Why is a tensor product used in local expression of Riemannian metric?

Question

$$g = g_{ij} \, du^i \otimes du^j $$ Why is a tensor product used here. Please explain.

Answer 1

The reason why is that the Riemannian metric is in fact a symmetric 2-tensor. Let $V$ be an $n$-dimensional vector space and consider the set of symmetric $2$-tensors $\Sigma^2(V)$ over $V$ (at least recall that the $2$-tensors are multilinear mappings $du^i\otimes du^j:V\times V \to \mathbb{R}$ with the property that $du^i\otimes du^j(v,w) = du^i(v) \, du^j(w)$; this is symmetric if you can interchange $v$ and $w$ without changing the value of the tensor product).

Now given the tangent space $T_pM$ at the point $p \in M$ we can consider the symmetric 2-tensors over this tangent space as $\Sigma^2(T_pM)$, and lastly we can "glue" these together to have the symmetric 2-tensor bundle over $M$:

$$ \Sigma^2(M) \;\; =\;\; \bigsqcup_{p \in M} \Sigma^2(T_pM) $$

where the square cup product indicates the smooth manifold structure endowed with the disjoint union topology. We can restrict our attention to those 2-tensors which are positive-definite and denote this tensor bundle as $\Sigma_+^2(M)$. Note that we have a canonical projection $\pi:\Sigma_+^2(M) \to M$ given by $\pi(p, \sigma_p) = p$. This admits smooth right inverses (though it doesn't admit a unique inverse) which we can write as maps of the form $g:M \to \Sigma_+^2(M)$ where we have $\pi \circ g = Id_M$. These maps $g$ can serve as Riemannian metrics, and therefore can be locally expressed as symmetric 2-tensors.