In Wald's General Relativity, an $n$-dimensional $C^{\infty}$ manifold $\mathit{M}$ is defined as a set, with subsets $\lbrace{O}_{\alpha}\rbrace$, which satisfies 3 properties. In particular, the third one (abbreviated) is:

If any two sets $\mathit{O}_{\alpha}$ and $\mathit{O}_{\beta}$ overlap, we can consider the map $\psi_{\beta}\circ \psi_{\alpha}^{-1}$, which takes points from subset $\psi_{\alpha}[\mathit{O}_{\alpha} \cap \mathit{O}_{\beta}]$ to points in subset $\psi_{\beta}[\mathit{O}_{\alpha} \cap \mathit{O}_{\beta}]$. We require these subsets of $\mathbb{R}^{n}$ to be open and this map to be $\mathit{C}^{\infty}$.

My question is: what is the purpose of this clause in the definition of a manifold? It appears to allow the existence of a transformation from different coordinate systems (from $\psi_{\alpha}$ to $\psi_{\beta}$). It requires the existence of an inverse. Other than that, is there any other motivation for defining the manifold in this way?

  • $\begingroup$ It does not require $\psi$ to be invertible: this is already assumed because, by definition (of an atlas on a manifold), $\psi$ is an homeomorphism. $\endgroup$
    – anderstood
    Aug 17, 2014 at 4:10
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    $\begingroup$ I guess this is required for the compatibility of several atlases. BTW, the title and the questions seem quite different to me. $\endgroup$
    – anderstood
    Aug 17, 2014 at 4:13

4 Answers 4


This axiom is a "smoothness" property of manifolds. You could define a manifold without this property simply as a space which is locally Euclidean. However, then something which looks smooth in one chart may be highly non-smooth in another chart. For example a smooth curve in one chart may be discontinuous in another chart (since the representation in the other chart is obtained by composing with $\psi_1^{-1} \circ \psi_2$). So one normally requires the chart transition maps to be smooth (or at least continuous, which then yields "topological manifolds" in contrast to "smooth manifolds"). This smoothness is used everywhere in differential geometry, for example in the definition of a tangent vector as a equivalence class of curves (where the equivalence relation is defined via a chart).


First, the definition above is a right definition to include things like (the surface of) a smooth torus – and any similar ... manifold ... with an arbitrary topology, but exclude objects that wouldn't be locally equivalent to smooth space.

If you changed something important in the definition, you would get a difference concept that doesn't agree with our intuitive understanding or other ways to define a manifold. Unless you would change one or several things in such a way that the definition would be different but still capturing the general spirit of what we want, or even equivalent.

To prove that one has a manifold by this definition, it must be fully covered by $O_\alpha$, open sets (patches) that overlap. Locally, each of the $O_\alpha$ is a piece of ${\mathbb R}$. Each $O_\alpha$ must be an open set so that with each point, it includes a vicinity of this point. The openness is needed to guarantee that there are no singular points included. The transition functions between the coordinates on the patches have to be invertible to avoid various pathological examples in which the required function would be pathological or discontinuous in some way.

Anderstood says that the functions have to be invertible because $\psi$ has to be a homomorphism. This isn't a real answer. It just creates one equally unanswered question – why homomorphisms – instead of the previous one.

At any rate, I think that I have explained why I think that the definition above is the right thing we need to describe the notion that may also be known intuitively etc. But in mathematics, one may define whatever she wants, of course. With most random mutations of the definition above, you won't be able to prove nice theorems etc. about the "mutated manifolds".

If you are really interested about some feature of the definition etc., you would have to formulate your question in a more focused way.

  • $\begingroup$ Thanks for the answer. Could you refer me to an example of a theorem that specifically utilizes this clause of the definition? $\endgroup$
    – yjc
    Aug 17, 2014 at 5:15
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    $\begingroup$ No but every theorem about manifolds does. $\endgroup$ Aug 18, 2014 at 4:45

A manifold should be thought of as a topological space that locally looks like a $\mathbb R^n$. This is made precise by saying that every point has a neighborhood that is homeomorphic to an open subset of $\mathbb R^n$. If you don't require anything else (except maybe Hausdorffness and the existence of a dense countable subset to make everything easier without sacrificing much) we speak of topological manifolds which are already very interesting, but you can't do calculus on them. If you want to do calculus, you have to say what differentiability means for these spaces. We would like to say that they are locally diffeomorphic to $\mathbb R^n$, but that won't lead anywhere if we haven't defined the differentiable structure on it yet. Rather, we define the differentiable structure to be the one that makes these local homeomorphisms $\psi_\alpha$ into diffeomorphisms, which implies that the transition homeomorphisms $ \psi_\beta \circ \psi_\alpha^{-1} $ are smooth, and actually is equivalent to it.


I realize now that you start out with an abstract set with subsets (I have always seen it defined as a topological space). Essentially the same is done in your case to define the topology: by forcing the $\psi_\alpha$ to be homeomorphisms.


This is the way the different "pieces" (i.e subsets/neighborhoods/atlases $\lbrace{O}_{\alpha}\rbrace$) can be "stitched" together to provide a smooth/uniform cover of the whole manifold.

Also this gives a way to measure/compare a function or vector on one subset (neighborhood/atlas) with a copy of it on a neighboring atlas. A necessary condition to transport vectors parallely on the manifold


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