I can't verify a relationship between cofactor and determinant

[Kisok]

TL;DR Summary

In "General Relativity" by Hobson, et. al., in p. 66, we find a sentence starting "If we denote the value of the determinant ... " in the lower part of the page. I can not verify this assertion. Please help me the verification.

On that sentence, cofactor of an element of a metric is derived. But I can not verify it. Here I attached the copy of the page.

 

[stevendaryl]

I think it's a typo. You can expand a determinant using cofactors as described here: https://en.wikipedia.org/wiki/Minor_(linear_algebra)#Cofactor_expansion_of_the_determinant

The key facts are that if $C_{ij}$ is the cofactor of $g_{ij}$, then

1. $\sum_{j} g_{ij} C_{jk} = 0$, if $k \neq i$
2. $\sum_{j} g_{ij} C_{ji} = det(g)$

Those two facts together imply that $C_{jk} = det(g) g^{jk}$

(where $g^{jk}$ is the inverse of the metric)

 

[TeethWhitener]

It’s because the inverse of $g_{\mu\nu}$ is $g^{\mu\nu}$, and the inverse of a matrix is given by the matrix of cofactors divided by the determinant.

 

[Cryo]

Rund and Lovelock ("Tensors, Differential Forms, and Variational Principles") have a good section on this in chapter 4

I hope you are familiar with Levi-Civita pseudo-tensors (https://en.wikipedia.org/wiki/Levi-Civita_symbol)

The determinant of the n-by-n matrix $g_{\alpha\beta}$ is given by:

$g = \epsilon^{\mu_1,\dots,\mu_n}g_{1\mu_1}\dots g_{n\mu_n}$

Now consider the following object:

$\omega^{\alpha\beta}=\sum_s\delta^\beta_s \epsilon^{\mu_1,\dots,\mu_{s-1},\alpha,\mu_{s+1},\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}g_{s+1,\mu_{s+1}}\dots g_{n\mu_n}$

$g_{\kappa,\alpha}\omega^{\alpha\beta}=\sum_s \delta^\beta_s \epsilon^{\mu_1,\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}\: g_{\kappa,\mu_s} \:g_{s+1,\mu_{s+1}}\dots g_{n\mu_n}$

Due to anti-symmetry of Levi-Civita, the only non-zero term in the $\sum_s$ is the one with $s=\kappa$, but thet term is only non-zero if $s=\beta=\kappa$. Now if all of this works, then we get the earlier expression for the determinant, so:

$g_{\kappa,\alpha}\omega^{\alpha\beta}=g\delta^\beta_\kappa$

So $\omega^{\alpha\beta}=g g^{\alpha\beta}$, the inverse times the determinant. The co-factor matrix is here somewhere, but you don't need it to proceed. Note that from the above definition of the determinant

$\partial_c g = \partial_c g_{\beta \alpha}\sum_s\delta^\beta_s \epsilon^{\mu_1,\dots,\mu_{s-1},\alpha,\mu_{s+1},\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}g_{s+1,\mu_{s+1}}\dots g_{n\mu_n} = g g^{\beta\alpha} \partial_c g_{\alpha\beta}$

Which is what you were after (once we use $g_{\alpha\beta}=g_{\beta\alpha}$)

 

[Kisok]

Thank you for all the replies there. I will study the precious information and suggestions given by all of you.

 

[matrixbud]

gαβ

Rund and Lovelock ("Tensors, Differential Forms, and Variational Principles") have a good section on this in chapter 4

I hope you are familiar with Levi-Civita pseudo-tensors (https://en.wikipedia.org/wiki/Levi-Civita_symbol)

The determinant of the n-by-n matrix $g_{\alpha\beta}$ is given by:

$g = \epsilon^{\mu_1,\dots,\mu_n}g_{1\mu_1}\dots g_{n\mu_n}$

Now consider the following object:

$\omega^{\alpha\beta}=\sum_s\delta^\beta_s \epsilon^{\mu_1,\dots,\mu_{s-1},\alpha,\mu_{s+1},\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}g_{s+1,\mu_{s+1}}\dots g_{n\mu_n}$

$g_{\kappa,\alpha}\omega^{\alpha\beta}=\sum_s \delta^\beta_s \epsilon^{\mu_1,\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}\: g_{\kappa,\mu_s} \:g_{s+1,\mu_{s+1}}\dots g_{n\mu_n}$

Due to anti-symmetry of Levi-Civita, the only non-zero term in the $\sum_s$ is the one with $s=\kappa$, but thet term is only non-zero if $s=\beta=\kappa$. Now if all of this works, then we get the earlier expression for the determinant, so:

$g_{\kappa,\alpha}\omega^{\alpha\beta}=g\delta^\beta_\kappa$

So $\omega^{\alpha\beta}=g g^{\alpha\beta}$, the inverse times the determinant. The co-factor matrix is here somewhere, but you don't need it to proceed. Note that from the above definition of the determinant

$\partial_c g = \partial_c g_{\beta \alpha}\sum_s\delta^\beta_s \epsilon^{\mu_1,\dots,\mu_{s-1},\alpha,\mu_{s+1},\dots,\mu_n}g_{1\mu_1}\dots g_{s-1,\mu_{s-1}}g_{s+1,\mu_{s+1}}\dots g_{n\mu_n} = g g^{\beta\alpha} \partial_c g_{\alpha\beta}$

Which is what you were after (once we use $g_{\alpha\beta}=g_{\beta\alpha}$)

Thank you so much. Not being able to prove this was driving me crazy.

Oh, and I believe that $\omega^{\alpha\beta}$ is the co-factor. (I am new to this forum and may not have yet figured out how to express omega to the alpha, beta correctly.

 

[PeroK]

(I am new to this forum and may not have yet figured out how to express omega to the alpha, beta correctly.

You have figured it out!