# Section of a vector bundle as a submanifold

I am currently working on one part of a problem surrounding sections and submanifolds.

Given a real vector bundle $\pi: E\rightarrow M$ of rank k, with a smooth global section $s:M\rightarrow E$, can the section be considered a submanifold of $E$ under its graph parametrization? Namely, for an open set $U\subseteq M$, consider $\Gamma(s)=\{(p,v)\in M\times \mathbb{R}^k|p\in U, v=s(p)\}$, which is a properly embedded m-dimensional submanifold of the product space $M\times \mathbb{R}^k$.

But the vector bundle itself has a local trivialization $h:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^k$ which is a diffeomorphism from its open sets ($\pi^{-1}(U)$) to open sets in the product space($U\times \mathbb{R}^k$). Can one piece together a submanifold of $E$ using these local trivializations?

I am asking all this because, the problem I am working on involves showing that for a compact manifold, where the rank of the bundle is greater than the dimension of the space, has a non-vanishing section. Since the bundle $E$ has dimension $m+k$, and the zero section is a submanifold of dimension $m$, my plan was to try and use transversality to show the two sections can not intersect due to dimensional considerations. Also unclear on where I might use compactness so I'm still struggling.

1. Yes. The image of a (sufficiently smooth) section is generally a submanifold. But it might not be nice. Consider the trivial line bundle over the reals, which I'll visualize as $\mathbb R^2$, with projection $(x, y) \mapsto x$. Then a section really is just the graph of a function on the reals. When you think about things like $y = |x|$ and $y = x^3$ and $y = x^{3/2}$, you realize how awkward things can be. You clearly need some smoothness conditions.
2. Your proof sounds as if it's headed in exactly the right direction. There's a small difficulty: if you start wtih the zero-section, $z$, and another section, $s$, and make them transverse by adjusting $s$, will the perturbed $s$ still be a section? The perturbation might move it so that more than one point of the image of $s$ happens to lie in a single fiber. Back to the trivial $R^2 \to R$ example: suppose we have the section $s(x) = x^{(1/3)}$, i.e., $x = y^3$. You can perturb that to $x = y^3 - \epsilon y$, and now it's no longer a section.
• Why would $x=y^3-\epsilon y$ not be a section in the above case? Is it because composing the the projection map with this function is not the identity on that fibre (due to the graph intersecting the x-axis multiple times)? – user90242 Jan 21 '14 at 1:59
• right: if you graph y = x^3 - x, the $x$-axis intersects the graph at $x =-1, 0, 1$; rotate that 90 degrees, and you have the situation I described. (The epsilon is in there to show that you can get this problem with a small perturbation). – John Hughes Jan 21 '14 at 2:39