# Connection between the Euclidean metric $\overline{g}=\sum_{i}dx^idx^i$ and $\left<v, w\right>_\overline{g}=\sum_{i}v^iw^i, v, w\in T_x\mathbb{R}^n$

This is a quite elementary question, but I am aware that there are two equivalent ways of characterizing the Euclidean metric in a Riemannian manifold: 1.) $$\overline{g}=\sum_{i}dx^idx^i$$ and 2.) $$\left_\overline{g}=\sum_{i}v^iw^i, v, w\in T_x\mathbb{R}^n$$. What I have trouble with is making the connection between the two. That is, if $$v = \sum_iv^i\partial_i|_x, w = \sum_iw^i\partial_i|_x$$, then how do we operate on $$v$$ and $$w$$ with $$\overline{g}=\sum_{i}dx^idx^i$$ such that we obtain the second characterization?

• By $dx^i\,dx^i$ I assume you mean $dx^i\otimes dx^i$ (or the 'symmetrized tensor product' which here reduces to the same thing $\frac{dx^i\otimes dx^i+dx^i\otimes dx^i}{2}=dx^i\otimes dx^i$). In this case, you just evaluate $g(v,w)$, by unwinding the definition of the tensor product, and see it equals $\sum_{i=1}^nv^iw^i$. Jun 23 at 8:36

By definition, $$\mathrm{d}x^i(v) = v^i , \text{ and } \mathrm{d} x^i (w) = w^i$$
$$\overline{g}(v, w) = \sum_{i}(\mathrm{d}x^i \otimes \mathrm{d}x^i)(v, w) = \sum_{i} v^i w^i$$