# Confusion regarding lattices

However, I have a confusion understanding the lattice itself. It says that $$H_d= \{{\vec{x} | \vec{x}.\vec{1} = 0}\} \subset R^{d+1}$$

The root lattice

$$A_d=\{\vec{x}=(x_0,...x_d) \epsilon Z^{d+1}| \vec{x}.\vec{1}=0\}$$

$$A_d=Z^{d+1}\cap H_d, A_d\subset H_d$$

Well my confusion is how come $$A_d\subset H_d$$ I mean

$$Z^{d+1}\cap H_d= H_d$$

itself isn't it? since $H_d$ itself is in $R^{d+1}$. Any insights?

Also it mentions $$A_d^{*} = \{{\vec{x}\epsilon H_d | \forall \vec{y} \epsilon A_d, \vec{x}.\vec{y} \epsilon Z}\}$$

Also I didn't understand what the above definition of permutohedral lattice means. I am beginner so I am finding it difficult to visualize. Any help? 

Also I didn't understand what they mean by

x is a remainder-k point while defining the lattice

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No, ${\mathbb Z}^{d+1} \cap H_d \ne H_d$. For example, if $d=2$, $(1/3,1/3,-2/3) \in H_2$ but is not in ${\mathbb Z}^{d+1}$: its entries are not integers.
$A^*_d$ is the set of vectors $\vec{x}$ such that $\vec{x}\cdot \vec{1} = 0$ and $\vec{x} \cdot \vec{y}$ is an integer for all $\vec{y} \in A_d$. Since the entries of $\vec{y}$ are integers, every member of $A_d$ is in $A^*_d$, but there are others. For example, if $d=2$, $(1/3,1/3,-2/3) \in A^*_d$. To see this, note that $(1/3,1/3,-2/3) \cdot (y_1,y_2,y_3) = (y_1 + y_2 - 2 y_3)/3 = - y_3$.
I didn't get it why (1/3,1/3,−2/3) is not in $Z^{d+1}$. Isn't $Z^{d+1}$ equivalent to $R^{d+1}$ –  user31820 Aug 24 '12 at 21:57
$\mathbb Z$ is the integers, $\mathbb R$ is the reals. –  Robert Israel Aug 24 '12 at 22:44