# Minkowski's Linear Forms Theorem With Complex Coefficients

Minkowski's Linear Forms Theorem is often stated about linear forms with real coefficients. However, in Narkiewicz's Elementary and Analytic Theory of Algebraic Numbers, the following generalization of Minkowski's Linear Forms Theorem is stated (the theorem is actually stated for a general lattice in real space, but I'm only interested in the case where the lattice is $\mathbb{Z}^N$):

Let $U = \{ L_j(x_1, \ldots, x_N) = \sum_{i=1}^N a_{ij} x_i \mid 1 \leq j \leq N \}$ be a system of $N$ linear forms with complex coefficients satisfying the following condition: if $L_j \in U$, then the conjugate form, $\overline{L_j} \in U$. Let $M = [a_{jk}]_{1\leq j,k \leq N}$. Let $\epsilon_1, \ldots, \epsilon_N \in \mathbb{R}^+$ such that $\prod_{i=1}^N \epsilon_i \geq | \det M |$ and $L_i = \overline{L_j}$ implies $\epsilon_i = \epsilon_j$. Then there exists a nonzero point $(x_1, \ldots, x_n) \in \mathbb{Z}^N$ such that $|L_j(x_1,\ldots,x_N)| \leq \epsilon_j$ for each $1 \leq j \leq N$ with strict inequality holding for all but one $j$.

However, instead of providing a proof, Narkiewicz refers to two books by Cassels, both of which only prove the case where the coefficients of the forms are real. The proof of the real version---which uses Minkowski's Convex Body Theorem on $M^{-1} ([-1,1]^N)$---does not easily generalize to the complex version. Does anyone have an idea of how the proof of the generalization might go? I've tried translating the problem to real-space (using the fact that $N$-dimensional complex space is isomorphic to $2N$-dimensional real space), but the set that I want to apply Minkowski's Convex Body Theorem to does not have large enough volume.

EDIT: A proof may appear in the 39th "chapter" in Minkowski's Geometrie der Zahlen (on page 113). However, I cannot read German. Does anyone know of an English translation of the work?

## migrated from math.stackexchange.comMar 17 '17 at 4:15

This question came from our site for people studying math at any level and professionals in related fields.

• A proof can be found, e.g., in Hecke's Lectures on the theory of algebraic numbers (Theorem 95, pp. 104--105 in the English translation). I can currently access this on Google Books: books.google.com/… – so-called friend Don Mar 21 '17 at 0:36
• @so-calledfriendDon Thank you very much! Of course, I'm just now seeing this after I made it through the proof myself with a big hint from Minkowski and the aid of Google Translate. On the bright side, I understand the proof very thoroughly now. – Greg K Mar 21 '17 at 3:37

First, note that we can assume that all $\epsilon_j = 1$. If not, replace $L_j$ by $\frac{L_j}{\epsilon_j}$ and note that all of the hypotheses pertaining to forms and their conjugates still hold.
Suppose there are $r$ forms which have real coefficients and $s$ pairs of complex conjugate forms so that $r + 2s = N$. Without loss of generality, $L_1, \ldots, L_r$ have real coefficients and $\bar{L}_{r+j} = L_{r+s+j}$ for each $1 \leq j \leq s$. Then we can write $$L_{r+j} = \frac{R_j + i I_j}{\sqrt{2}} \\ L_{r+s+j} = \frac{R_j - i I_j}{\sqrt{2}}$$ for linear forms with real coefficients $R_j$ and $I_j$ and $1 \leq j \leq s$. Observe that if we look at the set of forms $L_j$ for $1 \leq j \leq r$ and $R_k, I_k$ for $1 \leq k \leq s$, we have $r + 2s = N$ linear forms in $N$ variables, where all the linear forms have real coefficients. In particular, using the fact that $\bar{a}_{r+j,k} = a_{r+s+j,k}$ for all $1 \leq j \leq s$ and $1 \leq k \leq N$ (this is simply restating that $\bar{L}_{r+j} = L_{r+s+j}$), we can explicitly describe $R_j$ and $I_j$: $$R_j(x_1, \ldots, x_N) = \sqrt{2} \text{Re}[a_{r+j,1}]x_1 + \cdots + \sqrt{2} \text{Re}[a_{r+j,N}] x_N \\ I_j(x_1, \ldots, x_N) = \sqrt{2} \text{Im}[a_{r+j,1}]x_1 + \cdots + \sqrt{2} \text{Im}[a_{r+j,N}] x_N$$
Let $A$ be the matrix whose entries are the coefficients of the forms $L_j$ for $1 \leq j \leq r$ and $R_k, I_k$ for $1 \leq k \leq s$. I won't go into the gory details, but using elementary properties of determinants, you can show that $|\det(A)| = |\det(M)| \leq 1$. Now you can apply the version of Minkowski's Linear Forms Theorem where the coefficients of the linear forms are real (proofs of this theorem are plentiful) to find a nonzero point $\mathbf{x} = (x_1, \ldots, x_N) \in \mathbb{Z}^N$ where $|L_1(\mathbf{x})| \leq 1$, $|L_j(\mathbf{x})| < 1$ for $2 \leq j \leq r$ and $|R_k(\mathbf{x})|, |I_k(\mathbf{x})| < 1$ for each $1 \leq k \leq s$.
But then we have, for each $1 \leq j \leq s$ $$|L_{r+j}(\mathbf{x})|^2 = \left|\frac{R_j(\mathbf{x}) + i I_j(\mathbf{x})}{\sqrt{2}} \right|^2 = \frac{R_j(\mathbf{x})^2 + I_j(\mathbf{x})^2}{2} < 1 \\ |L_{r+s+j}(\mathbf{x})|^2 = \left|\frac{R_j(\mathbf{x}) - i I_j(\mathbf{x})}{\sqrt{2}} \right|^2 = \frac{R_j(\mathbf{x})^2 + I_j(\mathbf{x})^2}{2} < 1$$.
This shows that $|L_1(\mathbf{x})| \leq 1$ and for each $2 \leq j \leq N$, we have $|L_j(\mathbf{x})| < 1$, which is exactly what we wanted.