Suppose that $M$ is a Riemannian manifold $M$ in $\mathbb{R}^2$ and we are interested in what kind of metric does $g = e^xdx^2 + dy^2$ induce. Assuming that $g$ indeed is a Riemannian metric, should $g$'s precise definition be $g_p = e^{p_1}dx^2 + dy^2$ (where $p_k$ is the $k$th component of $p\in M$)? I have seen similarish objects as $g$ in couple of posts on this site and even in some textbooks, but I have trouble understanding how such a $g$ acts on the points, or point pairs, of our manifold $M$.
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3A metric does not act on points of the manifold. It provides at every point $p$ of $M$ a scalar product for the tangent space $T_pM,.$ Put this together with the fact that $dx^2=dx\otimes dx$ and $dy^2=dy\otimes dy$ are bilinear forms on $T_pM$ and you are done. – Kurt G. May 30 '23 at 16:14
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@KurtG. I see. But was I correct about the replacing $x$ by $p_1$ in $e^x$ of $e^xdx^2$? – Cartesian Bear May 31 '23 at 07:08
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This was correct. However, $g_p=e^{p_1},dx^2+dy^2$ is not an inch more precise than $g=e^{x},dx^2+dy^2,.$ – Kurt G. May 31 '23 at 08:21