Timeline for Hurwitz integers represented as sums of two squares of Hurwitz integers
Current License: CC BY-SA 3.0
18 events
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| Dec 16, 2013 at 2:23 | comment | added | Guillermo Pineda-Villavicencio | @few_reps Thanks a lot for your comments. | |
| Dec 13, 2013 at 7:54 | comment | added | few_reps | (... cont.) Here it how it goes. In an isotropic quaternion algebra $A$ you can always find an isotropic quaternion $\varepsilon$ whose trace is $1$. Then write $\varepsilon'=1-\varepsilon$. The subalgebra generated by $\varepsilon$ is $<\varepsilon,\varepsilon'>$. Then decompose the algebra as $A=\varepsilon A \varepsilon+\varepsilon A \varepsilon'+\varepsilon' A \varepsilon+\varepsilon' A \varepsilon'$ : you're done. | |
| Dec 13, 2013 at 7:42 | comment | added | few_reps | @GuillermoPineda-Villavicencio The point is the following : $\mathbf H$ (the Hurewitz integers) is a maximal order in $\mathbf H\otimes\mathbf Q$. This implies that it is locally so : $\mathbf H\otimes\mathbf Z_p$ is a maximal order in $\mathbf H\otimes\mathbf Q_p$. Now the norm is isotropic over $\mathbf H\otimes\mathbf Q_p$, and isotropic quaternions algebras over a field $k$ are (easily seen to be) isomorphic to $\mathrm{M}_2(k)$. But if you want to write things explicitely, the isomorphism will heavily depend on the prime $p$ you localize at ... (to be cont.) | |
| Dec 13, 2013 at 7:30 | comment | added | Guillermo Pineda-Villavicencio | @few_reps Thank you very much for your comments. Sorry but I still don't understand how you are representing an arbitrary Hurwitz integer $h=a +bi+cj+dk$ with $a, b, c, d$ integers or half-integers via a $2\times 2$ matrix. It seems all your matrices have integer coefficients. Specifically, what is the matrix representation for the aforementioned $h$? I am only aware of a representation over complex numbers. It is my impression that for every $h\in M_2(\mathbb{Z}[i])$ such that $h=x(y^2+z^2)$ with $x$ a unit you cannot assume that $y,z\in M_2(\mathbb{Z}[i])$. | |
| Dec 12, 2013 at 22:09 | history | edited | few_reps | CC BY-SA 3.0 |
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| Dec 12, 2013 at 17:37 | history | edited | few_reps | CC BY-SA 3.0 |
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| Dec 12, 2013 at 17:30 | history | edited | few_reps | CC BY-SA 3.0 |
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| Dec 12, 2013 at 15:59 | history | edited | few_reps | CC BY-SA 3.0 |
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| Dec 12, 2013 at 15:32 | history | edited | few_reps | CC BY-SA 3.0 |
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| Dec 12, 2013 at 14:47 | comment | added | few_reps | @GuillermoPineda-Villavicencio : I made an edit to explain the link ... | |
| Dec 12, 2013 at 14:46 | history | edited | few_reps | CC BY-SA 3.0 |
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| Dec 12, 2013 at 5:14 | comment | added | Guillermo Pineda-Villavicencio | @few_reps Thanks for your comments. How are you representing a Hurtwiz integer via matrices? Are you using $$1=\begin{bmatrix} 1&0\\0&1\end{bmatrix}, i=\begin{bmatrix} 1&1\\1&-1\end{bmatrix}, j=\begin{bmatrix} -1&1\\1&1\end{bmatrix}, k=\begin{bmatrix} 0&-1\\1&0\end{bmatrix}$$?. | |
| Dec 11, 2013 at 20:19 | comment | added | Will Jagy | When you put it that way, I guess this is just the worst percentage among my experiments; so, with distinguished real part, and doubling so as not to typeset fractions, we are looking for eight proofs of impossibility: (-7;5,3,1) (-5;7,3,1) (-3; 7,5,1) (-1; 7,5,3) (1;7,5,3) (3;7,5,1) (5;7,3,1) (7;5,3,1) | |
| Dec 11, 2013 at 17:57 | comment | added | few_reps | @WillJagy : One point to note is that a quaternion $q$ is the sum of two squares iff all $aqa^{-1}$ are sums of two squares ($a$ one of the $24$ units) ... and iff its conjugate is the sum of two squares ... this explains the symetries observed in the imaginary part (since we can change places and signs in the imaginary part by these operations) and the special role of the real part. | |
| Dec 11, 2013 at 2:15 | comment | added | Will Jagy | few, I don't think mod 8 is enough here, although there are some simple obstructions about the real coefficient alone.. In comparison mod 8, it did represent all the arrangements of $$\frac{3}{2} + \frac{3}{2} i + \frac{1}{2} j + \frac{1}{2} k $$ with fixed real part (so 3 * 8 = 24 with +- signs), also got $$\frac{-1}{2} + \frac{3}{2} i + \frac{3}{2} j + \frac{1}{2} k$$ and $$\frac{-3}{2} + \frac{3}{2} i + \frac{1}{2} j + \frac{1}{2} k,$$ but NOT any of the 24 $$\frac{1}{2} + \frac{3}{2} i + \frac{3}{2} j + \frac{1}{2} k.$$ | |
| Dec 11, 2013 at 1:03 | comment | added | few_reps | @WillJagy Did you try to run your program mod 4 or 8 ? This might give obstructions explaining this phenomenon ... | |
| Dec 11, 2013 at 0:05 | comment | added | Will Jagy | A worthwhile exercise, i think; there are 384 signed permutations of the quadruple $(1,3,5,7).$ My computer is convinced that not one of the 384 Hurwitz quaternions gotten from $$ \frac{1}{2} + \frac{3}{2} i + \frac{5}{2} j + \frac{7}{2} k $$ is the sum of two squares. I'd like to see a proof of that. | |
| Dec 10, 2013 at 10:44 | history | answered | few_reps | CC BY-SA 3.0 |