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

[Peter_Newman]

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?

 

[Office_Shredder]

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.

 

[Peter_Newman]

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?

 

[Office_Shredder]

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.

 

[Peter_Newman]

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.

 

[pasmith]

Given $x \notin \mathbb{Z}$, we have $\langle xe_i, e_i \rangle = x \notin \mathbb{Z}$. So a dual lattice vector cannot have a non-integer component.

 

[Peter_Newman]

@pasmith thanks! This goes in the direction of my post #3. But it's more general.

So this was everything?

 

[WWGD]

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?

 

[Peter_Newman]

It's the (n-dim.) vector space over $\mathbb{Z}^n$.

 

[WWGD]

It's the (n-dim.) vector space over $\mathbb{Z}^n$.

$\mathbb Z^n$ is not a field.

 

[pasmith]

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 $\{ \sum_{i=1}^n m_i e_i : m_i \in \mathbb{Z} \} \subset \mathbb{R}^n$, the $\mathbb{Z}$-lattice generated by the standard basis vectors $e_i \in \mathbb{R}^n$.

The dual lattice is then $\{ \sum_{i=1}^n m_i f_i : m_i \in\mathbb{Z}\}$ where $f_i \in \mathbb{R}^n$ is defined by $\langle f_i, e_j \rangle = \delta_{ij}$, which gives $f_i = e_i$. (This is the definition of "dual lattice" with which I am familiar, which is hopefully equivalent to that stated in the Wikipedia article.)

 

[Office_Shredder]

The dual lattice is then $\{ \sum_{i=1}^n m_i f_i : m_i \in\mathbb{Z}\}$ where $f_i \in \mathbb{R}^n$ is defined by $\langle f_i, e_j \rangle = \delta_{ij}$, which gives $f_i = e_i$. (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.

 

[WWGD]

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.

 

[Office_Shredder]

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.