Local representation of a riemannian metric (change of coordinates) Let's say I have a manifold $M$ and an atlas $\mathcal{A} = \{ (U,\varphi), (V, \phi)\}$ of $M$. I am asked to show that there is a unique riemannian metric $g$ such that its representations in the coordinates given by $\varphi$ is something of the like
$$g^{\varphi} = g_{ij}^{\varphi}\mathop{dx^i}\mathop{dx^j}$$
and to find the local representation in the coordinates given by $\phi$.
I'm not sure I completely understand the theory behind and I'd like to know whether what I was doing was okay. 
For the first part I really don't know how to prove the existence and uniqueness part, really. Some clue would be highly appreciated. 
For the other part, as I understand, I need to compute 
$$g^{\phi} = (D h)^t (g_{ij}^{\varphi})(Dh)$$
where $h = \varphi \circ \phi^{-1}$. Is this correct? Thank you.
 A: To give you some help...when you have a smooth manifold, you usually ask it to be T$^{2}$ and have a countable open covering. This is somewhat technical but it permits you to "glue" constructions you do in a chart, as long as such construction coincides in intersections of charts.
In that way, using a chart you may define a metric on the coordinate open set, through the metric of $R^n $(say dim M = n). That is, the metric on rn is inherited by the open set on the manifold, since they are diffeomorphic.
Then, as said bebore, you glue them all together.(they do coincide on an intersection)
Another thing you may see is that in general, given a basis of a vector space, say {$v_1$, $v_2$,...$v_n$} and a metric on that space, there is a matrix A representing the metric on such basis, so that g(X;Y) = $X^t$ A Y
where A = ($a_{ij}$) is indeed $a_{ij} $= g($v_i$,$v_j$)
(your $g_{ij}$)
But I would say that instead of d$x_i$, you should put $ \partial{}/\partial{x}_{i}$, since it is the basis of a tangent space you are considering. (the vector space)
