I'm reading Gallot-Hulin-Lafontaine, and in section 2.7 they say they following:

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I wanted to check that the second $v_g,$ given in a local oriented chart, satisfied the first property. So I took an oriented basis $(e_1,\ldots,e_n)$ satisfying $g(e_i,e_j) = \delta^i_j,$ compared it to the frame $(\partial_1,\ldots,\partial_n)$ (the coordinate vector fields in the $x^i$ coordinates) by writing $e_j = a^i_j \partial_i$ (using summation notation) for some invertible matrix $(a^i_j)$, and computed

$$ v_g(e_1,\ldots,e_n) = v_g(a^i_1 \partial_i, \ldots, a^i_n \partial_i) = \sqrt{\det(g_{i,j})} a^1_1 \cdots a^n_n. $$

This is supposed to equal 1, but I don't see how it follows from the relations

$$\delta^i_j = g(e_i,e_j) = a^k_i a^l_j g_{k,l}.$$

I think I made an error somewhere, but I don't know where. Thanks!


1 Answer 1


The problem is your initial calculation. You have $v_g=\lambda\,dx^1\wedge\cdots\wedge dx^n$ for some function $\lambda$. Then $$1=v_g(e_1,\dots,e_n)=\lambda\,dx^1\wedge\cdots\wedge dx^n\big(a^i_1\partial_i,\dots,a^i_n\partial_i\big)=\lambda\,\det(a^i_j)$$ The equation $\delta_{i,j}=g(e_i,e_j)=a_i^ka_j^lg_{kl}$ tells you that $AGA^t=I_n$ where $A=(a_i^j)$ and $G=(g_{ij})$. Since the charts are oriented, we get that $\det(A)$ is positive, and so from $\det(G)\det(A)^{2}=1$ it follows that $\det(A)=\det(G)^{-\frac12}$, and finally, $\lambda\det(A)=1$ implies $\lambda=\sqrt{\det(g_{ij})}$


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