Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:57:47Z http://mathoverflow.net/feeds/question/81235 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold Gunnar Magnusson 2011-11-18T08:51:15Z 2011-11-21T08:41:43Z <p>Let $X$ be a Kahler manifold. Associated to any hermitian metric $h$ on $X$ is a smooth real $(1,1)$-form $\omega = -\text{Im } h$, called the Kahler form of $h$. One of several equivalent conditions for the metric $h$ to be a Kahler metric is that $\omega$ is symplectic, or that $\text{d} \omega = 0$.</p> <p>Morally speaking, this implies that Kahler metrics are rare. In a sense, they are contained in a proper subspace of the cone of hermitian metrics on $X$.</p> <p><strong>Question:</strong> Does anyone know an example of a manifold $X$ and a natural metric $h$ on $X$ which is not Kahler?</p> <p><strong>For extra credit:</strong> Can we find such an example where $X$ is compact?</p> <p>I'll elaborate a bit on what I mean by "natural" and then provide some motivation for the questions. The following paragraphs are not meant to be mathematically rigorous, but rather heuristic, so please be gentle when you see any inaccuracies.</p> <p><em>"Naturality":</em> It is easy to give explicit examples of non-Kahler metrics on a Kahler manifold. Just take any Kahler metric $h$ and multiply it by a positive non-constant function $f$: the Kahler form of the new metric will then satisfy $\text{d} (f \omega) = \text{d} f \wedge \omega + f \text{d} \omega = \text{d} f \wedge \omega$. As $\omega$ is symplectic the wedge product $\text d f \wedge \omega$ can only be zero if $\text d f$ is zero, so the new metric is not Kahler.</p> <p>This feels like cheating to me. It's like starting a book on linear algebra by defining vector spaces axiomatically and then only giving the trivial space as an example. The example does not advance our understanding in any significant way.</p> <p>I would like to see an example where the metric $h$ arises in a geometric way or is somehow an obvious candidate for a metric on $X$. For example, consider a Hopf surface $X$, which arises as a quotient of $\mathbb C^2 \setminus {0}$ by a group $G$. The naive way to give an example of a metric on $X$ is to find a metric on $\mathbb C^2 \setminus {0}$ which is invariant under the action of $G$, and it is perfectly possible to give an explicit example of such a metric by some calculations (see [1] for an example). If only the Hopf surface were Kahler I would accept this as a "real" example.</p> <p><em>Motivation:</em> Given a hermitian metric $h$ there are several equivalent definitions of it being Kahler. One can say that its Kahler form is closed, that one can approximate the euclidean metric to the second degree in local coordinates, or that the Chern and Levi-Civita connections of $h$ are the same. This last condition is the one I like the most, because with good will one can interpret it as saying that the complex and Riemannian geometries defined by the metric are the same.</p> <p>This is all well and good, and I feel I understand the different definitions and the links between them. However, given an explicit metric, I have absolutely no intuition for if it is Kahler or not. I can't look at a metric and just go "Aha!", I have to "fly blind" and calculate.</p> <p>For example, take the Fubini-Study metric on $\mathbb P^n$. It can be obtained by considering a scalar product on $\mathbb C^{n+1}$ and saying that the scalar product of lines in that space is the "angle" between the lines (-ish). This is a very pretty and geometric way of obtaining a metric. Now, the only way I know to show that the metric obtained in this beautiful way is Kahler is by long and violent calculations. I can't give you an a priori plausibility argument for it being Kahler. The same is true for any explicit example of a Kahler metric on any manifold.</p> <p>I see this as failure on my part, and a sign that I have not really understood Kahler metrics. I think that an explicit example of a natural non-Kahler metric would help me understand complex geometry better.</p> <hr> <p>[1] <a href="http://mathoverflow.net/questions/57535/examples-of-non-kahler-surfaces-with-explicit-non-kahler-metric" rel="nofollow">http://mathoverflow.net/questions/57535/examples-of-non-kahler-surfaces-with-explicit-non-kahler-metric</a></p> http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold/81237#81237 Answer by Francesco Polizzi for Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold Francesco Polizzi 2011-11-18T09:43:09Z 2011-11-18T09:50:57Z <p>In his paper <a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;aid=4787044" rel="nofollow">Invariant Kahler metrics and projective embeddings of the flag manifold</a>, Bull. Austal. Math. Soc. <strong>49</strong> (1994), K. Yang considers the flag manifold $$F_{1,2,3}(\mathbb{C}^3):=SU(3)/S(U(1)^3)$$ and determines the space of invariant Hermitian and Kahler metrics on it.</p> <p>In particular, he shows that a Killing metric is <strong>not</strong> Kahler.</p> <p>On the other hand, by applying Kodaira embedding theorem, he proves that $F_{1,2,3}(\mathbb{C}^3)$ is projective algebraic and provides an explicit projective embedding of it.</p> <p>The computations are made quite explicitly in terms of the Maurer-Cartan form of $SU(3)$. </p> <p>So this could be one of the examples you are looking for.</p> http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold/81256#81256 Answer by Claudio Gorodski for Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold Claudio Gorodski 2011-11-18T15:57:24Z 2011-11-18T15:57:24Z <p>In <a href="http://www.jstage.jst.go.jp/article/kyushumfs/35/1/65/_pdf" rel="nofollow">Left invariant Riemannian metrics on complex Lie groups</a>, M. Goto and K. Uesu prove that if a complex analytic group $G$ admits a left-invariant Kahlerian metric, then $G$ is Abelian. I think this provides lots of natural examples on non-Abelian complex groups, but they are not compact. </p> http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold/81280#81280 Answer by José Figueroa-O'Farrill for Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold José Figueroa-O'Farrill 2011-11-18T20:46:14Z 2011-11-18T20:46:14Z <p>The (real) six-dimensional complex projective space $\mathbb{CP}^3$ is Kähler relative to the Fubini-Study metric; however it has another natural (at least in my opinion) metric which is <em>nearly Kähler</em> but non-Kähler. Moreover, just like the Fubini-Study metric, it is Einstein. It has the property that its metric cone has $G_2$ holonomy. Equivalently, $\mathbb{CP}^3$ with the nearly Kähler admits real Killing spinors. A good place to read about this is this paper by Moroianu, Nagy and Semmelmann: <a href="http://arxiv.org/abs/math/0611223" rel="nofollow">arXiv:math.DG/0611223</a> and references therein.</p> http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold/81485#81485 Answer by Gunnar Magnusson for Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold Gunnar Magnusson 2011-11-21T08:41:43Z 2011-11-21T08:41:43Z <p>Thanks Claudio, Francesco and José for your interesting answers.</p> <p>This is an "answer in absence" of Demailly; he doesn't use this site, but I thought his remark was nice enough to share. What follows is only his sketch, there are details to fill in that I haven't yet had the time to take care of.</p> <p>Let $X$ be a smooth projective variety, embedded in $\mathbb P^N$. Take $k$ big enough so that the vector bundle $T_X \otimes \mathcal O(k)$ is generated by its global sections. The Fubini-Study metric on $\mathbb P^N$ now gives a hermitian metric on $X$ by restriction, and on $\mathcal O(k)$ by taking powers of the determinant metric. These induce an $L^2$ metric on the global sections of $T_X \otimes \mathcal O(k)$.</p> <p>If we tensor by $\mathcal O(-k)$, then we have a surjective bundle map</p> <p>$$H^0(X, T_X \otimes \mathcal O(k)) \otimes \mathcal O(-k) \to T_X \to 0.$$</p> <p>The $L^2$ metric and the metric induced by Fubini-Study on $\mathcal O(-k)$ now gives a metric on the tensor product on the left. This induces a quotient metric $h$ on $T_X$. Despite its algebraic origins, this metric $h$ should (almost) never be a Kahler metric.</p> <p>Moreover, by applying approximation theorems of Tian, Demailly and others, one should be able to prove that these non-Kahler metrics are dense in the cone of hermitian metrics on $X$ -- i.e. starting from any metric on $X$ and using that to define the $L^2$ metric, it should be possible to fabricate a series of non-Kahler metrics as above which converges to the given metric. The process should in fact generalize and yield similar metrics on any holomorphic vector bundle over a projective manifold.</p> <p>This should imply that any heuristic of the form "my metric is geometric/algebraic, thus Kahler" is doomed to failure.</p>