##### # # This file is ~steger/donald/SU21/C20p2/D.txt # ##### CASE: $(C_{20},p=2)$ The BASE FIELDS: k = \QQ[\sqrt{7}] \ell = k[i] = k[\sqrt{-7}] = \QQ[\sqrt{7},i] $k$~ramifies over~$\QQ$ at the $2$-adic and $7$-adic places. $\ell$ doesn't ramify over~$k$ at any places. The EXTENSION FIELD used to present the division algebra: m = \QQ[\zeta_7,i] = \QQ[\zeta_{28}] = \ell[Z]/(Z^3 + (-\sqrt{-7}+1)/2 Z^2 + (-\sqrt{-7}-1) Z - 1) Here~$Z=\zeta_7$ is a primitive $7$'th root of unity. $m$~is a cylic Galois extension of degree~$6$ over~$k$ with Galois group \{ Z\mapsto Z, \mapsto Z^2, \mapsto Z^3, \mapsto Z^4, \mapsto Z^5, \mapsto Z^6 \} The Galois group of~$m$ over~$\ell$ is generated by~$\phi$: \phi(Z) = Z^2 \phi(Z^j) = Z^{2j} \phi^2(Z) = Z^{4} \phi^2(Z^j) = Z^{4j} Specifically, it's \{ Z\mapsto Z, \mapsto Z^2, \mapsto Z^4 \} which is the unique subgroup of order~$3$ of the Galois group of $m$~over~$k$. The intersection of~$m$ with the reals is generated by~$\sqrt{7}$ and W = Z + Z^{-1} whose irreducible polynomial over~$\QQ$ (or over~$\ell$) is W^3 + W^2 - 2W - 1 The discriminant of $W^3+W^2-2W-1$ is~$49$. \phi(W) = W^2-2 \phi(W^2) = -W^2-W+3 \phi^2(W) = -W^2-W+1 \phi^2(W^2) = W+2 Over~$\QQ$, $m$~ramifies at the $7$-adic place with ramification index $e=6$; at the $2$-adic place it ramifies with ramification index~$e=2$ Over~$k$, $m$~ramifies (only) at the $7$-adic place with ramification index $e=3$. Likewise, over~$\ell$, $m$ ramifies (only) at the $7$-adic places with ramification index $e=3$. The ramification at the $7$-adic place is \emph{tame} ramification. PRESENTATION of $\D$, degree three division algebra, central over~$\ell$: \D$ is generated by~$m$ and~$\sigma$ where \sigma x \sigma^{-1} = \phi(x) (for $x\in m$) \sigma^3 = (3 + \sqrt{-7}) / 4 Then (1') $\D\otimes\ell_{2\pm}$ has invariant $\pm 1$; (2') $\D\otimes\ell_v$ splits at every other place~$v$. For a place~$v$ which is neither $2$-adic nor $7$-adic, we know $\D\otimes\ell_v$ splits because $(3+\sqrt{-7})/4$ has valuation zero at~$v$ and because $m$~doesn't ramify over~$\ell$ at~$v$. Since there is no $\sqrt{7}$ in~$\QQ_2$, there is only one $2$-adic place of~$k$, which we continue to denote by~$2$. The completion~$k_2$ is a ramified degree~2 extension of~$\QQ_2$. Since \ell = k[i] = \QQ[\sqrt{7},i] = \QQ[\sqrt{7},\sqrt{-7}] = k[\sqrt{-7}] = k[S]/(S^2+7) and since there is a square root of~$-7$ in~$\QQ_2$, there are two 2-adic places of~$\ell$, denoted~$2+$ (respectively ~$2-$) and so that $S=\sqrt{-7}=5 \mod 16$ (respectively $11 \mod 16$). Since $\ell_{2\pm}$ is the same ramified degree~2 extension of~$\QQ_2$ as $k_2$, it doesn't contain any primitive seventh roots of unity. Therefore~$2\pm$ lifts to a unique place of~$m$, still denoted~$2\pm$. The completion~$m_{2\pm}$ is the unramified degree~3 extension of~$\ell_{2\pm}$. The residue field of~$m_{2\pm}$ is~$F_8$, which does have a full set of seventh roots of unity. Now let us show that $D\otimes\ell_{2+}$ is nontrivial. Since~$m_{2+}$ is unramified of degree~3 over~$\ell_{2+}$ the image of the norm map, $N_{m_{2+}/\ell_{2+}} (\ell_{2+}^\times)$, consists of elements with valuation divisible by~3. Since $\sqrt{-7}=5 \mod 16$ (3+\sqrt{-7}})/4 = (3+5)/4 = 2 \mod 4 Doing the same calculation for the place~$2-$, we use $\sqrt{-7}=11 \mod 16$, and get (3+\sqrt{-7}})/4 = (3+11)/4 = 7/2 \mod 4 The $2$-adic valuations of~$2$ and~$7/2$ are nonzero and opposite modulo~3, so the Hasse invariants of~$\D\otimes\ell_{2\pm}$ are likewise nonzero and opposite. There is a unique $7$-adic place of~$\ell$, denoted simply by~$7$. Using the fact that the sum of the Hasse invariants over all places of~$\ell$ comes to zero, one sees that this last Hasse invariant must be zero. To do the calculation directly, observe first that $\ell_7=\ell\otimes\QQ_7$ is a field, of degree~4 over~$\QQ_7$, with unramified degree~$f=2$ and ramification index $e=2$. Then note that $m_7=m\otimes\QQ_7$ is again a field, totally and tamely ramified of degree~3 over $\ell_7$. Consequently, a norm~1 element of~$\ell_7$ is a norm from~$m_7$ if and only if its reduction to the residue field, $\FF_{49}$, is a cube. Since (3+\sqrt{-7})/4 = 3/4 = -1 = (-1)^3 \mod 7 it follows that~$(3+\sqrt{-7})/4$ is a norm from~$m_7$ to~$\ell_7$. 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 (3+\sqrt{-7})/4 0 0 For COMPLEX calculation: \sqrt{7} = 2.6457513110645905905016158 Z = 0.6234898018587335305250049 + 0.7818314824680298087084445 i W = Z + Z^{-1} = 1.2469796037174670610500098 (3+\sqrt{-7})/4 = 0.75 + 0.6614378277661476476254039 i PRELIMINARY INVOLUTION of the second kind: \bar Z = Z^{-1} \bar W = W \overline{\sqrt{7}} = \sqrt{7} \iota_0(x) = \bar x (for $x\in m$) \iota_0(\sigma) = \sigma^{-1} On complex matrices: \iota_0(A) = A^* INVOLUTION of the second kind: Because $\iota_0$ gives the compact group $PU(3)\times PU(3)$ at the real place, we need to adjust it. We use T = (2-4\sqrt{7}/7) + (1-3\sqrt{7}/7)W + (-\sqrt{7}/7)W^2 \phi(T) = (-\sqrt{7}/7) + (\sqrt{7}/7)W + (1-2\sqrt{7}/7)W^2 \phi^2(T) = (3-9\sqrt{7}/7) + (-1+2\sqrt{7}/7)W + (-1+3\sqrt{7}/7)W^2 The adjusted involution is: \iota(x) = T^-1 \iota_0(x) T \iota(x) = \bar x (for $x\in m$) \iota(\sigma) = T^{-1} \sigma^{-1} T = \phi^2(T)/T \sigma^{-1} = ((1-3\sqrt(7)/7) + (3\sqrt(7)/7)W + (\sqrt(7)/7)W^2) \sigma^{-1} \iota(\sigma^{-1}) = T^-1 \sigma T = \phi(T)/T \sigma = ((-1-\sqrt(7)/7) + (-1+\sqrt(7)/7)W + (1-2\sqrt(7)/7)W^2) \sigma On complex matrices: \iota(A) = F^{-1} A^* F for T 0 0 F = 0 \phi(T) 0 0 0 \phi^2(T) which means unitary matrices ought to preserve the form: T |v_1|^2 + \phi(T) |v_2|^2 + \phi^2(T) |v_3|^2 INTEGRALITY: We want elements of the form \sum_{n=0,1} \sum_{j=0,1,2,3,4,5} \sum_{l=-1,0,1} c_{njl} i^n Z^j \sigma^l where the coefficients c_{njl}\in\QQ satisfy integrality conditions as follows: (.) For any prime~$v$ except~2, 3, or~7 the $c_{njl}$ are integral in $\QQ_v$, i.e. no factors of~$v$ occur in the denominators. (.) Factors of~2 may appear in the denominators. However, the power of~2 which appears is bounded above. Almost certainly all such powers disappear if one multiplies the~$c_{njl}$ by~256. (.) At the special prime~7, one must use a special condition, as calculated in ~/donald/SU21/C20p2/padic.gap. (.) At the special prime~3 one must use special conditions, as calculated in ~/donald/SU21/C20p2/padic.gap. There are two 3-adic places for~$k$, denoted~$3\pm$, depending on whether $\sqrt{7}=+1$ or $\sqrt{7}=-1 \mod 3$. One must either use (a) Type~1 conditions at both 3-adic places (b) the Type~1 condition for $3+$ and the Type~2 condition for $3-$, or (c) vice versa. These three possibilities give rise to three maximal subgroups~$\bar\Gamma$, with respective covolumes~$1/21$, $1/3$, and~$1/3$. The Type~1 condition amounts to requiring that an element of~$U(\D,\iota)$ fix a certain lattice in~$\ell_{3\pm}^3$ which is self-dual relative to the sesquilinear form associated to~$\iota$. The Type~2 condition amounts to requiring that an element of~$U(\D,\iota)$ fix a certain pair of mutually dual lattices in~$\ell_{3\pm}^3$, a pair which is also adjacent in the building of~$PGL(3,\ell_{3\pm})$. DETERMINANTS: An element of $U(\D,\iota)$ can have a determinant which is any power of~$(3+\sqrt{-7})/4$. In~$PU$, one can always choose a representative where the power is~$0$ or~$\pm 1$.