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In chapter $1$ page $44$ there's the following equation

$$ Vol(X_1(p),\ldots, X_n(p)) = \det(a_{ij}) = \sqrt{\det(g_{ij})}(p) (1) $$

Where $X_i(p) = \frac{\partial}{\partial x_i}(p)$, $i = 1 ,\ldots, n$; $X_i = \sum_{j} a_{ij} e_j$; $\left\{e_i, \ldots, e_j \right\}$ orthonormal basis for $T_pM$, where $M$ is a differentiable manifold and $T_p M$ tangent space at $p$, $g_{ij}(p) = \left\langle \frac{\partial}{\partial x_i}(p), \frac{\partial}{\partial x_j}(p) \right\rangle_{p}$

The question is where does the square root come from?

I'm sure this is a basic fact from linear algebra... I've tried to prove it by myself using the definition of the determinant

$$ \det(a_{ij}) = \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} \prod_{i=1}^n a_{i,\sigma(i)} $$

I didn't end up with nothing. I'm sure is something really simple though.

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By definition we have $$g_{ij}(p)=\left\langle X_i(p),X_j(p)\right\rangle = \left\langle \sum_k a_{ik}e_k,\sum_l a_{jl}e_l\right\rangle=\sum_k a_{ik}a_{jk},$$ so $(g_{ij}(p))=(a_{ij})\cdot (a_{ij})^T$. Now use multiplicativity of the determinant.

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  • $\begingroup$ As I said... it was simple... thank you! $\endgroup$ – user8469759 Jun 27 at 14:16

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