I Showing Determinant of Metric Tensor is a Tensor Density

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The discussion focuses on demonstrating that the determinant of the metric tensor, denoted as g, is a tensor density. The transformation of the metric tensor under a change of basis is expressed, and the participant seeks to relate this transformation to the determinant properties. A key equation is provided, showing how the determinant of the transformed metric tensor can be expressed in terms of the change-of-basis matrix C. The conversation highlights the importance of matrix multiplication properties and cyclic identities in linear algebra for this demonstration. The participant expresses gratitude for the hints that clarify their approach.
AndersF
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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?
 
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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.
 
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!
 
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haushofer said:
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).
 
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vanhees71 said:
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 said:
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!
 
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