# What's the difference between a real manifold and a smooth variety?

I am teaching a course on Riemann Surfaces next term, and would like a list of facts illustrating the difference between the theory of real (differentiable) manifolds and the theory non-singular varieties (over, say, $\mathbb{C}$). I am looking for examples that would be meaningful to 2nd year US graduate students who has taken 1 year of topology and 1 semester of complex analysis.

Here are some examples that I thought of:

1. Every $n$-dimensional real manifold embeds in $\mathbb{R}^{2n}$. By contrast, a projective variety does not embed in $\mathbb{A}^n$ for any $n$. Every $n$-dimensional non-singular, projective variety embeds in $\mathbb{P}^{2n+1}$, but there are non-singular, proper varieties that do not embed in any projective space.

2. Suppose that $X$ is a real manifold and $f$ is a smooth function on an open subset $U$. Given $V \subset U$ compactly contained in $U$, there exists a global function $\tilde{g}$ that agrees with $f$ on $V$ and is identically zero outside of $U$.

By contrast, consider the same set-up when $X$ is a non-singular variety and $f$ is a regular function. It may be impossible find a global regular function $g$ that agrees with $f$ on $V$. When $g$ exists, it is unique and (when $f$ is non-zero) is not identically zero on outside of $U$.

3. If $X$ is a real manifold and $p \in X$ is a point, then the ring of germs at $p$ is non-noetherian. The local ring of a variety at a point is always noetherian.

What are some more examples?

Answers illustrating the difference between real manifolds and complex manifolds are also welcome.

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2 is not quite correct, since f might blow up at the boundary of U. However, if V is any open set compactly contained in U, then there is a globally defined function g that agrees with f on V. We can furthermore take g to be identically zero outside U. –  Charles Staats May 8 '10 at 4:02
@Charles Staats, thanks for catching that! This has been fixed. –  jlk May 8 '10 at 4:27
It would be more striking to contrast complex manifolds with real-analytic manifolds (and likewise for vector bundles on them), such as the Grauert-Morrey theorem, or things like the lack of distinction between diffeomorphism and analytic isomorphism for real-analytic manifolds. –  BCnrd May 8 '10 at 4:33
Don't these all illustrate the differences between real and complex manifolds, rather than between real manifolds and smooth complex varieties? –  Rebecca Bellovin May 8 '10 at 4:38
@Rebecca Bellovin, sure. A better phrasing of the question would be give facts illustrating the difference between A) real manifolds, B) complex manifolds, and C) smooth complex varieties. I was, however, trying to keep the phrasing as simple as possible. –  jlk May 8 '10 at 4:58

Here is a list biased towards what is remarkable in the complex case. (To the potential peeved real manifold: I love you too.) By "complex" I mean holomorphic manifolds and holomorphic maps; by "real" I mean $\mathcal{C}^{\infty}$ manifolds and $\mathcal{C}^{\infty}$ maps.

• Consider a map $f$ between manifolds of equal dimension. In the complex case: if $f$ is injective then it is an isomorphism onto its image. In the real case, $x\mapsto x^3$ is not invertible.

• Consider a holomorphic $f: U-K \rightarrow \mathbb{C}$, where $U\subset \mathbb{C}^n$ is open and $K$ is a compact s.t. $U-K$ is connected. When $n\geq 2$, $f$ extends to $U$. This so-called Hartogs phenomenon has no counterpart in the real case.

• If a complex manifold is compact or is a bounded open subset of $\mathbb{C}^n$, then its group of automorphisms is a Lie group. In the smooth case it is always infinite dimensional.

• The space of sections of a vector bundle over a compact complex manifold is finite dimensional. In the real case it is always infinite dimensional.

• To expand on Charles Staats's excellent answer: few smooth atlases happen to be holomorphic, but even fewer diffeomorphisms happen to be holomorphic. Considering manifolds up to isomorphism, the net result is that many complex manifolds come in continuous families, whereas real manifolds rarely do (in dimension other than $4$: a compact topological manifold has at most finitely many smooth structures; $\mathbb{R}^n$ has exactly one).

On the theme of zero subsets (i.e., subsets defined locally by the vanishing of one or several functions):

• One equation always defines a codimension one subset in the complex case, but {$x_1^2+\dots+x_n^2=0$} is reduced to one point in $\mathbb{R}^n$.

• In the complex case, a zero subset isn't necessarily a submanifold, but is amenable to manifold theory by Hironaka desingularization. In the real case, any closed subset is a zero set.

• The image of a proper map between two complex manifolds is a zero subset, so isn't too bad by the previous point. Such a direct image is hard to deal with in the real case.

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I'm slightly confused. The equation $x_1^2 + ... + x_n^2 = 0$ defines a codimension one subset (complex submanifold) of $\mathbb{C}^n$ namely the complex $(n-1)$ dimensional unit sphere $S^n$. The unit sphere is compact and obviusly not a point. But any compact submanifold of $\mathbb{C}$ is necessarily a point (the coordinate function would have to be bounded on $S^n$ and therefore constant). Where does my reasoning go wrong? –  bavajee May 8 '10 at 11:58
@bajavee: the set $X=$ {$(x_1,\dots,x_n) \in \mathbb{C}^n\ \vert\ x_1^2+\dots+ x_n^2=0$}, called a complex quadric, is not compact (note that it is invariant by multiplication by a complex number). –  Dan Zaffran May 8 '10 at 12:18
Dan, thank you for your reply. I was thinking of $x_1^2 + ... x_n^2 = 1$. Anyway my confusion has cleared up: the complex unit sphere is not a complex submanifold of $\mathbb{C}^n$ because it is the zero set of $|x_1|^2 + ... + |x_n|^2 - 1$ (which is not a holomorphic function) and not of $x_1^2 + ... x_n^2 - 1$ (which is holomorphic). My mind was stuck in the real variable world. Embarrassing. –  bavajee May 8 '10 at 12:44

Any two compact surfaces (without boundary) of the same genus are diffeomorphic. However, if S is a surface of genus g > 0, there are uncountably many non-isomorphic complex (or, equivalently, algebraic) structures on S.

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A proper variety doesn't have (non-constant) global sections. A real manifold, compact or not, has lots of global sections.

There are lots of maps between real manifolds. Maps between varieties are much more restricted (e.g. by Riemann-Hurwitz in the case of curves).

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you mean "nonconstant" instead of "nonzero". –  Martin Brandenburg May 8 '10 at 10:15
oops, yeah, thanks, fixed. –  Kevin H. Lin May 8 '10 at 18:31

Some embedding statements.

A compact complex subvariety of ${\mathbb{C}}^n$ is a point. However, every compact real manifold of dimension $n$ can be realized as a submanifold of some ${\mathbb{R}}^{2n}$.

There are compact complex manifolds that cannot be embedded into complex projective space. An example most often quoted in textbooks is the Hopf manifold, which is not even Kahler. On the other hand, I heard that embedding into real projective space is not often considered in differential geometry.

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Some of these properties are local, and distinguish analytic and algebraic functions, from smooth functions.

Others are global and distinguish compact manifolds from projective manifolds by their difference in containment of many subvarieties.

some are as simple as contrasting the dimension, and homology of say projective Real space from that of projective Complex space, as noted.

As a deep contrast between smooth and analytic structure I like the answer pointing out that analytic riemann surfaces can have many non isomorphic structures on the same smooth manifold. this is interesting already for genus one manifolds. and it is not trivial to show that the sphere has only one complex analytic structure.

that might be a fun challenge to a class, to show two Riemann surfaces both diffeomorphic to the 2-sphere, are holomorphically isomorphic.

You might also be interested in some of the articles by Kolla'r on the Nash conjecture contrasting real varieties and real manifolds. such as "What are the simplest varieties?", Bulletin, vol 38. I like the pair of theorems 54, 51, subtitled respectively: "the Nash conjecture is true in dim 3", and "The Nash conjecture is false in dim 3".

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@mathwonk, which article of Kolla'r are you referring to? –  jlk May 8 '10 at 21:52

A connected real manifold can be disconnected by the removal of a submanifold but the complement of a subvariety on an irreducible variety is still connected.

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This is really just a consequence of the fact that there are two real dimensions per one complex dimension. Thus, when you remove a subvariety, you are removing a submanifold of real codimension at least two, which cannot possibly disconnect the original manifold. Incidentally, a smooth variety is irreducible iff it's connected, since the intersection of two irreducible components would be singular –  Charles Staats May 8 '10 at 4:11

For a closed analytic subset Z ⊂ S of a (say compact) complex manifold with complement U=S-Z one has additivity of the (topological) Euler characteristic:

Χ(S)=Χ(Z)+Χ(U).

This is wrong for if S and Z are topological spaces or smooth manifolds. Indeed, take for Z a point on a circle S. This (surprising) difference was recently pointed out to me by Manfred Lehn.

Of course there is also no additivity of Poincare-Polynomials or other "motivic" invariants of complex varieties.

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I think there's some big difference concerning the metric approach too.

In fact, the Gram-Schmidt process (which is real analytic) enables us -in real differential geometry- to find some local orthonormal frames (for any hermitian bundle, and in particular for the tangent bundle), whereas in the holomorphic case, very subtle differences may occur there.

For example, in the Kähler case, we can find "orthonormal" frames for the tangent bundle at order 2, which is the key for the Kähler identities, leading to fundamental results like the equality of all Laplacians and thus the Hodge decomposition theorem in the compact case.

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