Some Identities Presented in Peskin and Schroeder Problem 3.5 (Supersymmetry) I am using the solutions for Introduction to Quantum Field Theory by Dr. Zhong-Zhi Xianyu. Right under equation $(3.41)$, the author states that $$(\sigma ^2)^T = -\sigma^2.$$ Why is this true? Shouldn't there not be a minus sign?. Here was my logic. We know from Peskin and Schroeder that $\sigma = (1, \vec{\sigma})$, where $\vec{\sigma} = \sigma_x \hat{x} + \sigma_y \hat{y} + \sigma_z \hat{z}$. Here, the components of $\sigma$ are denoted as $\sigma^{\mu}$. Thus $\sigma^2 = \sigma \cdot \sigma = 1 - \vec{\sigma} \cdot \vec{\sigma} = 1 - 1 - 1- 1 = -2$. Hence $\sigma^2$ is $-2$ times the identity matrix. From this shouldn't we conclude that $(\sigma^2)^T = \sigma^2$?
Next to the previous identity we have $$\epsilon^T \sigma^2 \chi = \chi^T \sigma^2 \epsilon.$$ How do you prove / show this? I as assuming that $\chi$ is a vector of left and right handed Weyl spinors.
Equation $(3.46)$ says that $(\epsilon^T \sigma^2 \chi_k)(\chi_i^T \sigma^2 \chi_j) + (\epsilon^T \sigma^2 \chi_i)(\chi_j^T \sigma^2 \chi_k) + (\epsilon^T \sigma^2 \chi_j)(\chi_k^T \sigma^2 \chi_i) = 0$. Using the previous Weyl Spinor Identity I see that:
$(\epsilon^T \sigma^2 \chi_k)(\chi_i^T \sigma^2 \chi_j) + (\epsilon^T \sigma^2 \chi_i)(\chi_j^T \sigma^2 \chi_k) + (\epsilon^T \sigma^2 \chi_j)(\chi_k^T \sigma^2 \chi_i) = (\chi_k^T \sigma^2 \epsilon)(\chi_i^T \sigma^2 \chi_j) + (\chi_k^T \sigma^2 \epsilon)(\chi_j^T \sigma^2 \chi_k) + (\chi_k^T \sigma^2 \epsilon)(\chi_k^T \sigma^2 \chi_i)$ However, I am not sure why this is zero. The author of the solutions manual said that this can be proven by brute force, but since I do not know the form of the Weyl Spinors $\chi$ I am not sure how to prove this. Can any show me a away to prove this?
PS: I can see how the solution to this exercise follows from these identities, but I do not know why this identities are true. Without knowing the proofs of these identities, I will not be able to completely understand and appreciate this problem. Help will be appreciated.
 A: For the first identity, $\sigma^2$ is the second Pauli matrix, NOT $\sigma^\mu \sigma_\mu$.
The second identity is very easy to prove
\begin{align}
\epsilon^T \sigma^2 \chi &= \epsilon^a (\sigma^2)_{ab} \chi^b \\
&= - \chi^b (\sigma^2)_{ab} \epsilon^a \\
&= - \chi^b [ - (\sigma^2)_{ba} ] \epsilon^a \\
&= \chi^T \sigma^2 \epsilon
\end{align}
The first sign comes from exchanging the spinors since they are Grassmannian. The second minus uses the first identity.
I have no idea what calculation you are trying to do in your question though. That seems completely wrong.
A: Prahar gave a good answer to your first two questions so let me answer the third. If you let $\epsilon, \chi_i, \chi_j, \chi_k$ carry spinor indices $\alpha, \beta, \gamma, \delta$ respectively, then
\begin{equation}
(\epsilon^T \sigma^2 \chi_k)(\chi_i^T \sigma^2 \chi_j) + (\epsilon^T \sigma^2 \chi_i)(\chi_j^T \sigma^2 \chi_k) + (\epsilon^T \sigma^2 \chi_j)(\chi_k^T \sigma^2 \chi_i) = 0
\end{equation}
is the same as the identity
\begin{equation}
\varepsilon^{\alpha\beta}\varepsilon^{\gamma\delta} - \varepsilon^{\alpha\gamma}\varepsilon^{\beta\delta} + \varepsilon^{\alpha\delta}\varepsilon^{\beta\gamma} = 0.
\end{equation}
To go between them, we just use $\sigma^2 = i\varepsilon$ and the fact that spinor fields anti-commute.
When it comes to proving this identity, there are two possible answers. One is to learn about Fierz identities of which this is an example. However, sources on Fierz identities can often be misleading, giving readers the impression that the identities themselves should be learned rather than the algorithms for proving them. Memorizing identities is hopeless in invariant theory because if you want to build tensors with $N$ indices, the number of candidates built from the primitives will grow with $N$ much more quickly than the number of invariant tensors the group actually has.
So I'll explain why $SU(2)$ only has two invariant tensors with four spinor indices in a way that hopefully makes the generalization to other situations clear. What we do is write
\begin{equation}
x_1 \varepsilon^{\alpha\beta}\varepsilon^{\gamma\delta} + x_2 \varepsilon^{\alpha\gamma}\varepsilon^{\beta\delta} + x_3 \varepsilon^{\alpha\delta}\varepsilon^{\beta\gamma} = 0
\end{equation}
and see if there's a solution other than $x_1 = x_2 = x_3 = 0$. In other words, we are showing  that these three tensors are linearly dependent. To do so, we contract the above line three times. First with $\varepsilon_{\alpha\beta} \varepsilon_{\gamma\delta}$, next with $\varepsilon_{\alpha\gamma} \varepsilon_{\beta\delta}$ and finally with $\varepsilon_{\alpha\delta} \varepsilon_{\beta\gamma}$. Each contraction is easy to do once we use
\begin{equation}
\varepsilon_{\alpha\beta}\varepsilon^{\beta\gamma} = \delta^\gamma_\alpha.
\end{equation}
At the end of the day, we get three equations for the $x_i$ and they have a non-trivial kernel. Namely $\mathrm{span}(1, -1, 1)$ which proves the claim.
