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Any smooth manifold $M$ is locally diffeomorphic to an open set in $\mathbb{R}^N$. So the tangent space at each point $p \in M$ is also isomorphic to $R^N$ where $p$ is mapped into the origin.

So we have a orthonormal basis in $R^N$ if we take the inverse image of the same, we get a frame in $T_p(M)$ for which the metric can be just the Euclidean metric $g_E = (\delta_{ij})$ which is kind of the simplest metric we can handle. I believe it is also smoothly defined over the manifold.

So why are we worried about the metric in general ("A Riemannian manifold is a manifold along with a metric") ?

I don't understand how studying different metrics help us ? Probably I am not understanding some aspect of metric.

Sub-question : Is it possible to equip $\mathbb{R}^N$ with non-Euclidean metric so that it becomes a sphere $\mathbb{S}^N$ ? Does this question at all make sense ?

Any help will be appreciated. Thanks in advance.

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"I believe it is also smoothly defined over the manifold" this fails in general, because the metric you define on each chart domain by pulling back the metric on $\Bbb R^n$ will not paste together nicely. It will be different than that obtained by the same procedure using another chart (where they may differ : on the overlap of the chart domains.) –  Olivier Bégassat Jan 21 at 10:42
$\Bbb R^N$ and $S^N$ aren't homeomorphic or even homotopy equivalent, so the last part will fail. However, $\Bbb R^N$ and $S^N$ minus a point are homeomorphic via stereographic projection. –  Olivier Bégassat Jan 21 at 10:45
But the tangent space is changing when I change the point ? And the metric is kind of constant, is it not smooth then ? How does overlap cause any problem ? –  DiffeoR Jan 21 at 10:47
You can always define some metric on an arbitrary manifold, by a standard partitions of unity argument. But different metrics are … different! They produce different connections, for example. Without a connection, it is hard to differentiate vector fields in a meaningful fashion. There is a whole universe of mathematics on manifolds that requires a metric. –  Harald Hanche-Olsen Jan 21 at 10:49
@DiffeoR Just look at the easiest case : take $M=\Bbb R^n$ and consider the two charts $\phi:M\to\Bbb R^n, x\mapsto x$ and $\psi:M\to\Bbb R^n, x\mapsto 2x$. If you pull back the standard metric of $\Bbb R^n$ with these two charts you don't get the same metric on $M$! $\phi$ gives the standard metric $g$, while $\psi$ gives $\frac12 g$. This is the simplest case conceivable, but imagine how different the two metrics may be if $M$ is some general manifold and the charts are more complicated (in particular non-linear.) –  Olivier Bégassat Jan 21 at 12:41

1 Answer 1

up vote 5 down vote accepted

I think, I understand the cause of the confusion. Indeed, every smooth $n$-manifold $M$ is locally diffeomorphic to $R^n$. Hence, every point in $M$ has a (small) neighborhood which admits a flat, i.e., a metric which is isometric to the Euclidean Riemannian metric on (a domain in) $R^n$. Such metrics can be locally characterized y the property that they have zero sectional curvature. (This was Riemann's main discovery.) However, such flat metrics on open subsets of $M$ cannot be (in general) patched together to form a single (flat) Riemannian metric on $M$. The simplest obstruction for this comes from the Gauss-Bonnet formula: If a connected compact surface (without boundary) admits a metric of zero curvature, then this surface has zero Euler characteristic, i.e., is diffeomorphic to the torus or Klein bottle. In particular, the (topological) 2d sphere does not admit a flat metric. Similar things happen in higher dimensions (although the main obstruction is not the Euler characteristic). Much of modern Riemannian geometry revolves around the question of relation between curvature(s) (scalar, Ricci, sectional) and topology of the manifold, i.e., what restrictions on topology of $M$ are imposed by curvature properties of the metric.

Even if one is interested only in local aspects, i.e., Riemannian metrics on open subsets of $R^n$, considering non-flat metrics is still very useful; much of this motivation (going back to Einstein's work) comes from physics.

For your question about the sphere: There is nothing you can do to $R^n$ to make it diffeomorphic to the sphere. Still, you can ask if $R^n$ admits a (complete) metric of strictly positive sectional (or Ricci) curvature (as the round sphere has). The answer to this is negative, but the proof is not so easy, it comes from Myer's theorem: If a manifold $M$ admits a complete metric of strictly positive Ricci curvature, then $M$ is compact. (This is a nice example of interaction of curvature and topology!) Now, observe that $R^n$ is not compact.

Edit: By strictly positive Ricci curvature here I meant that there exists $k$ such that for all $x\in M$, $Ric(x)\ge k>0$. Otherwise, the conclusion is false.

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Good answer in the sense that you have weaved your answer with so many niceties in Riemannian Geometry which spark further interest. Do you have any suggestions for books on Riemannian Geometry for a good understanding of the subject in general ? –  DiffeoR Jan 22 at 9:48
The book I like most is do Carmo's Riemannian Geometry. –  studiosus Jan 22 at 14:09
@studiousus: I think your last paragraph is false. For example, $\mathbb{R}^n$ admits a complete rotational symmetric metric (think of a paraboloid) of positive curvature. Myer's (not Mayer's) theorem requires a that the Ricci curvature be bounded below by a positive number (which doesn't hold in the paraboloid picture). On the flip, $\mathbb{R}^n$ is the only noncompact space to admit a complete metric of positive curvature. This follows from the soul conjecture (which was proved by Perelman) –  Jason DeVito Nov 3 at 19:14
@JasonDeVito: By strictly positive, I meant $\ge k>0$, sorry for being sloppy here. Yes, I know about Perelman's solution of Cheeger-Gromoll conjecture. –  studiosus Nov 3 at 22:17
@studiousus: I didn't mean to imply you didn't know of Perelman's work - I intended that statement for other readers (sorry for not indicating that clearly). Thank you for the edit! –  Jason DeVito Nov 3 at 22:20

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