Why is the dual of Z^n again Z^n ?

In summary: An equivalent basis free definition is the dual is the set of all vectors whose inner product with every vector from the original lattice is an...integer?
  • #1
Peter_Newman
155
11
Hello,

how can one proof that the dual of ##\mathbb{Z}^n## is ##\mathbb{Z}^n##?

My idea:
The definition of a dual lattice says, that it is as set of all lattice vectors ##x \in span(\Lambda)## such that ##\langle x , y \rangle## is an integer. When we now consider ##\mathbb{Z}^n## we see that all Elements of ##\mathbb{Z}^n## are ##n##-dimensional vectors with integer coordinates. So the dot product of those vectors of a span of ##\Lambda## with all vectors ## y \in \Lambda## results in an sum of a multiplication of integer values for each vector component e.g. ##x_1 \cdot y_1 +... + x_n \cdot y_n##, where each ##x_i, y_i \in \mathbb{Z}##, so the sum is also in ##\mathbb{Z}## as required in the definition.

Is this ok for a "proof" or did I miss something?
 
Physics news on Phys.org
  • #2
That proves that ##\mathbb{Z}^n## is a subset of the dual. You also need to show that no vector outside of ##\mathbb{Z}^n## is in the dual.
 
  • Like
Likes WWGD and topsquark
  • #3
Office_Shredder said:
That proves that ##\mathbb{Z}^n## is a subset of the dual. You also need to show that no vector outside of ##\mathbb{Z}^n## is in the dual.
Hey @Office_Shredder thanks for your answer!

How can you show the last point exactly? With a vector that is not in ##\mathbb{Z}^n## e.g. ##(1/3, 2/3)^T##? How do you generalize that?
 
  • #4
Peter_Newman said:
Hey @Office_Shredder thanks for your answer!

How can you show the last point exactly? With a vector that is not in ##\mathbb{Z}^n## e.g. ##(1/3, 2/3)^T##? How do you generalize that?
Can you think of the simplest integer vector to inner product with to prove that isn't in the dual? You don't have to do anything clever here.

An alternate technique Is to use the determinant, the product of the determinant of the lattice and its dual equals 1.
 
  • Like
Likes topsquark and Peter_Newman
  • #5
Office_Shredder said:
Can you think of the simplest integer vector to inner product with to prove that isn't in the dual? You don't have to do anything clever here.
Hello @Office_Shredder, actually I can't find a simple vector here. This would then get back to the reasoning I gave in my first post. The scalar product of integer vectors is again an integer.
 
  • #6
Given [itex]x \notin \mathbb{Z}[/itex], we have [itex]\langle xe_i, e_i \rangle = x \notin \mathbb{Z}[/itex]. So a dual lattice vector cannot have a non-integer component.
 
  • Like
Likes topsquark and Peter_Newman
  • #7
@pasmith thanks! This goes in the direction of my post #3. But it's more general.

So this was everything?
 
Last edited:
  • #8
Peter_Newman said:
Hello,

how can one proof that the dual of ##\mathbb{Z}^n## is ##\mathbb{Z}^n##?

My idea:
The definition of a dual lattice says, that it is as set of all lattice vectors ##x \in span(\Lambda)## such that ##\langle x , y \rangle## is an integer. When we now consider ##\mathbb{Z}^n## we see that all Elements of ##\mathbb{Z}^n## are ##n##-dimensional vectors with integer coordinates. So the dot product of those vectors of a span of ##\Lambda## with all vectors ## y \in \Lambda## results in an sum of a multiplication of integer values for each vector component e.g. ##x_1 \cdot y_1 +... + x_n \cdot y_n##, where each ##x_i, y_i \in \mathbb{Z}##, so the sum is also in ##\mathbb{Z}## as required in the definition.

Is this ok for a "proof" or did I miss something?
What kind of Algebraic object is ## \mathbb Z^n ## ? A ## \mathbb Z ##- module ? A complicated construction. Or a Cartesian product ring of copies of ##\mathbb Z ##? A tensor product?
 
  • #9
It's the (n-dim.) vector space over ##\mathbb{Z}^n##.
 
  • Like
Likes WWGD
  • #10
Peter_Newman said:
It's the (n-dim.) vector space over ##\mathbb{Z}^n##.
##\mathbb Z^n ## is not a field.
 
  • #11
WWGD said:
What kind of Algebraic object is ## \mathbb Z^n ## ? A ## \mathbb Z ##- module ? A complicated construction. Or a Cartesian product ring of copies of ##\mathbb Z ##? A tensor product?

I think in context it's [itex]\{ \sum_{i=1}^n m_i e_i : m_i \in \mathbb{Z} \} \subset \mathbb{R}^n[/itex], the [itex]\mathbb{Z}[/itex]-lattice generated by the standard basis vectors [itex]e_i \in \mathbb{R}^n[/itex].

The dual lattice is then [itex]\{ \sum_{i=1}^n m_i f_i : m_i \in\mathbb{Z}\}[/itex] where [itex]f_i \in \mathbb{R}^n[/itex] is defined by [itex]\langle f_i, e_j \rangle = \delta_{ij}[/itex], which gives [itex]f_i = e_i[/itex]. (This is the definition of "dual lattice" with which I am familiar, which is hopefully equivalent to that stated in the Wikipedia article.)
 
  • Like
Likes Office_Shredder and WWGD
  • #12
pasmith said:
The dual lattice is then [itex]\{ \sum_{i=1}^n m_i f_i : m_i \in\mathbb{Z}\}[/itex] where [itex]f_i \in \mathbb{R}^n[/itex] is defined by [itex]\langle f_i, e_j \rangle = \delta_{ij}[/itex], which gives [itex]f_i = e_i[/itex]. (This is the definition of "dual lattice" with which I am familiar, which is hopefully equivalent to that stated in the Wikipedia article.)

An equivalent basis free definition is the dual is the set of all vectors whose inner product with every vector from the original lattice is an integer.
 
  • #13
Office_Shredder said:
An equivalent basis free definition is the dual is the set of all vectors whose inner product with every veclattice is an integer
Well, then usjng the ##e_i## basis, we can see that each component in a dual element must be an Integer. Since we had cincluded ##\mathbb Z^n## was in the dual, we're done.
 
  • #14
WWGD said:
Well, then usjng the ##e_i## basis, we can see that each component in a dual element must be an Integer. Since we had cincluded ##\mathbb Z^n## was in the dual, we're done.

Yes, this is what post #6 is doing.
 
  • Like
Likes WWGD

1. Why is the dual of Z^n again Z^n?

The dual of a vector space V is the set of all linear functionals on V. In the case of Z^n, the dual is the set of all linear maps from Z^n to the integers. Since Z^n is a discrete group, the only linear maps from Z^n to Z are those that send each element of Z^n to a single integer. Therefore, the dual of Z^n is again Z^n.

2. What is the significance of Z^n being its own dual?

The fact that Z^n is its own dual is significant because it means that the group has a self-dual structure. This can be useful in certain mathematical applications, such as in coding theory and cryptography.

3. How does the dual of Z^n relate to the dual of other groups?

The dual of Z^n is unique in that it is isomorphic to the original group. This is not the case for all groups. For example, the dual of the real numbers is the set of continuous functions on the real line, which is not isomorphic to the real numbers themselves.

4. Can the concept of duality be extended to other mathematical structures?

Yes, the concept of duality can be extended to other mathematical structures such as rings, modules, and algebras. In these cases, the dual is defined as the set of all homomorphisms from the structure to its underlying field or ring.

5. How is the dual of Z^n used in practical applications?

The concept of duality is used in various areas of mathematics, including linear algebra, functional analysis, and topology. In practical applications, the dual of Z^n can be used in coding theory to construct error-correcting codes, and in cryptography to generate secure keys.

Similar threads

  • Linear and Abstract Algebra
Replies
1
Views
152
  • Linear and Abstract Algebra
Replies
23
Views
2K
  • Linear and Abstract Algebra
Replies
11
Views
1K
  • Linear and Abstract Algebra
Replies
25
Views
1K
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
250
  • Linear and Abstract Algebra
Replies
1
Views
912
Replies
5
Views
863
  • Linear and Abstract Algebra
Replies
16
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
502
Back
Top