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Sarbeswar Pal - Rational curves and a conjecture of Drinfeld

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Abstract

Everyone is invited to the next talk of the Graduate Students’ Group seminar, whose details are as follows:

Speaker: Dr. Sarbeswar Pal

Topic: Rational curves and an application of it to a conjecture of Drinfeld.

Abstract: In this lecture I will introduce the basic notions of rational curves in a smooth projective variety over complex numbers. We will see some interesting properties of the variety of rational curves and see how this helps us to prove a conjecture of Drinfeld.

Date: 15 March, 2025 (Tomorrow)
Time: 6:00 PM - 7:00 PM
Venue: AB2 2A or 2B

Question

Let be a smooth projective variety. Does there exist a rational curve

In general, no.

is not nef.

We know

is a quasi-projective variety.

Suppose

is a rational curve and

is the projection from the tangent bundle.

Now the pullback

is a vector bundle on .

Grothendiek. Any vector bundle on is a direct sum of line bundle

for

As

Then

for

is free if and is non-free otherwise

is Fano if…

(canonical bundle)

Let is a globally generated line bundle?

Then …

is Fano if ample.

Fact: an open set such that any rational curve which intersects is free.

Question

is empty? is it proper?

We don’t know, in general.

Question

be a Fano projective manifold of (Picard rank 1) such that is not nef. Then is nonempty.

  1. What can we say about ?

Guess: is pure of co-dim 1.

Example

  • is Fano of Picard rank 1, and tangent bundle is…

Examples of non-nef

Let be a general hypersurface in of degree .

  • Intersection of two quadratics
  • is curve

be moduli of stable vec buns. It is smooth, Fano of Picard rank 1

is called wobbly if

such that .

Drinfeld's conjecture

Locus of wobbly bundles is pure of codim 1

A vector bundle is wobbly if a non-free rational curve passing through .

Conjecture

Let be Fano, Picard rank 1, different from . Then any non-constant endomorphism

is isomorphism.

All proved examples (Grassmanians, homogeneous spaces?) have nef.

Guess

Let Fano, Picard rank 1, , is not nef

then

This proves the conjecture.

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