5.3 Restrictions on Topology of Real Algebraic Surfaces

As in the case of real plane projective curves, the set of real points of a nonsingular spatial surface of degree $m$ is one-sided, if $m$ is odd, and two-sided, if $m$ is even. Indeed, by the Bézout theorem a generic line meets a surface of degree $m$ in a number of points congruent to $m$ modulo 2. On the other hand, whether a topological surface embedded in $\mathbb{R}P^3$ is one-sided or two-sided, can be detected by its intersection number modulo 2 with a generic line: a surface is one-sided, iff its intersection number with a generic line is odd.

There are some other restrictions on topology of a nonsingular surface of degree $m$ which can be deduced from the Bézout theorem.

5.3.A. (On Number of Cubic's Components).   The set of real points of a nonsingular surface of degree three consists of at most two components.

Proof. Assume that there are at least three components. Only one of them is one-sided, the other two are contractible. Connect with a line two contractible components. Since they are zero-homologous, the line should intersect each of them with even intersection number. Therefore the total number of intersection points (counted with multiplicities) of the line and the surface is at least four. This contradicts to the Bézout theorem, according to which it should be at most three. $\qedsymbol$

5.3.B. (On Two-Component Cubics).   If the set of real points of a nonsingular surface of degree 3 consists of two components, then the components are homeomorphic to the sphere and projective plane (i. e., this is $P\amalg S$).

Proof. Choose a point inside the contractible component. Any line passing through this point intersects the contractible component at least in two points. These points are geometrically distinct, since the line should intersect also the one-sided component. On the other hand, the total number of intersection points is at most three according to the Bézout theorem. Therefore any line passing through the selected point intersects one-sided component exactly in one point and two-sided component exactly in two points. The set of all real lines passing through the point is $\mathbb{R}P^2$. Drawing a line through the selected point and a real point of the surface defines a one-to-one map of the one-sided component onto $\mathbb{R}P^2$ and two-to-one map of the two-sided component onto $\mathbb{R}P^2$. Therefore the Euler characteristic of the one-sided component is equal to $\chi(\mathbb{R}P^2)=1$, and the Euler characteristic of the two-sided component is $2\chi(\mathbb{R}P^2)=2$. This determines the topological types of the components. $\qedsymbol$

5.3.C. (Estimate for Diameter of Region Tree).   The diameter of the region tree⁶ of a nonsingular surface of degree $m$ is at most $[m/2]$.

Proof. Choose two vertices of the region tree the most distant from each other. Choose a point in each of the coresponding regions and connect the points by a line. $\qedsymbol$

5.3.D   The set of real points of a nonsingular surface of degree 4 has at most two noncontractible components. If the number of noncontractible components is 2, then there is no other component.

Proof. First, assume that there are at least three noncontractible components. Consider the complement of the union of three noncontractible components. It consists of three domains, and at least two of them are not adjacent (cf. the previous subsection: the graph of adjacency of the domains should be a tree). Connect points of nonadjacent domains with a line. It has to intersect each of the three noncontractible components. Since they are zero-homologous, it intersects each of them at least in two points. Thus, the total number intersection points is at least 6, which contradicts to the Bézout theorem.

Now assume that there are two noncontractible components and some contractible component. Choose a point $p$ inside the contractible component. The noncontractible components divide $\mathbb{R}P^3$ into 3 domains. One of the domains is adjacent to the both noncontractible components, while each of the other two domains is bounded by a single noncontractible component. If the contractible component lies in a domain bounded by a single noncontractible component, then take a point $q$ in the other domain of the same sort, and connect $p$ and $q$ with a line. This line meets each of the three components at least twice, which contradicts to the Bézout theorem.

Otherwise (i. e. if the contractible component lies in the domain adjacent to both noncontractible components), choose inside each of the two other domains an embedded circle, which does not bound in $\mathbb{R}P^3$. Denote these circles by $L_1$ and $L_2$. Consider a surface $C_i$ swept by lines connecting $p$ with points of $L_i$. It realizes the nontrivial homology class. Indeed, take any line $L$ transversal to it. Each point of $L\cap C_i$ corresponds to a point of the intersection of $L_i$ and the plane consisting of lines joining $p$ with $L$. Since $L_i$ realizes the nonzero homology class, the intersection number of $L_i$ with a plane is odd. Therefore the intersection number of $L$ and $C_i$ is odd. Since both $C_1$ and $C_2$ realizes the nontrivial homology class, their intersection realizes the nontrivial one-dimensional homology class. This may happen only if there is a line passing through $p$ and meeting $L_1$ and $L_2$. Such a line has to intersect all three components of the quadric surface. Each of the components has to be met at least twice. This contradicts to the Bézout theorem. $\qedsymbol$

Remark 5.3.E   In fact, if a nonsingular quartic surface has two noncontractible components then each of them is homeomorphic to torus. It follows from an extremal property of the refined Arnold inequality 5.3.L. I do not know, if it can be deduced from the Bézout theorem. However, if to assume that one can draw lines in the domains of the complement which are not adjacent to both components, then it is not difficult to find homeomorphisms between the components of the surface and the torus, which is the product of these two lines. Cf. the proof of 5.3.B.

5.3.F. Generalization of 5.3.D.   Let $A$ be a nonsingular real algebraic surface of degree $m$ in the 3-dimensional projective space. Then the diameter of the adjacency tree of domains of the complement of $\mathbb{R}A$ is at most $[m/2]$. If the degree is even and the diameter of the adjacency tree of the connected components of the complement of the union of the noncontractible components is exactly $m/2$, then there is no contractible components.

The proof is a straightforward generalization of the proof of 5.3.D.$\qedsymbol$

Surprisingly, Bézout theorem gave much less restrictions in the case of surfaces than in the case of plane curves. In particular, it does not give anything like Harnack Inequality. Most of restrictions on topology of surfaces are analogous to the restrictions on flexible curves and were obtained using the same topological tools. Here is a list of the restrictions, though it is non-complete in any sense.

The restrictions are formulated below for a nonsingular real algebraic surface $A$ of degree $m$ in the 3-dimensional projective space. In these formulations and in what follows we shall denote the $i$-th Betti number of $X$ over field $\mathbb{Z}_2$ (which is nothing but $\dim_{\mathbb{Z}_2}H_i(X ; \mathbb{Z}_2)$) by $b_i(X)$. In particular, $b_0(X)$ is the number of components of $X$. By $b_*(X)$ we denote the total Betti number, i. e. $\sum_{i=0}^{infty} b_i(X)=\dim_{\mathbb{Z}_2}H_*(X ; \mathbb{Z}_2)$.

5.3.G. Generalized Harnack Inequality.  

$$b_*(\mathbb{R}A)\le m^3-4m^2+6m.$$

Remark 5.3.H   This is a special case of Smith-Floyd Theorem 3.2.B, which in the case of curves implies Harnack Inequality, see Subsections 3.2. It says that for any involution $i$ of a topological space $X$

$$b_*(fix(i))\le b_*(X).$$

Applying this to the complex conjugation involution of the complexification $\mathbb{C}A$ of $A$ and taking into account that $\dim_{\mathbb{Z}_2}H_*(\mathbb{C}A ; \mathbb{Z}_2)= m^3-4m^2+6m$ one gets 5.3.G.

5.3.I. Extremal Congruences of Generalized Harnack Inequality.   If

$$b_*(\mathbb{R}A)=m^3-4m^2+6m,$$

then

$$\chi(\mathbb{R}A)\equiv(4m-3m^2)/3 \mod 16.$$

If $b_*(\mathbb{R}A)= m^3-4m^2+6m-2$, then

$$\chi(\mathbb{R}A)\equiv(4m-m^3\pm 6)/3 \mod 16.$$

5.3.J. Petrovsky - Oleinik Inequalities.  

$$-(2m^3-6m^2+7m-6)/3\le\chi(\mathbb{R}A)\le(2m^3-6m^2+7m)/3.$$

Denote the numbers of orientable components of $\mathbb{R}A$ with positive, zero and negative Euler characteristic by $k^+$, $k^0$ and $k^-$ respectively.

5.3.K. Refined Petrovsky - Oleinik Inequality.   If $m\not= 2$ then

$$-(2m^3-6m^2+7m-6)/3\le\chi(\mathbb{R}A) -2k^+-2k^0.$$

5.3.L Refined Arnold Inequality.   Either $m$ is even, $k^+=k^-=0$ and

$$k^0=(m^3 -6m^2 +11m)/6,$$

or

$$k^0+k^-\le(m^3-6m^2+11m-6)/6.$$