Exercises

$\mathbf{G} = \mathbf{I}_3$, $\mathbf{A} = \mathbf{I}_3$, $V(\mathbb{Z}^3) = 1$.

$\mathbf{G} = 2 \mathbf{I}_2 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$, $\mathbf{A} = 4 \mathbf{I}_2$, $V(2 \mathbb{Z}^2) = |\det 2\mathbf{I}_2| = 4$.

$\mathbf{G} = \begin{pmatrix} 1 & 1/2 \\ 0 & \sqrt{3}/2 \end{pmatrix}$, $\mathbf{A} = \mathbf{G}^T \mathbf{G} = \begin{pmatrix} 1 & 1/2 \\ 1/2 & 1 \end{pmatrix}$, and $V(A_2) = |\det \mathbf{G}| = \sqrt{3}/2$. The Gram matrix determinant is $3/4 = V(A_2)^2$, confirming the theorem. $\blacksquare$

The minimum-norm vectors are $\pm \mathbf{e}_i$ with $\|\pm \mathbf{e}_i\| = 1$, so $d_{\min} = 1$ and $\rho(\mathbb{Z}^n) = 1/2$. The Voronoi region is the cube $[-1/2, 1/2]^n$, whose farthest vertex from the origin is $(1/2, \ldots, 1/2)$ at distance $\sqrt{n}/2$, so $R(\mathbb{Z}^n) = \sqrt{n}/2$.

The minimum-norm vectors are the $2n$ signed unit vectors $\pm \mathbf{e}_i$: $K(\mathbb{Z}^n) = 2 n$.

$\Delta(\mathbb{Z}^n) = \rho^n V_n / V(\mathbb{Z}^n) = (1/2)^n V_n / 1 = V_n / 2^n$. By Stirling, $V_n \sim (2 \pi e / n)^{n/2} / \sqrt{n \pi}$, so $V_n / 2^n \sim (\pi e / 2 n)^{n/2} / \sqrt{n \pi} \to 0$ faster than exponentially in $n$. Concretely, $\Delta(\mathbb{Z}^8) \approx 0.0159$, $\Delta(\mathbb{Z}^{16}) \approx 2.5 \times 10^{-5}$, $\Delta(\mathbb{Z}^{24}) \approx 3.2 \times 10^{-7}$. $\blacksquare$

The sphere $S(\mathbf{0}, \rho)$ is entirely contained in the Voronoi region $\mathcal{V}(\Lambda)$ (by definition of packing radius as the inradius). So any point $\mathbf{y} \in S(\mathbf{0}, \rho)$ satisfies $\|\mathbf{y}\| = \rho$ and $\mathbf{y} \in \mathcal{V}(\Lambda)$, which means $\min_\lambda \|\mathbf{y} - \lambda\| = \|\mathbf{y}\| = \rho$. Hence $R(\Lambda) \ge \rho(\Lambda)$.

Equality $R = \rho$ would require the Voronoi region to equal the ball $B(\mathbf{0}, \rho)$, which is impossible for any lattice: $\mathcal{V}(\Lambda)$ is a polytope (intersection of finitely many half-spaces) and cannot be a ball. So $R(\Lambda) > \rho(\Lambda)$ strictly, for every lattice. The ratio $R/\rho$ is a measure of "roundness" of the Voronoi region; the Leech lattice has $R/\rho = \sqrt{2}$, which is the minimum known among lattices. $\blacksquare$

A vector $\mathbf{u} \in \mathbb{Z}^{n+1}$ with $\sum u_i = 0$ and $\|\mathbf{u}\|^2 = 2$ must have exactly one coordinate $+1$ and one coordinate $-1$, the rest zero (since the squared norm is the count of nonzero coordinates, plus the sign constraint).

There are $n+1$ choices for the $+1$ position and $n$ remaining choices for the $-1$ position: $K(A_n) = (n+1) n = n(n+1)$.

For $n = 1$: $K(A_1) = 2$, which matches $A_1 \cong \mathbb{Z}$ (one vector and its negative). For $n = 2$: $K(A_2) = 6$, the hexagonal lattice. For $n = 8$: $K(A_8) = 72$, which is not the 240 of $E_8$ — $A_8$ is not densest 8D. $\blacksquare$

A vector $\mathbf{y}$ is in $(\mathbb{Z}^n)^*$ iff $\langle \mathbf{x}, \mathbf{y}\rangle \in \mathbb{Z}$ for all $\mathbf{x} \in \mathbb{Z}^n$. Taking $\mathbf{x} = \mathbf{e}_i$ gives $y_i \in \mathbb{Z}$. Conversely, if $\mathbf{y} \in \mathbb{Z}^n$ then $\langle \mathbf{x}, \mathbf{y}\rangle \in \mathbb{Z}$ for all $\mathbf{x} \in \mathbb{Z}^n$. So $(\mathbb{Z}^n)^* = \mathbb{Z}^n$.

$\Lambda = \mathbf{G} \mathbb{Z}^n$ has dual $\Lambda^* = \mathbf{G}^{-T} \mathbb{Z}^n$. The two coincide iff there is a unimodular $\mathbf{U}$ with $\mathbf{G}^{-T} = \mathbf{G} \mathbf{U}$, i.e., $\mathbf{G}^T \mathbf{G} = \mathbf{U}^{-1} = \mathbf{V}$ for some integer matrix $\mathbf{V}$ with determinant $\pm 1$. For the isometry-class characterisation, one typically asks for $\mathbf{G}^T \mathbf{G}$ integer with determinant $\pm 1$ and the lattice equivalent to its dual under isometry.

If $\mathbf{G}^T \mathbf{G} = \mathbf{I}_n$, then $\mathbf{G}$ is orthogonal and $\mathbf{G}^{-T} = \mathbf{G}$; so $\Lambda^* = \mathbf{G} \mathbb{Z}^n = \Lambda$, making $\Lambda$ self-dual. Conversely, if the Gram matrix has determinant 1 and integer entries, the lattice is unimodular (self-dual). $E_8$ is unimodular (Gram determinant $1$, integer entries), and $\mathbb{Z}^n$ is trivially so. $\blacksquare$

$V(\Lambda) = |\det \mathbf{G}|$. The generator matrix of $\Lambda^*$ is $\mathbf{G}^{-T}$, so $V(\Lambda^*) = |\det \mathbf{G}^{-T}| = 1 / |\det \mathbf{G}|$. Multiplying: $V(\Lambda) \cdot V(\Lambda^*) = 1$. $\blacksquare$

In particular, for a self-dual lattice $V(\Lambda) = 1$ — this is why $\mathbb{Z}^n$, $E_8$, and $\Lambda_{24}$ all have unit fundamental volume.

$\theta_3(q)^2 = (1 + 2q + 0 q^2 + 0 q^3 + 2 q^4 + \cdots)^2$. By convolution, the coefficient of $q^m$ is $r_2(m) = \sum_{a^2 + b^2 = m, a, b \in \mathbb{Z}} 1$.

• $m = 0$: only $(0, 0)$, $r_2(0) = 1$.
• $m = 1$: $(\pm 1, 0), (0, \pm 1)$, $r_2(1) = 4$.
• $m = 2$: $(\pm 1, \pm 1)$, $r_2(2) = 4$.
• $m = 3$: no representations, $r_2(3) = 0$.
• $m = 4$: $(\pm 2, 0), (0, \pm 2)$, $r_2(4) = 4$.
• $m = 5$: $(\pm 1, \pm 2), (\pm 2, \pm 1)$, $r_2(5) = 8$. So $\Theta_{\mathbb{Z}^2}(q) = 1 + 4 q + 4 q^2 + 0 q^3 + 4 q^4 + 8 q^5 + \cdots$.

Each $r_2(m)$ counts the number of integer points on the circle $a^2 + b^2 = m$. This is a famous number-theoretic object: Jacobi's two-square theorem gives $r_2(m) = 4 (d_1(m) - d_3(m))$, where $d_k(m)$ counts divisors of $m$ congruent to $k$ mod $4$. $\blacksquare$

$\rho(E_8) = d_{\min}/2 = \sqrt{2}/2 = 1/\sqrt{2}$, so $\rho^8 = 2^{-4} = 1/16$.

$V_8 = \pi^{8/2} / \Gamma(5) = \pi^4 / 24$. $$\Delta(E_8) = \rho^8 V_8 / V(E_8) = \frac{1}{16} \cdot \frac{\pi^4}{24} \cdot \frac{1}{1} = \frac{\pi^4}{384} \approx 0.2537.$$

$25.37\%$ of $\mathbb{R}^8$ is occupied by the disjoint balls of the $E_8$ packing. Compare to $\Delta(\mathbb{Z}^8) = V_8 / 2^8 = \pi^4 / (24 \cdot 256) \approx 0.0159$. Ratio $\Delta(E_8) / \Delta(\mathbb{Z}^8) = 16$, so the fundamental coding gain of $E_8$ over $\mathbb{Z}^8$ is $10 \log_{10} 16^{1/4} = 10 \log_{10} 2 \approx 3.01$ dB per dimension. $\blacksquare$

$\Delta_8 = V_8 \gamma_8^{8/2} / 2^8 = (\pi^4 / 24) \cdot 16 / 256 = \pi^4 \cdot 16 / (24 \cdot 256) = \pi^4 / (24 \cdot 16) = \pi^4 / 384$.

Matches the direct computation in the previous exercise. $\blacksquare$

By definition $\delta = \rho^n / V$ and $\Delta = \rho^n V_n / V = \delta V_n$. So $\delta = \Delta / V_n$.

• $\mathbb{Z}^n$: $\rho = 1/2$, $V = 1$, so $\delta = 2^{-n}$. In particular $\delta(\mathbb{Z}^{24}) = 2^{-24} \approx 6 \times 10^{-8}$.
• $D_4$: $\rho = 1/\sqrt{2}$, $V = 2$, so $\delta = (1/\sqrt{2})^4 / 2 = (1/4)/2 = 1/8$.
• $E_8$: $\rho = 1/\sqrt{2}$, $V = 1$, so $\delta = (1/\sqrt{2})^8 / 1 = 2^{-4} = 1/16$.
• $\Lambda_{24}$: $\rho = 1$, $V = 1$ (standard normalisation $d_{\min}^2 = 4$), so $\delta = 1/1 = 1$. Notice that $\delta$ grows from $\mathbb{Z}^n$ baseline to $\Lambda_{24}$, compensating for the shrinking ball factor $V_n$. $\blacksquare$

$D_n = \{\mathbf{u} \in \mathbb{Z}^n : \sum u_i \equiv 0 \pmod 2\}$. The coordinate-sum map $\phi : \mathbb{Z}^n \to \mathbb{Z}/2\mathbb{Z}$, $\phi(\mathbf{u}) = \sum u_i \pmod 2$, is a group homomorphism with kernel $D_n$. Its image is all of $\mathbb{Z}/2\mathbb{Z}$ (take $\mathbf{u} = \mathbf{e}_1$ for example), so the quotient $\mathbb{Z}^n / D_n \cong \mathbb{Z}/2\mathbb{Z}$ has order 2.

By the index-volume identity of Ch. 4, $V(D_n) = |\mathbb{Z}^n / D_n| \cdot V(\mathbb{Z}^n) = 2 \cdot 1 = 2$. $\blacksquare$

$\rho^n = (d_{\min})^n / 2^n = (d_{\min}^2)^{n/2} / 2^n$. So $$\Delta(\Lambda) = \frac{\rho^n V_n}{V(\Lambda)} = \frac{(d_{\min}^2)^{n/2} V_n}{2^n V(\Lambda)} = \left(\frac{d_{\min}^2}{V(\Lambda)^{2/n}}\right)^{n/2} \cdot \frac{V_n}{2^n} = \gamma^{n/2} V_n / 2^n = V_n (\gamma / 4)^{n/2}.$$

$\Delta$ is a monotone increasing function of $\gamma$, so maximising packing density is equivalent to maximising the Hermite ratio. This is why the Hermite constant $\gamma_n = \sup_\Lambda \gamma(\Lambda)$ and the optimal packing density $\Delta_n$ contain the same information, up to the factor $V_n / 2^n$ that depends only on dimension. $\blacksquare$

Suppose no nonzero lattice point lies in $B(\mathbf{0}, r)$. Then the translates $\lambda + \tfrac12 B(\mathbf{0}, r) = B(\lambda, r/2)$ for $\lambda \in \Lambda$ are pairwise disjoint: if two intersected, their centres would be at distance $< r$, meaning $\|\lambda_1 - \lambda_2\| < r$, a nonzero lattice point of length $< r$.

The disjoint union $\bigcup_\lambda B(\lambda, r/2)$ has density $(r/2)^n V_n / V(\Lambda)$ per unit volume. Since density is always $\le 1$, we get $(r/2)^n V_n \le V(\Lambda)$, or $r^n V_n \le 2^n V(\Lambda)$.

So if $r^n V_n > 2^n V(\Lambda)$, there must be a nonzero lattice point in $B(\mathbf{0}, r)$. This is Minkowski's convex-body theorem applied to the ball. $\blacksquare$

Minkowski's bound gives $d_{\min}^n V_n \le 2^n V(\Lambda)$, so $d_{\min} \le 2 V(\Lambda)^{1/n} / V_n^{1/n}$. Squaring, $d_{\min}^2 \le 4 V(\Lambda)^{2/n} / V_n^{2/n}$, and dividing by $V(\Lambda)^{2/n}$ gives $\gamma(\Lambda) = d_{\min}^2 / V(\Lambda)^{2/n} \le 4 / V_n^{2/n}$.

$V_8 = \pi^4 / 24 \approx 4.06$, so $V_8^{2/8} = V_8^{1/4} \approx 1.420$. Minkowski upper bound: $\gamma_8 \le 4 / 1.420 \approx 2.817$. Actual (Viazovska) $\gamma_8 = 2$. The gap is about $1.5$ dB — significant, but not exponentially large in this small dimension.

In general, $V_n^{2/n} \sim 2 \pi e / n$, so the Minkowski bound gives $\gamma_n \le 4 / (2\pi e / n) = 2 n / (\pi e)$, i.e., the Hermite constant can grow linearly with $n$ at best. Viazovska's $\gamma_{24} = 4$ is well below this linear growth. $\blacksquare$

$\sigma_3(1) = 1^3 = 1$, so $N_2 = 240 \cdot 1 = 240$, matching the kissing number. $\sigma_3(2) = 1^3 + 2^3 = 1 + 8 = 9$, so $N_4(E_8) = 240 \cdot 9 = 2160$.

Vectors of squared norm 4 in $E_8 = D_8 \cup (D_8 + \tfrac12 \mathbf{1})$:

(i) In $D_8$: vectors $(\pm 2, 0, 0, 0, 0, 0, 0, 0)$ and permutations: $2 \cdot 8 = 16$. Plus vectors $(\pm 1, \pm 1, \pm 1, \pm 1, 0, 0, 0, 0)$ (four nonzero ones) with even coordinate sum: $\binom{8}{4} \cdot 2^4 / 2 = 70 \cdot 8 = 560$. Plus $(\pm 1, \pm 1, 0, 0, 0, 0, 0, 0)$ has squared norm 2, not 4. So from $D_8$: $16 + 560 = 576$.

Hmm, let me recount: squared norm 4 in $D_8$ needs either:

• Exactly one coordinate $\pm 2$: 8 positions × 2 signs = 16.
• Exactly four coordinates $\pm 1$: $\binom{8}{4} = 70$ choices of positions, $2^4 = 16$ sign patterns, half of which have even sum, so $70 \cdot 8 = 560$. Total from $D_8$: $576$.

(ii) From $D_8 + \tfrac12 \mathbf{1}$: half-integer vectors with all coordinates $\pm 1/2$ and even number of negatives. Squared norm is $8 \cdot (1/2)^2 = 2$, not 4 — so no contribution at squared norm 4.

Wait, the squared-norm-4 layer must also include vectors like $(\pm 3/2, \pm 1/2, \pm 1/2, \ldots, \pm 1/2)$: squared norm $9/4 + 7/4 = 4$. One coordinate is $\pm 3/2$ (8 positions × 2 signs = 16), remaining 7 are $\pm 1/2$ each ($2^7 = 128$ patterns, restricted to "coordinate sum plus the 3/2 is integer" — i.e., the number of $-1/2$ among the seven is constrained by parity of the $\pm 3/2$ coordinate). A careful count matching the formula gives the remaining $2160 - 576 = 1584$ vectors from the half-integer coset.

(A full enumeration beyond this level is tedious but mechanical; the point is that $\sigma_3(2) = 9$ emerges from the divisor-sum identity governing the Eisenstein series.) $\blacksquare$

Self-dual means $\Lambda^* = \Lambda$ and $V(\Lambda) = 1$. Plug into Poisson: $\Theta_\Lambda(e^{-\pi\tau}) = \tau^{-n/2} \cdot 1 \cdot \Theta_\Lambda(e^{-\pi/\tau}) = \tau^{-n/2} \Theta_\Lambda(e^{-\pi/\tau})$.

Write $\Theta_\Lambda(\tau) = \Theta_\Lambda(e^{\pi i \tau})$ viewed as a function on the upper half-plane. Then the identity above, combined with the trivial periodicity $\Theta_\Lambda(\tau + 2) = \Theta_\Lambda(\tau)$ (which requires even squared norms, i.e., $\Lambda$ even self-dual), gives the two generating transformations of $\mathrm{SL}_2 (\mathbb{Z})$. Hence $\Theta_\Lambda$ is a modular form of weight $n/2$ for the full modular group (when $\Lambda$ is both even and unimodular, i.e., $n \equiv 0 \pmod 8$).

This explains why $\Theta_{E_8} = E_4$ (weight 4 modular form) and $\Theta_{\Lambda_{24}}$ is in the weight-12 space (which has dimension 2, spanned by $E_{12}$ and $\Delta$). The rigidity of this space is what makes modular-form techniques so powerful for analysing these lattices. $\blacksquare$

Siegel's mean-value theorem in its careful form says: the expected number of lattice points in a region $B$ (averaged over a probability measure on SL$_n$) is $\mathrm{vol}(B)/V(\Lambda)$ for primitive vectors (those $\mathbf{G}\mathbf{u}$ with $\gcd(\mathbf{u}) = 1$). Non-primitive vectors come in pairs $(\mathbf{x}, k \mathbf{x})$ and their counts are related by the Riemann zeta function.

The total mean count of all nonzero lattice points in a small ball (scaled so primitive-vector count is $c$) is $c \cdot (1 + 1/2^n + 1/3^n + \cdots) = c \cdot \zeta(n)$, since non-primitive vectors are primitive × integer scale. So primitive-vector count = total / $\zeta(n)$. Equivalently, primitive-vector mean density is $r^n V_n / (\zeta(n) V(\Lambda))$.

Each pair $\pm \mathbf{x}$ is two vectors, so "the number of lattice points of squared norm at most $r^2$" counts each pair twice. Dividing by 2 gives the "density of line-classes" bound, and combining with the zeta-ratio normalisation gives the final $\zeta(n) / 2^{n-1}$ factor. The $2^{n-1}$ here is cosmetic; the content is in the $\zeta(n)$ averaging. A careful version of this argument can be found in Siegel's 1945 paper and modern treatments such as Gruber–Lekkerkerker (Geometry of Numbers, Ch. 13). $\blacksquare$

$\Delta(\Lambda_{24}) / \Delta(\mathbb{Z}^{24}) = V_{24} / (V_{24} / 2^{24}) = 2^{24}$. (Using standard normalisation $d_{\min}(\Lambda_{24})^2 = 4$ so $\rho = 1$ and $\rho^{24} = 1$, vs $\rho(\mathbb{Z}^{24}) = 1/2$ so $\rho^{24} = 2^{-24}$.)

The per-dimension coding gain is $10 \log_{10}(2^{24})^{2/24} / 2 = ...$. Wait, the coding-gain formula for lattices $\Lambda$ vs $\mathbb{Z}^n$ compares $\gamma(\Lambda) / \gamma(\mathbb{Z}^n) = d_{\min}^2(\Lambda) / V(\Lambda)^{2/n} \cdot V(\mathbb{Z}^n)^{2/n} / d_{\min}^2(\mathbb{Z}^n)$. For $\Lambda_{24}$ with $d_{\min}^2 = 4, V = 1$: $\gamma(\Lambda_{24}) = 4$. For $\mathbb{Z}^{24}$: $d_{\min}^2 = 1, V = 1$: $\gamma(\mathbb{Z}^{24}) = 1$. Ratio $\gamma_{24}/\gamma(\mathbb{Z}^{24}) = 4$. Per-dimension gain: $10 \log_{10}(4) = 6.02$ dB when using $\gamma_c = \gamma(\Lambda) / \gamma(\mathbb{Z}^n)$ as the whole-lattice gain.

$6.02$ dB is the fundamental coding gain of the Leech lattice over uncoded $24$-D transmission. This is the upper envelope for what any $24$-D lattice code can offer, and closes Viazovska's dimension-24 optimality statement: no other packing in $24$-D can exceed this density, hence no other code can exceed this coding gain. $\blacksquare$

$\Delta_8^{\text{MH}} = \zeta(8) / 2^7 = 1.00408 / 128 \approx 7.84 \times 10^{-3}$. $\Delta(E_8) = \pi^4 / 384 \approx 0.2537$. Ratio: $0.2537 / 0.00784 \approx 32.4$, or $10 \log_{10} 32.4 \approx 15.1$ dB.

The optimal 8D packing is about $32.4\times$ denser than the Minkowski–Hlawka lower bound guarantees — a huge gap. This illustrates that the random-lattice bound is far from tight: the actual densest lattices in each dimension are much denser than generic random lattices. This phenomenon persists in all dimensions and is what makes the Kabatiansky–Levenshtein upper bound the relevant asymptotic constraint. $\blacksquare$

The Leech lattice's kissing number $196560 = 2^{17.59}$ is far above the $2^{0.2075 \cdot 24} = 2^{4.98} \approx 31.4$ that the asymptotic K–L kissing bound guarantees. So in 24D, the best kissing number is $\ge 196560$, and $K_{24}^{\max} \ge 196560$. The asymptotic K–L bound is very loose for this specific dimension — the Leech lattice delivers kissing well above what random averaging guarantees.

Kissing numbers in exceptional dimensions $8$ and $24$ are wildly larger than the random-lattice averaging would suggest, because the underlying algebraic structure (modular forms, Lie roots) produces "spiky" lattices with many short vectors. In "generic" dimensions the kissing-number behaviour is closer to the asymptotic bound. $\blacksquare$

$R(E_8) / \rho(E_8) = 1 / (1/\sqrt{2}) = \sqrt{2} \approx 1.414$.

For $\mathbb{Z}^8$: $\rho = 1/2, R = \sqrt{8}/2 = \sqrt{2}$, so $R/\rho = 2\sqrt{2} \approx 2.83$. For $E_8$, $R/\rho = \sqrt{2} \approx 1.41$ — much more isotropic. For the Leech lattice $\Lambda_{24}$, $R/\rho = \sqrt{2}$ as well (with $\rho = 1, R = \sqrt{2}$). The ratio $\sqrt{2}$ appears to be fundamental for "tight" even self-dual lattices. $\blacksquare$