What does "carry a representation" mean (in SUSY algebra)? 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 


*

*give me a general mathematical definition of this term

*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?
 A: 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).
A: 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.
A: 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.
A: 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.
