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I have two quesitons about a book by Nakahara: Geometry,topology and physics In the snippet below how do I compute that $$|\det(\frac{\partial x^\mu}{\partial y^\kappa}\frac{\partial x^\nu}{\partial y^\lambda}g_{\mu\nu})|^{\frac{1}{2}}dy^1\wedge...\wedge dy^m$$ equals to $$=|\det (\partial x^\mu/\partial y^\kappa)|\sqrt{|g|}\det(\partial y^\lambda/\partial x^\nu)dx^1...dx^m=\pm\sqrt{|g|}dx^1...dx^m,$$ by inserting $dy^\lambda=\frac{\partial y^\lambda}{\partial x^\mu}dx^\mu$.

Here, both euqalities are unclear to me.

EDIT I think now that I understand the second =.

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Related may be form the same book below $(8.81)$ why:

$$\nabla \Omega=\frac{1}{2}\partial_\lambda\Omega_{\mu\nu}dx^\lambda\wedge dx^\mu\wedge dx^\nu=d\Omega.$$

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user122424
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1 Answers1

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You need know that $\det(AB) = \det A \det B$. In addition you need to know that $$ dy^{1}\wedge \ldots\wedge dy^m = \frac{\partial y^1}{\partial x^{k_1}}dx^{k_1} \wedge \ldots \wedge \frac{\partial y^m}{\partial x^{k_m}}dx^{k_m} = \det\left(\frac{\partial y^\mu}{\partial x^{\nu}} \right) dx^1\wedge\ldots\wedge dx^m$$ In this last statement the first equality comes from inserting the relation between $dx^i$ and $dy^k$, while the second one is a known fact about the alternating product

Now note that $$\det\left(\frac{\partial y^\mu}{\partial x^{\kappa}} \frac{\partial y^\nu}{\partial x^{\lambda}}g_{\mu\nu}\right) $$

is just the determinant of the product of the three matrices with components $\left(\frac{\partial y^\mu}{\partial x^{\kappa}}\right)$, $\left(\frac{\partial y^\nu}{\partial x^{\lambda}}\right)$ and $(g_{\mu\nu})$, so it's the product of the determinants of the individual matrices. Just write down what this means (don't forget to write down the square root, also).

Then note that the matrix with components $\left(\frac{\partial y^\mu}{\partial x^{\kappa}}\right)$ is the inverse of the matrix with components $\left(\frac{\partial x^\sigma}{\partial y^{\tau}}\right)$ (by the chain rule), so that $$ |\det\left( \frac{\partial y^\mu}{\partial x^{\kappa}} \right)|\det\left( \frac{\partial x^\mu}{\partial y^{\kappa}} \right)= \pm 1$$

(I hope I did not produce too many typos with all the indices).

Thomas
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