# Differentiation of the determinant $g$

Let $$g$$ be the determinant of the metric tensor.

I want to derive the following equation $$g_{,\nu}=gg^{\lambda \mu}g_{\lambda \mu,\nu}$$. It is said that $$gg^{\lambda \mu}$$ is a cofactor, but I can't understand why. To begin with, I'm not familiar with how to express the determinant of the metric tensor i.e. $$g$$. I know that $$g$$, the determinant is the sum of the cofactors multiplied with corresponding matrix element.

Can I get some illuminations on this?

• I don't have time to check myself, but usually this always come into play with identities like these: en.wikipedia.org/wiki/Jacobi%27s_formula – JamalS Oct 17 '19 at 0:23
• I think the matrix identity $\ln{\det{M}}=\text{tr}\ln{M}$ is the key. – G. Smith Oct 17 '19 at 0:24
• @G.Smith It's taking me some time to understand. Can you elaborate? – Nuri Oct 17 '19 at 13:28
• @JamalS I have followed your link, and with that I've written in my own way below. – Nuri Oct 17 '19 at 13:29
• @YeonwookJung Differentiating gives $(\det{M})^{-1}\partial\det{M}=\text{tr}M^{-1}\partial M$. Now let $M$ be the metric tensor, and write this using index notation. – G. Smith Oct 17 '19 at 16:37

Let $$A$$ be a $$n\times n$$ matrix.
Recall that $$A \text{adj}(A)=\det(A)\cdot1$$ where $$1$$ is the identity matrix, $$\text{adj}(A)=C^T$$ is the adjoint matrix, $$C_{ij}=(-1)^{i+j}M_{ij}$$ is the cofactor matrix, $$M_{ij}$$ is the minor of the $$A_{ij}$$. This is justified because $$\det{A}=\sum_j A_{ij}C_{ij}=\sum_j A_{ij}adj_{ji}=(A\text{adj}(A))_{ii}.$$
Now consider $$d(\det(A))=\sum_{i,j}\frac{\partial\det(A)}{\partial A_{ij}}dA_{ij}$$ where $$\frac{\partial\det(A)}{\partial A_{ij}}=\frac{\partial\sum_k A_{ik} \text{adj}_{ki}}{\partial A_{ij}}=\sum_k\frac{\partial A_{ik}\text{adj}_{ki}}{\partial A_{ij}}=\sum_k \frac{\partial A_{ik}}{\partial A_{ij}}\text{adj}_{ki}+A_{ik}\frac{\partial \text{adj}_{ki}}{\partial A_{ij}}=\sum_k \delta_{jk}\text{adj}_{ki}=\text{adj}_{ji}$$ since $$\text{adj}_{ji}(A)=C_{ij}$$ is independent of $$A_{ij}$$ as it is the determinant of the matrix A with $$i^{th}$$ row and $$j^{th}$$ column removed. Thus we have $$d(\det(A))=\sum_{i,j}\text{adj}_{ji} dA_{ij}=\sum_{i,j} \text{adj}_{ij} A_{ji} = tr(\text{adj(A)} dA)=tr(\det(A)A^{-1}dA).$$ Finally putting $$(A)=(g)$$, we conclude $$g_{,\nu}=gg^{\mu\lambda}g_{\mu\lambda ,\nu}.$$