- #1

AndersF

- 27

- 4

- 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

##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?

##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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