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Context

The following is from the book "Ideas and methods in supersymmetry and supergravity" by I.L. Buchbinder and S.M Kuzenko, pg 56-60. It is about realizing the irreducible massive representations of the Poincare group as spin tensor fields which transform under certain representations of the homogeneous Lorentz group and are subject to some supplementary conditions.

Consider the linear space $\mathcal{H}_{(A,B)}$ of $(A/2,B/2)$ type spin tensor fields $\Phi_{\alpha_1\cdots\alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}(x)$ totally symmetric in their A undotted indices and independently in their B dotted indices, with $A+B=2s$ and which satisfy the following supplementary conditions: $$\begin{cases}\partial^{\dot{\alpha}\alpha}\Phi_{\alpha\alpha_1\cdots\alpha_{A-1}\dot{\alpha}\dot{\alpha}_1\cdots\dot{\alpha}_{B-1}}(x)=0 && (1)\\ (\partial^a\partial_a-m^2)\Phi_{\alpha_1\cdots\alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}(x)=0&&(2)\end{cases}$$ Here $\partial_{\alpha\dot{\alpha}}=(\sigma^a)_{\alpha\dot{\alpha}}\partial_a$ and $\partial^{\dot{\alpha}\alpha}=(\tilde{\sigma}^a)^{\dot{\alpha}\alpha}\partial_a=\varepsilon^{\dot{\alpha}\dot{\beta}}\varepsilon^{\alpha\beta}(\sigma^a)_{\beta\dot{\beta}}\partial_a$, $\sigma^a=(\text{Id},\vec{\sigma}) $ and $ \tilde{\sigma}^a=(\text{Id},-\vec{\sigma}).$ My metric convention is $\eta_{ab}=\text{Diag}(-1,1,1,1)$, spinor indices are greek letters whilst Lorentz indices are Latin.

Consider the following one-to-one map: $$\Delta_{{\alpha}_{A+1}}^{~~~\dot{\alpha}_B}: \mathcal{H}_{(A,B)}\rightarrow \mathcal{H}_{(A+1,B-1)}$$ $$\Phi_{\alpha_1\cdots\alpha_A\alpha_{A+1}\dot{\alpha}_1\cdots\dot{\alpha}_{B-1}}(x):=\Delta_{{\alpha}_{A+1}}^{~~\dot{\alpha}_B}\Phi_{\alpha_1\cdots\alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}(x) ~~\text{ where }~~ \Delta_{\alpha_{A+1}}^{~~\dot{\alpha_B}}=\frac{1}{m}\partial_{\alpha_{A+1}}^{~~\dot{\alpha_B}}$$ The map is one-to-one because it can be shown, using the mass shell condition (1), that it has an inverse defined by $\Delta_{\alpha}^{~~\dot{\alpha}}\Delta^{\beta}_{~~\dot{\alpha}}=\delta_{\alpha}^{\beta}$. Although for the purpose of this thread it is not necessary to look at the inverse map.

Question

To me, it is not obvious that after we have acted on an element of $\mathcal{H}_{(A,B)}$ with $\Delta_{{\alpha}_{A+1}}^{~~~\dot{\alpha}_B}$, the result is totally symmetric in its undotted indices, including the additional one created through the map. However, this is a claim of the author's. So I would like to prove the following: $$\Delta_{({\alpha}_{A+1}}^{~~\dot{\alpha}_B}\Phi_{\alpha_1\cdots\alpha_A)\dot{\alpha}_1\cdots\dot{\alpha}_B}(x)=\Delta_{{\alpha}_{A+1}}^{~~\dot{\alpha}_B}\Phi_{\alpha_1\cdots\alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}(x)$$

Attempt

I'm almost certain that the result should follow from supplementary condition (2) (which is sometimes reffered to as the 'spin selection' condition) along with the fact that $\Phi_{\alpha_1\cdots\alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}(x)$ is totally symmetric in its undotted indices. However, again, it isn't at all obvious as to why.

To gain some intuition for the problem I tried a simple case: $$\Delta_{\gamma}^{~~\dot{\beta}}:\mathcal{H}_{(2,2)}\rightarrow \mathcal{H}_{(3,1)}$$ $$X_{\alpha\beta\dot{\alpha}\dot{\beta}}\rightarrow X_{\alpha\beta\gamma\dot{\alpha}}:=\Delta_{\gamma}^{~~\dot{\beta}}X_{\alpha\beta\dot{\alpha}\dot{\beta}} ~~\text{ where } X_{(\alpha\beta)(\dot{\alpha}\dot{\beta})}=X_{\alpha\beta\dot{\alpha}\dot{\beta}}$$

The aim of this intuitive exercise would then be to show that $X_{\alpha\beta\gamma\dot{\alpha}}=X_{\gamma\beta\alpha\dot{\alpha}}$, for example.

If I explicitly plug in some numbers for the indices, I can see that, for example, the case $(\alpha=1,\gamma=2)$ is equal to the case $(\alpha=2,\gamma=1)$ as a result of supplementary condition (2). However I am struggling to generalize this and am certainly not content with leaving the argument here. Does anyone have a suggestion or hint for me? Thank you.

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Since $\Phi_{\alpha_1\cdots \alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}$ is completely symmetric in the undotted indices, it is enough to show that $$\partial^{\dot{\alpha}_B}_\alpha\Phi_{\alpha_1\cdots \alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B} = \partial^{\dot{\alpha}_B}_{\alpha_1}\Phi_{\alpha\alpha_2\cdots \alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B}.$$ Using the property of the antisymmetric tensor $\varepsilon^{\alpha\beta}$, $$\varepsilon^{\alpha\beta} = -\varepsilon^{\beta\alpha}$$ this equation is the same as requiring $$ \varepsilon^{\alpha\alpha_1}\partial^{\dot{\alpha}_B}_\alpha\Phi_{\alpha_1\cdots \alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B} = 0.$$ This holds by property (1) since $$ \varepsilon^{\alpha\alpha_1}\partial^{\dot{\alpha}_B}_\alpha\Phi_{\alpha_1\cdots \alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B} = \partial^{\dot{\alpha}_B\alpha_1}\Phi_{\alpha_1\cdots \alpha_A\dot{\alpha}_1\cdots\dot{\alpha}_B} = 0. $$

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  • $\begingroup$ Could you please explain why the second equation follows from the first? $\endgroup$ – NormalsNotFar Jul 1 '17 at 3:51
  • $\begingroup$ This is from the (anti)symmetry properties of $\varepsilon$. I have updated the answer. $\endgroup$ – Sean Pohorence Jul 1 '17 at 18:52

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