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
.