7
$\begingroup$

Carmo2 Carmo3

The above two pages are from Carmo's Riemannian Geometry.

According to 2.1 Definition and 2.3 Remark, is a differentiable manifold Hausdorff?

$\endgroup$
3
  • 3
    $\begingroup$ I think it's interesting that in Do Carmo's definition of a differentiable manifold, he does not assume a priori that $M$ is a topological space. That seems potentially simpler than most definitions of a differentiable manifold that I've seen. $\endgroup$
    – littleO
    Commented Jun 8, 2019 at 18:06
  • 1
    $\begingroup$ Follow Adam Chalumeau's deleted comment, it seems as though $X =$ the line with two origins $0_1,0_2$ may satisfy this definition. There is an obvious differentiable structure consisting of two parametrizations $x_1, x_2 : \mathbb{R} \to \mathbb{X}$ with $x_i(0)=0_i$ and $x_i(t) = t$ for $t \ne 0$; then $x_1^{-1} \circ x_2$ is simply the identity on the open set $\mathbb{R} \setminus \{0\}$, which is differentiable, and $x_2^{-1} \circ x_1$ likewise. This structure is not maximal, but if we believe do Carmo's comment, it should be possible to extend it to one that is maximal. $\endgroup$ Commented Jun 8, 2019 at 21:17
  • $\begingroup$ Why is this question still open? $\endgroup$ Commented Feb 2, 2021 at 1:11

3 Answers 3

5
$\begingroup$

As pointed out in littleO's comment, $M$ is not a topological space, but only a set. How can we define a reasonable topology on $M$? Certainly we want that all $U_\alpha$ are open in $M$ and that all $x_\alpha : U_\alpha \to x_\alpha(U_\alpha)$ are homeomorphisms. So let us consider the set $\mathcal{B}$ of all $V_\alpha = x_\alpha(U_\alpha) \subset M$. It has the following properties:

(1) For every $x \in M$ there exists $V_\alpha \in \mathcal{B}$ such that $x \in V_\alpha$.

(2) For any two $V_\alpha,V_\beta \in \mathcal{B}$ and any $x \in V_\alpha \cap V_\beta$ there exists $V_\gamma \in \mathcal{B}$ such that $x \in V_\gamma \subset V_\alpha \cap V_\beta$. In fact, we even have $V_\alpha \cap V_\beta \in \mathcal{B}$. This follows from do Carmo's (2) and (3).

Now it is a well-known fact that there exists a unique topology on $M$ having $\mathcal{B}$ as a base. By definition the $V_\alpha$ are open in $M$. Moreover, the $x_\alpha : U_\alpha \to V_\alpha$ are homeomorphisms. In fact, for each open $U \subset U_\alpha$ the restriction of $x_\alpha$ to $U$ is a bijection $U \to x_\alpha(U)$ and by (3) it belongs to the given family of injective mappings.

There is no reason why this topology should be Hausdorff. A non-Hausdorff example is the line with two origins. See Nate Eldredge's comment and The Line with two origins and https://en.wikipedia.org/wiki/Non-Hausdorff_manifold.

$\endgroup$
1
$\begingroup$

As others have noted here, it is common to start by defining M to be a topological space and then to define a smooth structure on top of this. As others have noted, do Carmo starts only with a set; as a historical note, this follows (for instance) Whitney in his famous Ann. Math. paper "Differentiable manifolds" from 1936 (probably one of the earliest references for smooth manifolds). This perspective is nice if one wishes to view a smooth structure as a mere refinement of the notion of topological (manifold) structure, and not as something constitutionally different.

(and I agree with the other answers which say "no" to your main question. According to the standard definitions, this is a definition of a differentiable atlas, not of a differentiable manifold)

$\endgroup$
0
$\begingroup$

Sadly, do Carmo is sloppy here. He does not assume manifolds to be Hausdorff (and 2nd countable). In Proposition 2.10 (Chapter 1) he specifically mentions Hausdorff and 2nd countable as assumptions for the existence of a Riemannian metric. Then, in Chapter 7, he "proves" that every connected Riemannian manifold is metrizable without assuming Hausdorffness, which is, of course, nonsense. My take on all this is that as far as local calculations are concerned, in do Carmo (and in general) Hausdorfness is not needed, but as soon as you try to deal with a global statement, you should assume that manifolds are Hausdorff. My personal preference is to assume Hausdorfness from the beginning.

$\endgroup$
2
  • $\begingroup$ On page 29 do Carmo specifically mentions that his definition does not restrict the topology of the manifold to Hausdorff or second countable. He then states them as two axioms/ $\endgroup$
    – mxnoqwerty
    Commented Jun 11, 2019 at 0:31
  • $\begingroup$ @mxnoqwerty: Right, but then he does not always assume these axioms even when they are clearly needed, e.g. in Chapter 7. $\endgroup$ Commented Jun 11, 2019 at 1:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .