##### # # This file is ~steger/donald/SU21/C10p2/D.txt # ##### CASE: $(C_{10},p=2)$ The BASE FIELDS: k = \QQ[\sqrt{2}] \ell = k[\sqrt{-5+2\sqrt{2}}] = \QQ[\sqrt{2},\sqrt{-5+2\sqrt{2}}] = \QQ[\sqrt{2},U] / (U^2-(1+\sqrt{2})U+2) = k[U] / (U^2-(1+\sqrt{2})U+2) where U = ((1+\sqrt{2}) + \sqrt{-5+2\sqrt{2}}) / 2 $k$~ramifies over~$\QQ$ only at the $2$-adic place. $k$~has two different $17$-adic places, denoted $17\pm$. $\ell$ ramifies over~$k$ only at~$17-$. $\ell$ is a Galois extension of~$k$, but is not Galois over~$\QQ$. Since $(-5+2\sqrt{2})(1-\sqrt{2})^2 = -7-4\sqrt{2}$, $\ell=k[\sqrt{-7-4\sqrt{2}}]$. In [Prasad--Yeung] $\ell$~is given as $k[\sqrt{-7+4\sqrt{2}]$ which is isomorphic to our~$\ell$ via an isomorphism which exchanges $\pm\sqrt{2}$. The EXTENSION FIELD used to present the division algebra: m = \QQ[\sqrt{2},U,\zeta_9+\zeta_9^-1] = \ell[\zeta_9+\zeta_9^-1] = \ell[W]/(W^3-3W+1) Here~$\zeta_9$ is a primitive 9'th root of unity, and $W=2\Re\zeta_9$ generates the real subfield of~$\QQ[\zeta_9]$. The discriminant of $W^3-3W+1$ is~$81$. $m$~is a cylic Galois extension of degree~$3$ over~$\ell$ with Galois group induced from the maps \{ \zeta_9\mapsto\zeta_9, \mapsto\zeta_9^4, \mapsto\zeta_9^2 \} The Galois group of~$m$ over~$\ell$ is generated by~$\phi$: \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; The intersection of~$m$ with the reals is generated by~$\sqrt{2}$ and~$W$. Over~$k$, $m$~ramifies at the~$3$-adic place with ramification index $e=3$ and at the place~$17-$ with ramification index~$e=2$ 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{2}) + \sqrt{-5+2\sqrt{2}}) / (2\sqrt{2}) = U / \sqrt{2} Then (1') $\D\otimes\ell_{2\pm}$ has Hasse 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 $((1+\sqrt{2})+\sqrt{-5+2\sqrt{2}})/(2\sqrt{2})$ has valuation zero at~$v$ and because $m$~doesn't ramify over~$\ell$ at~$v$. Since there is no $\sqrt{2}$ in~$\QQ_2$, there is only one $2$-adic place of~$k$, which we continue to denote by~$2$. The completion~$k_2=\QQ_2[\sqrt{2}]$ is totally ramified of degree~2 over~$\QQ_2$. There is a square root of $-5+2\sqrt{2}$ in~$\k_2$. Indeed: (3-5\sqrt{2})^2 - (-5+2\sqrt{2}) = 64-32\sqrt{2} and this can be extended to an exact square root using Hensel's Lemma. Thus, there are two 2-adic places of~$\ell$ denoted~ $2\pm$ and characterized by: \sqrt{-5+2\sqrt{2}) = \pm(3 - 5\sqrt{2}) \mod 16\sqrt{2} Each of~$\ell_{2\pm}$ is isomorphic to $k_2$. Adjoining a primitive 9'th root of unity to~$k_2$, we obtain $\QQ_2[\sqrt{2},\zeta_9]$, which is the unramified extension of degree~6 of~$k_2$. Since $[\QQ_2[\sqrt{2},\zeta_9]:\QQ_2[\sqrt{2},W]\leq 2$, it follows that $m_{2\pm}=\ell_{2\pm}[W]=k_2[W]=k_2[\zeta_9]$ is unramified of degree~3 over $\ell_{2\pm}$. Thus an element of $\ell_{2\pm}^\times$ is a norm from $m_{2\pm}^\times$ if and only if its valuation is a multiple of~3. For $\ell_{2+}$ we find: \sigma^3 = ((1+\sqrt{2}) + \sqrt{-5+2\sqrt{2}})/(2\sqrt{2}) = ((1+\sqrt{2})+(3-5\sqrt{2}))/(2\sqrt{2}) \mod 16\sqrt{2} = (4-4\sqrt{2})/(2\sqrt{2}) \mod 16\sqrt{2} = \sqrt{2}-2 \mod 8 which has valuation~$1$. Similarly, for $\ell_{2-}$ we find that $\sigma^3$ has valuation~$-1$ Thus, the Hasse invariants of~$\D$ at~$\ell_{2\pm}$ are nonzero and opposite. Because there is no~$\sqrt{2}$ modulo~$3$, there is a unique 3-adic place of~$k$, and $k_3=\QQ_3[\sqrt{2}]$ is the degree~2 unramified extension of~$\QQ_3$. One can check that in $\FF_9$ there is no square root of $-5+2\sqrt{2}$. Consequently, there is a unique 3-adic place of~$\ell$ and $\ell_3=\QQ_3[\sqrt{2},\sqrt{-5+2\sqrt{2}}]$ is the degree~4 unramified extension of~$\QQ_3$. Since we have calculated the Hasse invariants at all the other places, and since those sum to zero, the Hasse invariant at the 3-adic place must also be zero. In ~donald/SU21/C10p2/padic.gap there is an explicit calculation showing that~$\sigma^3\in\ell_3$ is a norm from~$m_3$. 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{2})+\sqrt{-5+2\sqrt{2}))/(2\sqrt{2}) 0 0 0 1 0 = 0 0 1 U/\sqrt{2} 0 0 For COMPLEX calculation: \sqrt{2} = 1.414213562373095048801688724209 \sqrt{-5+2\sqrt{2}} = 1.4736257582079005931550091 i \sqrt{-5-2\sqrt{2}} = 4.3020528829540906907583866 i U = (1+\sqrt{2}+\sqrt{-5+2\sqrt{2}})/2 = 1.2071067811865475244008444 + 0.7368128791039502965775046 i U' = (1-\sqrt{2}+\sqrt{-5-2\sqrt{2}})/2 = -0.2071067811865475244008444 + 1.3989663259659067020315405 i U/\sqrt{2} = ((1+\sqrt{2})+\sqrt{-5+2\sqrt{2}))/(2\sqrt{2}) = 0.8535533905932737622004222 + 0.5210053832799870771404151 i U'/(-\sqrt{2}) = ((1-\sqrt{2})+\sqrt{-5-2\sqrt{2}))/(-2\sqrt{2}) = 0.1464466094067262377995778 - 0.9892185757421227086489334 i W = 1.5320888862379560704047853 \phi(W) = -1.8793852415718167681082186 \phi^2(W) = 0.3472963553338606977034332 PRELIMINARY INVOLUTION of the second kind: \overline{\sqrt{2}} = \sqrt{2} \overline{\sqrt{-5+2\sqrt{2}}} = -\sqrt{-5+2\sqrt{2}} \bar W = W \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 = -\sqrt{2}+(1-\sqrt{2})W+W^2 \phi(T) = (4-3\sqrt{2})+\sqrt{2}W+(-1+\sqrt{2})W^2 \phi^2(T) = (2+\sqrt{2})-W-\sqrt{2}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} = (-\sqrt{2}+(-3+\sqrt{2})W+(1-\sqrt{2})W^2) \sigma^{-1} \iota(\sigma^{-1}) = T^-1 \sigma T = \phi(T)/T \sigma = ((1-2\sqrt{2})+(1+2\sqrt{2})W+(1+\sqrt{2})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: Recall that U = ((1+\sqrt{2}) + \sqrt{-5+2\sqrt{2}}) / 2 We want elements of the form \sum_{i=0,1} \sum_{j=0,1} \sum_{k=0,1,2} \sum_{l=-1,0,1} c_{ijkl} \sqrt{2}^i U^j W^k \sigma^l where the coefficients c_{ijkl}\in\QQ satisfy integrality conditions as follows: (.) For any prime~$v$ except~$2$, $3$, or~$17$ the $c_{ijkl}$ 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 primes~$3$ and~$17$ one must use special conditions, as calculated in ~/donald/SU21/C10p2/padic.gap. (.) For the prime~$3$, one always uses a Type~1, hyperspecial condition. (.) $\ell$~has two $17$-adic places, denoted $17\pm$. At~$17+$ we always wish to use a Type~1, hyperspecial condition. If we wish to use a Type~1 condition for~$17-$ too, then the $17$-adic condition on the $c_{ijkl}$ is the simple one --- no factors of~$17$ should appear in the denominators. (.) If, on the other hand, we wish to use a Type~2 condition for~$17-$, then we must use a pair of condition matrices. Each matrix of the pair is $18\times 36$. The first matrix corresponds to the Type~1 condition at $17+$ and the second to the Type~2 condition at~$17-$. The Type~1 conditions (correctly called the hyperspecial conditions for~$3$ and for $17+$) amount to requiring that an element of~$U(\D,\iota)$ fix a certain lattice in~$\ell_3^3 or~$\ell_{17\pm}^3$, a lattice which is self-dual relative to the sesquilinear form associated to~$\iota$. The Type~2 conditions amount to requiring that an element of~$U(\D,\iota)$ fix a certain pair of mutually dual lattices in~$\ell_{17-}^3$, a pair which is also adjacent in the building of~$PGL(3,\ell_{17-})$. DETERMINANTS: Elements of $U(\D,\iota)$ can have determinant which is any power of~$U/\sqrt{2} = ((1+\sqrt{2})+\sqrt{-5+2\sqrt{2}})/(2\sqrt{2})$. In~$PU$, one can always choose a representative where the power is~$0$ or~$\pm 1$.