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Next: 2.4 Further Examples Up: 2 A Real Algebraic Previous: 2.2 The First Examples

2.3 Classical Small Perturbations from the Complex Point of View

To consider further examples, it would be useful to understand what is going on in the complex domain, when one makes a classical small perturbation (see Section 1.5).

First, consider the simplest special case: a small perturbation of the union of two real lines. Denote the lines by $ L_1$ and $ L_2$ and the result by $ C$. As we saw above, $ \mathbb{C}L_i$ and $ \mathbb{C}C$ are homeomorphic to $ S^2$. The spheres $ \mathbb{C}L_1$ and $ \mathbb{C}L_2$ intersect each other at a single point. By the complex version of the implicit function theorem, $ \mathbb{C}C$ approximates $ \mathbb{C}L_1\cup\mathbb{C}L_2$ outside a neighborhood $ U_0$ of this point in the sense that $ \mathbb{C}C\smallsetminus U_0$ is a section of a tulubular neighborhood $ U_1$ of $ (\mathbb{C}L_1\cup\mathbb{C}
L_1)\smallsetminus U_0$, cf. 1.5.A. Thus $ \mathbb{C}C$ may be presented as the union of two discs and a part contained in a small neighborhood of $ \mathbb{C}L_1\cap\mathbb{C}L_2$. Since the whole $ \mathbb{C}C$ is homeomorphic to $ S^2$ and the complement of two disjoint discs embedded into $ S^2$ is homeomorphic to the annulus, the third part of $ \mathbb{C}C$ is an annulus. The discs are the complements of a neighborhood of $ \mathbb{C}L_1\cap\mathbb{C}L_2$ in $ \mathbb{C}L_1$ and $ \mathbb{C}L_2$, respectively, slightly perturbed in $ \mathbb{C}P^2$, and the annulus connects the discs through the neighborhood $ U_0$ of $ \mathbb{C}L_1\cap\mathbb{C}L_2$.

This is the complex view of the picture. Up to this point it does not matter whether the curves are defined by real equations or not.

To relate this to the real view presented in Section 1.5, one needs to describe the position of the real parts of the curves in their complexifications and the action of $ conj$. It can be recovered by rough topological agruments. The whole complex picture above is invariant under $ conj$. This means that the intersection point of $ \mathbb{C}L_1$ and $ \mathbb{C}L_2$ is real, its neighborhood $ U_0$ can be chosen to be invariant under $ conj$. Thus each half of $ \mathbb{C}C$ is presented as the union of two half-discs and a half of the annulus: the half-discs approximate the halves of $ \mathbb{C}L_1$ and $ \mathbb{C}L_2$ and a half of annulus is contained in $ U_0$. See Figure 15.

Figure 15:
\begin{figure}\centerline{\epsffile{f2_1n.eps}}\end{figure}

This is almost complete description. It misses only one point: one has to specify which half-discs are connected with each other by a half-annulus.

First, observe, that the halves of the complex point set of any curve of type I can be distinguished by the orientations of the real part. Each of the halves has the canonical orientation defined by the complex structure, and this orientation induces an orientation on the boundary of the half. This is one of the complex orientations. The other complex orientation comes from the other half. Hence the halves of the complexification are in one-to-one correspondence to the complex orientations.

Now we have an easy answer to the question above. The halves of $ \mathbb{C}L_i$ which are connected with each other after the perturbation correspond to the complex orientations of $ \mathbb{R}L_i$ which agree with some orientation of $ \mathbb{R}C$. Indeed, the perturbed union $ C$ of the lines $ L_i$ is a curve of type I (since this is a nonempty conic, see Section 2.2). Each orientation of its real part $ \mathbb{R}C$ is a complex orientation. Choose one of the orientations. It is induced by the canonical orientation of a half of the complex point set $ \mathbb{C}C$. Its restriction to the part of the $ \mathbb{R}C$ obtained from $ \mathbb{R}L_i$ is induced by the orientation of the corresponding part of this half.

The union of two lines can be perturbed in two different ways. On the other hand, there are two ways to connect the halves of their complexifications. It is easy to see that different connections correspond to different perturbations. See Figure 16.

Figure 16:
\begin{figure}\centerline{\epsffile{f2_2n.eps}}\end{figure}

The special classical small perturbation considered above is a key for understanding what happens in the complex domain at an arbitrary classical small perturbation. First, look at the complex picture, forgetting about the real part. Take a plane projective curve, which has only nondegenerate double points. Near such a point it is organized as a union of two lines intersecting at the point. This means that there are a neighborhood $ U$ of the point in $ \mathbb{C}P^2$ and a diffeomorphism of $ U$ onto $ \mathbb{C}^2$ mapping the intersection of $ U$ and the curve onto a union of two complex lines, which meet each other in 0. This follows from the complex version of the Morse lemma. By the same Morse lemma, near each double point the classical small perturbation is organized as a small perturbation of the union of two lines: the union of two transversal disks is replaced by an annulus.

For example, take the union of $ m$ projective lines, no three of which have a common point. Its complex point set is the union of $ m$ copies of $ S^2$ such that any two of them have exactly one common point. A perturbation can be thought of as removal from each sphere $ m-1$ disjoint discs and insertion $ \frac{m(m-1)}2$ tubes connecting the boundary circles of the disks removed. The result is orientable (since it is a complex manifold). It is easy to realize that this is a sphere with $ \frac{(m-1)(m-2)}2$ handles. One may prove this counting the Euler characteristic, but it may be seen directly: first, by inserting the tubes which join one of the lines with all other lines we get a sphere, then each additional tube gives rise to a handle. The number of these handles is

$\displaystyle \left(\begin{matrix}m-1  2\end{matrix}\right)=\frac{(m-1)(m-2)}2.$

By the way, this description shows that the complex point set of a nonsingular plane projective curve of degree $ m$ realizes the same homology class as the union of $ m$ complex projective lines: the $ m$-fold generator of $ H_2(\mathbb{C}P^2)(=\mathbb{Z})$.

Now let us try to figure out what happens with the complex schemes in an arbitrary classical small perturbation of real algebraic curves. The general case requirs some technique. Therefore we restrict ourselves to the following intermediate assertion.

2.3.A   (Fiedler [Rok-78, Section 3.7] and Marin [Mar-80].) Let $ A_1, \dots, A_s$ be nonsingular curves of degrees $ m_1, \dots, m_s$ such that no three of them pass through the same point and $ A_i$ intersects transversally $ A_j$ in $ m_im_j$ real points for any $ i$, $ j$. Let $ A$ be a nonsingular curve obtained by a classical small perturbation of the union $ A_1\cup\dots A_s$. Then $ A$ is of type I if and only if all $ A_i$ are of type I and there exists an orientation of $ \mathbb{R}A$ which agrees with some complex orientations of $ A_1, \dots, A_s$ (it means that the deformation turning $ A_1\cup\dots A_s$ into $ A$ brings the complex orientations of $ A_i$ to the orientations of the corresponding pieces of $ \mathbb{R}A$ induced by a single orientation of the whole $ \mathbb{R}A$).

If it takes place, then the orientation of $ \mathbb{R}A$ is one of the complex orientations of $ A$.

Proof. If some of $ A_i$ is of type II, then it has a pair of complex conjugate imaginary points which can be connected by a path in $ \mathbb{C}A_i\smallsetminus \mathbb{R}A_i$. Under the perturbation this pair of points and the path survive (being only slightly shifted), since they are far from the intersection where the real changes happen. Therefore $ A$ in this case is also of type II.

Assume now that all $ A_i$ are of type I. If $ A$ is also of type I then a half of $ \mathbb{C}A$ is obtained from halves of $ \mathbb{C}A_i$ as in the case considered above. The orientation induced on $ \mathbb{R}A$ by the orientation of the half agrees with orientations induced from the halves of the corresponding pieces. Thus a complex orietation of $ A$ agrees with complex orientations of $ A_i$'s.

Again assume that all $ A_i$ are of type I. Let some complex orientations of $ A_i$ agree with a single orientation of $ \mathbb{R}A$. As it follows from the Morse Lemma, at each intersection point the perturbation is organized as the model perturbation considered above. Thus the halves of $ \mathbb{C}A_i$'s defining the complex orientations are connected. It cannot happen that some of the halves will be connected by a chain of halves to its image under $ conj$. But that would be the only chance to get a curve of type II, since in a curve of type II each imaginary point can be connected with its image under $ conj$ by a path disjoint from the real part. $ \qedsymbol$

Figure 17: Construction of nonsingular cubic curves. Cf. Figure 2.
\begin{figure}\centerline{\epsffile{f2-1s.eps}}\end{figure}

Figure 18: Construction of nonsingular quartic curves. Cf. Figure 3.
\begin{figure}\centerline{\epsffile{f3-1s.eps}}\end{figure}

Figure 19: Construction of nonsingular quintic curves. Cf. Figure 4.
\begin{figure}\centerline{\epsffile{f4-1s.eps}}\end{figure}


next up previous
Next: 2.4 Further Examples Up: 2 A Real Algebraic Previous: 2.2 The First Examples
Oleg Viro 2000-12-30