# Why doesn't the Superconformal Algebra close?

What I mean is, if I take the conformal algebra and the supersymmetry charges, how do I show that $[K, Q]$ is not in this algebra?

I see this fact stated everywhere but with no proof. Any references that show this explicitly would be greatly appreciated. A similar question has been answered before (The superconformal algebra) but there the same fact is also just stated with no justification.

• Well, this commutator has to have scaling dimension $-1+\frac{1}{2}=-\frac{1}{2}$, and there are no such operators in conformal algebra. Also, superconformal algebra does close, as it contains S. – Peter Kravchuk Sep 24 '17 at 4:34
• In other words, if you assume that it is one of the known operators, then the Jacobi identity with dilatation operator fails. – Peter Kravchuk Sep 24 '17 at 4:42
• Mathematical nitpick: The fact that $[K,Q]$ is not a linear combination of the "known operators", does not mean it "is not in this algebra", of course. It just means that the "known operators" do not form a basis of the vector space underlying the algebra---they still generate the algebra, by construction. – Danu Sep 24 '17 at 9:27
• @PeterKravchuk I tried your suggestion and for me the Jacobi identity seems to work like a charm anyway. Even if it didn't, it seems like that still couldn't rule out that $[K,Q]=0$ – AlexM Sep 25 '17 at 21:18

Let $C$ be the conformal algebra, cosisting of Lorentz generators $M$, translations $P$, special conformal transformations $K$ and dilatations $D$. Let $C_Q$ be the linear space spanned by elements of $C$ and some supersymmetry generators $Q^i_a$, where $a$ is a spinor index in whatever dimension we are considering and $i$ is an $R$-symmetry index. (In even dimensions there are two kinds of spinors, but we can ignore that in the arguments to follow.)
We are studying $[K_\mu,Q_a^i]$. It cannot be equal to zero. Indeed, we have the Jacobi identity, $$[P_\nu,[K_\mu,Q_a^i]]=[[P_\nu,K_\mu],Q_a^i]+[K_\mu,[P_\nu,Q_a^i]].$$ If $[K_\mu,Q_a^i]=0$ then we have, since $[P_\nu,Q_a^i]=0$, $$0=[[P_\nu,K_\mu],Q_a^i]=-2[D\delta_{\mu\nu}-M_{\mu\nu},Q_a^i].$$ The right hand side is non-zero, since $[D,Q^i_a]=\frac{1}{2}Q^i_a$ (this follows straightforwardly from $[Q,Q]=P$ and Jacobi identities if we assume that $Q^i_a$ has a definite scaling dimension) and $[M_{\mu\nu},Q_a^i]=-(\mathcal{M_{\mu\nu}})_a{}^b Q_b^i$, where $\mathcal{M_{\mu\nu}}$ is the matrix which furnishes the spinor representation of Lorentz group (this fact can be thought of either as definition of "$a$ is a spinor index" or as a part of SUSY algebra).
Having established that $[K_\mu,Q_a^i]\neq 0$, we can see that it cannot lie in $C_Q$. Indeed, on $C_Q$ the spectrum of $D$ consists of $+1$ (on translations $P$), $0$ (on $D$ and $M$) $-1$ (on $K$) and $\frac{1}{2}$ (on $Q$). On the other hand, we have the Jacobi identity $$[D,[K_\mu,Q_a^i]]=[[D,K_\mu],Q_a^i]+[K_\mu,[D,Q_a^i]]=-[K_\mu,Q_a^i]+\frac{1}{2}[K_\mu,Q_a^i]=-\frac{1}{2}[K_\mu,Q_a^i],$$ so $D$ takes eigenvalue $-\frac{1}{2}$ on $[K_\mu,Q_a^i]$. Thus this commutator is necessarily not in $C_Q$ i.e. a "new operator".