CASE: $(a=7,p=2)$ The BASE FIELDS: k=\QQ \ell=\QQ[\sqrt(-7)] $\ell$~ramifies over~$\QQ$ only at the $7$-adic place. PRESENTATION of $\D$, degree three division algebra, central over~$\ell$: m=\ell[Z]/(Z^3+3Z^2+3) This is a cylic Galois extension of~$\ell$ with Galois group \rangle 1, \phi, \phi^2 \rangle with $\phi$ given by: \phi(Z) = \phi^2(Z) = (-\sqrt{-7}/7) Z^2 (\sqrt{-7}/7) Z^2 + (-1/2 - 9\sqrt{-7}/14) Z + (-1/2 + 9\sqrt{-7}/14) Z + (-3/2 - 3\sqrt{-7}/14) + (-3/2 + 3\sqrt{-7}/14) \phi(Z^2) = \phi^2(Z^2) = (-1/2 + 9\sqrt{-7}/14) Z^2 (-1/2 - 9\sqrt{-7}/14) Z^2 + (15\sqrt{-7}/7) Z + (-15\sqrt{-7}/7) Z + (9/2 + 3\sqrt{-7}/14); + (9/2 - 3\sqrt{-7}/14); The discriminant of $Z^3+3Z^2+3$ is $(-7)81$. $m$~ramifies over~$\ell$ only at the $3$-adic place. $m$~splits at the $7$-adic place of~$\ell$. \D$ is generated by~$m$ and~$\sigma$ where \sigma x \sigma^{-1} = \phi(x) \sigma^3 = 2 Then (1') $\D\otimes\ell_{2\pm}$ has invariant $\pm 1$; (2') $\D\otimes\ell_v$ splits at every other place~$v$. EMBEDDING of $\D$ in $\Mat_{3\time 3}(\CC)$: x 0 0 x \mapsto 0 \phi(x) 0 (for $x\in m$) 0 0 \phi^2(x) 0 1 0 \sigma \mapsto 0 0 1 2 0 0 For REAL calculation: Z = -3.2790187861665935794914426057761899 INVOLUTION of the second kind: \bar Z = Z \overline{\sqrt{-7}} = -\sqrt{-7} \iota(x) = \bar x (for $x\in m$) \iota(\sigma) = \sigma On complex matrices: \iota(A) = F^{-1} A^* F for 1 0 0 F = 0 0 1/2 0 1/2 0 which means unitary matrices ought to preserve the form: |z_1|^2 + \Re (z_1 \bar z_2) INTEGRALITY: We want elements of the form \sum_{j=0,1} \sum_{l=0,1,2} c_{j0l} \sqrt{-7}^j \sigma^l + c_{j1l} \sqrt{-7}^j Z \sigma^l + c_{j2l} \sqrt{-7}^j (Z^2+Z-2)/\sqrt{-7} \sigma^l where the coefficients c_{jkl} \in \ZZ[1/2] The powers of~$2$ which occur in the denominators of unitary elements are bounded above, probably by~$2$ or~$4$, possibly a little bit higher. Action on BUILDINGS and TREES: Let $v$ be any place of~$\ell$ and let~$w$ be some place of~$m$ lying above~$v$. If the equation $Z^3+3Z+3=0$ is solvable in $\ell_v$, there are three such places corresponding to the three solutions; for each of the three places $m_w=\ell_v$. If there is no solution, then there is only one choice for~$w$, and $m_w$~is obtained from~$\ell_v$ by adjoining a solution. Now, consider the map $\D \to \Mat_{3\times 3}(m_w)$ given by x 0 0 x \mapsto 0 \phi(x) 0 (for $x\in m$) 0 0 \phi^2(x) 0 1 0 \phi \mapsto 0 0 1 2 0 0 The map to $\Mat_{3\times 3}(\CC)$ was given by identical formulas. Indeed, the map to $\Mat_{3\times 3}(\CC)$ is the special case of this one for the situation where $v=v_\infty$, the real place of~$\ell$. The integrality condition for an element $\gamma\in\D$ corresponds to the condition that at each place~$v$, except for the $2$-adic places, the matrix for $\gamma$ has entries which are integral in~$m_w$. Suppose $v$~lies over the rational prime~$p$. One must analyse separately: (a) the case where there exists a~$\sqrt{-7}$ in $\QQ_p$, and so $\ell_v=\QQ_p$. (b) the case where $\ell_v=\QQ_p[\sqrt{-7}]$ is of degree two over~$\QQ_p$. In case~(a), the building of the unitary group at the prime~$p$ is of type~$\tilde A_2$. In case~(b) the building is a tree. Case (b)~divides further into: (b1) $p\neq 7$, so $\ell_v$~is the unramified extension of~$\QQ_p$, and the tree has is bihomogeneous with degrees~$p^3+1$ (hyperspecial vertices) and~$p+1$ (nonhypspecial vertices). (b2) $p=7$, so $\ell_v=\QQ_7[\sqrt{-7}]$, and the tree is homogeneous of degree~$8$ at all vertices, hyperspecial and not. This division into cases is inherent in the algebraic group we want to study, namely $PU(\D,\iota)$, considered as an algebraic group over~$\QQ$. Vice versa, the specific presentation of~$\D$ which we're using is an arbitrary choice on our part. There are many fields which could be used in place of~$m$. The actual choice of~$m$ was the one that appeared ``simplest''. Nonetheless, having chosen~$m$, we must further divide each of cases~(a) and~(b) according to one of: (c) $z^3+3z^2+3$ has a root in~$\ell_v$, so $m_w=\ell_v$; (d) $z^3+3z^2+3$ has no roots in~$\ell_v$, so $m_w$~is of degree~3 over~$\ell_v$, obtained by adjoining a root. Nonetheless, considering separately cases~(ac), (ad), (bc), and~(bd), one sees that unitary elements $\gamma\in\D$ satisfying our integrality condition are those which fix a hyperspecial vertex in each of the sequence of buildings associated to the rational primes~$p$. This is right even for~$p=2$, because the $2$-adic building reduces to a single point. Actually, the division into four cases causes less trouble than you might think, while the single prime~$p=3$ requires extra effort, since in that case $m_w$~is ramified over $\ell_v=\ell_3$. The CANONICAL FORM of the arithmetic group, $\Lambda$: In summary, the integrality condition and~$\iota$ specified above lead to a group which, for every prime~$p$, fixes a hyperspecial vertex in the $p$-adic building. So the above choices give the \emph{canonical version} of the arithmetic group, $\Lambda$. The alternative form corresponds to the case where the group fixes a nonhyperspecial vertex in the tree at the $7$-adic place. Of course this same adjustment could be made at any other prime where the building is a tree, or for any finite number of them, but the resulting alternative arithmetic group would have larger covolume than the original. At the $7$-adic place we get the unique alternative form with the same covolume. So let us consider what's going on at the $7$-adic place. There is a single root of~$z^3+3z^2+3$ in $\QQ_7$. That root, call it~$Z$, satisfies~$Z=1 \mod 7$. In $\ell_v=\QQ_7[\sqrt{-7}]$ the polynomial~$z^3+3z^2+3$ splits. The two new roots are both $= -2 \mod 7$, and they do not belong to~$\QQ_7$. Corresponding to these 3~roots, there are 3~place of~$m$ lying over~$v$. We choose the one for which the image of $Z\in m$ under $m\to m_w$ is the first root, $Z\in\QQ_7$, with $Z=1\mod 7$. Thus: m_w=\ell_v=\QQ_7[\sqrt{-7}] With these choices made, we have a matrix representation of~$\D$ over $\ell_v$, given by the formulas above. In that matrix representation, which matrices correspond to unitary elements of~$\D$? This works out just as it did over the complexes. They must be unitary relative to the form: 1 0 0 F = 0 0 1/2 0 1/2 0 Which matrices correspond to integral elements of~$\D$? Those which have entries which are integral in~$\ell_v$, which is to say those which preserve the standard base lattice $L_0$: L_0 = \O_v^3 \subset \ell_v^3 Since $L_0$ is self-dual relative to~$F$, it gives a hyperspecial vertex of the $7$-adic tree. That vertex is fixed by our~$\Lambda$. The ALTERNATIVE FORM, $\Lambda'$: To get the alternative form of the arithmetic group, we must modify either $\iota$ or the integrality condition. It seems easier to modify~$\iota$: \iota'(\cdot) = (2+Z)^{-1} \iota(\cdot) (2+Z) Since $\iota$~is an antiautomorphism, so is~$\iota'$. Since $\iota$ is antilinear, so is~$\iota'$. Since $\iota(2+Z)=2+Z$, and since $\iota$~is an involution, $\iota'$ is an involution too. At the level of matrices over~$\CC$, we have: \iota(A) = F^{-1} A^* F \iota'(A) = (F(2+Z))^{-1} A^* (F(2+Z)) which means that $\iota'$-unitary elements $\gamma\in\D$, those elements satisfying $A \iota'(A) = \id$, map to matrices which are unitary relative to the form $2F(2+Z)$. 2 0 0 2+Z 0 0 2F (2+Z) = 0 0 1 0 \phi(2+Z) 0 0 1 0 0 0 \phi^2(2+Z) 4+2Z 0 0 = 0 0 2+\phi^2(Z) 0 2+phi(Z) 0 4+2Z 0 0 = 0 0 \bar W 0 W 0 where W = 2 + \phi(Z) = (-\sqrt{-7}/7) Z^2 + (-1/2 - 9\sqrt{-7}/14) Z + (1/2 - 3\sqrt{-7}/14) Exactly the same calculation works if we want to think about matrices with entries in $\QQ_7[\sqrt{-7}]$. Working in $\QQ_7[\sqrt{-7}] and using $Z=1 \mod 7$, we find: 4+2Z = 6 \mod 7 W = -\sqrt{-7} \mod 7 So, $\mod 7$, the hermitian form which $\iota'$-unitary elements preserve is: 6 0 0 0 0 \sqrt{-7} \mod 7 0 -\sqrt{-7} 0 The integral elements in~$\D$ are exactly as before. As before, they preserve the standard base lattice~$L_0$. Relative to the modified hermitian form, the lattice $L_0'$, dual to~$L_0$ has basis: 1 0 0 0 , 1/\sqrt{-7} , and 0 0 0 1/sqrt{-7} Hence $L_0$ and~$L_0'$ are adjacent vertices in the $\tilde A_2$-building of $PGL(3,\QQ_7[\sqrt{-7}]$. As such they correspond to a vertex of nonhyperspecial type in the tree of $PU(\QQ_7\otimes\D,\iota')$. This is the vertex fixed by the integral elements. Therefore, the arithmetic group $\Lambda'$ of integral, $\iota'$-unitary elements of~$\D$ fixes a nonhyperspecial vertex in the $7$-adic tree. At all the other places, the modification from~$\iota$ to~$\iota'$ changes nothing. This is because~$Z+2$ is a unit in~$\O_v$ if $v$~is any place of~$\ell$ except the $7$-adic place. And that is because $N_{\ell/\QQ} (Z+2) = 7$.