4.1. Output of the Newton–Puiseux Algorithm for Triple Points
In this section, we use the Newton–Puiseux algorithm to study triple points for a reduced algebraic plane curve. Let : be a plane curve of degree d and let P be a triple point for (i.e., ).
We recall that, by Remark 3, the curve
can have three 1-branches, or one 2-branch and one 1-branch, or a 3-branch at
P. Anyway, if the tangent cone is made up of three distinct tangent lines,
P is an ordinary singularity, and it is analytically equivalent to
, while if there are two distinct tangents, the singularity is analytically equivalent to
,
, see [
12].
Hence, we focus on the most interesting case, i.e., when
has a unique triple tangent
r at
P (for an analysis modulo analytic equivalence of the branches at a triple point, see [HH]). It is not restrictive to assume that
and that
r is given by
, hence,
The cases where
f is divisible by
y are trivial, so we can assume that
, that is,
f is Newton-convenient. Notice that if
has one 3-branch at
O, then its parameterization will be of type
(this follows by Definition 3, since the unique tangent to the branch is
).
We use the notation of
Section 3.2, with the unique exception that the
associated to
will be denoted by
, in order to specify which edge we refer to. Notice that if
is an edge of the Newton polygon, the height of
is denoted by
, while the height of the Newton polygon, denoted by
, is defined to be the sum of the heights of its edges.
We indicate with the coefficient in . We also denote by the sub-algorithm of Algorithm 1 starting inside step n with the choice of the root for .
In order to describe the output of the algorithm, we have to answer the following questions:
- 1.
Which are the possible choices of and in each case?
- 2.
When, for each choice of , does the algorithm stop?
The Stop Criterion 3.1 is an answer to question 2, but we can improve on it:
Remark 5. By Lemma 3, if has multiciplity 1 as a root of , then satisfies the hypothesis of Theorem 1, because in , there is the monomial . Hence, if has multiplicity 1 as a root of , stops at the step , i.e., without calculating . From now on, we use this new Stop criterion because it is more efficient.
To answer question 1, we observe that the following statements hold:
- (i)
Since the height of the Newton Polygon decreases at each step, i.e., , and , we have to consider Newton Polygons of height only;
- (ii)
In general, . However, since , there exists such that . Hence, for , the vertices of are in .
We start answering question 1 just for ; we shall see later (see Remark 7) that this is enough.
Remark 6. The possible shapes of the Newton Polygons arising when applying our algorithm in the case of a triple point with a triple tangent and for
are described in the
Figure 10, and each of the corresponding cases is discussed below. Namely, in each case, we write explicitly the
associated with each edge and its roots
with their relative multiplicities, so that we have a complete answer to question 1 first; after that, we examine the algorithms
for each
.
Case 1: ,
, and we have
has a unique root of multiplicity 1,
with
, which by Remark 5, gives a 1-branch of
, and the algorithm
stops here.
Case 2.1: ,
, and
, so we have
has two distinct roots of multiplicity 1 each:
By Remark 5, the algorithms
and
stop here; moreover, they give equivalent parameterizations of the same 2-branch of
, since, if that were not true, the sum of the multiplicities of the branches at
O would be
. The parameterization
, composed with the biolomorphic function
, is equal to
; in fact, since they are equivalent, by Remark 1, there is an
,
, such that
is obtained from
by substituting
; since
, we obtain
.
Case 2.2: ,
, and
, so we have
Let
denote the roots of
; we distinguish two cases:
Case 2.2.1: If , i.e., if has two roots of multiplicity 1, they give two distinct 1-branches of and the algorithms and stop here.
Notice that if
, we apparently are in the same situation as Case 2.1, but here
is a positive integer, so the two simple roots are
and the biolomorphic function
used in 2.1 does not work here: it does not give the equivalence of
and
. In fact, the two simple roots correspond to two non-equivalent parameterizations, each one for a different 1-branch.
Case 2.2.2: If , i.e., if has one double root, the algorithm goes on: it can yield either two 1-branches or one 2-branch of , hence, globally, we may find either three 1-branches or one 2-branch and one 1-branch. Notice that is still in , since .
Case 3.: ,
; we have
Both
and
have a unique root of multiplicity 1,
and
; each of them gives a different 1-branch of
, and the algorithms stops here.
Case 4.1: ,
, and
, so we have
has three distinct roots of multiplicity 1, namely:
and, as in Case 2.1, all of them give equivalent parameterizations of the same 3-branch of
, since otherwise, the sum of the multiplicities of the branches at
O would be
; for example the parameterization
composed with the biolomorphic function
is equal to
. Since the three roots have multiplicity 1, the algorithms stop.
Case 4.2: ,
, and
, so we have
Let
denote the roots of
; we distinguish three cases:
Case 4.2.1: If , i.e., if has three roots of multiplicity 1, they give three distinct 1-branches of so the algorithms stop here.
Notice that if
, we apparently are in the same situation as Case 4.1, but here
is a positive integer, the three simple roots are
and the biolomorphic function
used in 4.1 does not work here: it does not give the equivalence of
and
. In fact, the three simple roots correspond to three non-equivalent parameterizations (each of a different 1-branch).
Case 4.2.2: If , i.e., if has two roots, one double and one simple; the algorithm stops and gives a 1-branch of . On the other hand, does not stop because the relative root has multiplicity 2; continuing, it can give two 1-branches or one 2-branch of . Notice that in this case is still in , since .
Case 4.2.3: If , i.e., if has one triple root, the algorithm goes on. Notice that in this case, is still in , since .
Case 5.1: ,
and
, so we have
has two distinct roots of multiplicity 1:
, with
, which give equivalent parameterizations of the same 2-branch of
, while
has one root of multiplicity 1:
, giving one 1-branch. The algorithms stop here.
Case 5.2: ,
and
, so we have
has one root of multiplicity 1, which gives a 1-branch of
and the relative algorithm stops here, while, if
denote the roots of
, we distinguish two cases (notice that
may be 0, in which case, we are apparently in the same situation as Case 5.1):
Case 5.2.1: If , i.e., if has two roots of multiplicity 1, each of them gives a 1-branch of , so we have three 1-branches of and the algorithms stop here.
Case 5.2.2: If , i.e., if has the double root , the algorithm does not stop; continuing, it can give two 1-branches or one 2-branch of , hence, we may find three 1-branches or one 2-branch and one 1-branch, in all. Notice that in this case, is still in , since .
Case 6.1: Case 6.1:,
and
, so we have
Since
, the discussion is analogous to Case 5.1:
has one 1-branch and one 2-branch, and the algorithms stop here.
Case 6.2: ,
and
, so we have
Notice that the assumption
gives
, hence, the discussion is analogous to Case 5.2, namely, we distinguish two cases:
Case 6.2.1: if has two roots of multiplicity 1, each of them gives a 1-branch of , so we have globally three 1-branches of and the algorithms stop here. Notice that may be 0, in which case, we are apparently in the same situation as Case 6.1, but here .
Case 6.2.2: if has one double root , the relative algorithm does not stop; continuing, it can give two 1-branches or one 2-branch of , hence, globally, we may find three 1-branches or one 2-branch and one 1-branch. Notice that in this case, is still in , since .
Case 7: ,
and we have
,
and
have one simple root each and the three roots give a 1-branch of
each: the algorithms stop here.
Remark 7. Notice that, in all the cases we have just considered, we obtain a polynomial (for the following step of the Algorithm) which is always in , i.e., we are always in case .
4.2. A Theorem for Triple Points
Now we are ready to see what happens when we run the Newton-Puiseux Algorithm 1 in the case of a triple point with a triple tangent.
Theorem 2. Let be a plane curve of degree d, such that O is a triple point for and the x-axis is its triple tangent at O, so thatWith the previous notation, the output of the Newton–Puiseux Algorithm 1 is as follows: A finite sequence of n steps like case 4.2.3 with , and then the following four cases are possible:
step is a case 4.1; the algorithm stops: has one 3-branch at O, and a parameterization for it is of the form step is one of the following cases: 4.2.1, 5.2.1, 6.2.1, 7; the algorithm stops: has three 1-branches at O, each of which with a parameterization of the formwhere in case 4.2.1, in cases 5.2.1, 6.2.1; step is one of the following cases: 5.1, 6.1; the algorithm stops:
has one 1-branch and one 2-branch at O, with parameterizations of the form: step is one of the following cases: 4.2.2, 5.2.2, 6.2.2; in order to decide if has one 1-branch and one 2-branch or three 1-branches at O, the sub-algorithm relative to the double root must go on, so that we find a finite sequence of steps like case 2.2.2, and then: if the next step is a case 2.1, the algorithm stops: has one 1-branch and one 2-branch at O, with parameterizations of the formif the next step is a case 2.2.1 or 3, the algorithm stops: has three 1-branches at O, with parameterizations of the formwhere in case2.2.1.
Proof. First, we claim that, in order to compute the number and multiplicities of the branches of at a triple point with a triple tangent, it is enough to consider Newton polygons whose vertices are in ; that is, applying the Newton–Puiseux Algorithm to f, we encounter only the cases described in Remark 6 above.
Proof of the claim. This is true for , since . Now, we choose a path as described in Algorithm 1, and assume the claim is true for the steps ; then the Newton Polygon of is one of those described above. If the algorithm does not stop, we are in one of the following five cases: 2.2.2, 4.2.2, 4.2.3, 5.2.2, 6.2.2, and since in each of these cases the vertices of the Newton Polygon of are again in , the same is true for . On the other hand, if the algorithm stops, we do not need to go any further, so the claim is true.
Now, we are going to analyze what happens in the assumption that ; if, on the contrary, , there is a branch coming from an algebraic component of , and thus, the corresponding series is in fact a polynomial.
Assume that after n steps, we are in case 4.2.3. Since for , we have for , hence, each previous step was one of the cases between 4.1 and 7. If one of the previous steps was a case among 4.1, 4.2.1, 5.1, 5.2.1, 6.1, 6.2.1, 7, we would have already stopped. If one of the previous steps was a case among 4.2.2, 5.2.2, 6.2.2, only the sub-algorithm relative to the double root would have gone on, and in the next steps, the height of the Newton Polygon would have been 2. Therefore, all the previous steps were of type 4.2.3.
Let us start with the step 0 which is a case
4.2.3, and set
; since
,
k is a positive integer, and for a suitable coefficient
b, we have
Applying the algorithm, we obtain:
,
,
and since each power of
x has an exponent
, we find
If
is a case
4.2.3, again we have:
and we find
with
; hence, if the sequence of cases
4.2.3 goes on to step
n, we have a sequence of roots, setting
(notice that
, otherwise, the tangent cone of
C at
O would be different from
)
Let us assume that the algorithm does not stop and continutes with an infinite sequence of cases
4.2.3; then, the Puiseux series
is in fact a power series giving a unique 1-branch counted 3 times, but our curve
is reduced and this is not possible.
Hence, the only possible way in which 4.2.3 can appear is as follows: we have a finite sequence of n steps 4.2.3 and then one of the other cases.
If the step
is a case
4.1,
has one 3-branch at
O, we have found:
which gives us
and since the common lowest denominator of
is 3, a parameterization for the 3-branch is
with
integers, and
. Thus, we obtain
.
If the step is not a case 4.1, then it has to be one of the cases from 4.2.1 to 7 (except 4.2.3), since the height of the Newton Polygon is still 3. It is easily seen that cases 4.2.1, 5.2.1, 6.2.1, 5.1, 6.1 give and just following the discussion in Remark 6.
If the step is one of the cases 4.2.2, 5.2.2, 6.2.2, in order to decide if has one 1-branch and one 2-branch or three 1-branches at O, the sub-algorithm relative to the double root must go on; now, the height of the Newton Polygon is decreased to 2, so that we have to use cases from 2.1 to 3, and again, it is immediate to see that holds following the discussion in the previous Remark 6. □
We give two examples studied with the Algorithm as in the Theorem above.
Example 4. Let us consider the curve:
We have that
and the unique tangent of
at
O is
. By applying Algorithm 1, one finds that
has at
O one 2-branch given by
and one 1-branch given by
.
Example 5. Let us consider the curve:
We have that
and the unique tangent of
at
O is
. By applying the Algorithm 1 one finds that
has three 1-branches at
O given by
, and
.
Our last observation is on cuspidal triple points, i.e., case of the theorem above.
Definition 8. In case , we say that the triple point O is of type s. Notice that this definition is well-posed: if are two equivalent parameterization of the 3-branch, by Remark 1, we have the same value of s for the two parameterizations.
Remark 8. We point out this definition because in case
, the type
s is an invariant which determines the topological structure of the singularity; in fact, if a plane algebraic curve
has a triple point with one 3-branch at
O, of type
s, then it is locally topologically equivalent, see [
19], to one of type
, i.e., with equation
.
Notice that, in fact, a complete description of the analytical equivalence classes for this kind of triple points can be given, see ([page 4], [
20]).
Example 6. Let us consider the following curve:
We have that
and the unique tangent of
at
O is
. By applying Algorithm 1, one finds that
has one 3-branch at
O given by
; thus, we are in case
of the theorem, and the triple point of
at
O is of type 10, hence, it is topologically equivalent to
.