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A square is a topological manifold with boundary but not a smooth manifold with boundary because of its corners. But I am confused about it. I think since for a specific corner $p$, there is only one chart to cover $P$ (in order to be compatible), say $(U,f)$ , thus in the NBHD of $P$ the transition maps can only be $ff^{-1},f^{-1}f$, so the charts are compatible, so a square has a smooth structure. I feel confused about it. Could you tell me where I went wrong? Thank you!

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The topological space known as the square definitely does have a smooth structure, since it's homeomorphic to a disc. However, it does not have a smooth structure such that the inclusion map $i: \square \to \Bbb R^2$ is a smooth embedding.

Proof: Put a smooth structure on the square. Let $(U,\varphi)$ be a boundary chart about one of the corners $p$ such that $\varphi(p)=0$. If $i$ is a smooth embedding, then the differential $di_p$ is an isomorphism. The tangent cone of a point $q$ is the set of vectors $v \in T_qM$ such that $v = \gamma'(0)$, where $\gamma: [0,\varepsilon) \to M$ is smooth. For an interior point, the tangent cone is the whole of $T_qM$; for a boundary point of a smooth manifold with boundary, it's a half-space, precisely the half-space of "inward-pointing" tangent vectors. (This is a useful notion when considering other sorts of 'manifolds with corners'; see my answer here for another application.)

Now note that tangent cones are functorial: if $f: M \to N$ is a smooth map, then $df_p(C_p) \subset C_{f(p)}$. (If $f$ is a diffeomorphism we see that the linear isomorphism $df_p$ preserves the tangent cones.)

Now let's break our assumption on the smooth map $i$. Because it's a boundary point in a manifold with boundary, $C_p$ is a half-space. But $C_{(0,0)}$ is a quadrant of $\Bbb R^2$, and there is no linear isomorphism that sends a half-space into a quadrant.

Relatedly, it is worth thinking about the upper right quadrant in $\Bbb C = \Bbb R^2$, and why the map $z \mapsto z^2$ is a homeomorphism onto the upper half plane but not a diffeomorphism, and how the idea of this argument comes from that fact.

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  • $\begingroup$ Thank you very much! But I am still a little confused. On Lee's textbook, he said "squares are topological manifold with boundary but not smooth manifolds with boundary because of they have corners" without saying about the inclusion map you referred, so I wonder if he meant the same thing as you said or he meant something else? $\endgroup$ – 6666 Nov 28 '15 at 6:46
  • $\begingroup$ @6666: The rigorous phrasing is what I said. Ultimately what he means is that certain subsets of $\Bbb R^n$ can automatically be given a smooth manifold structure: these are those such that the inclusion map is a smooth embedding (or with certain charts, usually called slice charts). Take it as a moral statement instead of as a totally correct one. $\endgroup$ – user98602 Nov 28 '15 at 6:49

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