Showing that the determinant of the metric tensor is a tensor density

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AndersF
TL;DR Summary
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

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

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?

Last edited:

Gold Member
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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.

AndersF
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!

Last edited:
vanhees71 and AndersF
Homework Helper
Gold Member
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).

vanhees71, haushofer and AndersF
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!

vanhees71