I have following questions about the complexified octonion algebra $\mathbb C \otimes_{\mathbb R} \mathbb O$. Zero divisors are of shape $p+i\otimes q$ (shortly $p+iq$) where $p$, $q$ are perpendicular octonions of the same length. We denote octonion conjugation by an overbar, so $\overline{z \otimes x} = z \otimes \bar x$. An element $x$ of the algebra is invertible when $x\bar{x}$ is a non-zero complex number. Each element of the algebra is either a zero divisor or invertible. If we call the norm of $x$ the value $x\bar{x}$, then zero divisors are elements of norm equal to zero.

1) Is it true that, for $u$ a zero divisor, the span of $\{xu:\text{$x$ invertible}\}$ includes all zero divisors?

2) Is it true that, for each zero divisor $u$, there are zero divisors $a$, $b$ such that $u=a\bar{b}$?

3) For a given zero divisor $a$, what is the dimension of $\{b:ab=0\}$? Probably the set is just $\{\bar {a}x: \text{$x$ invertible}\}$.

4) What is the dimension of $\{(a,b):ab=0\}$? I discovered that elements $b=\bar{p}(qs)+i\otimes s$, for $s \in \mathbb O$, satisfy $ab=0$ for $a=p+iq$, $|p|=|q|=1$, $p\bar {q}$ imaginary.

5) What is the dimension of $\{(a,b,c):ab=0, bc=0, ca=0\}$ ?

The answers are needed for my analysis of reflections in $E_6$ Lie group. The definition of reflection is contained in my other question.


Here are some answers, but let me introduce some notation: What you are calling the norm of $x\in\mathbb{C}{\otimes}\mathbb{O}$, I will denote by $N(x) = x\bar x = \bar x x\in \mathbb{C}$. The complexified octonions satisfy $N(xy) = N(x)N(y)$, and the norm is a nondegenerate quadratic form on $\mathbb{C}{\otimes}\mathbb{O}$, considered as a complex vector space.

1) No. Given a zero divisor $u$, consider the quadratic form $Q_u(x) = N(xu)$ on $\mathbb{C}\otimes\mathbb{O}$. This quadratic form is identically zero, so this implies that the subspace $(\mathbb{C}{\otimes}\mathbb{O})u$ is a totally $N$-null linear subspace of $\mathbb{C}{\otimes}\mathbb{O}$, an $8$-dimensional vector space on which $N$ is nondegenerate. Thus, this subspace has dimension at most $4$ and it contains all of the elements $xu$ where $x$ is invertible. Thus, these latter elements can't span $\mathbb{C}{\otimes}\mathbb{O}$ all zero divisors because the cone $N(u) = 0$ of zero divisors does not lie in any proper linear subspace of $\mathbb{C}{\otimes}\mathbb{O}$. A similar argument applies to the elements $ux$ where $x$ is invertible.

2) Yes. To understand this, you need to realize that there are two kinds of zero divisors. The first kind (generic) are the zero divisors $u$ such that $u+\bar u = \lambda\not=0$, i.e., $u$ is not 'purely imaginary'. For such a $u$, we have $$ 0 = u\bar u = u(\lambda-u) = \lambda u - u^2, $$ so setting $a = u$ and $b = \bar u/\lambda$, we have $u = a\bar b$. The second kind has $\bar u = - u\not=0$, so we can write $u = p + iq$ where $p$ and $q$ are purely imaginary elements of $\mathbb{O}$ that have the same norm, say, $p\bar p = q\bar q = r>0$, and are orthogonal. Then $1,p,q,pq$ span an associative quaternion subalgebra of $\mathbb{O}$. We then compute, using that $p^2 = q^2 = -r$ and $pq=-qp$ that $$ (1+ipq)(p+iq) = (1+r)(p+iq). $$ Dividing by $1+r>0$, we have $p+iq$ written as a product of zero divisors.

3) For any (nonzero) zero-divisor $a$, the subspace consisting of those $b$ that satisfy $ab = 0$ is a (complex) vector space of dimension $4$. Hence the subspace consisting of (left) multiples of $a$ has dimension $8-4 = 4$ as well. The same dimension counts hold true for right multiplications. In particular, by dimension count, the subspace consisting of elements of the form $\bar a x$ is equal to the subspace consisting of elements $b$ that satisfy $ab=0$.

4) As a complex analytic variety, it has dimension 11. The reason is that, in the pair $(a,b)$, as a fixed $a\not=0$ ranges over the $N$-null cone (which has complex dimension $7$), $b$ is only required to lie in the $4$-dimensional subspace that is the kernel of left multiplication by $a$. When $a=0$, you get an $8$-plane, since there is then no restriction on $b$, but this doesn't affect the count.

5) (Corrected after the OP's comment.) The complex dimension of this singular variety is $14$. The reason is that the set of pairs $(a,c)$ that satisfy $ca=0$ has complex dimension $11$, and, for each such pair with neither $a$ nor $c$ equal to zero, the equations $ab=bc=0$ are linear equations for $b$ that, it turns out, have a (complex) $3$-dimensional space of solutions. Thus, the variety has dimension $11+3 = 14$. (The special cases when one of $a$, $b$, or $c$ vanishes only have dimension $11$, so they can't increase the count.)

  • $\begingroup$ Thank you for the answer ! I need some time to digest it. Regards, Marek $\endgroup$ – Marek Mitros Aug 17 '16 at 5:24
  • $\begingroup$ 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. $\endgroup$ – Marek Mitros Aug 18 '16 at 7:08
  • $\begingroup$ 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. $\endgroup$ – Marek Mitros Aug 18 '16 at 7:15
  • $\begingroup$ @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. $\endgroup$ – Robert Bryant Aug 18 '16 at 11:59
  • 1
    $\begingroup$ @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 $\mathbb{C}{\otimes}\mathbb{H}=M_2(\mathbb{C})$. $\endgroup$ – Robert Bryant Aug 18 '16 at 15:57

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