I'm learning special relativity and I am having a confusion on this mathematical point. Whenever any sort of motion or non motion happens in the world, it can only be perceived by the scientist in a chart (inertial frame). But, we say there exists a common manifold which is being charted by different observers when doing observation.

How can one from just seeing the charts with transition maps and metric (spacetime - metric) know the existence of a manifold?

In a more simplistic sense, could I reconstruct how the earth sits in space just from seeing pages the world map (with marked distances)?

  • $\begingroup$ Manifold construction lemma. Lemma 1.35 in in Lee's Introduction to Smooth Manifold book $\endgroup$
    – erolbarut
    May 20 at 12:28
  • 3
    $\begingroup$ Get a pot of glue, cut out the pictures of the charts from all the pages, and paste all the charts together using the transition maps. Literally, that's what one does, although it is formalized with the topological concept of quotient space. $\endgroup$
    – Lee Mosher
    May 20 at 13:06

3 Answers 3


Here are some thoughts on mathematical aspects of the question:

Can a smooth manifold be reconstructed from a collection of charts?


Assume you're given open sets $U_k\subset\mathbb{R}^d$ $(k=1,\dots, N)$ (think: pages in an atlas of the earth) and diffeomorphisms $\psi_{kl}\colon \operatorname{Domain}(\psi_{kl})\supset U_k\rightarrow U_l$ satisfying the obvious consistency condition (think: you keep track of certain locations on the earth appearing on different pages of the atlas). Then you can build from this a space $X=\coprod U_k/\sim$ by gluing the sets $U_k$ along the overlaps (think: you tear out the pages of the atlas and glue them together such the overlaps match up). Then there's a natural way to turn $X$ into a smooth manifold.

For any smooth manifold $X$ one could first consider an atlas (in the geometry sense), which gives rise to chart domains $U_k$ and transition functions $\psi_{kl}$ as above and then the preceding construction again yields $X$ (think: if you follow the instructions from above for an actual atlas of the earth, say with strechable paper such that you can nicely glue everything, then you end up with a sphere)

Can a Riemannian manifold be reconstructed from a collection of charts and marked distances?


Say in the setting above each chart domain $U_k$ comes with a distance function $d_k\colon U_k\times U_k\rightarrow [0,\infty)$, again being compatible with each other (think: the distance between Paris and London is the same no matter on which page of the atlas you measure it). Then on $X$ you get a function $d\colon \mathcal{D}\subset X\times X\rightarrow [0,\infty)$, defined on some neighbourhood $\mathcal D$ of the diagonal in $X\times X$ such that $d(x,y)=d_k(x,y)$ if $x,y\in U_k. $

  • Can $d$ be extended to all of $X\times X$? (think: there might be no page in the atlas in which you can measure the distance between Paris and Tokyo - can you still compute how far they are away from each other?) If you're able to comptue $d$ for points that are close to each other, you can define the length of a curve in $X$ (by marking $M$ points, very close to each other, adding up their distances and then taking the limit $M\rightarrow \infty$). If you can measure lenghts of curves, then you can measure the distance between any two points $x,y\in X$ by taking the infimum over the length of all curves between $x$ and $y$. This should give a metric $d^*$ on $X$ and if the metrics $d_k$ you started with were reasonable nice, it should be true that $d^*=d$ whenever the latter was already defined. Checking these claims should be possible after reading a bit about length spaces.

  • Next, if the original manifold $X$ that was mapped out in the atlas came with a Riemannian metric $g$, is it possible to recover this from $d$? (think: if you do this, then you can not only compute distance but also other intrinsic metric quantities, like the Gaussian curvature of the earth) The answer to this is yes, see here.

Can we reconstruct how the earth sits in space?

This question is of course a bit vague -- with the data available in a common atlas and specified above ($U_k,\psi_{kl},d_k$) -- the answer is no. From a mathematical point of view this is just saying that you only get the intrinsic geometry but no extrinsic features that depend on the precise embedding of $(X,g)$ in an ambient manifold. But for the earth this should be obvious -- after all, by looking at an atlas, how would you figure out the distance to the sun?

One reasonable question would be the following: Assume we are not given the intrinsic distances $d_k$, but the extrinsic distances $\tilde d_k$ (think: the direct line by going through the earth) - can we determine some geometric features beyond what is said above? This is related to the famous boundary rigidity problem, see e.g. here. Having this data and using the (very recent) theorem from the previous link, one could e.g. say that in a $3$-dimensional layer below the earth's surface the space has zero curvature.


It may depend on your philosophical standing, but most of us would probably agree that space-time is the manifold, regardless of whether any scientist has used a chart to measure anything. Charts came after the existence of the manifold.

In pure math, there are at least two different ways to define a manifold. One can start with a topological space, and then insists it has some nice properties including admitting a maximal family of compatible charts. We may also starts with a family of charts $U\subset\mathbb R^n$ and transition maps $\rho_{UV}$ that satisfy the compatibility property, then we glue the charts together and define the manifold to be $\bigsqcup U /\sim$ where $\sim$ is the equivalence relation induced by $\rho_{UV}$'s in the obvious way. The second approach seems closer to how science is developed from measurement, but the first approach is argubly better, for example, it's much harder to address Hausdorff-ness in the second definition. And from the point of view of algebraic geometry, the underlying topological space is essential and charts are just sections of the smooth function sheaf over the space. A chart/measure is just a function, a way to assgin numbers to events.

Put in another way, scientists use their local charts to understand the world around themselves. All the observers can patch together their charts to gain a global understanding, and the way they patch is to assume the world itself is a consistent manifold (whatever this means).

You have to be more specific about "how the earth sits in space" means, but basically if different observations are combined, we can recover the manifold through the second definition.


To my mind there are two ways to answer this question. From a mathematical point of view, a variety is a set with a set of maps which are bijections with open sets of $R^n$, subject to some conditions. So in principle, from the definition, we know what is the variety. From another point of view, we can also start from a set of charts and try to put two by two pieces together, as we do with road maps, or with an atlas. It is more intuitive, but it is not the usual definition.

  • $\begingroup$ Qhats a variety? I'm asking about manifolds $\endgroup$ May 20 at 12:16
  • $\begingroup$ variety= manifold $\endgroup$
    – Thomas
    May 20 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.