# Determinant as Skew-symmetric n-linear form

I'm reading Mathematics for Physics by Goldbart section on Skew-symmetric n-linear form and I'm not able to understand a particular step.

We can think of determinant as being skew-symmetric n-linear form i.e. map $$\omega : V\times V\times \cdots V\rightarrow \mathbb{F}$$ We have the following property:

$$\omega(\lambda \mathbf{a}+\mu \mathbf{b},\mathbf{c}_2,\cdots ,\mathbf{c}_n)=\lambda \omega(\mathbf{a},\mathbf{c}_2,\cdots ,\mathbf{c}_n)+\mu\omega(\mathbf{b},\mathbf{c}_2,\cdots ,\mathbf{c}_n)$$ $$\omega(\cdots \mathbf{a}_i\cdots \mathbf{a}_j\cdots)=-\omega(\cdots \mathbf{a}_j\cdots \mathbf{a}_i\cdots)$$

$$\wedge^n(V^*)$$: Space of skew-symmetric $$n$$-linear form of $$V$$.

Let $$\{\mathbf{e}_i\}$$ is basis of $$V$$. If $$\mathbf{a}_i=a_{ij}\mathbf{e}_j$$ is set of $$n$$ vector. We compute

$$\omega(\mathbf{a}_1,\cdots ,\mathbf{a}_2)=a_{1i_1}a_{2i_2}\cdots a_{ni_n}\omega(\mathbf{e}_{i_1},\cdots \mathbf{e}_{i_2},\cdots,\mathbf{e}_{i_n})$$ $$=a_{1i_1}a_{2i_2}\cdots a_{ni_n}\epsilon_{i_1,i_2,\cdots ,i_n}\omega(\mathbf{e}_1,\cdots,\mathbf{e}_n)$$

I don't understand How the last two lines came about? What's the difference between $$\mathbf{e}_{i_j}$$ and $$\mathbf{e}_j$$? How do they go from first to second? What is the expansion? Is it $$\mathbf{a}_1=a_{1i_1}\mathbf{e}_{i_1}$$ Or $$\mathbf{a}_1=a_{1i}\mathbf{e}_i$$?

• They hide the summation notation, do you understand that? Sep 7, 2021 at 13:58
• I understand the Einstein summation convention but not how they write the first and second lines. Sep 8, 2021 at 7:41

## 1 Answer

They are essentially using $$\mathbf{a_i} = a_{ij} \mathbf e_j$$ carefully. If you plug in the expression into $$\omega(\mathbf{a}_1,\cdots ,\mathbf{a}_n)$$, you get

$$\omega(\mathbf{a}_1,\cdots ,\mathbf{a}_n) = \omega (a_{1j} \mathbf e_j, \cdots a_{nj} \mathbf e_j),$$

but this is a very bad notation, since you used the same $$j$$ to $$n$$ different summation. To avoid this, you instead write $$\mathbf a_j= a_{j i_j} \mathbf e_{i_j}$$ and thus you get

$$\omega(\mathbf{a}_1,\cdots ,\mathbf{a}_n) = \omega (a_{1i_1} \mathbf e_{i_1}, \cdots a_{ni_n} \mathbf e_{i_n})= a_{1i_1} \cdots a_{n i_n}\ \omega (e_{i_1}, \cdots, e_{i_n}).$$

For the second equality, it is more or less the definition of $$\epsilon$$, which is $$\epsilon_{i_1\cdots i_n} = \begin{cases} 0 & \text{ if } i_j = i_k \text{ for some } j\neq k,\\ (-1)^{|\sigma|} &\text{ otherwise}. \end{cases},$$ where $$\sigma$$ is the permutation which send $$j$$ to $$i_j$$.