Tensor contraction with multiple indices

I think I missed a rule somewhere, because I can contract the following expression in multiple ways.

$\epsilon^{\alpha \beta} \sigma_{\dot{\alpha} \alpha} \epsilon^{\dot{\alpha} \dot{\beta}} = \sigma_{\dot{\alpha} \alpha} \epsilon^{\alpha \beta} \epsilon^{\dot{\alpha} \dot{\beta}}=\sigma_{\dot{\alpha}}^{~\beta}\epsilon^{\dot{\alpha} \dot{\beta}}=\sigma^{ \dot{\beta}\beta}$

I also know that $\epsilon^{\alpha \beta}=-\epsilon^{ \beta \alpha}$, so

$\epsilon^{\alpha \beta} \sigma_{\dot{\alpha} \alpha} \epsilon^{\dot{\alpha} \dot{\beta}} = -\epsilon^{ \dot{\beta} \dot{\alpha}}\sigma_{\dot{\alpha} \alpha} \epsilon^{\alpha \beta} =-\sigma^{\dot{\beta}}_{~\alpha}\epsilon^{\alpha \beta}=-\sigma^{ \dot{\beta}\beta}$

What is going wrong here?

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What is $\epsilon$ and what is $\sigma$? What are the dots denoting? Why did you introduce the negative sign in the second computation? –  Qiaochu Yuan Jul 1 '12 at 17:55
$\epsilon$ is a completely antisymmetric tensor, the negative comes from the exchange of a pair of indices. The $\sigma$ is a Pauli matrix and the dots indicate that the index refers to a conjugated spinor. The dots shouldn't be important however, they can just be regarded to be different indices than their dotless counterparts. –  Ben Ruijl Jul 1 '12 at 18:26
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1 Answer

Both calculations are wrong, I'm afraid. Since when does $\epsilon^{\lambda\eta}\sigma_{\eta\nu} = \sigma^\lambda{}_\nu$? This is valid only for metric tensor, but $\epsilon^{\lambda\eta}$ ($\begin{pmatrix}0&1\\-1&0\end{pmatrix}$) is not a metric tensor, assuming $\epsilon^{\lambda\eta}$ is the two-dimensional Levi-Civita symbol.

Suppose $T^{\beta\delta} = \epsilon^{\alpha\beta}\sigma_{\gamma\alpha}\epsilon^{\gamma\delta}$, and the metric is identity. Since there are only 4 distinct elements, we could carry out the computation directly:

• $T^{00} = \epsilon^{10}\sigma_{11}\epsilon^{10} = \sigma_{11} = -\sigma_{00}$
• $T^{01} = \epsilon^{10}\sigma_{01}\epsilon^{01} = -\sigma_{01}$
• $T^{10} = \epsilon^{01}\sigma_{10}\epsilon^{10} = -\sigma_{10}$
• $T^{11} = \epsilon^{01}\sigma_{00}\epsilon^{01} = \sigma_{00} = -\sigma_{11}$

or use the identities $\epsilon_{ab}=\epsilon^{ab}$ and $\epsilon_{ab}\epsilon^{cd} = \delta_a^c\delta_b^d - \delta_a^d\delta_b^c$ to arrive at

$$T^{\beta\delta} = (\delta^{\alpha\gamma}\delta^{\beta\delta} - \delta^{\alpha\delta}\delta^{\beta\gamma})\sigma_{\gamma\alpha} =\delta^{\beta\delta}\operatorname{tr}(\sigma) -\sigma^{\beta\delta} = -\sigma^{\beta\delta}.$$

Edit: If you define $\epsilon^{\mu\nu}$ to be the metric, then your first formula is wrong, because index raising is done by

$$\huge x^{\color{red}\mu} = g^{\color{red}\mu\color{green}\nu} x_{\color{green}\nu} \tag{Correct}$$

and not

$$\huge x^{\color{red}\mu} = g^{\color{#FF4000}\nu\color{#808000}\mu} x_{\color{green}\nu} \tag{Wrong}$$

especially when your metric $g^{\mu\nu}$ is asymmetric.

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$\epsilon$ behaves like a metric for Weyl spinors and so it has the same behaviour for Pauli matrices. I should have clarified this, sorry for that. –  Ben Ruijl Jul 1 '12 at 20:53
@BenRuijl: OK. See update. –  KennyTM Jul 1 '12 at 21:17
How does this work with multiple indices, should I contract the $\nu$ first: $x^{\mu \alpha}=g^{\mu \nu} g^{\alpha \beta} x_{\nu \beta}$, or doesn't it matter? That's where my confusion arose. –  Ben Ruijl Jul 2 '12 at 8:56
@BenRuijl: It doesn't matter whether you contract $\nu$ and $\beta$ first. For summation with finite terms, $\sum_a\sum_b f(a,b)=\sum_b\sum_a f(a,b)$. –  KennyTM Jul 2 '12 at 10:17
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