## “The complex version of Nash’s theorem is not true”

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The title quote is from p.221 of the 2010 book, The Shape of Inner Space: String Theory and the Geometry of the Universe's Hidden Dimensions by Shing-Tung Yau and Steve Nadis. "Nash's theorem" here refers to the Nash embedding theorem (discussed in an earlier MO question: "Nash embedding theorem for 2D manifolds"). 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!

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## 4 Answers

The failure is actually more profound than you might guess at first glance:

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.

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.

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 $M$.

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.

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I am quite willing to be more than a little heuristic under such brilliant guidance, Robert! Thanks for so much enlightenment in so few lines. – Georges Elencwajg Apr 2 2011 at 18:37

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i really like the hueristic argument too! – Colin Tan Apr 19 2011 at 12:22

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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 $\mathbb{C}^n$.

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Perfect---Thank you!! – Joseph O'Rourke Mar 27 2011 at 18:21

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Nitpick: Stein manifolds are those that admit a proper embedding into $\mathbb C^n$. An open subset of $\mathbb C^n$ is not necessarily Stein. – jvp Mar 29 2011 at 11:45

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Moreover, Nash theorem is about isometric embedding... – diverietti Apr 4 2011 at 19:31
jvp & diverietti: Thanks for the comments. I've removed my remark about Stein manifolds. – Faisal Apr 4 2011 at 21:48

As Faisal says, there is no hope to have a general Nash-type theorem for all 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.

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.

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 $H^2(X,\mathbb Z)\subset H^{2}(X,\mathbb R)$.

The celebrated Kodaira's embedding theorem states that the converse is in fact true:

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 a posteriori $X$ is indeed algebraic by Chow's theorem).

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 ```$$
\begin{aligned}
\varphi_{|L^{\otimes m}|}\colon & X\setminus\operatorname{Bs}(L^{\otimes m})\to\mathbb 
P(H^0(X,L^{\otimes m})^*) \\
& x\mapsto\{\sigma\in H^0(X,L^{\otimes m})\mid \sigma(x)=0\},
\end{aligned}
$$``` 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 $$\langle\langle\sigma,\tau{}\rangle\rangle_{L^2}=\int_X\langle\sigma(x),\tau(x)\rangle_{h^{\otimes m}}\frac{\omega^{n}}{n!},$$
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).

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).

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.

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.

(1) J.-P. Demailly, "Complex Analytic and Differential Geometry",

(2) C. Voisin, "Théorie de Hodge et géométrie algébrique complexe".

(3) R. O. Wells Jr., "Differential analysis on complex manifolds".

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

(4) L. Hörmander, "An introduction to complex analysis in several variables".

would do the job, at least for an introduction.

[edited after comments by Joel Fine and Robert Bryant]

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@diverletti: Check your statement of the Kodaira embedding theorem. The embedding $\iota:X\hookrightarrow\mathbb{P}^N$ that pulls back the Fubini-Study $(1,1)$-form to be the given form is NOT holomorphic in general; it's only symplectic. The Kodaira embedding theorem is about the space of sections of `positive' bundles, and you don't mention this. If $X$ is a Riemann surface of genus $g>1$ and $\omega$ is the positive integral $(1,1)$-form that has constant negative curvature, there is NO holomorphic embedding into $\mathbb{P}^N$ that induces an integral multiple of $\omega$. – Robert Bryant May 9 2011 at 15:19
Dear Robert, I think that my statement is definitely true. You just make some confusion. The Kähler metrics I consider, being integral, are already seen as curvature forms of some line bundle (whose first Chern class is represented by the given Kähler form and, as such is positive, so that the line bundle itself is, by definition, positive). Your counterexample is not true because of the following reason: you consider the Kähler metric as a metric on the tangent bundle (which, in the case of Riemann surfaces is incidentally a line bundle, too) and then you take the curvature (to be continued) – diverietti May 9 2011 at 23:14
which, in the cases your consider, cannot -of course- be positive. Thus, it does not satisfy the hypotheses of my statement since being a (1,1)-form not positive it cannot be a Kähler form. What your example says, instead, is that if you normalize your Kähler form in such a way that its total volume is for instance one, then you will find a (ample) hermitian line bundle on X whose curvature is exactly your form. – diverietti May 9 2011 at 23:19

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@diverletti: You need to look up a statement of the Kodaira embedding theorem. It does not say what you think it does. I stand by my claim that there is NO holomorphic curve (compact or not) in $\mathbb{CP}^n$ such that the Fubini-Study metric induces a metric (i.e., a positive (1,1)-form) of constant negative curvature on the curve. (It is a old theorem of Calabi, not mine.) As Joel points out below, the Kodaira embedding theorem is not about isometric embeddings. – Robert Bryant May 12 2011 at 3:59

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I think this version of your answer is much better! Just one small technical point: the convergence of the sequence you mention is in fact in the C-infinity topology. Tian originally proved C^2 convergence but this has subsequently been improved. (I guess this extension is due to Ruan, but I'm a little hazy on the exact history. There is also important work of Zelditch and Catlin on this.) – Joel Fine Dec 21 2011 at 11:41
show 2 more comments

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:

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 $\omega$?

As Robert says, taken literally the answer is no. However, there is a beautiful theorem of Tian which says that the answer is yes, provided one is willing to let $n$, the dimension of the projective space, tend to infinity.

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 $\omega$.

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Joel, I am not sure I can understand completely your answer. For Kodaira's embedding theorem already follows that if you rescale the restriction of the Fubini-Study metric by one over the power of the line bundle you need to embed it, then it is the original metric. When does the $L^2$ metric come into the picture? Which is exactely the theorem of Tian you are talking to? – diverietti Apr 4 2011 at 19:38
Kodaira's theorem tells you that for sufficiently large $k$, the map $f_k$ gives you an embedding for any choice of basis of hol sections of $L^k$ whatsoever. If you pull back the cohomology class of the Fubini-Study metric and divide by 1/k then you get the first Chern class of L. But Kodaira's theorem says nothing at all about the pull back of the actual metric itself. To get a well defined metric via an embedding, you must specify the basis (at least up to the action of the unitary group). That's where you need the $L^2$ inner-product. Continued... – Joel Fine Apr 4 2011 at 19:55
... If you want to recover the original metric you started with, you use an $L^2$-orthonormal basis of sections for each $k$, pull back the metric, rescale by $1/k$ and then take a limit as $k \to \infty$. This gives you a sequence of Kähler metrics on $X$ converging to the one you first thought of. This is proved by Tian in his article "On a set of polarized Kähler metrics on algebraic manifolds." J. Differential Geom. 32 (1990).  – Joel Fine Apr 4 2011 at 19:58
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