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.) $\left<v, w\right>_\overline{g}=\sum_{i}v^iw^i, v, w\in T_x\mathbb{R}^n$. 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?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ 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 $\sum_{i=1}^nv^iw^i$. $\endgroup$– peek-a-booJun 23 at 8:36
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
By definition, $$\mathrm{d}x^i(v) = v^i , \text{ and } \mathrm{d} x^i (w) = w^i$$
so that:
$$\overline{g}(v, w) = \sum_{i}(\mathrm{d}x^i \otimes \mathrm{d}x^i)(v, w) = \sum_{i} v^i w^i $$
