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I've been referring to Pg.36-Pg.38 in Introduction to Supersymmetry by Wiedamann. For understanding the precise origin of dotted, undotted indices on Spinors. He starts off my saying that $M$ acts on $\Psi$ as a representation on the spinor space.He says, that a spinor $\Psi_A$ transforms as $$\Psi_A = M_A^B\psi_B$$ under a rep $M$ of $SL(2,\mathbb{C})$. Then he proves that $M^{-1T}$ is an equivalent representation to $M$. He introduces the standard lowering and raising $\epsilon$ through this. And shows that, $\psi^A \equiv \epsilon^{AB}\psi_B$ is an object that transforms in the $M^{-1T}$ and therefore, we give it a new notation of contra variant index, $$\Psi^A = (M^{-1T})^{A}_B\Psi^A$$ So far everything is fine. But on Pg.39, he writes $$\psi^A \equiv \epsilon^{AB}\psi_B = -\psi_B\epsilon^{BA}$$ I'm unclear with how the second part come that is, $-\psi_B\epsilon^{BA}$. I know $\epsilon$ is anti-symmetric but, what is the need for changing the order? Why not $-\epsilon^{BA}\psi_B$?

Understanding the Spinors as Mathematical Objects

Now, I know that there is a very uncanny resemblance of a metric and $\epsilon$ here. I fail to see the exact connection. It does the same role as raising and lower indices. Namely, from a $\mathbb{R}^4$ manifold point of view, I know that the metric gives a notion of distance for vectors in the tangent space $T_p(\mathcal{M})$ of a manifold $\mathcal{M}$. Therefore, it makes sense to write $g_{\mu\nu} x^\nu x^\nu$, for vector which are labeled as $x_\nu$ and $x_\mu$ in a basis. FOR CONVIENCE (I hope I'm right with this), we choose to write $g_{\mu\nu}x^\nu \equiv x_\mu$ as column vector to have an easy way of having the action of duals on its space.

Similarly. I'm guessing the whole point of $\epsilon$ is a notion of distance on the spinor space with a spinor vector whose components labeled by $\psi_A$. And $\epsilon^{AB}\psi_A\psi_B$ gives a notion of distance on this space. In which case, I dont' see why not $-\epsilon^{BA}\psi_B =- \psi_B\epsilon^{BA}$. As they are just some numbers. I would like to know if this is what we do. Probably, someone with the whole picture of a supermanifold and metrics on it can explain me the relation between these.Because, the fact that we consider these labels $\Psi_A$ and $\Psi_B$ to be anti-commuting $\{\Psi_A,\Psi_B\} =0$ seems to suggest to be that these are Grassmann numbers on superspace if I'm not wrong.

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    $\begingroup$ I don't think you need to change the order, it just "looks better" when the contracted indices are next to each other. That is, aesthetically, $\psi_B\epsilon^{BA}$ is preferable to $\epsilon^{BA}\psi_B$. I'm not 100% sure, just keep this in mind. $\endgroup$ – Ryan Unger Mar 11 '16 at 14:22
  • $\begingroup$ Here $\epsilon^{BA}$ is a matrix acting on $\psi_B$. I think it is completely relevant to consider the order. $\endgroup$ – Sai krishna Deep Mar 11 '16 at 14:29
  • $\begingroup$ Ok, this is some strange convention. Normally, $\epsilon^{AB}$ is as much of a matrix as $T_{ab}$ or $R_{ab}$ is. $\endgroup$ – Ryan Unger Mar 11 '16 at 15:03
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This is purely for aesthetics.

Following Wald, we have a 2-dimensional vector space $W$ over $\mathbb{C}$. The spinors are elements of $W$ and furnish a rep of $\mathrm{SL}(2,\mathbb{C})$. The antisymmetric tensor $\epsilon_{AB}$ gives an isomorphism $W\to W^*$, the dual space. Note that nothing we do here depends on the special linear group or anticommuting variables. The indices $A,B,$ etc. are to be understood in the abstract sense.

Given $\psi^A\in W$ we associate to it $\psi_A\in W^*$ by $\psi_B=\epsilon_{AB}\psi^A=-\epsilon_{BA}\psi^A$. In the other direction, we assign $\psi^A$ to $\psi_A$ by $\psi^A=\epsilon^{AB}\psi_B=-\epsilon^{BA}\psi_B=-\psi_B\epsilon^{BA}$. Since these are just tensors, everything commutes in the abstract index notation. There are no "matrices" here.

In Riemannian geometry, we have a vector $v^a$, a metric $g_{ab}$ and an associated covector $v_a$. Then $v^a=g^{ab}v_b=g^{ba}v_b=v_bg^{ba}$. This process is entirely analogous, modulo a sign.

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  • $\begingroup$ Ok, thanks for this. I'll have a look at Wald on its reference to Spinors. But the other part of the question is still unanswered. $\endgroup$ – Sai krishna Deep Mar 12 '16 at 9:25

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