Cubic Shaping of Lattice Constellations from Multi-Level Constructions from Codes †
Abstract
:1. Introduction
2. Preliminary Information
2.1. Quadratic Ring of Integers
- Either and B is prime or are pairwisely coprime.
- .
- .
2.2. Constructions of Lattices from Linear Codes
- (Code Formula) is defined as , where
- (Construction D) is defined as , where
2.3. Tileability
3. Cubic Shaping of
3.1. Introductory Theorems in Tileability
- Suppose that is not bijective. This means that there exist two distinct elements such that . Suppose that a and b are with coefficients , respectively. We define , which will have coefficients where . Since a and b are distinct, it follows that ’s are not all zero.However, we have . By considering the real and imaginary parts of the equation, which are under modulo w and , respectively, together with the fact that , one obtains and as desired.
- Suppose that there exists an element d with coefficients , where ’s are not all zero and such that and . Next, we separate the coefficients of d into positive and negative ones and use them to construct such a and b. DefineSince the coefficients of a and b are in , . It is obvious that , and from the fact that together with the property that and , one may combine the real and imaginary parts into one equation, namely, . This concludes that is not bijective.
- 1.
- For any w, the bijectivity of for and is the same.
- 2.
- If m is even, the bijectivity of for , , , and is the same.
- Suppose that for is not bijective. According to Proposition 3, there exists an element d with coefficients , where ’s are not all zero and such that and .We construct a counterexample for by introducing an element with coefficients for all . From the definition, we can see that ’s are not all zero with . Furthermore,Therefore, and as well. From Proposition 3, it implies that is not bijective for .
- Suppose that m is even. It now suffices to show that the bijectivity of for and coincides.Suppose that is not bijective for . From Proposition 3, there exists an element d with coefficients , where ’s are not all zero and such that and are both divisible by .We construct a counterexample again, this time with employing the same coefficients. In other words, , implying that . Nonetheless, it follows from that . As a result,Since and are divisible by , so are and .
3.2. Totally Tiling Bases
- For m even, we use . Since is the componentwise modulo on the real and imaginary parts, for any , for some . Therefore,The equation shows that is the p-ary decomposition through base , which is unique.
- For m odd, we use and . Similarly, for any ,This p-ary decomposition base is unique as long as is also uniquely represented. Here, its real and imaginary parts are and , which are then both uniquely represented, since and .
- For , we have .
- For , we have .
- For , we have .
- For , we use . Because is componentwise modulo on both real and imaginary parts, for any , for some . Consequently,
- For , we use and . Therefore, for any ,From the uniqueness property of binary decomposition base , it is left to show that the term is uniquely represented. Here, its real and imaginary parts are and , and so the uniqueness of the expressions ensues from the fact that and .
- For , we use and . For any , we can write the tiling map asBy applying the same technique as in the previous case, we conclude that it is uniquely represented.
- For , we use . Hence, for any element , we have for some . It can be rewritten as
- For , we use and . For any element , we have for some . Therefore,We invoke unique decomposition base , and it remains to be shown that is uniquely represented. By considering its real part, we have , which is uniquely represented via division modulo and that . Furthermore, its imaginary part is , which is uniquely represented.
- For , the first possibility is and . For , we have for some . Therefore,Again, it is only left to show that the term is uniquely represented. Notice thatSince is a free -module which can be generated by 1 and , we have that the expressions and are uniquely represented. However, the first term is uniquely represented by and , since and . As a result, must be unique as well.
- For , the second possibility is and . For any , for some . Then,In this case, we are left to show that is uniquely represented. Since its real part is , where and , it is uniquely represented by and . From the uniqueness of , it follows that must be unique, in order to make unique.
- For , we use and . For any element , for some . Hence,Lastly, it is left to show that is uniquely represented by and . Notice thatWe once again use the fact that is a free -module generated by 1 and to conclude that the expressions and are unique. Since and , the first term is uniquely represented by and . Consequently, is unique as well.
3.3. Counterexamples for the Remaining Cases
- On , for with norm , we have .
- On , for with norm , we have .
- On , for with norm , we have .
- For , , and , it is obvious that they obey .
- For , suppose on the contrary that . The inequality can be written as a polynomial inequality in x as
- –
- When , we have the ring . The inequality of discriminant is . Therefore, , and thus . The only possible bases are , , which have already been given a counterexample.
- –
- When , we have the ring . The inequality of discriminant is . As a result, we have and or and . The resulting bases are . However, each of them either has norm or has been given a counterexample previously.
- For with norm , we use as a counterexample.
- For other bases, we take , , , and , which gives us . Again, we have yet to verify that the coefficients obey .
- –
- It is clear that and satisfy .
- –
- For , we have . The only case of equality is when . This would imply where a counterexample has already been constructed.
- –
- For , we have as desired.
- For and the base with norm , we use as a counterexample.
- Suppose that . According to Lemma 2, either or . Therefore, the only possible value for b is 1, and thus . This leads to the bases with norm in . For the base , we choose as a counterexample.
- Suppose that . We choose , , , and , which yields . It is left to show that the coefficients satisfy .
- –
- It is clear that and follow .
- –
- The inequality is equivalent to either , which is true by the assumption, or , which is always true for all and . Hence, .
- –
- For , we suppose on the contrary that , which is equivalent to a polynomial inequality in x as follows:The inequality has a solution if and only if the discriminant of the polynomial is non-negative, meaning that . The only solutions to this inequality where are . Here, contradicts Lemma 2 and implies with norm in , both of which are not in consideration.
- Suppose that . We choose , , , and , which gives . Again, we show next that for all ’s.
- –
- The statement is clear for and .
- –
- For , the inequality is equivalent to either , which is always the case for all and , or , which is true from the assumption. Thus, .
- –
- For , we suppose on the contrary that , which can be written as a polynomial inequality in x asSimilar to the previous case, the discriminant is non-negative when . For to satisfy the inequality, the only solutions are . Once again, contradicts Lemma 2. The solutions give the following bases.
- –
- For , in . Each of them either fails or .
- –
- For , in , where . Each of them either fails or or has already been given a counterexample.
- –
- For , in where . They both fail the assumption .
Lastly, the solutions give bases in , and they all fail to satisfy . We conclude that .
- First, we show that implies or , and that implies or .
- –
- Suppose that . Since t is an integer, must be divisible by p. However, , and so we have . As a result, . Because and , it follows that , which eventually implies that or p.
- –
- Suppose that . Because t is an integer, must be divisible by p. Due to the fact that and , we obtain , and thus or p.
Therefore, if or is equal to p, the other one must be either 0 or p. This takes us to the next step of the proof. - We show that are all impossible.
- –
- Suppose that and . It follows that . Unfortunately, and , meaning that t is not an integer, which is a contradiction.
- –
- Suppose that and . It follows that . Again, and , and so t is not an integer, which is a contradiction.
- –
- Suppose that . It follows that . Because and , is an integer if and only if is divisible by p. Since , it implies , or equivalently . The only solutions here are . Excluding those from the family of known tiling bases, we are left with the following bases, for which we now provide a counterexample.
- –
- For in with norm , we have .
- –
- For in with norm , we have .
We have shown that and , thus guaranteeing the existence of a counterexample, and the proof is now complete.
- If , we have , or equivalently . However, , and so . Since p is a prime, . Because , we have , or . The existence of solutions occurs when , and they are . These solutions only produce bases of norms or .
- If , the inequality implies that , and thus .
- Suppose that , or . Notice that . According to Theorem 1, it is sufficient to show the case , whose norm is . Here, we choose as a counterexample, which implies that is not bijective.
- Suppose that . Since and ,This inequality simplifies to either or . When x is considered a variable, the discriminant of these inequalities is either or . We can omit the case since an inequality with a negative discriminant cannot have a real solution. On the other hand, if , then a solution exists only when , and they are . These solutions give bases of norm .
- We show that implies or , and that implies or
- –
- Suppose that . Since t is an integer, is divisible by p. However, by the fact that , we have that . As a result, . From and , we have , which means that or p.
- –
- Suppose that . Since s and t are integers, and are divisible by p. Hence, is divisible by p. Since , must then be divisible by p. This, however, makes divisible by p. Since and , we conclude that , implying or p.
- We show that are impossible.
- –
- Suppose that and . It follows that and . Because s and t are integers and , and must both be divisible by p. This implies that p divides , contradicting the fact that and .
- –
- Suppose that and . It follows that . Since and , it contradicts the fact that t is an integer.
- –
- Suppose that . It follows thatBecause are integers and , we haveFrom these equations, we obtain . Since and , we have , which is a contradiction. This finishes the last piece of the theorem.
4. Summary and Application
- The trivial in where and B is prime,
- The trivial in , where and is prime,
- in ,
- in .
- For where and B is prime, the norm is , and .
- For where and is prime, the norm is , and .
- For , the norm is . Here, if , then , and if , then .
- For , the norm is . If , then , and if , then .
Author Contributions
Funding
Data Availability Statement
Conflicts of Interest
References
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Pooksombat, P.; Kositwattanarerk, W. Cubic Shaping of Lattice Constellations from Multi-Level Constructions from Codes. Mathematics 2025, 13, 1562. https://doi.org/10.3390/math13101562
Pooksombat P, Kositwattanarerk W. Cubic Shaping of Lattice Constellations from Multi-Level Constructions from Codes. Mathematics. 2025; 13(10):1562. https://doi.org/10.3390/math13101562
Chicago/Turabian StylePooksombat, Perathorn, and Wittawat Kositwattanarerk. 2025. "Cubic Shaping of Lattice Constellations from Multi-Level Constructions from Codes" Mathematics 13, no. 10: 1562. https://doi.org/10.3390/math13101562
APA StylePooksombat, P., & Kositwattanarerk, W. (2025). Cubic Shaping of Lattice Constellations from Multi-Level Constructions from Codes. Mathematics, 13(10), 1562. https://doi.org/10.3390/math13101562