# Antisymmetric part of a tensor with Levi-Civita symbol

How to write the antisymmetric part of a tensor in general? It has been asked here but I want to write it using the Levi Civita symbol. I have thought of:

$$A^{\mu_1} A^{\mu_2} - A^{\mu_2} A^{\mu_1} = A^{[\mu_1 \mu_2]} = \epsilon_{\alpha \beta} A^{\mu_{\alpha} \mu_{\beta}}$$

but I don't know if this is the best way. (Of course I want this for more than 2 indices).

## 2 Answers

The basic problem is that in $$n$$ dimensions you have $$n$$ indices on your orientation tensor so you have to map a $$[q, 0]$$-tensor to a $$[0, n-q]$$-tensor with one application.

You can probably get it with two, that is in $$n$$ dimensions we probably would have $$A^{[ab]} = \frac12 A^{ab} - \frac12 A^{ba} \propto \epsilon^{abcd\dots p} ~\epsilon_{cd\dots pqr}~A^{qr}$$ but there is a reason that this is a “weird thing to do” and it's that you can antisymmetrize in a non-orientable space so these notions must be formally separate, and we are here just describing a property that the orientation tensor should have when it exists.

• OK so you mean that, in a $n-$dimensional space (where indices go from $1$ to $n$), you can construct the antisymmetric part of a $\text{rank} \, p$ tensor $A^{a_1 ... a_p}$ only if $p \leq n$. You do it as follows: $A^{[a_1 ... a_p]} \propto \epsilon^{a_1 ... a_p b_1 ... b_{n-p} } \epsilon_{b_1 ... b_{n-p} c_1 ... c_p} A^{c_1 ... c_p}$ – MBolin Sep 23 '18 at 15:10
• (And if $p>n$ the antisymmetric part of the tensor is the zero tensor, trivially antisymmetric) – MBolin Sep 23 '18 at 15:14

You can define the map $$\overline{\text{Alt}}:\otimes^kV\otimes^lV^*\to\text{Alt}(\otimes^kV\otimes^lV^*),$$ called the alternating map, to create an alternating $$(k,l)$$-tensor out of a general $$(k,l)$$-tensor. The alternating map is defined as $$\overline{\text{Alt}}(\tilde T)\equiv \frac{1}{k!l!}\sum_{\substack{\sigma\in S_k\\\lambda\in S_l}}\text{sgn}(\sigma)\text{sgn}(\lambda){}_{\{\sigma,\lambda\}}\tilde T\hspace{0.1cm},$$ where the quantity $${}_{\{\sigma,\lambda\}}\tilde T$$ is defined by: $${}_{\{\sigma,\lambda\}}\tilde T(\omega_1,...,\omega_k,v_1,...,v_l)\equiv \tilde T(\omega_{\sigma(1)},...,\omega_{\sigma(k)},v_{\lambda(1)},...,v_{\lambda(l)}).$$

Then, using the alternating property of $$\overline{\text{Alt}}(\bar T)$$ we have that \begin{align} \hspace{-0.15cm}\overline{\text{Alt}}(\tilde T)(\omega_1,...,\omega_k,v_1,...,v_l)&=\frac{1}{k!l!}\sum_{\substack{\sigma\in S_k\\\lambda\in S_l}}\text{sgn}(\sigma)\text{sgn}(\lambda)\tilde T(\omega_{\sigma(1)},...,\omega_{\sigma(k)},v_{\lambda(1)},...,v_{\lambda(l)})\nonumber\\ &=\sum_{\substack{a_1,...,a_k=1 \\ b_1,...,b_l=1}}^n\frac{\omega_{a_1}...\omega_{a_k}v^{b_1}...v^{b_l}}{k!l!}\sum_{\substack{\sigma\in S_k\\\lambda\in S_l}}\text{sgn}(\sigma)\text{sgn}(\lambda)\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\tilde T(e^{a_{\sigma(1)}},...,e^{a_{\sigma(k)}},e_{b_{\lambda(1)}},...,e_{b_{\lambda(l)}})\nonumber\\ &=\sum_{\substack{a_1,...,a_k=1 \\ b_1,...,b_l=1}}^n\frac{\omega_{a_1}...\omega_{a_k}v^{b_1}...v^{b_l}}{k!l!}\sum_{\substack{\sigma\in S_k\\\lambda\in S_l}}\text{sgn}(\sigma)\text{sgn}(\lambda)\tilde T^{a_{\sigma(1)}...a_{\sigma(k)}}_{\hspace{0.2cm}b_{\lambda(1)}...b_{\lambda(l)}}\in\mathbb{F}, \end{align} where $$\mathbb{F}$$ is the field up on which the vector space $$V$$ is defined. It is obvious that the action of the alternating map on an already alternating tensor is equal to the alternating tensor itself. Thus, the alternating map just "picks out" the alternating part of a general tensor.

Finally, since $$\overline{\text{Alt}}(\tilde T)$$ is an alternating $$(k,l)$$-tensor it can be expanded in the standard bases of both $$\otimes^kV\otimes^lV^*$$ and $$\text{Alt}(\otimes^kV\otimes^lV^*)$$. This is the general procedure on how to find the alternating (or in your rank-2 case, antisymmetric) part of a general tensor and the sums over the possible permutations can be substituted with appropriate sums over generalised Levi-Civita symbols.

For example, $$\sum_{\substack{\sigma\in S_k\\\lambda\in S_l}}\text{sgn}(\sigma)\text{sgn}(\lambda)\tilde T^{a_{\sigma(1)}...a_{\sigma(k)}}_{\hspace{0.2cm}b_{\lambda(1)}...b_{\lambda(l)}}=\sum_{\substack{\sigma\in S_k\\\lambda\in S_l}}\epsilon_{a_{\sigma(1)}...a_{\sigma(k)}}\epsilon^{b_{\lambda(1)}...b_{\lambda(l)}}\bar T^{a_{\sigma(1)}...a_{\sigma(k)}}_{\hspace{0.2cm}b_{\lambda(1)}...b_{\lambda(l)}}.$$

I hope this helps.