Let $f: X\to A$ be a (flat) morphism from a smooth complex projective variety $X$ to an abelian variety $A$. Consider the following natural diagram: $$T^*X\overset{df}{\longleftarrow}X\times H^0(A, \Omega_A^1)\overset{f\times id}{\longrightarrow}A\times H^0(A, \Omega_A^1)$$ where $T^*A$ and $T^*X$ are the cotangent bundles, and $df$ is the usual differential. Notice that $$(f\times id)(df^{-1}(o_X))=\{(a, \omega)| \ \text{there is a point}\ x\in f^{-1}(a)\ \text{such that}\ f^*\omega(x)=0\},$$ where $o_X$ is the zero section of $T^*X.$

My question is:

Is there an example that there is an irreducible component $Z$ of $(f\times id)(df^{-1}(o_X))$ such that $\dim Z<\dim A$? I am curious about this question, because I noticed in Schnell's notes (Lemma in page 3), we always have $\dim (f\times id)(df^{-1}(o_X))\leq\dim A$.

  • 1
    $\begingroup$ I think such an example will come from a projective example of your previous question. $\endgroup$
    – AG learner
    Aug 14 '20 at 17:46
  • 1
    $\begingroup$ It should be easy to make local examples global and then generically finite over an AV, see for example the proof of Prop 3.13 of arxiv.org/pdf/1212.5105.pdf $\endgroup$
    – Hacon
    Aug 14 '20 at 19:25
  • $\begingroup$ @AGlearner Sorry I messed up example and counterexample. I think an projective example of my previous question, i.e., a proper map $f: U\to V$ with $\dim f(Z)<r$ cannot give a example of above question, since the image of the tangent map at singular points of a singular fiber may move around downstairs. This a priori can make each irreducible of $(f\times id)(df^{-1}(o_X))$ is of the right dimension $\dim A$. However, if one can prove the previous question is a right statement for $f$ being proper and flat. Then each irreducible component has dimension $\dim A$ if $f: X\to A$ is flat. $\endgroup$
    – Feng Hao
    Aug 14 '20 at 20:29
  • $\begingroup$ @Hacon Thank you very much for your comment! Right now I don't think I have a local example. For the example of my previous question $f: \mathbb{A}^5\to \mathbb{A}^3; (x_1, x_2, x_3, x_4, x_5)\mapsto (x_1, x_2, x_1x_3+x_2x_4+x_5^2)$. The ``$(f\times id)(df^{-1}(o_X))$'' has dimension 3. $\endgroup$
    – Feng Hao
    Aug 14 '20 at 20:41
  • $\begingroup$ One local example is a modification of the earlier example, namely $f(x,y,z)=(x,x^2z+y^2).$ $\endgroup$ Aug 15 '20 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.