"The complex version of Nash's theorem is not true" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:32:53Z http://mathoverflow.net/feeds/question/59738 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true "The complex version of Nash's theorem is not true" Joseph O'Rourke 2011-03-27T16:55:21Z 2011-12-21T12:47:39Z <p>The title quote is from p.221 of the 2010 book, <em>The Shape of Inner Space: String Theory and the Geometry of the Universe's Hidden Dimensions</em> by Shing-Tung Yau and Steve Nadis. "Nash's theorem" here refers to the Nash embedding theorem (discussed in an earlier MO question: "<a href="http://mathoverflow.net/questions/37708/" rel="nofollow">Nash embedding theorem for 2D manifolds</a>"). I would appreciate any pointer to literature that explains where and why embedding fails. This is well-known in the right circles, but I am having difficulty locating sources. Thanks!</p> http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true/59740#59740 Answer by Faisal for "The complex version of Nash's theorem is not true" Faisal 2011-03-27T17:05:30Z 2011-04-04T21:48:11Z <p>Using the maximum modulus principle you can show that $\mathbb{C}^n$ doesn't have any compact complex submanifolds of positive dimension. It follows that lots of complex manifolds, such as complex grassmannians and projective spaces for example, do not embed into $\mathbb{C}^n$.</p> http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true/59771#59771 Answer by diverietti for "The complex version of Nash's theorem is not true" diverietti 2011-03-27T22:34:50Z 2011-12-21T12:47:39Z <p>As Faisal says, there is no hope to have a general Nash-type theorem for <em>all</em> complex manifold, when the ambient space considered for the (isometric) embedding is some $\mathbb C^N$: no compact complex manifold of positive dimension could admit it. On the other hand there are a lot of compact complex manifold of positive dimension.</p> <p>Restricting the attention to the compact case then, one may guess if there is a natural analogous of the Nash embedding theorem. For instance, one may wonder if given any compact hermitian manifold $(X,\omega)$ one can embed it isometrically into some projective space, endowed with its natural Fubini-Study metric.</p> <p>It turns out that there are several restrictions both of analytic and geometrical nature. For instance, the starting metric $\omega$ must then be a Kähler form (as a restriction of the Fubini-Study Kähler form). It then gives a nonzero cohomology class in $H^{1,1}(X,\mathbb R)\subset H^2(X,\mathbb R)$. Moreover, being the Fubini-Study metric the Chern curvature form of the (anti)tautological line bundle, its cohomology class must be integral, that is in the image $H^{1,1}(X,\mathbb Z)$ in $H^{1,1}(X,\mathbb R)$ of the natural inclusion $H^2(X,\mathbb Z)\subset H^{2}(X,\mathbb R)$.</p> <p>The celebrated Kodaira's embedding theorem states that the converse is in fact true: </p> <p>Let $X$ be a compact complex manifold of dimension $\dim X=n$ possessing an integral $(1,1)$-cohomology class $[\omega]\in H^{1,1}(X,\mathbb Z)$ such that $[\omega]$ can be represented by a closed positive smooth $(1,1)$-form $\omega$. Then, there is an embedding $\iota\colon X\hookrightarrow \mathbb P^N$ to some complex projective space (and <em>a posteriori</em> $X$ is indeed algebraic by Chow's theorem).</p> <p>This embedding is obtained as follow. Given $[\omega]$, there exists a holomorphic hermitian line bundle $L\to X$ with hermitian metric $h$ such that $c_1(L)=[\omega]$ and moreover the Chern curvature form is such that $\frac i{2\pi}\Theta(L,h)=\omega$. Then, for all $m\in\mathbb N$, one considers the holomorphic map <code>\begin{aligned} \varphi_{|L^{\otimes m}|}\colon &amp; X\setminus\operatorname{Bs}(L^{\otimes m})\to\mathbb P(H^0(X,L^{\otimes m})^*) \\ &amp; x\mapsto\{\sigma\in H^0(X,L^{\otimes m})\mid \sigma(x)=0\}, \end{aligned}</code> where $\operatorname{Bs}(L^{\otimes m})$ is the set of points of $X$ where all sections in $H^0(X,L^{\otimes m})$ vanish simultaneously. What can be shown is in fact that for all sufficiently big $m$, one has that $\operatorname{Bs}(L^{\otimes m})=\emptyset$ and $\varphi_{|L^{\otimes m}|}$ is an immersive homeomorphism onto its image. Moreover, denoting by $\mathcal O(1)$ the (anti)tautological line bundle on $\mathbb P(H^0(X,L^{\otimes m})^*)$, one has that $$\varphi_{|L^{\otimes m}|}^*\mathcal O(1)\simeq L^{\otimes m}.$$ Now, $H^0(X,L^{\otimes m})$ has a natural inner product, namely the $L^2$-product given by $$\langle\langle\sigma,\tau{}\rangle\rangle_{L^2}=\int_X\langle\sigma(x),\tau(x)\rangle_{h^{\otimes m}}\frac{\omega^{n}}{n!},$$<br> and thus we have a corresponding Fubini-Studi metric on $\mathbb P(H^0(X,L^{\otimes m})^*)$; call it $\omega_{FS}$. Then, $\varphi_{|L^{\otimes m}|}^*\omega_{FS}$ lies in the same cohomology class of $c_1(L^{\otimes m})=m\cdot c_1(L)$. In particular, $\omega$ and $\frac 1m\varphi_{|L^{\otimes m}|}^*\omega_{FS}$ are cohomologous and since $X$ is Kähler by the $\partial\bar\partial$-lemma we have that $$\omega-\frac 1m\varphi_{|L^{\otimes m}|}^*\omega_{FS}=\frac i{2\pi}\partial\bar\partial f$$ for some globally defined smooth function $f\colon X\to\mathbb R$. This means that the new hermitian metric $\tilde h=he^{f}$ on $L$ has Chern curvature equal to $\frac 1m\varphi_{|L^{\otimes m}|}^*\omega_{FS}$ and with this new metric $\varphi_{|L^{\otimes m}|}$ becomes (after rescaling by the factor $m$) an isometric embedding (of course this is almost tautological: the only point here is that we modify the original metric just by rescaling the hermitian metric on the line bundle whose curvature was our original metric).</p> <p>Note that we didn't obtain an isometric embedding for the original metric. The best you can do we the original metric is to approximate it in the $C^2$-topology by the sequence of metrics $(\frac 1m\varphi_{|L^{\otimes m}|}^*\omega_{FS})_{m\in\mathbb N}$ by a result contained in the PhD thesis of G. Tian (please see the answer below by Joel Fine for more on that) (the convergence is now known to be in the $C^\infty$ topology on the space of symmetric covariant $2$-tensors, cf. the comment of Joel Fine here below). </p> <p>Remark also that not any compact Kähler manifold admits such a cohomology class, thus the theorem "really give a criterion". For instance a generic compact complex torus of complex dimension greater than or equal to two is a compact Kähler manifold (with its natural flat metric) which does not admit any embedding in some projective space.</p> <p>Turing to the literature, regarding the compact case there are several wonderful books in the literature. I'll give you two or three names, which are my favorites.</p> <p>(1) J.-P. Demailly, "Complex Analytic and Differential Geometry", </p> <p>(2) C. Voisin, "Théorie de Hodge et géométrie algébrique complexe".</p> <p>(3) R. O. Wells Jr., "Differential analysis on complex manifolds".</p> <p>For the non-compact case, there are also plenty of books of course. If you want to know more on the theory of Stein manifolds (precisely the analytic close submanifold of some $\mathbb C^N$), then for example </p> <p>(4) L. Hörmander, "An introduction to complex analysis in several variables".</p> <p>would do the job, at least for an introduction.</p> <p><strong>[edited after comments by Joel Fine and Robert Bryant]</strong></p> http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true/60207#60207 Answer by Robert Bryant for "The complex version of Nash's theorem is not true" Robert Bryant 2011-03-31T17:08:09Z 2011-03-31T17:08:09Z <p>The failure is actually more profound than you might guess at first glance: </p> <p>There are conformal metrics on the Poincare disk that cannot (even locally) be isometrically induced by embedding in $\mathbb{C}^n$ by any holomorphic mapping. For example, there is no complex curve in $\mathbb{C}^n$ for which the induced metric has either curvature that is positive somewhere or constant negative curvature.</p> <p>You can get around the positivity problem by looking for complex curves in $\mathbb{P}^n$ (with the Fubini-Study metric, say), but even there, there are no complex curves with constant negative curvature.</p> <p>More generally, for any Kahler metric $g$ on an $n$-dimensional complex manifold $M$, there always exist many metrics on the Poincare disk that cannot be isometrically induced on the disk via a holomorphic embedding into $M$. </p> <p>This should not be surprising if you are willing to be a little heuristic: Holomorphic mappings of a disk into an $n$-dimensional complex manifold essentially depend on choosing $n$ holomorphic functions of one complex variable and each such holomorphic function essentially depends on choosing two (analytic) real functions of a single real variable. However, the conformal metrics on the disk depend essentially on one (positive) smooth function of two variables, which is too much `generality' for any finite number of functions of a single variable to provide. </p> http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true/60520#60520 Answer by Joel Fine for "The complex version of Nash's theorem is not true" Joel Fine 2011-04-04T03:45:55Z 2011-04-04T03:45:55Z <p>This is not so much an answer to the original question, more an addition to the answer of diverietti and part of the answer Robert Bryant. Both mention the following analogue of Nash's embedding problem:</p> <p>Given a Kähler manifold $(X, \omega)$, is there a projective embedding $X \to \mathbb{CP}^n$ for which the Fubini-Study metric pulls back to give $\omega$?</p> <p>As Robert says, taken literally the answer is no. However, there is a beautiful theorem of Tian which says that the answer is yes, <i>provided one is willing to let $n$, the dimension of the projective space, tend to infinity.</i> </p> <p>More precisely, let $L \to X$ be a positive holomorphic line bundle on $X$. This means there is a Hermitian metric $h$ in $L$ whose curvature is a Kähler form $\omega$ in $c_1(L)$. (Moreover, all Kähler forms in $c_1(L)$ arise this way.) With this metric $h$ and the volume form $\omega^n$ you can define an $L^2$-inner-product on the space of holomorphic sections of $L^k$, where $k$ is a large integer. Let $s_0, \ldots , s_n$ be an orthonormal basis of holomorphic sections (where $n$ depends on $k$ roughly $n \sim k^m$ where $m$ is the dimension of $X$). Then (for large $k$) the map $$f_k(x) = [s_0(x) : \cdots : s_n(x) ]$$ defines an embedding to $\mathbb{CP}^n$ which has the following property: if we restrict the Fubini-Study metric from projective space to $X$ via the map $f_k$, rescale by $1/k$ (to keep the total volume fixed) and then take the limit as $k \to \infty$ we get the original metric $\omega$.</p>