Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:27:16Z http://mathoverflow.net/feeds/question/53185 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53185/noncompact-kahler-manifolds-with-nonzero-ricci-tensor-but-vanishing-scalar-curvat Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature Peter 2011-01-25T04:21:51Z 2012-09-28T07:08:47Z <p>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</p> <p>$K = \bar{X} X + \bar{Y} Y + \log(\bar{X} X + \bar{Y} Y)$</p> <p>e.g. this is a 2D complex manifold. I claim that its Ricci form is nonzero, whereas its scalar curvature is identically zero.</p> <p>I'm wondering if such manifolds possess any interesting properties and how can we classify them.</p> <p><strong>UPD</strong>.</p> <p>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'' <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.cmp/1104162166" rel="nofollow">paper</a>. 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.</p> <p>$K = \bar{X}X+\bar{Y}Y+a\log(\bar{X}X+\bar{Y}Y)$</p> <p>where $a>0$. </p> <p>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</p> <p>$K = \sum\limits_{i=1}^N \sum\limits_{i=1}^{\tilde N}|X^i Y^j|^2 + a \log \sum\limits_{i=1}^N|X^i|^2$</p> http://mathoverflow.net/questions/53185/noncompact-kahler-manifolds-with-nonzero-ricci-tensor-but-vanishing-scalar-curvat/53186#53186 Answer by Zatrapilla for Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature Zatrapilla 2011-01-25T04:25:05Z 2011-01-25T04:25:05Z <p>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'.</p> http://mathoverflow.net/questions/53185/noncompact-kahler-manifolds-with-nonzero-ricci-tensor-but-vanishing-scalar-curvat/53273#53273 Answer by diverietti for Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature diverietti 2011-01-25T20:46:46Z 2011-01-25T20:46:46Z <p>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 $SL(n,\mathbb C)$. </p> <p>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.</p> <p>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:</p> <p>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. </p> http://mathoverflow.net/questions/53185/noncompact-kahler-manifolds-with-nonzero-ricci-tensor-but-vanishing-scalar-curvat/53379#53379 Answer by Peter for Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature Peter 2011-01-26T17:40:22Z 2011-02-16T17:03:31Z <p>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'' <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.cmp/1104162166" rel="nofollow">paper</a>. 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.</p> <p>$K = \bar{X}X+\bar{Y}Y+a\log(\bar{X}X+\bar{Y}Y)$</p> <p>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.</p> http://mathoverflow.net/questions/53185/noncompact-kahler-manifolds-with-nonzero-ricci-tensor-but-vanishing-scalar-curvat/108321#108321 Answer by Gunnar Magnusson for Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature Gunnar Magnusson 2012-09-28T07:08:47Z 2012-09-28T07:08:47Z <p>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.</p> <p>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 $X$.</p> <p><strong>Claim.</strong> The space $X$ has non-zero Ricci curvature but zero scalar curvature.</p> <p><em>Proof.</em> 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 $s = 0$.</p>