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Problem 5-6 in Michael Spivak's Calculus on Manifolds reads:

If $f:\mathbb R^n\to\mathbb R^m$, the graph of $f$ is $\{(x,y):y=f(x)\}$. Show that the graph of $f$ is an $n$-dimensional manifold if and only if $f$ is differentiable.

(here manifold means really submanifold, &c, and generally the statement has to be read in the context of the book, of course)

Now, this statement suffers from counterexamples: for example, the graph of the function $f:t\in\mathbb R\mapsto t^{1/3}\in\mathbb R$ is a submanifold.

  • I only have access to a printing of the original edition of the book: anyone happens to know if newer editions have the statement changed?

  • I can prove the statement if I change it to read «the graph $\Gamma$ of $f$ is an $n$-dimensional manifold and the differential of map $\Gamma\to\mathbb R^n$ given by projection on the first $n$ components has maximal rank everywhere if and only if &c». This seems like a rather strong hypothesis: can you think of a weaker one which will still give a sensible true statement? (I don't like this hypothesis because at that point in the book one does nto yet have the differential available)

Later. MathSciNet tells me there is a russian translation. Maybe one of our Russian-enabled friends on the site can tell me if the problem is there too? Translation into Russian traditionally includes some fixing, irrc :)

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    The problem statement is identical in my copy (doesn't say edition number; 1965 copyri$g$ht; 27th printin$g$ in January 1998; preface dated 1968)2011-10-03

2 Answers 2

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In the very next section (same page) he gives the tangent space for one of his submanifolds, meaning you can define the normal space, and add as a condition to problem 5-6 that the normal space contains no horizontal vector.

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    Or the tangent space no vertical ones, which is slightly more economical :)2011-10-03
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One statement that gets rid of the differential but which works out well is:

  • The graph of $f$ is a smooth manifold such that $\mathbb R^n \ni x \longmapsto (x,f(x)) \in graph(f)$ is a diffeomorphism, if and only if $f$ is smooth.

I suspect that's what Spivak was thinking. i.e. the graph of $f$ is not just a smooth manifold, but the function itself is a chart.

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    His definition of charts includes differentiability (as a function into $R^{n+m}$, in this case), so this reading does turn the problem into a simple one.2011-10-03