Prove basis independence of determinant with Levi-Civita symbol

Question

Let $V$ be an $N$-dimensional vector space over $\mathbb{C}$, and $A$ a linear operator on it. Choose a basis $\{e_i\}$ of $V$, under which $A$ is represented by the matrix ${A^i}_j$ (upper/lower indices transform contravariantly/covariantly with basis changes), i.e. the image of $e_j$ is

$$ A e_j = e_i {A^i}_j $$

We define the determinant of $A$ as

$$ \det A = \frac{1}{N!} \epsilon_{i_1 ... i_N} \epsilon^{j_1 ... j_N} {A^{i_1}}_{j_1} \cdots {A^{i_N}}_{j_N} $$

Here both $\epsilon_{i_1 ... i_N}$ (with all lower indices) and $\epsilon^{j_1 ... j_N}, $ (with all upper indices) are Levi-Civita symbols. But these "symbols" does not change with basis changes. The index positions of $\epsilon$ does not have special meanings; they are written simply to comply with tensor analysis (Ricci calculus) rules: we sum over paired upper/lower indices.

In usual tensor analysis, if an expression does not contain unpaired (free) upper/lower indices, then it is invariant under a basis change. But here the Levi-Civita symbols are, strictly speaking, not tensors (it only behaves like a tensor under orthogonal basis changes, see this Math.SE question). How to start from this definition of $\det A$ to show that it is basis independent?

As shown by Sal in his answer to this question, you can make the Levi-Civita symbol tensorial by adjusting it with a prefactor. Letting $\epsilon$ be the Levi-Civita symbol, you can define the Levi-Civita tensor $\varepsilon$ as $$\varepsilon_{i_1\dots i_n}=\sqrt{|g|}\epsilon_{i_1\dots i_n}$$ Where $g$ is the metric. — K.defaoite, Oct 19 '23 at 15:56
@K.defaoite What if the vector space $V$ does not have a metric? — Zhengyuan Yue, Oct 19 '23 at 16:06
Correction: $Ae_j=A^i{}_ke^k_j$, not $e_iA^i{}_j$. Note that $e_iA^i{}_j=e_1A^1{}_j+\dots +e_nA^n{}_j$ which is not the same thing as $Ae_j$. — K.defaoite, Oct 19 '23 at 19:19
@K.defaoite Indices on LHS in your correction do not match RHS. My $A e_j$ is actually $A(e_j)$. — Zhengyuan Yue, Oct 20 '23 at 00:18

score 0 · Answer 1 · answered Dec 01 '24 at 00:44

Although I'm a bit late, and this is my first answer, the product of two permutation symbols is in fact an absolute tensor. Recall that the Levi-Civita symbol (with subscripts) is a tensor density of weight 1, while the Levi-Civita symbol (with superscripts) a tensor density of weight -1. Thus their tensor product is a tensor density of rank $(n,n)$ and weight $w=+1-1=0$, i.e. it's an absolute tensor. Here, n are the dimensions and the number of indices of each permutation symbol. Since I just figured out you can write equations: \begin{equation} \tag{1} \tilde{\epsilon}_{i'_{1} i'_{2} \dots i'_{n}}= \frac{\partial x^{i_{1}}}{\partial \tilde{x}^{i'_{1}}} \frac{\partial x^{i_{2}}}{\partial \tilde{x}^{i'_{2}}} \dots \frac{\partial x^{i_{n}}}{\partial \tilde{x}^{i'_{n}}} \epsilon_{i_{1} i_{2} \dots i_{n}} D \end{equation} D is the Jacobi determinant.Also \begin{equation} \tag{2} \tilde{\epsilon}^{i'_{1} i'_{2} \dots i'_{n}}= \frac{\partial \tilde{x}^{i'_{1}}}{\partial x^{i_{1}}} \frac{\partial \tilde{x}^{i'_{2}}}{\partial x^{i_{2}}} \dots \frac{\partial \tilde{x}^{i'_{n}}}{\partial x^{i_{n}}} \epsilon^{i_{1} i_{2} \dots i_{n}} \frac{1}{D} \end{equation} Their product is: \begin{equation} \tag{3} \tilde{\epsilon}^{i'_{1} i'_{2} \dots i'_{n}} \tilde{\epsilon}_{j'_{1} j'_{2} \dots j'_{n}}= \frac{\partial \tilde{x}^{i'_{1}}}{\partial x^{i_{1}}} \frac{\partial \tilde{x}^{i'_{2}}}{\partial x^{i_{2}}} \dots \frac{\partial \tilde{x}^{i'_{n}}}{\partial x^{i_{n}}} \frac{\partial x^{j_{1}}}{\partial \tilde{x}^{j'_{1}}} \frac{\partial x^{j_{2}}}{\partial \tilde{x}^{j'_{2}}} \dots \frac{\partial x^{j_{n}}}{\partial \tilde{x}^{j'_{n}}} \epsilon_{j_{1} j_{2} \dots j_{n}} \epsilon^{i_{1} i_{2} \dots i_{n}} \frac{D}{D} \end{equation} which is the transformation rule for a rank $(n,n)$ tensor. Furthermore, the metric tensor isn't mentioned here. Recall that we first define tensor densities using the Jacobi determinant and then, usually we write the Jacobi determinant in terms of the determinant of the metric tensor. So, I'm assuming that we can still define a Levi-Civita tensor, even if we don't equip our vector space with an inner product.

Prove basis independence of determinant with Levi-Civita symbol

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