I am working through a paper https://arxiv.org/pdf/1011.1690.pdf on $J$-holomorphic surves and I don't understand the part in the theorem
where it says "smooth section transverse to the zero section".
I could interpret this as $s(t) \neq 0_{section}(t)=0$ but this doesn't make sense to me..
Can somebody explain this to me?
On
The above claim is not different from the usual transversality theorem which says that if $f: N \to M$ is transversal to $Z$, $f \pitchfork Z, Z$ being a submanifold, then $f^{-1}(Z)$ is a submanifold. The zero section as a set is a submanifold, which happens to be diffeomorphic to $B$, and the section $s$ being transversal to $0$-section is same as $s$ being transversal to the submanifold $0$-section. Hence the conclusion.
First I will assume that $k\leq n$ so the statement makes more sense.
Let $z$ denote the zero section of $\pi \colon E\to B$. We have that $s$ and $z$ are transverse at $y\in s(B)\cap z(B)$ if for $x = \pi(y)$ $$Ds_x(T_xB) + Dz_x(T_xB) = T_yE.$$
We say that the sections are transverse if they are transverse at every point of intersection of their images (which can be empty).
Take a point $p\in B$ and a small neighborhood $V$ of $p$ in $B$ such that $E|_V \simeq V \times \mathbb{R}^k$ is trivial. Then in this trivialization we have $$ z\colon V \rightarrow V \times \mathbb{R}^k, \quad z(x) = (x,0) \\ s\colon V \rightarrow V \times \mathbb{R}^k, \quad s(x) = (x,f(x)) \\ $$ for some (smooth) function $f\colon V \rightarrow \mathbb{R}^k$. Now $s(p)= z(p)$ if and only if $f(p) =0$ and the transversality condition translates to $Df_p$ having maximal rank ($=k$).
Note that $s^{-1}(0) = s^{-1}\left(s(B)\cap z(B) \right)$ is (locally) given by the system of equations $f(x)=0$ and the transversality condition allows us to apply the Implicit function theorem to prove that $s^{-1}(0)$ is a submanifold of dimension $n-k$.