Degree 3. By Harnack's inequality, the number of components is at most 2. By 1.3.B a curve of degree 3 is one-sided, thereby the number of components is at least 1. In the case of 1 component the real scheme is , and the type is II by Klein's congruence 2.6.C. In the case of 2 components the type is I by 2.6.B. The real scheme is . Thus we have 2 possible complex schemes: (realized above) and . For the first one and for the second . Since the right hand side of the complex orientation formula is and , only the first possibility is realizable.
Degree 4. By Harnack's inequality the number of components is at most 4. We know (see 1.4) that only real schemes , , , , and are realized by nonsingular algebraic curves of degree 4. From Klein's congruence 2.6.C it follows that the schemes and are of type II. The scheme is of type II by 2.6.A. By 2.6.B is of type I.
The scheme is of type II, since it admits no orientation satisfying the complex orientation formula. In fact, for any orientation while the right hand side is .
By 2.6.D the scheme is of type I. A calculation similar to the calculation above on the scheme , shows that only one of the two semiorientations of the scheme satisfies the complex orientation formula. Namely, . It was realized in Figure 18.
Degree 5. By Harnack's inequality the number of components is at most 7. We know (see 1.4) that only real schemes , , , , , , , are realized by nonsingular algebraic curves of degree 5. From Klein's congruence 2.6.C it follows that the schemes , , are of type II. By 2.7.F and are of type II.
By 2.6.B is of type I. The complex orientation formula gives the value of (cf. Proof of 2.7.F): . This determines the complex scheme. It is .
By 2.6.D is of type I. The complex orientation formula allows only the semiorientation with . Cf. Figure 19.
The real scheme is of indefinite type, as follows from the construction shown in Figure 19. In the case of type I only one semiorientation is allowed by the the complex orientation formula. It is .