We know that Calabi constructed some extremal metrics on $\mathbb{C}P^n\#\overline{\mathbb{C}P^n}$ which are not constant scalar curvature ones.

I just want to know given a Kähler class in $\mathbb{C}P^n\#\overline{\mathbb{C}P^n}$ ($n\geq 3$), is there any criterion to detemine whether or not there contains a constant scalar curvature metric? Or are there some known obstructions which can be used to rule out the existence of csc metrics in some Kähler classes in $\mathbb{C}P^n\#\overline{\mathbb{C}P^n}$? If there do have, what Kähler classes in $\mathbb{C}P^n\#\overline{\mathbb{C}P^n}$ can be ruled out up to now? Futaki invariant of course is an obstrution, but it involves the holomorphic vector field.

  • 1
    $\begingroup$ The Futaki invariant is indeed an obstruction. Your manifold is toric, so for torus-invariant Kahler classes it can be computed explicitly from the moment polytope (see e.g. arxiv.org/pdf/0803.0985 ) $\endgroup$
    – YangMills
    May 1, 2012 at 13:43
  • $\begingroup$ Thanks for your reply. What do torus-invariant Kahler classes mean? Given a Kahler class on a toric manifold, it doesnot change under this toric action, isn't it? OK. Maybe I should ask my question more explicitly. Let $x$ and $y$ be the two generators of $H^2(CP^n\sharp\bar{CP^n})$ corresponding to the $CP^n$ and $\bar{CP^n}$ respectively. We assume $\int x^n=-\int y^n=1$. Now any Kahler class should be written in the form $ax+by$ $(a^n-b^n>0)$. My question is, for what pairs $(a,b)$(a^n-b^n)$, the correponding Kahler classes don't contain csc metrics. I want to know the concrete results. $\endgroup$
    – Ping
    May 2, 2012 at 0:33
  • $\begingroup$ Sorry. In the last sentence, $(a,b)$(a^n-b^n)& should be $(a,b)$ $(a^n-b^n>0)$ $\endgroup$
    – Ping
    May 2, 2012 at 0:36
  • $\begingroup$ On a compact toric manifold there are in general non-torus invariant Kahler classes. $\endgroup$
    – YangMills
    May 2, 2012 at 15:09
  • $\begingroup$ Thanks. But I just want to know what kinds of $(a,b)$ cannot admit CSC metric:-) $\endgroup$
    – Ping
    May 2, 2012 at 23:16

1 Answer 1


There are no constant scalar curvature Kähler metrics on $M$, the blowup of $\mathbb{CP}^n$ at one point in any Kähler class and for any $n>1$.

This is because of the Lichnerowicz-Matsushima obstruction, which in this case says that if any such metric existed then $Aut^0(M)$ the connected component of the identity of the automorphism group of $M$ would be reductive (this is because $M$ is Fano so all holomorphic vector fields have a zero somewhere).

But $Aut^0(M)$ is readily seen to be isomorphic to the subgroup of $PGL(n+1,\mathbb{C})$ of matrices (modulo multiples of the identity) with arbitrary entries except for the first column which looks like $(*,0,0,\dots,0)$ where * is any nonzero complex number. This group is not reductive, so $M$ does not admit any constant scalar curvature Kähler metric in any class.

There are of course extremal Kähler metrics in some classes, which have nonconstant scalar curvature, the first example was constructed by Calabi in 1982 in his paper "Extremal Kähler metrics".

  • $\begingroup$ Yes, you are right:-) This is the most important obstruction before Futaki's integral invariant, which I suddenly forgot:-) We can also use it to show that the blowup of $\mathbb{CP}^n$ at two genric points also cannot any csc metric. $\endgroup$
    – Ping
    May 8, 2012 at 0:55
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    $\begingroup$ A Kähler metric is called extremal if it gives a critical point of $Φ=\int R_g^2dv$. We denote by $R_g$ the scalar curvature on $X$. In particular, if $R_g$ is constant, $g$ is extremal. The converse is also true (due to Fujiki)if $\dim L(X)=0$, where $L(X)$ is the maximal connected linear algebraic subgroup of $Aut^0X$. Note also that any Kähler-Einstein metric is of constant scalar curvature. $\endgroup$
    – user21574
    Jul 28, 2017 at 10:27

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