Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let us consider a noncompact K\"{a}hler manifold with vanishing scalar curvature but nonzero Ricci tensor. I'm wondering what can it tell us about the manifold. The example (coming from physics) has the following K\"{a}hler form

$K = \bar{X} X + \bar{Y} Y + \log(\bar{X} X + \bar{Y} Y)$

e.g. this is a 2D complex manifold. I claim that its Ricci form is nonzero, whereas its scalar curvature is identically zero.

I'm wondering if such manifolds possess any interesting properties and how can we classify them.

UPD.

Partly the answer for 4 manifolds (2d complex manifolds) is given in the paper by C Lebrun ``Counter-examples to the generalized positive action conjecture'' paper. The author considers vanishing scalar curvature and derives the most generic form of the Kahler potential such that it vanishes. There are several integration constants in the final answer, playing with them we can get different manifolds including the one I was talking above. For that case the Kahler metric is the metric of a standard blow-up in the origin.

$K = \bar{X}X+\bar{Y}Y+a\log(\bar{X}X+\bar{Y}Y)$

where $a>0$.

Now one can ask the same question about manifolds of higher dimension if they all with vanishing scalar curvature (but nonvanishing Ricci tensor) are described by the blow-ups of $\mathbb{C}^n$'s. In particular, I'm interested in the following Kahler potential

$K = \sum\limits_{i=1}^N \sum\limits_{i=1}^{\tilde N}|X^i Y^j|^2 + a \log \sum\limits_{i=1}^N|X^i|^2$

share|improve this question
1  
I'd suggest that you change the title of your question to mention your additional requirement that the Ricci tensor is non-zero; generally it is good to have as specific a title as possible, so that someone wondering about the same question will be able to find it. –  Zev Chonoles Jan 25 '11 at 18:21
1  
Frankly, Peter, from your originally question it is impossible to understand that you actually wanted to have information about non-compact Kähler manifolds with zero Ricci curvature... You should edit your question, indeed... –  diverietti Jan 25 '11 at 20:28
2  
The use of "Ricci curvature" to mean "scalar curvature" is intolerable : it is already the name of what you call the "Ricci tensor". "Ricci scalar" is rarely used anymore. Read any modern text in riemannian/differential geometry to learn vocabulary. –  BS. Jan 26 '11 at 10:14
1  
Peter, this is a math site, so you really should make an effort to follow the terminology mathematicians use. Otherwise, you're creating unneeded confusion. Your question uses the phrase "Ricci curvature", which for any mathematician means the same as "Ricci curvature tensor" and definitely not "Ricci scalar curvature". And I would avoid "Ricci scalar curvature" here. Just say "scalar curvature". –  Deane Yang Jan 27 '11 at 15:07
1  
Wald, "General Relativity" calls it scalar curvature, Misner, Thorne and Wheeler, "Gravitation" calls it scalar curvature. It's scalar curvature whether you're a mathematician or a physicist and calling it the Ricci something is just going to cause confusion in either camp and will not increase understanding. –  Jeff Harvey Jan 27 '11 at 20:40

4 Answers 4

On a $n$-dimensional Kähler manifold $(X,\omega)$, the Ricci form is (minus) the curvature of the canonical bundle $K_X$ endowed with the induced metric. Thus, if $X$ has zero Ricci curvature then its canonical bundle is flat. Thus, the structure group can be reduced to a subgroup of the special linear group $SL(n,\mathbb C)$.

However, Kähler manifolds already possess holonomy in $U(n)$, and so the (restricted) holonomy of a Ricci flat Kähler manifold is contained in $SU(n)$. Conversely, if the (restricted) holonomy of a $2n$-dimensional Riemannian manifold is contained in $SU(n)$, then the manifold is a Ricci-flat Kähler manifold.

In the case when $X$ is compact the celebrated solution of Yau to the Calabi problem asserts that if $c_1(X)=0$ then $X$ posses a metric with vanishing Ricci curvature. For the non compact case, there are some (among others) results by Tian and Yau which concerns the existence of complete Ricci-flat Kähler metrics on quasiprojective varieties. One of their main theorems is the following:

Suppose that $X$ is a smooth complex projective variety with ample anticanonical line bundle (i.e. a Fano manifold), and that $D\subset X$ is a smooth anticanonical divisor. Then $X\setminus D$ admits a complete Ricci-flat Kähler metric.

share|improve this answer
    
Thanks a lot, it is very instructive. I did mean noncompact K\"{a}hler manifolds and nonvanishing Ricci tensor, while Ricci curvature is zero. I guess I should look in the paper, thanks. –  Peter Jan 26 '11 at 6:44
1  
You meant that but you wrote "My question was if noncompact Kahler manifolds with zero Ricci curvature possess any interesting properties"... So you want non-compact Kähler manifolds with non identically vanishing Ricci curvature but identically zero scalar curvature, right? Then, you should edit your question consequently... –  diverietti Jan 26 '11 at 9:27

The Ricci scalar is the average gaussian curvature in all the two-dimensional subspaces passing through the point, I believe. Whence you can derive the 'meaning'.

share|improve this answer
    
Well, first of all this is a 2d complex manifold, second of all it's just an example, I can give a manifold of higher dimension. I'd like to know if a generic statement exists –  Peter Jan 25 '11 at 5:09

Partly the answer for 4 manifolds (2d complex manifolds) is given in the paper by C Lebrun ``Counter-examples to the generalized positive action conjecture'' paper. The author considers vanishing scalar curvature and derives the most generic form of the Kahler potential such that it vanishes. There are several integration constants in the final answer, playing with them we can get different manifolds including the one I was talking above. For that case the Kahler metric is the metric of a standard blow-up in the origin.

$K = \bar{X}X+\bar{Y}Y+a\log(\bar{X}X+\bar{Y}Y)$

where $a>0$. Now one can ask the same question about manifolds of higher dimension if they all with vanishing scalar curvature (but nonvanishing Ricci tensor) are described by the blow-ups of $\mathbb{C}^n$'s.

share|improve this answer

Dear Peter, I don't think one can say anything about such manifolds because the scalar curvature is too weak an invariant to be of use. Here is an infinite family of non-diffeomorphic compact examples to support my claim; for non-compact ones, remove a subvariety.

Let $Y$ be a projective manifold of dimension $n$ with ample canonical bundle. By the Calabi-Yau theorem, $Y$ admits a Kahler-Einstein metric $\omega_Y$ with $Ric \omega_Y = - \omega_Y$. Recall that the projective space $\mathbb P^n$ admits the Fubini-Study metric $\omega_{FS}$ that has $Ric \omega_{FS} = \omega_{FS}$. We set $X = \mathbb P^n \times Y$ and equip this space with the product metric $\omega = \omega_{FS} \oplus \omega_Y$. (Here and everywhere we should write $pr_1^\ast\omega_{FS} \oplus pr_2^*\omega_Y$ for the appropriate projection maps.) By varying $Y$ among projective manifold with ample bundle (which are legion) we get non-diffeomorphic $X$.

Claim. The space $X$ has non-zero Ricci curvature but zero scalar curvature.

Proof. The dimension of $X$ is $2n$. We have $\omega^{2n} = \binom{2n}{n} \omega_{FS}^n \wedge \omega_Y^n$. A calculation in local coordinates then gives that $$Ric \omega = Ric \omega_{FS} + Ric \omega_Y = \omega_{FS} - \omega_Y \not= 0.$$ The scalar curvature $s$ of $\omega$ satisfies $$ 2n s dV = Ric \omega \wedge \omega^{2n-1} / (2n-1)!, $$ where $dV = \omega^{2n}/(2n)!$ is the volume form of $\omega$. Since $$ \omega^{2n-1} = \binom{2n-1}{n} \bigl( \omega_{FS}^{n-1}\wedge \omega_Y^n + \omega_{FS}^n \wedge \omega_Y^{n-1} \bigr)$$ we get $$ 2n s dV = \frac{1}{(2n-1)!}\binom{2n-1}{n}\bigl( (2n)! dV - (2n)! dV\bigr) = 0, $$ whence $s = 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.