Undergrad Properties of Lattice Projections

[Peter_Newman]

Hello,

let's say that we have a lattice $\Lambda$. Now I take one of it's basis vectors say $b_1$ and project all the lattice points onto the line which is orthogonal to $b_1$, this line could be calculated via Gram-Schmidt for instance. But does this then form a lattice again? It is possible to formalize this a bit more? With my definition of a lattice $\Lambda = \{\sum \lambda_ib_i, \lambda_i \in \mathbb{Z}\}$, however, I cannot prove that this projection on the line again represents a lattice.

 

[Paul Colby]

Some questions. Do the $b_i$ span a finite or infinite dimensional space? Are all the $b_i$ linearly independent?

 

[Office_Shredder]

Is your lattice 2 dimensional, or is there more than one choice of line to pick, or did you mean you would project onto the orthogonal hyperspace?

 

[Peter_Newman]

Thank you for your questions. So this is the question I asked myself for an $n$-dimensional lattice. The $b_i$ are all linearly independent according to definition of an $n$-dimensional lattice. Then I would project the $n$-dimensional lattice $\Lambda$ onto the hyperplane orthogonal to $b_1$ (of the dimension $n-1$), if this is thought correctly. In the $2$-dimensional it is a line like this:

I tried to represent this graphically in 2D. The blacks are the lattice points and the greens their projection onto the plane/space/... orthogonal to $b_1$, the point in the origin belongs to both...

 

[Office_Shredder]

In the 2d case it's not hard to see it's a lattice. Let $c$ be orthogonal to $b_1$ with unit length.

$\alpha b_1 + \beta b_2\to \beta (<b_2,c> c)$ so the image is a 1 dimensional lattice generated by $<b_2,c> c$.

 

[Peter_Newman]

In the 2d case it's not hard to see it's a lattice. Let $c$ be orthogonal to $b_1$ with unit length.

$\alpha b_1 + \beta b_2\to \beta (<b_2,c> c)$ so the image is a 1 dimensional lattice generated by $<b_2,c> c$.

How you came up with this ($\alpha b_1 + \beta b_2\to \beta (<b_2,c> c)$ ...) so quickly, I have yet to understand... If I understand this right, $<b_2,c>$ expresses the $b_2$ component of a lattice point, now I just think what $<b_2,c> c$ means then...

How is that then generally in the n-dimensional case?

 

[Office_Shredder]

How you came up with this ($\alpha b_1 + \beta b_2\to \beta (<b_2,c> c)$ ...) so quickly, I have yet to understand... If I understand this right, $<b_2,c>$ expresses the $b_2$ component of a lattice point, now I just think what $<b_2,c> c$ means then...

How is that then generally in the n-dimensional case?

The inner product measures the length of the projection, and the multiplication by c measures the direction of the new post- projection vector. It probably looks too obvious because you already knew it was in the c direction.

 

[Peter_Newman]

The inner product measures the length of the projection

Exactly, that's what I wanted to say in my previous post :)

 

[Office_Shredder]

For the $n$ dimensional case, you basically get the same result. Let $P$ be the projection. Obviously the projected lattice is spanned by integer combinations of $Pb_2,...,Pb_n$. We need to prove these are linearly independent.

Suppose $\sum_{i\geq 2} \alpha_i Pb_i=0$. Then $\sum_{i\geq 2} \alpha_i b_i=\alpha_1 b_1$ since it has to live in the kernel of $P$
But the original basis is linearly independent, so all the coefficients must be zero.

 

[Peter_Newman]

Hey, so I would have started like this now. So the projection of a vector from the lattice onto the span orthogonal to $b_1 = b_1^*$ would be, to my knowledge:

$\pi_{i=2}(x) = \sum_{j\geq i=2}^n \frac{\langle x, b_j^*\rangle}{\langle b_j^*, b_j^*\rangle}b_j^* \quad\text{(1)}$

But now you would have to show that this is a lattice, which I think is difficult here, because the fraction would have to be an integer, compare the lattice definition from above:

$\Lambda = \{\sum \lambda_ib_i, \lambda_i \in \mathbb{Z}\}$

now I am not sure how to show that equation (1) is or builds a lattice... The $b_j^*$ are not a problem because they are in $R^n$ but the factor has to be in $Z$ to build a lattice...

 

[Office_Shredder]

What is your definition of $b_2^*$? For that matter, what is $b_1^*$?

I suspect your issue is just that those denominators are absorbed into your new lattice basis which you haven't correctly identified, but I'm not really sure what's going on.

 

[Peter_Newman]

Hey, sorry I forgot to name this, with the asterisks I express Gram-Schmidt vectors. Then e.g. $\pi_{i=2}(x)$ is composed of all vectors orthogonal to $b_1 = b_1^*$.

I believe, without having checked or proved this now, that what you say concerning the denominator could also fit, that the denominator together with the $b_j^*$ is my new basis lattice vector. It then remains to show, that $\langle x, b_j^* \rangle$ is an integer value, but I'am not sure how to proof this. We could express $x$ in the Gram-Schmidt basis, something like $x = \sum a_i b_i^*$. Then $\langle x, b_j^* \rangle = \langle \sum a_i b_i^*, b_j^* \rangle = a_j \langle b_j^*, b_j^* \rangle$ remains and here would be the question, whether $a_j \langle b_j^*, b_j^* \rangle$ is an integer...

 

[Office_Shredder]

Got it, I think those are not a basis. In general there is no reason for your projected lattice to contain an orthogonal basis.

 

[Peter_Newman]

Ok and yes I would agree with that. But I actually meant with my question rather whether $\pi_{i}(\Lambda)$ forms again a lattice.

 

[Office_Shredder]

Yes, the projection forms a lattice, and my proof gives the right basis.

$\pi_i(\Lambda)$ is spanned by integer combinations of $\pi_i(b_j)$ for $i\neq j$. These vectors are also linearly independent - if $\Sigma \alpha_j \pi_i(b_j)=0$, then $\pi_i(\Sigma \alpha_j b_j)=0$. But this means $\Sigma \alpha_j b_j$ lies in the span of $b_i$, which by linear independence of the original basis is impossible unless all the coefficients are zero. Hence they are linearly independent.

And any integer span of linearly independent vectors forms a lattice (note the integer span of linearly dependent vectors may not form a lattice, unless you happen to get lucky)

 

[Peter_Newman]

Suppose $\sum_{i\geq 2} \alpha_i Pb_i=0$. Then $\sum_{i\geq 2} \alpha_i b_i=\alpha_1 b_1$ since it has to live in the kernel of $P$ But the original basis is linearly independent, so all the coefficients must be zero.

In reverse order, I understand that. But not quite as it stands there.

the original basis is linearly independent, so all the coefficients must be zero

Works! This is exactly the definition of a basis.

$\sum_{i\geq 2} \alpha_i b_i=\alpha_1 b_1$ since it has to live in the kernel of $P$

The kernel of $P$ is or better satisfies $\sum_{i=1}^n \alpha_i b_i = 0$ so, you have simply rewritten this statement?

Suppose $\sum_{i\geq 2} \alpha_i Pb_i=0$

Why can we assume that this is true? I would rather conclude that by reading the proof backwards?!

 

[Office_Shredder]

So the goal is to prove that $\pi_i(\Lambda)$ is a lattice. Let's start real slow. Do you agree that $\pi_i(\Lambda)$ is spanned by integer combinations of $\pi_i(b_j)$ for $j\neq i$, and also that if those are linearly independent, you have proven that this is an n-1 dimensional lattice?

 

[Peter_Newman]

Do you agree that $\pi_i(\Lambda)$ is spanned by integer combinations of $\pi_i(b_j)$ for $j\neq i$

Yes, that could fit. $\Lambda$ is a linear combination of lattice basis vectors and a scalar (in front of each basis vector) that is an integer by definition. The span of the projection could then actually come down to integer linear combinations of $\pi_i(b_j)$. The projection subtracts a component of $\Lambda$, so we can't have a full rank anymore, right? So only components remain, which are on the space of the projection, on which we project.

 

[Office_Shredder]

So all that's left is to prove that they are linearly independent. Suppose a linear combination $\sum_j \alpha_j \pi_i(b_j)=0$. Use linearity of $\pi_i$ to write down a vector in the kernel of $\pi_i$. But what is the kernel of $\pi_i$?

To give a non example to see what the point is, suppose I have a "lattice" with basis vectors $(1,0,0)$, $(1,1,0)$ and $(1,\pi,0)$. I project the x axis to 0. $\pi_1(b_2)=(0,1,0)$ and $\pi_1(b_3)=(0,\pi,0)$. This does *not* form a lattice, because integer combinations of those two vectors form a dense subset. It's a non example because the original thing also was not a lattice, but this is the kind of situation you need to watch out for.

 

[Peter_Newman]

But what is the kernel of $\pi_i$?

So the image of $\pi_i$ looks like this $\pi_{i}(x) = \sum_{j\geq i}^n \frac{\langle x, b_j^*\rangle}{\langle b_j^*, b_j^*\rangle}b_j^*$. Now we know that by the projection thus the components from $b_1^*$ to $b_{i-1}^*$ respectively $b_1$ to $b_{i-1}$ are lost.

The kernel (ker f := {a $\in$ A | f(a) = 0}) of a linear mapping f are the solutions of the system f(a) = 0. So if I pick any vector $(a_1,...,a_n)^T$ then what is left after the projection $\pi_{i}$ is of the form $(0_{1},...,0_{i-1},\beta_{i},...,\beta_n)^T$ (only the indices are important here). So that means in the kernel of $\pi_i$ are $\sum_{i=1}^{i-1}a_i b_i = 0$. Is this possible to say?

 

[Office_Shredder]

So that means in the kernel of $\pi_i$ are $\sum_{i=1}^{i-1}a_i b_i = 0$. Is this possible to say?

Why did you write =0? The kernel is a set of vectors, not an equation.

 

[Peter_Newman]

With this I wanted to express that the kernel consists of the solutions of the homogeneous system of equations. But was the rest right so far?

 

[Office_Shredder]

With this I wanted to express that the kernel consists of the solutions of the homogeneous system of equations. But was the rest right so far?

The equation $\sum_{j<i} a_j b_j =0$ has only one solution, which is every $a_j=0$
We know this because the b's are linearly independent.

So I'm not sure what you mean here.

 

[Peter_Newman]

Hey @Office_Shredder the "rest" was related to my post #20

Yes the linear independence of the vectors implies then that there can be only this solution, that makes sense!