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Question 1: Does the metric define the inner product or does the inner product define the metric? I have $$(u,v) := \delta_{ij} u^i v^j$$ and I don't know if the $\delta$ is the metric or just the delta function. In the definition of the metric tensor one also uses the inner product but in that case it is always $$(u,v) := u^i v^i$$ So I guess the metric defines the inner product. Is that correct?

Question 2: In the definition of an isometry, $\phi$ we have $$(u,v)_{g_1} = (\phi(u), \phi(v))_{g_2}$$ What does this notation mean? Simpily $$ g_{1_{ij}} u^i v^i = g_{2_{ij}} \phi(u)^i \phi(u)^j $$ or something else?

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  • $\begingroup$ An inner product is an "example" of a metric. Not all metrics are inner products, since for example positive definiteness may fail (for example, the Minkowski metric). I prefer to think of a metric as being a parametrized "inner product", where the parametrization space is, say, your manifold $M$. The "inner product" is on the fiber of some vector bundle $E \to M$. $\endgroup$ – Rellek Jun 21 '18 at 12:12
  • $\begingroup$ Ponder this as well: Every inner product induces a norm induces a metric induces a topology. $\endgroup$ – Jo Mo Jun 21 '18 at 13:26
  • $\begingroup$ Also, try to prove that if $\phi$ is an isometry then it is injective. $\endgroup$ – Jo Mo Jun 21 '18 at 13:29
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If $c:(I,|\ |)\rightarrow (X,d)$ is a continous curve in metric space, then define a length of $c$ to be $$ \lim_n\ \sum_{i=1}^n\ d(c(t_i),c(t_{i+1})) $$ where $t_i$ is a partition on $I$.

Here if $p=c(0),\ q=c(1)$, then infimum of lengths of curves between $p$ and $q$ is intrinsic metric. One of example of intrinsic metric is norm on $\mathbb{R}^n$. Here if norm satisfies $\| v-w\|^2 + \|v+w\|^2 =2(\|v\|^2+\|w\|^2)$, then the norm is in fact inner product $g$. Further, in inner product space we can define a length of curve as followes : $$ {\rm length}\ c =\int_0^1\ g(c'(t),c'(t))^\frac{1}{2}\ dt \ \ast$$

Consider inner product space $(X,g)$. From $\ast$, we have a metric $d_g$ which is a infimum of length of curves. If $f :(X,g)\rightarrow (X',g')$ is bijective isometry, then $f$ is isometry between $d_g,\ d_{g'}$, i.e. $$ d_g(x,y)=d_{g'}(f(x),f(y))$$

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