In this section we define the notion of hyperhomography on a Krasner hyperfield, as a quotient structure of a classical field by a normal subgroup. Using it, we introduce some hyperoperations and investigate the properties of the associated hypergroups.
Let be the Krasner hyperfield associated with the field F and . Define the generalized homography transformation on F as on , and call it the hyperhomography relation. We call the set hyperhomography, while is a homography, for all and .
Notice that the hyperhomography is a generalization of a homography , because is equivalent with , i.e., . The classical operations on the field F have been extended to the hyperoperation ⊕ and operation ⊙ on , where by we denote the hyperaddition between and the opposite of with respect to the hyperoperation ⊕. Besides, since the result of a hyperoperation is a set, the equality relation in the definition of a homography is substitute by a "belongingness" relation in the definition of a hyperhomography.
Moreover denote and , for all It follows that .
The relation between homographies and hyperhomographies is given by the following identity
thus there exists
, such that
and the following implications hold:
. Conversely, suppose that
then the following implications hold, too:
Thanks to Theorem 2, we call the set the hyperhomography on F, while the set is a homography on .
Example 2. Let be the field of all integers modulo and . Thus the quotient set is and the hyperaddition and the multiplication are defined on as follows:where , and . Consider now the hyperhomography
and the homographies
Then it follows that
as stated by Theorem 2.
A hyperhomography in is called nondegenerate, if the following conditions hold, respectively:
for all , if , then can be omitted from ,
for all and there is
for all the element can be added to where ∞ is an element outside of
By consequence, under the same conditions, also is called a nondegenerate hyperhomography in .
For a nondegenerate hyperhomography in , we fix some new notations: , and , for any .
Example 3. If we go back to Example 2 and use the concepts in Definition 6, then we can omit from and from , respectively, and add in both sets the element . Then and are nondegenerate homographies, while is a nondegenerate hyperhomography with the property where,
Definition 7. Let be a nondegenerate hyperhomography in . Setting , we obtain that . Define
for all and , where .
Moreover set , where , for all and .
Let be a nondegenerate hyperhomography in and G be a normal subgroup of . If and , then
According to Definition 7, if and , then or . In the second case, solving the equation we get . □
In the following, for a nondegenerate hyperhomography in , we define the lines passing through two points.
Definition 8. Let be a nondegenerate hyperhomography in . For all and , define and
where means the formal derivative of . In addition we call the line passing through the points and Intuitively, for each , the line passing through is a vertical line. In other words, plays an asymptotic extension role for .
Taking two arbitrary points
on the homography
Using the definition of the lines
are equal to
according to Definition 8, we have
is well defined, therefore we have
We will better illustrate the above defined notions in the following example.
Example 4. Consider the field and the homography transformation over F, so its graph is the hyperbola represented below in Figure 4. Taking on two arbitrary points and , we draw the line passing through and . Then , where is the x-axis. Then we obtain the point on the hyperbola . Proposition 1. Let be a nondegenerate hyperhomography in and . Then, it follows that
Based on Definition 8 and on the fact that
, by simple computations, we obtain
is a homography group, for all . Moreover, notice that "" is the group operation on the homography
On a nondegenerate hyperhomography
we introduce the equivalence relation “∼” by considering
and denote the set of the equivalence classes of
, respectively. It follows that
Furthermore, if we introduce the notation and We will have , and
Thus and are called the equipped hyperhomographies in and , respectively. Besides, if we admit that for all and , then the bijective map defined by , where equip the quotient with a group structure and gives us a group isomorphism
In addition, the concepts in Definition (8) can be similarly defined on , only by substituting a with .
Let be an equipped hyperhomography. We define the hyperoperation "∘" on as follows.
Let . If and for some and , then
If is an equipped hyperhomography, then has a hypergroup structure.
Suppose that such that and where, . First we notice that because implies that , i.e is a non-empty set and belongs to Besides, if and , then and , meaning that and . Hence we have and therefore , equivalently with . By consequence, the hyperoperation “∘” is well defined.
If or or then the associativity is obvious. If not, we have the following cases.
Case 1: .
This means that
and we have
Similarly, it holds that
On the other hand we have
In other words, for the six points
on the curve we have
and therefore, by Pascal’s theorem (see Theorem 1), it follows also that
, equivalently with
By Definition 8 we know that
where, by the associativity of the group operation "
", it holds
This leads to the equality
Case 2: .
, then we have
On the other hand
(ii) If , then the associativity holds, similarly as in the case (i).
, then we have
Case 3: .
In this case we have
On the other hand
Therefore the hyperoperation "∘" is associative.
In order to prove the reproduction axiom, we consider two cases as below:
Case 1. If , then and where . It follows that is a homography group, so the reproduction axiom holds.
Case 2. If
consider an arbitrary element
Similarly, and thus the reproduction axiom is proved. Therefore, is a hypergroup. □
If , then the hyperhomography and the associated hypergroup are the classical homography and the homography group, respectively.
Example 5. Let us consider again Example 3, where we deal with the nondegenerate hyperhomography as a subset of , having the form where, and , while the associated equipped hyperhomography is where,
Now let and Then and are reversible subhypergroups of , which are defined by the following Cayley tables, respectively
For a better understanding, we will explain all details in computing, for example, in the table of T the hyperproduct . For doing this, since , we use the function and the field . Based on Corollary 1, we obtain
Based on Proposition 1, we have
which imply that
Similarly, all the other hyperproducts in both tables can be obtained.
The next result gives a characterization of the subhypergroups of the equipped hyperhomographies in .
Let H be a non-empty subset of the hypergroup . Then H is a subhypergroup of the equipped hyperhomography if and only if it can be written as , where , or H is a subhypergroup of , for some .
. Suppose that H is a subhypergroup of and , for every in . There exist in such that . Now let . Thus we have . Hence .
. It is obvious. □
Let H be a subhypergroup of the hypergroup . Then H is reversible if and only if H is a subhypergroup of , for some .
. First we prove that any subhypergroup H of is a regular reversible hypergroup, for any . The regularity is clear, because is an identity and each element is an inverse for itself. In order to prove the reversibility, let and be arbitrary elements in . We distinguish three different situations.
Case 1. If
Case 2. If , then , with . Thus . It follows that .
Case 3. If , then , implying that and . Notice that , for all (i.e. every element is one of its inverses).
. Suppose that H is a reversible subhypergroup of such that it is not a subhypergroup of any , with . Based on Theorem 4, we have , where . Let be arbitrary elements in and , respectively, that are not equal to , with . If , then, based on the reversibility, we have , where , hence . Thus , which is in contradiction with the supposition that . Therefore , for some . □
In the following we will present two new hypergroup structures isomorphic with the equipped homography in the case when and , respectively.
Theorem 6. Consider the field F and define on the hyperoperation
Then, for every , there is the homomorphism .
It is easy to see that is a hypergroup. Now, taking , consider the bijective function defined by and the function defined by , where and . Geometrically, is the graph of the function , thus it is the line passing through the points of and , while is the map that projects the points of the hyperhomography on the above mentioned line.
Thus, using Proposition 1, for all
Now suppose that
are arbitrary elements in
. It follows that
Take now with as the projection map on the second component and define by .
, for all
is a bijective map and also a homomorphism because
Therefore is isomorphic to □
Theorem 7. Consider the field F and define on the hyperoperation
Then, if , there is the homomorphism .
is a hypergroup. Consider the bijective function
its graph. Besides define
. Therefore, for all
and for all
As in the previous theorem, let , be the projection map on second component and define by .
Therefore, for all
is a bijective map. We claim that
is a homomorphism, too, because
Therefore is isomorphic with . □