Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 20 / 2

Use the following skew-symmetric matrix, whose columns reduce modulo three to four of the specified nonzero tetracode words:Set $B_{+}=3I+S$, $B_{-}=-3I+S$, and define the full Euclidean lattices $L_{\pm }^0=B_{\pm }{\mathbb{Z}}^4$. Their determinants have absolute value $144$, sinceFor arbitrary positive real numbers $a_1,\dots ,a_4$, put $D=\mathrm{diag}(\sqrt{a_1},\dots ,\sqrt{a_4})$ and $L_{\pm }=D{L}_{\pm }^0$. This is the tetracode length-isospectral lattice construction. We prove equality of the length multisets by an explicit bijection, not just equality of the shortest lengths.

Modulo three, both lattices have image $\mathrm{im}S=T$. The matrix $S$ has rank two over ${\mathbb{F}}_3$; its four column lines are precisely the four lines in $T$. Also $S^2=0$ modulo three, so $\ker S=\mathrm{im}S=T$. Each nonzero tetracode word has exactly one zero coordinate. Every $v\in L_{\pm }^0$ therefore has some coordinate divisible by three. Moreover, all four coordinates of $v$ have the same parity, because every entry of either $B_{\pm }$ is odd.

For $v\in L_{+}^0$, choose the least index $i$ with $3\mid v_i$ and set $w=v-2v_i{e}_i$. The coordinate reflection bijection for tetracode lattices will show that $w\in L_{-}^0$. First, $w\equiv v(\mathrm{mod}3)$, so $Sw\equiv 0(\mathrm{mod}3)$. Second, positive-lattice membership implies $Sv\equiv 3v\equiv -v(\mathrm{mod}4)$. HenceThe expression in parentheses has $i$th coordinate zero and every other coordinate $v_j+v_i{S}_{ji}$ even, by the common parity and $S_{ji}=\pm 1$. Thus $(I-S)w\equiv 0(\mathrm{mod}4)$. Together the congruences modulo three and four say $(-3I-S)w\equiv 0(\mathrm{mod}12)$, which is precisely $B_{-}^{-1}w\in {\mathbb{Z}}^4$.

Conversely, if $v\in L_{-}^0$, the same reflection has unchanged reduction modulo three, and now $Sv\equiv v(\mathrm{mod}4)$. ThusIts $i$th coordinate is $4v_i$ and its other coordinates are even multiples of two, so it vanishes modulo four; it vanishes modulo three as well. The reflected vector is in $L_{+}^0$. The chosen index is unchanged by reflection, including for vectors reducing to zero, so this map is an involution between the two lattices. It preserves every $|v_j|$ and therefore preserves $\Vert Dv{\Vert }^2={\sum }_j{a}_j{v}_j^2$. The two lattices have the same length spectrum, including every multiplicity, for every positive choice of weights.

To prove a genuine failure of isometry, chooseThese four numbers are linearly independent over $\mathbb{Q}$: changing the signs of the independent square roots in the multiquadratic field $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ isolates each coefficient of a rational relation. If an orthogonal transformation $O$ maps $L_{+}$ to $L_{-}$, thenand $P^T\mathrm{diag}(a_j)P=\mathrm{diag}(a_j)$. Comparing rational coefficients of the independent $a_j$ in the diagonal entries gives $P_{ji}^2={\delta }_{ji}$. Consequently $P$ is diagonal with entries $\pm 1$.

Such a coordinate sign change must preserve $T$ modulo three. For each zero coordinate $i$, there are exactly two opposite nonzero tetracode words. Preserving their line requires the other three signs to be equal. Taking two different zero coordinates forces all four signs to be equal. Thus $P=I$ or $P=-I$, either of which would imply $L_{+}^0=L_{-}^0$. ButThis contradiction proves $L_{+}$ and $L_{-}$ are not isometric. Allowing affine isometries changes nothing: translating the image of zero back to zero reduces them to the orthogonal case.

For the flat tori, take period lattices equal to the dual lattices:The standard spectrum of a flat torus lemma states that its characters of a real torus are indexed by the dual of its period lattice. Here they are $e^{2\pi i⟨v,x⟩}$ for $v\in L_{\pm }$, with Laplacian eigenvalue $4{\pi }^2\Vert v{\Vert }^2$. These characters give a complete orthogonal basis, so our length-preserving bijection proves that $X_{+}$ and $X_{-}$ are isospectral manifolds. The second standard lemma is that an isometry of flat tori lifts to an affine isometry of their Euclidean universal covers. Its orthogonal part maps the period lattices to one another, and therefore maps their dual lattices to one another. This would give the impossible isometry $L_{+}\cong L_{-}$. Hence $X_{+}$ and $X_{-}$ are isospectral non-isometric flat four-dimensional tori.