Showing Determinant of Metric Tensor is a Tensor Density

[AndersF]

TL;DR

I'm trying to show that the determinant $g \equiv \det(g_{ij})$ of the metric tensor is a tensor density.

I'm trying to show that the determinant $g \equiv \det(g_{ij})$ of the metric tensor is a tensor density. Therefore, in order to do that, I need to show that the determinant of the metric tensor in the new basis, $g'$, would be given by

$g'=\operatorname{sgn}\bigg(\big(\det(C)\big)\bigg)\big(\det(C)\big)^wg \quad \quad \quad (1)$

With $C=(C^a_b)_{n \times n}$ the change-of-basis matrix.

I know that the metric tensor transforms under a change of basis in this way

$\tilde{g}_{i j}=C_{i}^{\alpha} C_{j}^{\beta} g_{\alpha \beta} \quad \quad \quad (2)$

I see that if I could identify in this last equation (2) a matrix multiplication, then I could use the properties of the determinants to get something similar to equation (1). But I'm stuck here, since these terms don't have the form of "classical" matrix multiplications, $P^i_j=M^i_k N^k_j$.

Could somebody give me a hint on how to accomplish this demonstration?

 

[vanhees71]

In matrix notation you can write $\mathrm{det} \tilde{g}=\mathrm{det}(C^{\text{T}} g C)=(\mathrm{det} C)^2 \mathrm{det} g$. So $g_{\mu \nu}$ is a tensor density of weight 2.

 

[haushofer]

To add: in linear algebra one has the cyclic identity det(ABC)=det(BCA)=det(CAB), and det(A^T)=det(A) ;)

Edit: typo corrected, thnx robphy!

 

[robphy]

To add: in linear algebra one has the cyclic identity det(ABC)=det(BCA)=det(CBA), and det(A^T)=det(A) ;)

Typo: The third expression in this cyclic identity should be det(CAB).

 

[AndersF]

In matrix notation you can write $\mathrm{det} \tilde{g}=\mathrm{det}(C^{\text{T}} g C)=(\mathrm{det} C)^2 \mathrm{det} g$. So $g_{\mu \nu}$ is a tensor density of weight 2.

To add: in linear algebra one has the cyclic identity det(ABC)=det(BCA)=det(CAB), and det(A^T)=det(A) ;)

Edit: typo corrected, thnx robphy!

Ok, these were just the "tricks" I was looking for, thank you very much!