Lie algebra $L$, and vice-versa.

In particular, note that a Lie group representation $V$ of the Lie group $G$ is also a Lie algebra representation $V$ of the corresponding Lie algebra $L$, while the opposite is not necessarily the case.

  • $\begingroup$ +1: This answer is good to future visitors, even though the OP, and other mathematicians, may not be happy with this, . $\endgroup$ Sep 6 '13 at 13:07
  • $\begingroup$ This answer (v1) was written in an attempt to answer and clarify OP's question. It might not be what OP is actually looking for, only OP can tell, but really, there is nothing mathematically controversial or wrong about the claims it makes. $\endgroup$
    – Qmechanic
    Sep 16 '13 at 21:20

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