# Periodic functions over different lattices in $\mathbb R^d$ are linearly independent [closed]

I have the following claim that I think have been proved by someone, but I can not find the reference, hence I would like to ask for help. Here is the claim:

Let $f_1, \ldots, f_n$ be continuous functions from $\mathbb R^d$ to $\mathbb R$ and $L_1, \ldots, L_n$ are $d$-dimensional lattices in $\mathbb R^d$ with the property that $$f_i(x)=f_i(x+l),$$ for all $x$ in $\mathbb R^d$ and all $l$ in $L_i$.

Suppose that $L_i$ and $L_j$ are not in a bigger lattice $L$ for all distinct $i$ and $j$, $f_1, \ldots, f_n$ are linearly independent, or one of them is a constant function.

I am thinking of something along the line of Fourier Transform to solve this problem. Any ideas? Thanks in advance.

## closed as off-topic by Qiaochu Yuan, Ryan Budney, S. Carnahan♦May 4 '14 at 22:28

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• Well Fourier transform won't help you here, because periodic continuous functions are not $L^{1}$, if you want to go along this lines, you would need to go to Fourier series along the related tori. Anyways, an elementary solution would be to project the linear combination into say a fundemental domain for ome $L_i$ and then use Kronecker lemma/Weyl's theorem to ensure that the $L_i$ periodic function must be constant. But notice that in the settings you've described, the thm does not hold. From the Kronecker argument I've described, you get that the function must be constant along leafs of – Asaf May 4 '14 at 17:17
• foliation of the torus by sub-tori (notice that Kronecker system is semi-simple, meaning it decomposes to a disjoint union of minimal ones, it is not necessarily minimal itself). The exact condition needed here is that any two sub-groups (not necessarily lattices (i.e. finite co-volume)) are not commensurable. – Asaf May 4 '14 at 17:19
• Thanks for your comment. Do you have reference for the Kronecker argument that you mention? – sweehong May 4 '14 at 17:28
• It was discussed recently here - mathoverflow.net/questions/162875/… Loosely speaking, this is some multi-dimensional generalization of the fact that irrational multiples are equidistributed in the torus, this appears in every basic book about ergodic theory, for example in Furstenberg's book or the recent book by Einsiedler-Ward book (volume I). I believe that the Fourier approach (in general LCA groups) appears in Katznelson book. – Asaf May 4 '14 at 17:40

## 1 Answer

Although it surely accomplishes nothing more of substance than more elementary-sounding arguments, taking Fourier transforms in the sense of tempered distributions gives a quick outcome: the Fourier transform of $e^{i\xi x}$ is a constant multiple of Dirac $\delta$ at $\xi$, so the Fourier transform of an $L$-periodic functions is supported on (edit: thx @Yoav Kallus) the dual lattice $\check{L}$, as tempered distribution. Thus, for $f_i$ $L_i$-periodic, $0=\sum_i f_i$ gives $0=\sum_i \widehat{f}_i$ as tempered distribution, and we look at supports. Depending what hypotheses you really want/need on the lattices, we easily reach conclusions: for example, if $L_i\cap L_j=\{0\}$ for any two $i\not=j$, (edit) then the same is true of the duals and the support of every $\widehat{f}_i$ appearing must be (contained in) $\{0\}$.

The pairwise intersections can be intermediate discrete subgroups of $\mathbb R^n$, of course, without being lattices or $\{0\}$.

• Thanks for the answer. My hypothesis is that the additive subgroup generated by L_i and L_j is not discrete, and I think we may need heavier machinery such as Weyl's criterion for that route, or is there an easier argument? – sweehong May 4 '14 at 18:31
• @sweehong, I think that hypothesis is insufficient: if two lattices in $\mathbb R^2$ are commensurate in one direction, but incommensurate in another, non-trivial linear combinations can be $0$. – paul garrett May 4 '14 at 18:39