What is the simplest way to show that a section of a vector bundle is transverse to the zero set - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:41:23Z http://mathoverflow.net/feeds/question/79723 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79723/what-is-the-simplest-way-to-show-that-a-section-of-a-vector-bundle-is-transverse What is the simplest way to show that a section of a vector bundle is transverse to the zero set Ritwik 2011-11-01T15:31:45Z 2011-11-04T07:54:10Z <p>Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$, be the space of homogeneous degree $d$ polynomials in three variables $[X,Y,Z] \in \mathbb{P}^2$ upto scaling, where $\delta_d = \frac{d(d+3)}{2}$. Note that we have two tautological line bundles $$\gamma_{\mathcal{D}} \rightarrow \mathcal{D}, \qquad \gamma_{\mathbb{P}^2} \rightarrow \mathbb{P}^2.$$<br> A section of the line bundle $$\mathcal{O}(d) = \gamma_{\mathbb{P}^2}^{* d} \rightarrow \mathbb{P}^2$$ is a homogeneous degree $d$ polynomial. This gives us a section of the rank $3$ vector bundle $$\psi_1: \mathcal{D} \times \mathbb{P}^2\rightarrow \gamma_{\mathcal{D}}^* \otimes \gamma_{\mathbb{P}^2}^{*d} \oplus \gamma_{\mathcal{D}}^* \otimes \gamma_{\mathbb{P}^2}^{*d} \otimes T_{}^* \mathbb{P}^2$$ given by $$\psi_1( [s], p) = s(p), \nabla s|_p$$ It is basically the evaluation map and the derivative'' at that point. We can take $\nabla$ to be any connection. It is a fact that if $d$ is large (in fact in this case we only need $d>1$), then $\psi_1$ is transverse to the zero set. Note that if $s(p)=0$ then $\nabla s|_p$ is the same for every connection. So it doesn't matter what connection we chose. My question is the following: Consider the following section $$\psi_2: \mathcal{D} \times (\mathbb{P}^2)^2 \rightarrow \gamma_{\mathcal{D}}^* \otimes \gamma_{\mathbb{P}^2}^{*d} \oplus \gamma_{\mathcal{D}}^* \otimes \gamma_{\mathbb{P}^2}^{*d} \otimes T_{}^* \mathbb{P}^2 \oplus \gamma_{\mathcal{D}}^* \otimes \gamma_{\mathbb{P}^2}^{*d} \oplus \gamma_{\mathcal{D}}^* \otimes \gamma_{\mathbb{P}^2}^{*d} \otimes T_{}^* \mathbb{P}^2$$ given by $$\psi_2( [s], p_1, p_2) = s(p_1), \nabla s|_{p_1}, s(p_2), \nabla s|_{p_2}$$ Is it true that the section $\psi_2$ is transverse to the zero set when $p_1 \neq p_2$, provided $d$ is large? Is there some reference someone can point out where they prove either this or some similar statement? My idea for proving this statement is as follows: </p> <p>Choose any two distinct points you like say $[1,0,0]$ and $[0,0,1]$ and show transversality at those points. After that, I want to argue that there is no loss of generality'' in assuming that those two points were $[1,0,0]$ and $[0,0,1]$, because you can always change coordinates and bring your points to these two points. </p> <p>I essentially want to prove the statement for $k$ distinct points (the vector bundle there will be of rank $3k$ and the base space will be $\mathcal{D} \times (\mathbb{P}^2)^k$ (provided $d$ is large). Everything is over the complex numbers. </p> http://mathoverflow.net/questions/79723/what-is-the-simplest-way-to-show-that-a-section-of-a-vector-bundle-is-transverse/80018#80018 Answer by Nikita Kalinin for What is the simplest way to show that a section of a vector bundle is transverse to the zero set Nikita Kalinin 2011-11-04T07:54:10Z 2011-11-04T07:54:10Z <p>It sounds like a multijet transversality theorem (see for example M. Golubitsky, V. Guillemin, Stable mappings and their singularities) in context of algebraic geometry. So, the answer is true --- a proof of this theorem uses only polynoms for perturbations which achieves general position.</p>