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First let me make a definition. Let $M$ be a smooth manifold and $S \subset M $ a topological subspace of $M$. We say that $S$ has "dimenion" at most $k$ if $S$ is a subset of
$$ X_1 \cup X_2 \ldots \cup X_n $$ such that each of the $X_i$ is a smooth manifold of dimension less than or equal to $k$. Of course $n$ is finite. Note that as per my definition if $S$ has dimension at most $4$ then it also has dimension at most $5$. Also note that the $X_i$ need not be closed.

Suppose $\pi: E \rightarrow M $ be a compact fiber bundle over a compact manifold $M$ (everything is smooth). Suppose $S \subset E$ is a topological subspace that has dimension at most $k$. Is it true that $\pi(S)$ also has dimension at most $k$?

It seems to me that this should follow from the fact that there can not be a smooth surjective map from $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ if $n > m$, but I am not sure.

In case the answer is no, suppose everything was in the complex setting, ie $\pi: E \rightarrow M$ is a compact complex fiber bundle, $M$ is complex manifold, $\pi$ is a homolomorphic and $S$ has complex dimension at most $k$ (there is an obvious definition for having complex dimension at most $k$). Is my claim true in that case, ie $\pi(S)$ has complex dimsnion at most $k$?

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  • $\begingroup$ Is there some reason why you want to use this particular definition of dimension? If you use Hausdorff dimension instead, for example, then there is a well-developed theory for this sort of thing. $\endgroup$
    – Neil Strickland
    Commented Jun 30, 2013 at 10:42
  • $\begingroup$ My reason is the following fact: Let $V \rightarrow M$ be a rank $k+1$ vector bundle over $M$ and $X$ a smooth submanifold of $M$ of dimension $k$. Then the zero set of a generic smooth section $s: M \rightarrow V$ does not intersect $X$. Would such a statement be true if $X$ was a space with hausdorf dimension $k$? $\endgroup$
    – Ritwik
    Commented Jun 30, 2013 at 11:48
  • $\begingroup$ I have heard that one can deduce this kind of thing from the results in Section 4.3.20 of "Geometric measure theory" by Federer. However, I have never tried to work out the details myself. $\endgroup$
    – Neil Strickland
    Commented Jun 30, 2013 at 18:50

2 Answers 2

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I guess "smooth manifold" means "smooth submanifold". In particular a curve which pass through a given point infinitely many times is not a smooth submanifold.

If yes then it is easy to construct an example. Take a smooth curve $\gamma\colon(0,1)\to \mathbb R^2$ which pass through $(0,0)$ infinitely many times under different angles. Then the curve $t\mapsto (t,\gamma(t))\in\mathbb R^3$ has dimension 1 in your sense while its projection $\gamma(0,1)$ on $\mathbb R^2$ has dimension 2.

You can do the same construction in the 3-torus, so everything can be made to be compact.

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  • $\begingroup$ I see what you are saying. What about in the complex setting (the second question I asked)? Is there still a counter example? $\endgroup$
    – Ritwik
    Commented Jun 30, 2013 at 11:52
  • $\begingroup$ Yes, you can complexity the example. $\endgroup$
    – Anton Petrunin
    Commented Jun 30, 2013 at 12:10
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Apply Sard's theorem to $\pi$ restricted to each $S_i$; thus $\pi$ cannot increase the dimension of any $X_i$.

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    $\begingroup$ I am not seeing why $\pi(X_i)$ is contained inside a FINITE union of manifolds of dimension less than or equal to the dimension of $X_i$. The image of $X_i$ need not be smooth. $\endgroup$
    – Ritwik
    Commented Jun 30, 2013 at 9:33

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