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Consider the graphs of the functions $f_1(x) = |x|$, and $f_2(x) = x$ under the subspace topology of $\mathbb{R}^2$.

Both of these graphs are smooth manifolds, just pick coordinate charts to be $(x, f_i(x)) \leftrightarrow x$.

Moreover, they are diffeomorphic via the map $(x, f_1(x)) \rightarrow (x, f_2(x))$.

This seems to clash with my intuition. For example, the graph of $f_1$ has a corner, so it "shouldn't" be smooth, much less diffeomorphic to $f_2$, which is just a straight line.

Can someone explain what's going on here?

In light of these examples, how should I visualize smooth manifolds and diffeomorphisms?

Chill2Macht
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Student1235
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    There's an old question on the first manifold. I think Prof Wong's answer could help a lot, here. – Dylan Moreland Sep 04 '11 at 02:17
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    I think maybe your intuition isn't fully thought-out. You're giving a set a smooth structure based on a bijection with $\mathbb R$ -- this is not a very natural thing to do. You can make the Cantor set a smooth manifold diffeomorphic to $S^n$ or $\mathbb R^n$ for any $n \geq 2$ using this technique, so it's not particularly interesting. – Ryan Budney Sep 04 '11 at 02:37

2 Answers2

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You could do the same thing with the set $T=\{(x,g(x))\}$ for any continuous $g$. The reason this seems non-intuitive is that you haven't used the smooth structure of $\mathbb{R}^2$ at all in defining the smooth structure of $T$; you've just taken the smooth structure on $\mathbb{R}$ and "transported" it onto $T$. Another way to say it: The intuitive non-smoothness of $T$ (for, say, $g(x)=|x|$) comes from looking at the way that $T$ is sitting in $\mathbb{R}^2$.

Abstractly, it's very much the same as the following situation, which may be clearer. The integers $\mathbb{Z}$ form a group under addition. The set $T = \{17,59\} \subset \mathbb{Z}$ is not a subgroup of $\mathbb{Z}$ under addition. It is true that $T$ can be made into group by transporting the structure of a 2-element group onto $T$, but you don't expect that this group will have anything to do with $\mathbb{Z}$ as a group anymore since you didn't use the group structure on $\mathbb{Z}$ to define the group structure on $T$.

Ted
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Your intuition is broken because the inclusion of the graph of $f_1$ (with the smooth structure you describe) into $\mathbb{R}^2$ isn't smooth. The graph of $f_1$, as a subset of $\mathbb{R}^2$ with its usual smooth structure, is not a smooth manifold for exactly the intuitive reason.

Qiaochu Yuan
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