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Let $c_i,d_j <n$, be a set of integers and define $$ M=\prod Gr(c_i,n),\quad N=\prod Gr(d_j,n),$$ where $Gr(k,n)$ is the grassmannian of k-planes in $\mathbb{C}^n$. Let $E_M=\oplus E_i^*$, where $E_i$ is the pullback of tautological bundle to $M$ from the i-th component. Similarly, let $E_R=\oplus E_j^*$, where $E_j$ is the pullback of tautological bundle to $N$ from the j-th component.

I am intending to show that the bundle $$E_M\boxtimes E_N = \oplus_{i,j} E_i^* \boxtimes E_j^* $$ over $M\times N$ has a transverse section. Note that each component of the dirsct sum has lots of sections. In fact, the set of sections correspond to $n\times n$ matrices.

Even if you don't know how to answer this particular question, what are the general methods of approaching such question? How can we find whether a given bundle has transverse sections? May be some sort of numerical criterion?

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    $\begingroup$ What is a transverse section? $\endgroup$
    – abx
    Dec 14, 2013 at 17:03
  • $\begingroup$ You probably mean transverse to the zero section. $\endgroup$
    – abx
    Dec 14, 2013 at 17:19
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    $\begingroup$ what else could it mean? $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Dec 14, 2013 at 17:31

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If a vector bundle on a smooth variety is globally generated then its generic section is transversal to zero. This is Bertini Theorem, so of course you need the base field to be infinite. In your special case note that the dual tautological bundles are globally generated and that pullbacks, tensor products, and direct sums of globally generated bundles are globally generated.

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  • $\begingroup$ That sounds very good then. Thanks. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Dec 14, 2013 at 18:54
  • $\begingroup$ just to make sure, you said tensor product, pull backs, and direct sums of globally generated bundles are globally generated. Does your list also include product (more precisely \boxtimes in this case). $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Dec 14, 2013 at 18:58
  • $\begingroup$ Exterior tensor product (i.e. \boxtimes) is the tensor product of pullbacks. $\endgroup$
    – Sasha
    Dec 14, 2013 at 19:01
  • $\begingroup$ @Sasha isn't the above only true for line bundles? Why is Bertini's theorem true for general vector bundles? $\endgroup$
    – Tobias Shin
    Oct 9, 2018 at 4:36
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    $\begingroup$ @TobiasShin: Bertini is true for any globally generated vector bundle. Just note that the zero locus of a particular section of a vector bundle is the same as the fiber of the zero locus of the universal section over the corresponding point on the projective space of all sections. When the bundle is globally generated, the universal zero locus is just a projective bundle over the original variety hence smooth. So, it remains to note that a general fiber of a morphism of smooth varieties is smooth. $\endgroup$
    – Sasha
    Oct 9, 2018 at 14:12

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