# Projection of a lattice onto a subspace

Let $$G$$ be a $$n \times n$$ matrix with real entries and let $$\Lambda = \{x^n \colon \exists i^n \in \mathbb{Z}^n \text{ such that } x^n = G \cdot i^n\}$$ define a lattice. I am interested in projecting the lattice points onto a $$k$$-dimensional subspace $$U$$ with $$k < n$$. Let $$A$$ be a $$n \times k$$ matrix of vectors that span $$U$$. Then, the projection of $$\Lambda$$ onto $$U$$ has the generator matrix $$P_A \cdot G$$ where $$P_A = A (A^TA)^{-1}A^T$$ is the projection operator. Lets also assume that the entries of $$G$$ have no structure between them - i.e., they are generally chosen from the reals.

I am interested in finding out when the projection of $$\Lambda$$ onto $$U$$ (call it $$\Lambda_U$$) is also a $$k$$-dimensional lattice. It seems to me that in most cases, $$\Lambda_U$$ will fill out the $$k$$-dimensional space $$U$$ completely, i.e., the shortest distance between neighboring points in $$\Lambda_U$$ will be arbitrarily small (obs. 1). For certain subspaces $$U$$ that seem to be "aligned" correctly with $$\Lambda$$, we get a well-defined $$\Lambda_U$$. By well-defined, I mean that the spacing between the nearest neighbors of $$\Lambda_U$$ is bounded away from $$0$$. As far as I can tell, this happens when $$A$$ is chosen such that the columns of $$P_A \cdot G$$ are rational multiples of one another (obs. 2).

Questions:

1. Are the observations (obs. 1) and (obs. 2) correct?
2. If so, is this a well-studied concept? I didn't have much success googling with the obvious keywords such as lattice "projection" or "matrix column rational multiples" etc.
3. Assuming the observations are correct, given $$G$$, is there a way to choose $$A$$ such that the columns of $$P_A \cdot G$$ are rational multiples of one another?
• I have this hunch that you can, without loss of generality, replace one of the matrices $A$ or $G$ with the identity matrix. In other words, the question doesn't change if you apply a linear transformation to the whole vector space. – Greg Graviton Dec 17 '10 at 18:20

## 2 Answers

I disagree with observation 2. It gives a sufficient condition that is not necessary unless you only consider projections onto a 1-dimensional subspace.

A weaker sufficient condition, where I am not sure whether its also necessary is the following: if there is a decomposition $P_A G = BC$, where $B$ is any regular matrix and $C$ is a matrix with only integer entries then $\Lambda_U$ also must be a lattice. This must be because $C$ represents a map $\mathbb{Z}^n \rightarrow \mathbb{Z}^n$, and $B$ represents a vector space isomorphism. (note that it does not matter if $C$ can be integer or rational, for any common denominator can be moved into $B$.)

It seems clear to me for geometric reasons that for any regular $G$ a suitable 1-dimensional $U$ can be found. Consider the basis vectors of the lattice, i.e. the columns of $G$. There must be a hyperplane through them and a line through the origin perpendicular to that hyperplane. An orthogonal projection onto that line will move points parallel to the hyperplane. In particular, the basis vectors will all be projected to the intersection of the line with the hyperplane. It follows that integer combinations of the basis vectors will be mapped to integer multiples of that.

I am not sure whether there is a simple way to find a base for this $U$, but it should always possible to take differences of basis vectors as a basis for the hyperplane and use Gramm-Schmidt orthonogonalization to find a normal vector.

If you are interested, it can be proved easily, at least in $\mathbb{R}^2$, that there is in fact an (countable) infinity of suitable $U$s. This can be done by setting up a sine/cosine parametrization of the unit vectors and showing that the ratio of the projection of the basis vectors of the lattice is a continuous non-constant function of the angle.

Addendum: Actually, it is algebraically not as complicated as I imagined. Supposing that $G$ is invertible, let $H = G^{-1}$ and take for $D$ an $n \times 1$ matrix of all ones.

Now if we let $A = H^TD$ then $A^TA$ becomes scalar and we can simplify $$P_A = \frac{H^TDD^TH}{D^THH^TD}$$ This means that $$P_AG = \frac{H^TD}{D^THH^TD} D^T$$ which means that the columns are all equal. Alternatively, you could of course replace the ones in D with any rationals.

• @ Niels: Thanks for your answer. I am still working my way through your response. – Dinesh Dec 23 '10 at 21:10

The technique you suggest is sometimes used to construct quasi-crystaline structures or in mathematical parlance, aperiodic tilings. If you take a high dimensional lattice and you project on an appropriate subspace U, the result can be the lattice structure of a quasi-crystal.

• Thanks for the reference. It should come in handy for googling about this. – Dinesh Dec 23 '10 at 21:10