# Chapter 1 - Do Carmo's Riemennian Geometry Definition of volume

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.

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.