Timeline for Properties of complexified octonions
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 19, 2016 at 12:42 | comment | added | user21230 | Yes, that sounds as a simple thing. Maybe my brain needs some oil :) I am trying to imagine 2x2 matrices as 4-dim vector space. I prefer to see geometrically why something happens, not just rely on formulas. Say for matrix $\pmatrix{1 & 0 \\ 0 & 0 }$ the kernal is generated by $\pmatrix{0 & 0 \\ 1 & 0 }$ and $\pmatrix{0 & 0 \\ 0 & 1 }$. | |
| Aug 19, 2016 at 12:01 | comment | added | Robert Bryant | @MarekMitros: I'm not sure why this isn't clear to you. If $a$ is a $2$-by-$2$ complex matrix of rank $1$, then the set of $2$-by-$2$ complex matrices $c$ such that $ac = 0$ is clearly a $2$-dimensional vector space, as is the set of $2$-by-$2$ complex matrices $d$ such that $da = 0$. (This is just linear algebra.) Thus, by the Cayley-Dickson formula, the kernel of left multiplication by $a$ in the complexified octonions has dimension $2+2=4$. I must be missing what is worrying you. | |
| Aug 19, 2016 at 9:17 | comment | added | user21230 | $L_a=\begin{pmatrix} L_a & \\ & R_a \end{pmatrix} $ where in the matrix we have left and right multiplication in $M_2 \mathbb C$. We should somehow conclude that $L_a$ and $R_a$ both has rank 2. How we can conlude it ? In this comment $a$ is again the octonion - sorry for bad notation. | |
| Aug 19, 2016 at 9:12 | comment | added | user21230 | Thank you again ! And again I need some time to analyze it. Surprisingly (for me) $\mathbb C\otimes\mathbb O=\mathbb C\otimes O'$ where $\mathbb O'$ are split octonions. Octonions $p$, $q$ lies in quaternion subalgebra $\mathbb H$ of octonions. That's clear ! Now let's verify this: "$a=p+iq$ can be viewed as 2-by-2 complex matrix of rank 1" ? Let's check for perpendicular quaternions $p=a+bj$, $q=c+dj$ we have matrix $$\begin{pmatrix} a & -\bar b \\ b & \bar a \end{pmatrix} $$ + $$\begin{pmatrix} ic & -i\bar d \\ id & i\bar c \end{pmatrix} $$ which has determinant 0. Next from Cayley-Dickson | |
| Aug 18, 2016 at 15:57 | comment | added | Robert Bryant | @მამუკაჯიბლაძე: Actually $\epsilon$ is any element in $\mathbb{O}$ such that $\epsilon^2=1$ and $\epsilon$ is perpendicular to $\mathbb{H}\subset\mathbb{O}$, so that $\mathbb{O}=\mathbb{H}\oplus \mathbb{H}\epsilon$ (orthogonal direct sum). Then the Cayley-Dickson formula (see F. Reese Harvey's Spinors and Calibrations, pg. 105, for example) says that $$ (a + b\epsilon)(c+d\epsilon)=(ac - \bar d b) + (da + b\bar c)\epsilon$$ for all $a,b,c,d\in\mathbb{H}$. Now, complexify this formula and use the fact that $\bar a = \mathrm{tr}(a) I_2 - a$ in $\mathbb{C}{\otimes}\mathbb{H}=M_2(\mathbb{C})$. | |
| Aug 18, 2016 at 15:27 | comment | added | მამუკა ჯიბლაძე | Can you give a reference for the last formula in your last comment? Is $\epsilon$ a square zero element? Can it be written down explicitly? | |
| Aug 18, 2016 at 12:21 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Fixed my erroneous calculation of the dimension in question 5.
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| Aug 18, 2016 at 12:17 | comment | added | Robert Bryant | @MarekMitros: Let the zero divisor be $a = p+iq$. By Artin's Theorem, the subalgebra of $\mathbb{O}$ generated by $p$ and $q$ is associative and, hence, lies in a quaternion subalgebra $\mathbb{H}\subset\mathbb{O}$. Now $\mathbb{C}{\otimes}\mathbb{H}\simeq M_2(\mathbb{C})$, the complex $2$-by-$2$ matrices. Thus, $a$ can be viewed as a $2$-by-$2$ complex matrix of rank $1$. Now use the Cayley-Dickson formula for multiplication in $$\mathbb{C}\otimes\mathbb{O}=\mathbb{C}{\otimes}\mathbb{H} \oplus\bigl(\mathbb{C}{\otimes}\mathbb{H}\bigr)\epsilon =M_2(\mathbb{C})\oplus M_2(\mathbb{C})\epsilon $$ | |
| Aug 18, 2016 at 11:59 | comment | added | Robert Bryant | @MarekMitros: To your first comment about point 5: I'm sorry, I was careless in the final count and forgot to notice that a 1-dimensional and a 2-dimensional space of solutions were actually transverse, so you are right that the dimension of the solution space of $ab = bc = 0$ is $3$ when $a$ and $c$ are nonzero. I'll fix that. | |
| Aug 18, 2016 at 7:15 | comment | added | user21230 | Additional question. I tested in GAP and indeed for picked two perpendicular unit octonions $p$,$q$ the 8x8 matrix $L_p+iL_q$ has rank 4. Can you provide some basic arguments why this is so ? I do not follow your arguments in point 1) using quadratic forms. | |
| Aug 18, 2016 at 7:08 | comment | added | user21230 | Regarding point 5). I tested in GAP below code for given zero divisors a,b satisfying ab=0 and it returned dimension 3. So I believe that answer should be 7+4+3=14 complex dimension. What do you say ? V:=Intersection(VectorSpace(CF(20),NullspaceMat(TransposedMat(left(b)))), VectorSpace(CF(20),NullspaceMat(TransposedMat(right(a))))); I should add that the functions "left" and "right" return matrices 8x8 of left and right multiplications in octonions. | |
| Aug 17, 2016 at 5:24 | comment | added | user21230 | Thank you for the answer ! I need some time to digest it. Regards, Marek | |
| Aug 17, 2016 at 5:21 | vote | accept | CommunityBot | ||
| Aug 16, 2016 at 20:44 | history | answered | Robert Bryant | CC BY-SA 3.0 |