6
$\begingroup$

If the holomorphic vector bundle of $E$ on a Kähler manifold $M$ admit Hermitian-Yang-Mills metric then we have the following known universal inequality of Yau-Uhlenbeck $$\int_Mc_2(End E)\wedge \omega^{n-2}\geq 0$$

Is it known for other chern classes $c_{2n}$?

$$\int_Mc_{2k}(End E)\wedge \omega^{n-2k}\geq 0$$

We know

$ch(E) = r + c_1(E) + 1/2(c_1(E)^2 − 2c_2(E)) + · · · ,$ and $$ch(End(E)) = ch(E ⊗ E^∗ ) = ch(E)ch(E^∗ ) = (r + ch_1(E) + ch_2(E) + · · ·)(r − ch_1(E) + ch_2(E) − · · ·) = (r^2 + 2r · ch_2(E) − (ch_1(E))^2 + · · ·).$$

We obtain,

$$c_1(End(E)) = 0, c_2(End(E)) = 2rc_2(E) − (r − 1)c_1(E)^ 2,...$$

$\endgroup$
2
  • 1
    $\begingroup$ Your first formula seems to be for an $n$-dimensional Kahler manifold $M$, but in the second formula you're integrating the $(6n-4)$-form $c_{2n}(\operatorname{End}(E))\wedge\omega^{n-2}$ on $M$. I think you want the second formula to be $\int_Mc_{2k}(\operatorname{End}(E))\wedge\omega^{n-2k} \geq 0$. $\endgroup$ Commented May 1, 2017 at 16:30
  • 1
    $\begingroup$ I was thinking some time ago about the same type of question. Maybe Remark 4.3 of this paper of mine arxiv.org/pdf/1503.02512.pdf might be of some interest for you. $\endgroup$
    – diverietti
    Commented May 11, 2017 at 9:00

1 Answer 1

2
$\begingroup$

This is not true, as we can check by calculating the intersection number $\int_X c_4(\operatorname{End}(T_X)) \cup \omega^{n-4}$ for some easy examples of Kahler-Einstein spaces:

This intersection number seems to be always positive for $\operatorname{End}(T_{\mathbb{P}^n})$ by numerical calculations. There it is given by a degree-$5$ polynomial in $n$ that has a positive leading coefficient and is postive for the first 10 million $n$.

For a Calabi-Yau hypersurface $X \subset \mathbb{P}^{n+1}$, that is, a hypersurface of degree $n+2$, and the Kahler class of the restriction of the Fubini-Study metric, the intersection number is negative for the first few thousand dimensions $n$ and seems to decrease with $n$.

For a hypersurface $X \subset \mathbb{P}^{n+1}$ with ample canonical bundle, that is, a hypersurface of degree $> n+2$, this intersection number can be either positive or negative. For example, for hypersurfaces in $5$-dimensional projective space, it is negatives in degrees $6, 7, 8, 9$ and positive in degrees $10$-$1000$.

This follows by noting that we have $$ 12 c_4(\operatorname{End} E) = (r-1) c_1(E)^4 -(r-1) c_1(E)^2c_2(E) +(r+1) c_2(E)^2 +(r-1) c_1(E)c_3(E) -r c_4(E) $$ for a holomorphic vector bundle $E \to X$ of rank $r$, and recalling that the Chern classes of projective space are $$ c_k(\mathbb{P}^n) = \binom{n}{k} h^k, $$ where $h$ is the hyperplane class, and the Chern classes of a hypersurface $X$ of degree $d$ in $(n+1)$-dimensional projective space satisfy $$ c_{k+1}(X) = \binom{n+1}{k+1} h^{k+1} - d c_k(X) \cup h. $$ We can then simply have a computer crunch through the intersection numbers $$ p(n,d) := 12 \int_X c_4(\operatorname{End} T_X) \cup h^{n-4} \Bigm/ \int_X h^n $$ and tell us when they are positive or negative. This is done here.

$\endgroup$
4
  • $\begingroup$ If your couter-example is really a quartic surface in $\mathbb{P}^3$, it has no $c_3$ (and a different $c_2$). $\endgroup$
    – abx
    Commented May 11, 2017 at 8:12
  • $\begingroup$ I needed it to check other invariants ,In fact the difference between Yang Mills functional and Hermitian Yang Mills functional is such discriminant when $k=1$ $\endgroup$
    – user21574
    Commented May 11, 2017 at 9:57
  • 1
    $\begingroup$ @GunnarÞórMagnússon ,Thank you, I appreciate your effort for solving this question. $\endgroup$
    – user21574
    Commented May 13, 2017 at 14:06
  • 3
    $\begingroup$ Just as complementary comment: Let $X ⊂ \mathbb P^n$ be a smooth hypersurface of degree $d$ by zero locus of non-singular $f$. If $\mathcal O_X(1)$ is the restriction of$ \mathcal O_{\mathbb P^n }(1)$, we have by adjunction formula $\omega_X\cong \mathcal O_X(d- n-1)$. If $d = n + 1$, then $\omega_X \cong O_X$ $\endgroup$
    – user21574
    Commented Jul 20, 2017 at 19:03

You must log in to answer this question.