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EDIT, Tuesday, July 26. I have convinced myself, with my own C++ programs, that the sum of five squares does satisfy the "easier" Borcherds-Allcock condition, although it fails Pete L. Clark's criterion. Four squares works, I believe that six and seven squares will work, while eight squares will be "borderline," with the worst behavior happening at the point with all coordinates being 1/2. Now, as Prof. Nebe has designed a new algorithm for covering radius, and Magma uses this now, I can at least hope that she gets interested enough to design an algorithm for this hybrid condition. Otherwise it is going to take me forever to find all of the "odd" lattices that succeed. Not at all by the way, the original impetus for Prof. Nebe's involvement was a request by Richard Parker for the positive integral lattices P for which it would be possible to calculate Aut(P + U) where U is the unimodular two-dimensional "even" Lorentzian lattice...

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I finally figured out that Nebe gets every lattice to be "even" by doubling all coefficients, this particularly happening to the "odd" lattices such as the sum of $k$ squares. If we were to leave the odd lattices as they are, her condition (and Pete's) would be that the covering radius be strictly less than 1! The reality check is that both Nebe and Pete include the sum of three squares, but both exclude the sum of five squares. For comparison, see LINK Before continuing, let me point out that this strong condition happens only a few times, $$x^2, 3 x^2, \; x^2 + y^2, x^2 + 2 y^2, 2 x^2 + 2 x y + 3 y^2, \; x^2 + y^2 + z^2, x^2 + 2 y^2 + 2 y z + 2 z^2,$$ in dimension no larger than 3.

Note that the earlier proof for "even" lattices does, in fact, apply to the "odd" lattices.

Both Richard Borcherds and his student, Daniel Allcock of U. T. Austin, pointed out that I did not have quite the "correct" condition. In the Duke Math. J., vol. 103 (2000) pp. 303-333, see LINK section 6, Allcock discusses what he calls a "well-covering," which is a covering by balls of varied radius, where the radius of the ball around a given lattice point is decided by the norm of the lattice point. In particular, he says (but not in the paper) that the correct condition for odd'' lattices is that $\mathbb R^n$ is covered by {\bf open} balls of radius 1 around all lattice points of odd norm, then radius $\sqrt 2$ around lattice points of even norm, where the norm of some $x \in \Lambda$ is $x \cdot x.$ This is called a "strict well-covering." This condition is milder than Pete's.

Given such a positive odd lattice $\Lambda$ with a strict well-covering as described, form the integral Lorentzian lattice $L = \Lambda \oplus U.$ Elements are of the form $(\lambda, m,n)$ where $\lambda \in \Lambda, \; m,n \in \mathbb Z.$ The norm on $L$ is given by $$(\lambda, m,n)^2 = \lambda^2 + 2 mn.$$ We infer the inner product $$(\lambda_1, m_1,n_1) \cdot (\lambda_2, m_2,n_2) = \lambda_1 \cdot \lambda_2 + m_1 n_2 + m_2 n_1.$$

Note that the sublattice of all points with even norm has index 2, and that norm 0 vectors in $L = \Lambda \oplus U$ can only be created from such points with even norm.

We choose a particular set of roots (elements of norm 2) beginning with any $\lambda \in \Lambda$ of even norm by $$\tilde{\tilde{\lambda}} = \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$

However, starting with points $\beta \in \Lambda$ of odd norm, we create root-ones by $$\tilde{\beta} = \left( \beta, 1, \frac{ 1 - \beta^2}{2} \right),$$ which are elements of norm 1 in $L.$ Note that when an English soccer goalkeeper boots the ball down the center of the pitch and attack remains central, this is referred to as root-one football. I'm sure I have that right.

Let the group $R \subseteq \mbox{Aut}(L)$ be generated by reflections in all the $\tilde{\tilde{\lambda}}$ and all the $\tilde{\beta}$ and by $\pm 1.$

Given any root $r \in L,$ meaning $r^2 = 2,$ we get the reflection $$s_r (z) = z - ( r \cdot z) r.$$ Also $s_r^2(z) = z.$

Given any root-one $\tilde{\beta} \in L,$ meaning $\tilde{\beta}^2 = 1,$ we get the reflection $$s_{\tilde{\beta}} (z) = z - 2( \tilde{\beta} \cdot z) \tilde{\beta}.$$ Also $s_{\tilde{\beta}}^2(z) = z.$

Given some primitive null vector $$z = (\xi, a,b),$$ so that $2 a b = - \xi^2,$ and $b = \frac{- \xi^2}{2 a} \in \mathbb Z.$ Note that, in order to have a primitive null vector, if one of $a,b$ is 0, then $\xi = 0$ and the other one of $a,b$ is $\pm 1.$

In the first case, suppose $|b| < |a|.$ Then $| 2 a b| = \xi^2 < 2 a^2,$ so in fact $$\left( \frac{\xi}{a} \right)^2 < 2.$$ We choose the root $\tilde{\lambda} = \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and $$s_{\tilde{\lambda}} (z) = z - ( \tilde{\lambda} \cdot z) \tilde{\lambda} = (\xi, -b,-a).$$

Therefore, we may always force the second case, which is $|a| \leq |b|.$ We assume that $a \neq 0,$ so that $b = \frac{- \xi^2}{2a}.$ Now we have $2 a^2 \leq | 2 a b| = \xi^2,$ so $\left( \frac{\xi}{a} \right)^2 \geq 2.$ From the covering radius condition, there is then some nonzero vector $\lambda \in \Lambda$ such that the rational number $$\left( \frac{\xi}{a} - \lambda \right)^2 < 2.$$

If $\lambda$ has even norm, everything proceeds as before.

However, it is possible the only point close enough is $\beta \in \Lambda$ of odd norm. We create the root-one $\tilde{\beta} = \left( \beta, 1, \frac{ 1 - \beta^2}{2} \right).$

By Allcock's strict well-covering condition, $$\left( \frac{\xi}{a} - \beta \right)^2 < 1.$$ For convenience we write $$a' = a \left( \frac{\xi}{a} - \lambda beta \right)^2$$

From the equation $2 z \cdot \tilde{\beta} = a - a',$ we see that $a' \in \mathbf Z.$

Well, we have $$s_{\tilde{\beta}} (z) = (\xi - ( a - a') \lambda, beta, a',b'),$$ where $$b' = b - ( a - a') \left( \frac{1 - \beta^2}{2} \right) = \frac{- \xi^2}{2a} - ( a - a') \left( \frac{1 - \beta^2}{2} \right) .$$

Now, if $a'=0,$ then $s_{\tilde{\beta}} (z) = (0,0,\pm 1),$ and an application of $\pm 1 \in R$ takes us to $w = (0, 0,1),$ with $E(w) = \Lambda.$

If, instead, $a' \neq 0,$ note that our use of the strict well-covering condition shows that $| a'| < | a|,$ while $a,a'$ share the same $\pm$ sign, that is their product is positive. So $a - a'$ shares the same sign, and $| a - a'| < |a|.$ From $2 a b = - \xi^2$ we know that $b$ has the opposite sign. But $2 a' b' = - (\xi - ( a - a') \beta)^2,$ so $b'$ has the opposite sign to $a'$ and $a-a'$ and the same sign as $b.$ Now, $\beta^2 \geq 1,$ so $$\frac{1 - \beta^2}{2} \leq 0,$$ and $( a - a') \left( \frac{1 - \lambda^2}{2} beta^2}{2} \right)$ has the same sign as $b$ and $b'.$ From $$b' = b - ( a - a') \left( \frac{ 1 -\lambda^2}{2} \beta^2}{2} \right)$$ we conclude that $| b'| \leq |b|.$

So, these steps also have $|a| + |b|$ strictly decreasing, until such time that one of them becomes 0, and we have arrived at $w = (0, 0,1).$

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I finally figured out that Nebe gets every lattice to be "even" by doubling all coefficients, this particularly happening to the "odd" lattices such as the sum of $k$ squares. If we were to leave the odd lattices as they are, her condition (and Pete's) would be that the covering radius be strictly less than 1! The reality check is that both Nebe and Pete include the sum of three squares, but both exclude the sum of five squares. For comparison, see LINK Before continuing, let me point out that this strong condition happens only a few times, $$x^2, 3 x^2, \; x^2 + y^2, x^2 + 2 y^2, 2 x^2 + 2 x y + 3 y^2, \; x^2 + y^2 + z^2, x^2 + 2 y^2 + 2 y z + 2 z^2,$$ in dimension no larger than 3.

Note that the earlier proof for "even" lattices does, in fact, apply to the "odd" lattices.

Both Richard Borcherds and his student, Daniel Allcock of U. T. Austin, pointed out that I did not have quite the "correct" condition. In the Duke Math. J., vol. 103 (2000) pp. 303-333, see LINK section 6, Allcock discusses what he calls a "well-covering," which is a covering by balls of varied radius, where the radius of the ball around a given lattice point is decided by the norm of the lattice point. In particular, he says (but not in the paper) that the correct condition for odd'' lattices is that $\mathbb R^n$ is covered by {\bf open} balls of radius 1 around all lattice points of odd norm, then radius $\sqrt 2$ around lattice points of even norm, where the norm of some $x \in \Lambda$ is $x \cdot x.$ This is called a "strict well-covering." This condition is milder than Pete's.

Given such a positive odd lattice $\Lambda$ with a strict well-covering as described, form the integral Lorentzian lattice $L = \Lambda \oplus U.$ Elements are of the form $(\lambda, m,n)$ where $\lambda \in \Lambda, \; m,n \in \mathbb Z.$ The norm on $L$ is given by $$(\lambda, m,n)^2 = \lambda^2 + 2 mn.$$ We infer the inner product $$(\lambda_1, m_1,n_1) \cdot (\lambda_2, m_2,n_2) = \lambda_1 \cdot \lambda_2 + m_1 n_2 + m_2 n_1.$$

Note that the sublattice of all points with even norm has index 2, and that norm 0 vectors in $L = \Lambda \oplus U$ can only be created from such points with even norm.

We choose a particular set of roots (elements of norm 2) beginning with any $\lambda \in \Lambda$ of even norm by $$\tilde{\tilde{\lambda}} = \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$

However, starting with points $\beta \in \Lambda$ of odd norm, we create root-ones by $$\tilde{\beta} = \left( \beta, 1, \frac{ 1 - \beta^2}{2} \right),$$ which are elements of norm 1 in $L.$ Note that when an English soccer goalkeeper boots the ball down the center of the pitch and attack remains central, this is referred to as root-one football. I'm sure I have that right.

Let the group $R \subseteq \mbox{Aut}(L)$ be generated by reflections in all the $\tilde{\tilde{\lambda}}$ and all the $\tilde{\beta}$ and by $\pm 1.$

Given any root $r \in L,$ meaning $r^2 = 2,$ we get the reflection $$s_r (z) = z - ( r \cdot z) r.$$ Also $s_r^2(z) = z.$

Given any root-one $\tilde{\beta} \in L,$ meaning $\tilde{\beta}^2 = 1,$ we get the reflection $$s_{\tilde{\beta}} (z) = z - 2( \tilde{\beta} \cdot z) \tilde{\beta}.$$ Also $s_{\tilde{\beta}}^2(z) = z.$

Given some primitive null vector $$z = (\xi, a,b),$$ so that $2 a b = - \xi^2,$ and $b = \frac{- \xi^2}{2 a} \in \mathbb Z.$ Note that, in order to have a primitive null vector, if one of $a,b$ is 0, then $\xi = 0$ and the other one of $a,b$ is $\pm 1.$

In the first case, suppose $|b| < |a|.$ Then $| 2 a b| = \xi^2 < 2 a^2,$ so in fact $$\left( \frac{\xi}{a} \right)^2 < 2.$$ We choose the root $\tilde{\lambda} = \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and $$s_{\tilde{\lambda}} (z) = z - ( \tilde{\lambda} \cdot z) \tilde{\lambda} = (\xi, -b,-a).$$

Therefore, we may always force the second case, which is $|a| \leq |b|.$ We assume that $a \neq 0,$ so that $b = \frac{- \xi^2}{2a}.$ Now we have $2 a^2 \leq | 2 a b| = \xi^2,$ so $\left( \frac{\xi}{a} \right)^2 \geq 2.$ From the covering radius condition, there is then some nonzero vector $\lambda \in \Lambda$ such that the rational number $$\left( \frac{\xi}{a} - \lambda \right)^2 < 2.$$

If $\lambda$ has even norm, everything proceeds as before.

However, it is possible the only point close enough is $\beta \in \Lambda$ of odd norm. We create the root-one $\tilde{\beta} = \left( \beta, 1, \frac{ 1 - \beta^2}{2} \right).$

By Allcock's strict well-covering condition, $$\left( \frac{\xi}{a} - \beta \right)^2 < 1.$$ For convenience we write $$a' = a \left( \frac{\xi}{a} - \lambda \right)^2$$

From the equation $2 z \cdot \tilde{\beta} = a - a',$ we see that $a' \in \mathbf Z.$

Well, we have $$s_{\tilde{\beta}} (z) = (\xi - ( a - a') \lambda, a',b'),$$ where $$b' = b - ( a - a') \left( \frac{1 - \beta^2}{2} \right) = \frac{- \xi^2}{2a} - ( a - a') \left( \frac{1 - \beta^2}{2} \right) .$$

Now, if $a'=0,$ then $s_{\tilde{\beta}} (z) = (0,0,\pm 1),$ and an application of $\pm 1 \in R$ takes us to $w = (0, 0,1),$ with $E(w) = \Lambda.$

If, instead, $a' \neq 0,$ note that our use of the strict well-covering condition shows that $| a'| < | a|,$ while $a,a'$ share the same $\pm$ sign, that is their product is positive. So $a - a'$ shares the same sign, and $| a - a'| < |a|.$ From $2 a b = - \xi^2$ we know that $b$ has the opposite sign. But $2 a' b' = - (\xi - ( a - a') \beta)^2,$ so $b'$ has the opposite sign to $a'$ and $a-a'$ and the same sign as $b.$ Now, $\beta^2 \geq 1,$ so $$\frac{1 - \beta^2}{2} \leq 0,$$ and $( a - a') \left( \frac{1 - \lambda^2}{2} \right)$ has the same sign as $b$ and $b'.$ From $$b' = b - ( a - a') \left( \frac{ 1 -\lambda^2}{2} \right)$$ we conclude that $| b'| \leq |b|.$

So, these steps also have $|a| + |b|$ strictly decreasing, until such time that one of them becomes 0, and we have arrived at $w = (0, 0,1).$