- #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?
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?