Is $\left|\frac{\det(A_{\mu\nu})}{\det(B_{\mu\nu})}\right|$ an invariant for two tensors $A_{\mu\nu}$ and $B_{\mu\nu}$ in a manifold?

Question

I already posted this question on mathoverflow.com. Yet I post it here too, since the community is bigger.

I was doing some math around the determinant of 2nd-order covariant tensors. In a general $n$-dimensional manifold, I deduced that the determinant of a tensor $A_{\mu\nu}$ can be defined as $$ \det(A_{\mu\nu})=\tilde\epsilon^{\mu_1\cdots \mu_n}\tilde\epsilon^{\nu_1\cdots \nu_n}A_{\mu_1\nu_1}\cdots A_{\mu_n\nu_n}, $$ based on a generalization of 2-dimensional tensor determinants, where $\tilde\epsilon^{\mu_1\cdots \mu_n}$ is the Levi-Civita symbol with $\tilde\epsilon^{1\cdots n}=1$ for a general metric.

Using the definition of the Levi-Civita symbol, I proved that the Levi-Civita tensor, if being antisymmetric w.r.t. any pair of its superscripts, must have the following form: $$ \epsilon^{\mu_1\cdots \mu_n}=s\cdot \tilde\epsilon^{\mu_1\cdots \mu_n}, $$ where $\epsilon^{\mu_1\cdots \mu_n}$ is the Levi-Civita tensor and $s$ is a (non-invariant) scalar. Using the tensor transformation law and assuming $s$ to be non-negative, I deduced that a sufficient condition for $\epsilon^{\mu_1\cdots \mu_n}$ to be a tensor, is to have $s=I\cdot \frac{1}{\sqrt{|\det(A_{\mu\nu})|}}$ for an arbitrary non-singular tensor $A_{\mu\nu}$ and arbitrary non-negative invariant $I$. My question originates from here.

If both $I\cdot \frac{1}{\sqrt{|\det(A_{\mu\nu})|}}\tilde\epsilon^{\mu_1\cdots \mu_n}$ and $J\cdot \frac{1}{\sqrt{|\det(B_{\mu\nu})|}}\tilde\epsilon^{\mu_1\cdots \mu_n}$ are tensors for arbitrary non-negative invariants $I$ and $J$ and non-singular tensors $A_{\mu\nu}$ and $B_{\mu\nu}$, does it mean that $\frac{|\det(A_{\mu\nu})|}{|\det(B_{\mu\nu})|}$ is an invariant? If not, where did I go wrong?

Edit

Here is my proof attempt (the tensor transformation law happens between two coordinates systems, $x^\mu$ and $\hat x^\mu$; a tensor $T$ in $x^\mu$-system has an equivalent for of $\hat T$ in $\hat x^\mu$-system):

$$ {\hat\epsilon^{\hat\mu_1\cdots\hat\mu_n}=\epsilon^{\mu_1\cdots \mu_n} \frac{\partial \hat x^{\hat\mu_1}}{\partial x^{\mu_1}}\cdots\frac{\partial \hat x^{\hat\mu_n}}{\partial x^{\mu_n}} \\\implies \hat s\cdot \hat{\tilde\epsilon}^{\hat\mu_1\cdots\hat\mu_n} = s\cdot \tilde\epsilon^{\mu_1\cdots \mu_n} \frac{\partial \hat x^{\hat\mu_1}}{\partial x^{\mu_1}}\cdots\frac{\partial \hat x^{\hat\mu_n}}{\partial x^{\mu_n}}. } $$ If we apply $\hat\epsilon^{\hat\nu_1\cdots\hat\nu_n}\hat T_{\hat \mu_1\hat\nu_1}\cdots \hat T_{\hat \mu_n\hat\nu_n}$ to the both sides of the latter, we obtain $$ { \hat s\cdot\hat{\tilde\epsilon}^{\hat\mu_1\cdots\hat\mu_n} \hat\epsilon^{\hat\nu_1\cdots\hat\nu_n}\hat T_{\hat \mu_1\hat\nu_1}\cdots\hat T_{\hat \mu_n\hat\nu_n} = s\cdot\tilde\epsilon^{\mu_1\cdots \mu_n} \hat\epsilon^{\hat\nu_1\cdots\hat\nu_n}\hat T_{\hat \mu_1\hat\nu_1}\cdots \hat T_{\hat \mu_n\hat\nu_n} \frac{\partial \hat x^{\hat\mu_1}}{\partial x^{\mu_1}}\cdots\frac{\partial \hat x^{\hat\mu_n}}{\partial x^{\mu_n}} \\\implies |\hat s|^2\cdot\det(\hat T) = |s|^2\cdot\tilde\epsilon^{\mu_1\cdots \mu_n} \tilde\epsilon^{\hat\nu_1\cdots\hat\nu_n} \frac{\partial \hat x^{\hat\nu_1}}{\partial x^{\nu_1}}\cdots\frac{\partial \hat x^{\hat\nu_n}}{\partial x^{\nu_n}} \hat T_{\hat \mu_1\hat\nu_1}\cdots \hat T_{\hat \mu_n\hat\nu_n} \frac{\partial \hat x^{\hat\mu_1}}{\partial x^{\mu_1}}\cdots\frac{\partial \hat x^{\hat\mu_n}}{\partial x^{\mu_n}} \\\implies |\hat s|^2\cdot\det(\hat T) = |s|^2\cdot\tilde\epsilon^{\mu_1\cdots \mu_n} \tilde\epsilon^{\hat\nu_1\cdots\hat\nu_n} T_{ \mu_1\nu_1}\cdots T_{\mu_n\nu_n} \\\implies |\hat s|^2\cdot\det(\hat T)=|s|^2\cdot\det(T), } $$ from which, I deduced that $|s|^2\cdot\det(T)$ must be an invariant, and so forth.

Answer 1

We will be using ricci calculus notation.

For a $(1,1)$-type tensor $W_{a}^{b}$, $\;$ the value $\;\det(W_{a}^{b})$ is an invariant tensor formula depending only on the dimension of the tensor, and not on the choice of coordinates.

Eg. for dimension $1$ it is $W_{a}^{a}$, the trace of the tensor.

For dimension $2$ it is $\;(W_{a}^{a} W_{b}^{b} - W_{a}^{b} W_{b}^{a})/2 = W^{a_1}_{[a_1}W^{a_2}_{a_2]}$ .

For an $n$-dimensional tensor the formula is: $W_{[a_1}^{a_1}W_{a_2\phantom{]}}^{a_2}\dots W_{a_n]}^{a_n}$
The determinant of an $n$-dimensional tensor of type $(1,1)$ is the contraction of the antisymmetric part of the tensor product of $n$ copies of itself.

As it uses only "valid" tensor operations: swapping of indices of the same type, tensor products, and contractions between indices of opposite types (covariant with contra-variant) it is coordinate-invariant.

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Now to prove $\frac{|\det(A_{\mu\nu})|}{|\det(B_{\mu\nu})|}$ is invariant.

In itself, the determinant of a type $(0,2)$-type tensor is not an invariant. However, the ratio of two such values is. Let $g^{ab}$ be an arbitrarily tensor of non-zero determinant (in Riemannian geometry we commonly use the contravariant metric tensor). We have:

$$\frac{|\det(A_{\mu\nu})|}{|\det(B_{\mu\nu})|} =\frac{|\det(A_{\mu\nu}) \det(g^{\nu\tau}) |}{|\det(B_{\mu\nu}) \det(g^{\nu\tau})|} = \frac{|\det(A_{\mu\nu}g^{\nu\tau}) |}{|\det(B_{\mu\nu} g^{\nu\tau})|} = \frac{|\det(A_{\mu}^{\tau})|}{|\det(B_{\mu}^{\tau})|} $$

The left-hand side shows that the quantity does not depend on choice of $g^{ab}$.
The right-hand side shows it is invariant to change of of coordinates, as, for a given fixed $g^{ab}$, it is the ratio of two coordinate-invariant quantities.

Answer 2

Without the indices of tensor notation — and without turning a bilinear form into a linear map by raising an index as in the accepted answer — note that a second order covariant tensor transforms by $\tilde A = PAP^\top$ when we change basis by $P$. Then $\det\tilde A = (\det P)^2\det A$. Thus, if we have two such tensors $A$ and $B$, we have $\det\tilde B = (\det P)^2\det B$ as well, and so $$\frac{\det \tilde A}{\det\tilde B} = \frac{\det A}{\det B},$$ as desired. No absolute values required.