##### # # This file is ~steger/donald/SU21/C2p2/D.txt # ##### CASE: $(C_2,p=2)$ The BASE FIELDS: k = \QQ[\sqrt{5}] \ell = k[\zeta_3] = k[\sqrt{-3}] = \QQ[\sqrt{5},\sqrt{-3}] $k$~ramifies over~$\QQ$ only at the $5$-adic place. $\ell$ ramifies over~$k$ only at the $3$-adic place. In place of $\sqrt{5}$ one can use the Golden~Ratio, \GR = (1+\sqrt{5})/2 The EXTENSION FIELD used to present the division algebra: m = \QQ[\sqrt{5},\zeta_9] = k[\zeta_9] = k[Z]/(Z^6+Z^3+1) Here~$Z=\zeta_9$ is a primitive $9$'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^4, \mapsto Z^8, \mapsto Z^7, \mapsto Z^5 \} $m$~is also an extension of $\ell$, where we take: \sqrt{-3} = 1 + 2\zeta_3 = 1 + 2Z^3 The Galois group of~$m$ over~$\ell$ is generated by~$\phi$: \phi(Z) = Z^4 = \zeta_3 Z \phi(Z^j) = Z^{4j} = \zeta_3^j Z^j \phi^2(Z) = Z^{16} = \zeta_3^2 Z \phi^2(Z^j) = Z^{16j} = \zeta_3^{2j} Z^j Specifically, it's \{ Z\mapsto Z, \mapsto Z^4, \mapsto Z^16 \} which is the unique subgroup of order~$3$ of the Galois group of $m$~over~$\ell$. The intersection of~$m$ with the reals is generated by~$\sqrt{5}$ and W = Z + Z^{-1} whose irreducible polynomial over~$\QQ$ (or over~$\ell$) is W^3 - 3W + 1 The discriminant of $W^3-3W+1$ is~$81$. \phi(W) = -W^2-W+2 \phi(W^2) = W+2 \phi^2(W) = W^2-2 \phi^2(W^2) = -W^2-W+4 Over~$\QQ$, $m$~ramifies at the $3$-adic place with ramification index $e=6$; at the $5$-adic place it ramifies with ramification index~$e=2$ Over~$k$, $m$~ramifies (only) at the $3$-adic place with ramification index $e=6$. Over~$\ell$, $m$ ramifies (only) at the $3$-adic places with ramification index $e=3$. The ramification at the $3$-adic place is \emph{wild} 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 = (1 + \sqrt{-15}) / 4 = 1/4 + \sqrt{5}/4 + \sqrt{5}Z^3/2 = \GR/2 + (-1/2+\GR)Z^3 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 $3$-adic, we know $\D\otimes\ell_v$ splits because $(1+\sqrt{-15})/4$ has valuation zero at~$v$ and because $m$~doesn't ramify over~$\ell$ at~$v$. Since there is no $\sqrt{5}$ in~$\QQ_2$, there is only one $2$-adic place of~$k$, which we continue to denote by~$2$. The completion~$k_2$ is the unramified degree~2 extension of~$\QQ_2$. Since \ell = k[\sqrt{-3}] = \QQ[\sqrt{5},\sqrt{-3}] = \QQ[\sqrt{5},\sqrt{-15}] = k[\sqrt{-15}] = k[S]/(S^2+15) and since there is a square root of~$-15$ in~$\QQ_2$, there are two 2-adic places of~$\ell$, denoted~$2+$ (respectively ~$2-$) and so that $S=\sqrt{-15}=9 \mod 16$ (respectively $7 \mod 16$). Since $\ell_{2\pm}$ is still the unramified degree~2 extension of~$\QQ_2$, it doesn't contain any primitive ninth 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~$\ell_{2\pm}$, therefore the unramified degree~6 extension of~$\QQ_2$. The residue field of~$m_{2\pm}$ is~$F_{64}$, which does have a full set of ninth 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 $\sqrt{-15}=9 \mod 16$ (1+\sqrt{-15})/4 = 10/4 = 5/2 \mod 4 Doing the same calculation for the place~$2-$, we use $\sqrt{-15}=7 \mod 16$, and get (1+\sqrt{-15})/4 = 8/4 = 2 \mod 4 The $2$-adic valuations of~$5/2$ and~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 $3$-adic place of~$\ell$, denoted simply by~$3$. 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. Moreover in ~steger/donald/SU21/C18p3/padic.gap there is a direct calculation showing that $(1+\sqrt{-15})/4$ is a norm from~$m_3$ to~$\ell_3$. 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 (1+\sqrt{-15})/4 0 0 For COMPLEX calculation: Z = 0.7660444431189780352023927 + 0.6427876096865393263226434 i W = Z + Z^{-1} = 1.5320888862379560704047853 (1+\sqrt(-15))/4 = 0.25 + 0.9428090415820633658677925 i \sqrt{5} = 2.2360679774997896964091737 \GR = 1.6180339887498948482045868 PRELIMINARY INVOLUTION of the second kind: \bar Z = Z^{-1} \bar W = W \overline{\sqrt{5}} = \sqrt{5} \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\sqrt{5} + (-1+\sqrt{5})W + 2W^2 = 2 ((1-2*\GR) + (-1+\GR)W + W^2) = (4-2\sqrt{5})+(-3+\sqrt{5})Z+(3-\sqrt{5})Z^2 -2Z^4+(1-\sqrt{5})Z^5 = 2 ((3-2\GR)+(-2+\GR)Z+(2-\GR)Z^2-Z^4+(1-\GR)Z^5) \phi(T) = 2 + (3-\sqrt{5})W + (1-\sqrt{5})W^2 = 2 (1 + (2-\GR)W + (1-\GR)W^2) = (4-2\sqrt{5})+2Z-2Z^2+(-1+\sqrt{5})Z^4+(-3+\sqrt{5})Z^5 = 2 ((3-2\GR)+Z-Z^2+(-1+\GR)Z^4+(-2+\GR)Z^5 \phi^2(T) = (10-4\sqrt{5}) -2W + (-3+\sqrt{5})W^2 = 2 ((7-4\GR) - W + (-2+\GR)W^2) = (4-2\sqrt{5})+(1-\sqrt{5})Z+(-1+\sqrt{5})Z^2 +(3-\sqrt{5})Z^4+2Z^5 = 2 ((3-2\GR)+(1-\GR)Z+(-1+\GR)Z^2+(2-\GR)Z^4+Z^5) 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} = ((3-\sqrt{5}) + (-3-\sqrt{5})W/2 + (-1+\sqrt{5})W^2/2) \sigma^{-1} = ((4-2\GR) + (-1-\GR)W + (-1+\GR)W^2) \sigma^{-1} = (2 + (-1-\sqrt{5})Z + (1+\sqrt{5})Z^2 + (1-\sqrt{5})Z^4/2 + (3+\sqrt{5})Z^5/2) \sigma^{-1} = (2 - 2\GR Z + 2\GR Z^2 + (1-\GR)Z^4 + (1+\GR)Z^5) \sigma^{-1} \iota(\sigma^{-1}) = T^-1 \sigma T = \phi(T)/T \sigma = ((-1-\sqrt{5})/2 + (5+\sqrt{5})W/2 + (-1-\sqrt{5})W^2/2) \sigma = (-\GR + (2+\GR)W - \GR W^2) \sigma = ((-3-3\sqrt{5})/2 + (3+\sqrt{5})Z + (-3-\sqrt{5})Z^2 + (1+\sqrt{5})Z^4/2 + (-5-\sqrt{5})Z^5/2) \sigma = (-3\GR + (2+2\GR)Z + (-2-2\GR)Z^2 + \GR Z^4 + (-2-\GR)Z^5) \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_{i=0,1} \sum_{j=0,1,2,3,4,5} \sum_{l=-1,0,1} c_{ijl} \GR^i Z^j \sigma^l where the coefficients c_{ijl}\in\QQ satisfy integrality conditions as follows: (.) For any prime~$v$ except~2 or~3 the $c_{ijl}$ 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~$c_{ijl}$ by~256. (.) At the special prime~3 one must use special conditions, as calculated in ~/donald/SU21/C2p2/padic.gap. There are two different conditions, Type~1 and Type~2. The Type~1 condition amounts to requiring that an element of~$U(\D,\iota)$ fix a certain lattice in~$\ell_3^3$ 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~$\ell_3^3$, a pair which is also adjacent in the building of~$PGL(3,\ell_3)$. DETERMINANTS: Elements of $U(\D,\iota)$ can have determinant which is $1$, $\zeta_3$, or $\zeta_3^2$ multiplied by any power of~$(1+\sqrt{-15})/4$. In~$PU$, one can always choose a representative where the power is~$0$ or~$\pm 1$.