most recent 30 from http://mathoverflow.net 2013-06-19T01:41:36Z http://mathoverflow.net/feeds/question/25866 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25866/smooth-sections-of-smooth-fiber-bundles Roberto Frigerio 2010-05-25T12:15:57Z 2011-01-30T14:06:50Z

<p>A maybe trivial question about fiber bundles (I'm not an expert, and I didn't find quickly an answer looking here and there). Suppose you are given fiber bundle $p\colon E\to M$, where $E,M$ are smooth manifolds and $p$ is smoothly locally trivial. Also suppose that the fiber $F$ of the bundle is smoothly contractible. Is that true that $p$ admits a smooth section? </p>

http://mathoverflow.net/questions/25866/smooth-sections-of-smooth-fiber-bundles/25884#25884 Andrew Stacey 2010-05-25T14:25:14Z 2010-05-25T14:25:14Z

<p>Yes. This follows from smooth paracompactness of $M$. Basically, you choose local trivialisations over a suitable cover and then on the intersections you smoothly interpolate between the different choices. By choosing your cover carefully, at each stage you are interpolating between a finite number of choices so the interpolation amounts to smoothly filling in a simplex, which by assumption you can do. (In slightly more detail, but only slightly, once you've filled in the simplex, you use a partition of unity to specify where in the simplex you should be. Pictorially, if you imagine a line at $y = 0$ and one at $y = 1$, then between $x=0$ and $x=1$ you must get from one line to the other. "Filling in the simplex" corresponds to filling in the rectangle $(0,0) -- (1,0) -- (1,1) -- (0,1)$. A partition of unity in this case is just a smooth function $[0,1] \to [0,1]$ which is $0$ near $0$ and $1$ near $1$. The graph of this defines the interpolation between the two lines and thus the way to get from the line $y = 0$ to the one at $y= 1$.)</p>

http://mathoverflow.net/questions/25866/smooth-sections-of-smooth-fiber-bundles/25885#25885 Thomas Kragh 2010-05-25T14:29:28Z 2010-05-25T14:29:28Z

<p>The answer is: Yes (at least for finite dimensional manifolds).</p> <p>In fact you only need that the fiber is contractible not smoothly contractible. Take any continuous section $s_0 \colon B \to E$. cover $B$ by open sets $U_i$ such that the bundle is trivializable over each $U_i$, also make sure that the closure of each $U_i$ is compact and that the cover is locally finite.</p> <p>Furthermore, give $E$ any complete Riemannian structure. This provides each fiber $E_x$ with an induced Riemannian structure, which is also complete. We use this to define the obvious supremum distance between any two sections of $E$ over any sub-space of $B$. We may also construct a continuous map $r \colon E \to \mathbb{R}_+$ such that the ball of radius $r(x)$ and center $x$ in each fiber is geodesically convex. </p> <p>The construction now goes in 2 steps:</p> <p>1) local construction: for each $i$ we may find a smooth section $s \colon U_i \to E$ such that the supremum distance defined above is smaller than $r$ on all the points of $S_0$ restricted to $U_i$. This is easy and follows from smooth approximation of any function from $U_i \to F$ defining a section $U_i \to U_i \times F$.</p> <p>2) global construction: use a partion of unity to get a global construction. This is now possible because we were carefull enough to create the smooth local sections such that they lie in a geodesically convex neighborhood.</p> <p>While finishing this Andrew Stacey posted a similar answer, but it seemed a waste not to poste this also. Especially since we are focussing on different details.</p>

http://mathoverflow.net/questions/25866/smooth-sections-of-smooth-fiber-bundles/53780#53780 Giuseppe Tortorella 2011-01-30T14:06:50Z 2011-01-30T14:06:50Z

<p>A proof of the existence of this section appears in Steenrod "The Topology of fibre bundles". The first edition is about 1950, at your knowledge, are there older proofs? </p>