# I am confused by the different definitions of manifolds.

I'm currently learning manifold from Do Carmo's Riemannian Geometry. This is his definition of differentiable manifold:

But this is different from what I saw in wiki:

A differentiable manifold is a topological manifold equipped with an equivalence class of atlases whose transition maps are all differentiable. In broader terms, a Ck-manifold is a topological manifold with an atlas whose transition maps are all k-times continuously differentiable.

Carmo defines differentiable structure on a set, so nothing about continuity of the mappings can be said. While wiki's definition is on topological space in the first place, therefore it make sense to require the charts to be homeomorphism. Moreover, the mapping directions of Carmo's $\bf{x_\alpha}$ and the charts are just the opposite.

After defining differentiable structure on set $M$, Carmo then define a $A\subset M$ be open iff $\textbf{x}_\alpha^{-1}(A\cap\textbf{x}_\alpha(U_\alpha))$ is open in $\mathbb R^n$ for all $\alpha$. From this I can see that $\textbf{x}_\alpha$ are continuous.

My question is that, are these two definition talking about the same thing? Why do they seems so different? Also, I don't understand why Carmo requires $\textbf{x}_\alpha^{-1}(W)$ to be open. And how come there are homeomorphism in wiki's definition, but not in Carmo's?

• I'll let others who know more answer this question, but this is a subject that I am interested in as well. I have found that Do Carmo's book to not be the best book because, for example, he defines the chart functions as going from a subset of $\mathbb{R_n}$ to a subset of $M$ which is the opposite of the way most other authors do it. I like John Lee's "Introduction to Smooth Manifolds" and Spivak's book on differential geometry. – jeo15 Aug 8 '15 at 15:44
• The definition of a smooth manifold shouldn't be complicated since it's a fairly intuitive object. Practically speaking if you wanna see whether a topological space is a manifold or not, you first find a suitable open covering, map each set in the open covering homeomorphically into R^n and check whether the "transition functions" on the areas of overlap are smooth. – Ahsan Aug 8 '15 at 16:38

I am not absolutely sure but I believe the point of do Carmo's (2) is to "import" the topology from $\mathbb{R}^n$. If the preimage of $W$ is open, then the map is continuous if $W$ is open. Defining this $W$ to be open basically gives a topology on $M$.

The fact that do Carmo's coordinate maps go the opposite way is irrelevant, since they are bijective. This can be seen, because they are defined to be injective, so if their codomain is restricted to the ranges, then they are also bijective, and since their domain is open, and we defined their range to be open, so they are continuous both ways, ergo they are homeomorphisms. And for a homeomorphism the "initial direction" of the mapping is irrelevant.

To help you see that these are the same, I will give you a complete definition of smooth manifolds as it is usually done (and also as it is done in Lee's book):

We first define topological manifolds.

Definition:

A real, topological manifold of dimension $n$ is a set $M$ for which it is true that

a) $M$is equipped with a topology, $\tau$;

b) $\tau$ is Hausdorff and second countable;

c) $M$ is locally euclidean.

By c) I mean, that for any $p\in M$ point there is an open set $U\in\tau$ that contains $p$ and there exists a homeomorphism $\varphi:U\rightarrow\mathbb{R}^n$.

Obviously, since $p$ is arbitrary, this means that there must be enough of these $U$s that they cover $M$. The $(U,\varphi)$ pair is called a chart. A set of charts $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in\mathbb{A}}$ that cover $M$, eg. $\cup_\alpha U_\alpha=M$ is called an atlas.

We say that an atlas is $C^k$, if for any two $U_\alpha$, $U_\beta$ that have nonzero intersections, the map $\varphi_\beta\circ\varphi^{-1}_\alpha:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is $C^k$.

We call two atlases, $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ $C^k$-compatible, if their unification is a $C^k$ atlas.

An atlas $\mathcal{A}$ is maximal if it contains all possible $C^k$-compatible atlases. The definition of a maximal atlas is needed so that two manifolds with different atlases, but which are $C^k$-compatible will not be considered different manifolds. A maximal $C^k$ atlas is what we call a $C^k$ differentiable structure.

Then we define $M$ to be a real, $n$-dimensional $C^k$ manifold if $M$ is a real, $n$-dimensional topological manifold, with a maximal $C^k$ atlas on it.

As you can see, this definition is the same as wikipedia's, but has been broadened and clarified. It is not hard to see that do Carmo's definition is also the same, his (1) is the requirement that charts cover the manifold, his (2) defines topology, atlases, and their differentiable property, and his (3) extends the atlas maximally.

• Do Carmo does indeed give $M$ the topology "imported" from $\mathbb{R}^n$—the coarsest topology in which all of the charts $\mathbf{x}_\alpha^{-1}$ are continuous. The OP mentions this in the sentence that begins, "After defining differentiable structure on set $M$...." – Vectornaut Aug 9 '15 at 21:29
• @Vectornaut Aye, but he also asked why do Carmo wants $\mathbb{x}^{-1}_\alpha[W]$ to be open. This condition is needed for the "importing", hence my explanation. I do admit I could have phrased that better though. – Bence Racskó Aug 9 '15 at 21:31
• Whoops, I missed that! There may be multiple reasons for the open-preimage clause in (2). I suspect the main one is that most authors only define differentiability for maps from open subsets of $\mathbb{R}^n$, so the open-preimage clause is necessary for the differentiability clause to make sense. – Vectornaut Aug 9 '15 at 21:41
• @Vectornaut At this point, I am really not sure, this was also a "gut feeling" for me. What I am not sure about is if this is a requirement after all, I have a feeling that with the rest given, this is automatically satisfied. On a second thought, if we define the chart domains to be open sets, then should not this clause enforce the finite intersection of open sets to be also open? – Bence Racskó Aug 9 '15 at 22:20
• Thank you for your detailed answer. Your definition given is just like what I usually see, and I can start smelling the homeomorphism in Carmo's definition. But there one place left unclear to me. I understand $x_\alpha$ are bijection. Also, since for all open $A\subset \textbf{x}_\alpha(U_\alpha)$, $\textbf{x}_\alpha ^{-1}(A)$ is open, so $\textbf{x}_\alpha$ are continuous. However, how can we show $\textbf{x} _\alpha^{-1}$ are also continuous? – A. Chu Aug 10 '15 at 13:26