Let $M$ be a compact connected manifold. Is there a chart $\Psi:U \to \mathbb{R}^n$ such that the closure of $U$ is $M$? This is true for $S^n, T^n, K$, all compact surfaces, etc. If it is not true in general, what is the obstruction?
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9$\begingroup$ How about this: the compact manifold admits a Morse function, the gradient flow as usual provides a CW-structure. The top cell is diffeomorphic to $\mathbb{R}^n$, and the complement is a closed subset. Choosing a diffeomorphism from the top cell to $\mathbb{R}^n$ should provide a chart whose closure is $M$. $\endgroup$– Matthias WendtAug 2, 2014 at 15:19
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1$\begingroup$ The tag is "differential geometry", I would say that the author of the question is assuming the smooth structure... $\endgroup$– Ilias A.Aug 2, 2014 at 16:03
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4$\begingroup$ The answer is yes. Start with a handle decomposition of the manifold and take a maximal forest in the 1-skeleton. That's a collection of discs in the manifold, and you can inflate its interior to be dense in the manifold, since its complement is a regular neighbourhood of the dual (n-1)-skeleton. This is a pretty common observation in courses where you study handle decompositions, the h-cobordism theorem and such. $\endgroup$– Ryan BudneyAug 2, 2014 at 18:24
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1$\begingroup$ @MatthiasWendt Possibly dumb question: You seem to be assuming just one top cell. It seems obvious to me that a compact connected manifold has a Morse function with only one local maximum, but how do you prove it? $\endgroup$– David E SpeyerAug 3, 2014 at 2:00
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5$\begingroup$ As valeri says in the comment below every $n$-manifold contains an open dense subset diffeomorphic to an $n$-disk: equip the manifold with a complete Riemannian metric, fix any point, and note that the complement to the cut locus at the point is diffeomorphic to a disk. The cut locus is nowhere dense. See e.g. Sakai's "Riemannian geometry''. $\endgroup$– Igor BelegradekAug 3, 2014 at 2:53
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2 Answers
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The exponential map for any Riemannian metric on your compact manifold $M$, based at any point $p$ of $M$, maps the tangent space $T_p M$, an ${\mathbb R}^n$, onto $M$ and is a diffeo inside the cut locus. Back on the tangent space, this `inside' of the cut locus is a star shaped domain relative to the origin, so defines a domain $V$ in ${\mathbb R}^n$ which is mapped diffeomorphically onto an open set $U$ whose closure is $M$.
(The closure of the domain $V$ is homeomorphic to the closed ball in the tangent space, so this same argument shows that every compact manifold is the quotient of the n-ball by some identification of points on its boundary, the n-sphere. )
answered Aug 6, 2014 at 4:47
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Take a covering $\mathcal{U}_0 = \{U_0^\alpha \,|\, \alpha < \kappa \}$ of $M_0 = M$ by some charts. Define $V_0 = U_0^0$ and consider $M_1 = M_0 \setminus \overline{U_0}$. Then $U_1^\alpha = U^\alpha_0 \cap M_1$ is a covering of $M_1$. Proceed by (transfinite) induction to obtain $V_\alpha$. (If $M$ is compact then you can assume that the covering $\mathcal{U}_0$ was finite and hence you have a finite set $\{V_\alpha\}$ of open subsets of $M$.) Now $M\setminus \bigcup_\alpha V_\alpha$ is a collection of boundaries of $U_\alpha$ which are manifolds of dimension $n-1$. Hence the closure of $\bigcup_\alpha V_\alpha$ is the whole $M$. If $U_\alpha$ were domains of charts $\varphi_\alpha$, then one obtains, translating the image of $\varphi_\alpha$ if necessary, a well defined chart $\varphi$ on $\bigcup_\alpha V_\alpha$ just by restriction $\varphi|_{V_\alpha} = \varphi_\alpha$, since the sets $V_\alpha$ are disjoint.
I haven't thought about the noncompact case so I'm not sure the transfinite induction will go through the limit ordinals. If it is even true, can one use a partition of unity to obtain a uniform proof?
Finally, a better notion suited for studying this kind of problems is the Lusternik Schnirelmann category.
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$\begingroup$ Yes, for a sphere I'd get two hemispheres and I can map them e.g. to two disjoint open disks such as $B(0,1)$ and $B(3,1)$. $\endgroup$ Aug 2, 2014 at 17:07
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3$\begingroup$ since the tag is differential geometry, how about this argument: take a point and issue from this point a geodesic g(t) in every direction v until (some t<t_0) it is minimal. Then the subset of all tv in the tangent space is homeomorphic to R^n and its image under exponential map is almost all M^n as required. $\endgroup$– valeriAug 2, 2014 at 17:16
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$\begingroup$ I did not notice your comment Valeri. It really says the same as my answer above-r.m. $\endgroup$ Aug 7, 2014 at 13:12
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$\begingroup$ And Igor Belegradek provided a reference for proof of the fact that cut locus is nowhere dense in another comment. For example, I've never even heard that term during 3 semesters of differential & Riemannian geometry. $\endgroup$ Aug 7, 2014 at 20:16