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I found that related to the Kähler cone there are many discussions on MathOverflow. Recently I am interested in the very special manifold of the one-point blow up of $\mathbb{C}P^n$ and just want to see what the general results on Fano Kähler manifolds look like when it comes to this special manifold.

My question is very concrete. Let $x$ and $y$ be the two generators of $H^2(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{Z})$ corresponding to the two components respectively. Moreover we assume $\int x^n=-\int y^n=1$. $H^2(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})=H^{1,1}(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})$ as $b_2(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})=2$. Thus every element in $H^{1,1}(\mathbb{C}P^n\sharp\bar{\mathbb{C}P^n};\mathbb{R})$ can be written in the form $ax+by$, $a,b\in\mathbb{R}$. So my question is $$ \{(a,b)\in\mathbb{R}^2~|~ax+by>0\}=?$$ Of course this set is contained in $\{(a,b)~|~a^n-b^n>0\}$.

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  • $\begingroup$ Since people felt the need to correct the spelling of K\"ahler's name (ironically the tag seems not to support the umlaut), I took the liberty of tweaking some other minor points $\endgroup$
    – Yemon Choi
    Commented Jul 25, 2013 at 8:29
  • $\begingroup$ Perhaps we can also enclose all the sharp signs with mathbin if we are going to be approaching MO like copy-editors... $\endgroup$
    – Yemon Choi
    Commented Jul 25, 2013 at 8:34

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The Kahler cone of any compact manifold is described by a theorem of Demailly and Paun. If $X$ is a compact Kahler manifold, then its Kahler cone is one of the connected components of the set $$ \mathcal P = \lbrace \alpha \in H^{1,1}(X,\mathbb R) \mid \int_Z \alpha^p > 0 \rbrace $$ where $Z$ runs through all the $p$-dimensional closed complex subspaces of $X$. If $X$ is projective, then the Kahler cone is actually this set.

Since the blowup of a projective variety in a point is projective, and the cohomology ring of $\mathbb P^n$ blown up in a point is pretty explicit, this lets us calculate the Kahler cone. Indeed, by some fun manipulations one gets $$ \mathcal P \simeq \lbrace (a,b) \in \mathbb R^2 \mid a > 0, \quad b > 0, \quad a > b \rbrace $$ where $(a,b) \mapsto aH - bE$. Here $H$ is the divisor of a general hyperplane in $\mathbb P^n$, pulled back to the blowup, and $E$ is the exceptional divisor of the blowup. You seem to be missing the $a, b > 0$ conditions, since $aH - bE > 0$ is equivalent to $a^n - b^n = (a-b) \cdot (a^{n-1} + \ldots + b^{n-1}) > 0$, and one can fulfill this condition with zero or negative $a$ or $b$, which would place us outside of the nef, or even pseudoeffective, cones.

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  • $\begingroup$ Gunnar, thanks very much. But suddenly I am a little confused with the exceptional divisor $E$. Is $\int E^n=(-1)^n$ ? Another question is why $b$ should be positive. For example, why is the pull back of $H$ to the blowup not a Kahler class? (in this case $a=1$ and $b=0$) $\endgroup$
    – Ping
    Commented May 6, 2012 at 11:35
  • $\begingroup$ Sorry. I think $\int E^n=(-1)^{n+1}$. Am I right?:-) $\endgroup$
    – Ping
    Commented May 6, 2012 at 11:38
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    $\begingroup$ I think your $E^n = (-1)^{n+1}$ is right (I'm not sure, so I should go revise my blowups), so then there should be a $(-1)^{n+1}$ in front of all $b$'s above. The pullback of the Kahler class $H$ to the blowup is unfortunately not Kahler, since it is degenerate on the exceptional divisor. $\endgroup$
    – Gunnar Þór Magnússon
    Commented May 6, 2012 at 12:41
  • $\begingroup$ Yes. I am sure $\int E^n=(-1)^{n+1}$ as I used Hirzebruch $\chi_y$-genus and Chern classes to verify it. My knowledge on complex and algebraic geometry is very limited, so what all the $p$-dimensional closed complex subspaces of $CP^n\sharp\bar{CP^n}$ look like? For $CP^n$ I can imagine is must be the unique $p$-dimenisonal subspace up to automorphisms. $\endgroup$
    – Ping
    Commented May 7, 2012 at 1:05
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    $\begingroup$ Your last sentence is wrong, for example a smooth cubic in $\mathbb{CP}^2$ is diffeomorphic to a 2-torus, while lines are diffeomorphic to a 2-sphere. However, the homology class of a smooth cubic equals 3 times the homology class of a line, which is all you care about when computing intersection numbers $\int_Z\alpha^p$. $\endgroup$
    – YangMills
    Commented May 7, 2012 at 4:35

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