# Is an associative division algebra required for this phenomenon?

For which integers $d \geq 1$ can we find real matrices $R_1, \dotsc, R_d$ of size $d \times d$ such that for any unit vector $v \in \mathbb{R}^d$, $$R_1 v, \dotsc, R_d v$$ is an orthonormal basis? Note that the chosen set of $R_i$s has to work simultaneously for all $v$. (Does this phenomenon have a name?)

It may not be obvious at first, but this question seems to be related to associative division algebras (i.e., reals, complex numbers, and quaternions). Indeed, I know that such matrices $R_i$ can be found when $d=1,2,4$:

• $d=1$ is trivial—simply take $$R_1 = \begin{pmatrix}1\end{pmatrix}$$
• $d=2$ is based on complex numbers—we represent $1$ and $i$ by $2 \times 2$ real matrices as follows: $$R_1 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\qquad R_2 = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$$
• $d=4$ is based on quaternions—we represent $1$, $i$, $j$, $k$ by $4 \times 4$ matrices as follows: $$R_1 = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\qquad R_2 = \begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}\qquad\\ R_3 = \begin{pmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{pmatrix}\qquad R_4 = \begin{pmatrix}0&0&0&1\\0&0&-1&0\\0&1&0&0\\-1&0&0&0\end{pmatrix}\qquad$$

Is this possible for any other $d$? I.e., can we find $R_i$s even when there is no associative division algebra of dimension $d$?

• I think associativity is irrelevant. What you want is probably closer to en.wikipedia.org/wiki/Composition_algebra. – Qiaochu Yuan Aug 19 '15 at 9:30
• I disagree. I can't use octonions for $d=8$ since they are not associative and thus can't be represented by matrices (because matrix multiplication is associative). – Māris Ozols Aug 19 '15 at 10:07
• I have to take this back - octonions indeed work (see the answer by Adam Przeździecki). – Māris Ozols Aug 19 '15 at 15:07
• This phenomenon has some interesting applications in quantum information which I was not aware of at the time I asked the question (see dx.doi.org/10.1103/PhysRevA.65.022316). – Māris Ozols Feb 2 '16 at 20:26

1. Consider the matrix $\sum_i a_iR_i$. One can show that it sends every vector of length 1 to a vector of length $\sqrt{\sum_i a_i^2}$. It follows that if the norm of $a=(a_i)$ is 1, then the matrix $\sum_i a_iR_i$ is orthogonal. In particular, all the matrices $R_i$ are orthogonal.
2. By multiplication from the left or from the right with an orthogonal matrix, we can assume without loss of generality that $R_1=Id$ (the collection $\{SR_iT\}$ where $S$ and $T$ are orthogonal matrices will still satisfy the condition.
3. Consider now $R_2$. Then $R_2$ cannot have a real eigenvalue $\lambda$, because then $R_2-\lambda R_1$ will have nontrivial kernel, which contradicts 1. It follows that, since $R_2$ is orthogonal, it can be written as the direct sum of $2\times 2$ rotation matrices. If one of these rotation angles is $\theta$, then by considering 1. for the matrix $R_2+R_1$ we reach the conclusion that $\cos(\theta)=0$. So $\theta=90$ or $270$. In any case, it follows that $R_2^2 = -1$, and the same is true for $R_i$ $i=3,\ldots, d$ of course. We also get that $d$ is even.
4. We can now multiply the collection $\{R_i\}$ from the left by $-R_2$. Again we will receive a collection of matrices which satisfies the condition, and contains the matrix $-R_2^2=Id$. But then it follows that all the non-unital matrices in this collection square to -1. In other words, we get that $(R_iR_j)^2=-1$ for every $i\neq j$, $i,j>1$.
5. Consider now the algebra $\mathbb{R}\langle X_1,\ldots X_{d-1}\rangle/(X_i^2=-1, X_iX_j + X_jX_i=0 (i\neq j))$. This algebra is a twisted group algebra with the group $G=(\mathbb{Z}/2)^{d-1}$. This algebra is semisimple, and one can prove directly that its center is 2 dimensional and generated by $X:=X_1\cdots X_{d-1}$.
If $r=2 \mod 4$, then $X^2=-1$. The center is then isomorphic with $\mathbb{C}$, so we get a central simple algebra of dimension $2^{d-2}$ over $\mathbb{C}$, which must be the matrix algebra $M_n(\mathbb{C})$, where $n= 2^{(d-2)/2}$. This algebra has pnly one irreducible representation, and it is of dimension $2^{(d-2)/2}$ over $\mathbb{C}$, and therefore of dimension $2^{d/2}$ over $\mathbb{R}$. This rules out the possibility that $r=6,10,14...$.
If $r=0 \mod 4$, then $X^2 = 1$, and the center is isomorphic with $\mathbb{R}\oplus\mathbb{R}$, and the algebra splits as the direct sum of two central simple $\mathbb{R}$-algebras, each of dimension $2^{(d-2)}$. Such an algebra will either be isomorphic with $M_n(\mathbb{R})$ where $n = 2^{(d-2)/2}$, or with $M_n(D)$ where $D$ is the quaternion algebra and $n= 2^{(d-6)/2}$. In the first case the algebra has a unique simple representation of dimension $2^{(d-2)/2}$, and in the second case it has a unique irreducible representation of dimension $2^{(d+2)/2}$. For $d>8$ both dimensions are bigger than $d$ so there is no solution. For $d=8$ we can study carefully the algebra and find a solution, the one given by Adam. In order to analyze the algebra we can, for example, consider the subalgebra generated by $X_1,X_2$ (in the proper quotient) which is isomorphic with the quaternion algebra. Then the algebra will be a tensor product of the quaternion algebra and its centralizer. This reduces the question to studying an algebra of a smaller dimension, and we can find out that at the end we get a matrix algebra over the reals, and so we have a solution for $d=8$.
• I'm not following the last point entirely. Is this the $\mathcal{C}\ell_{0,d-1}(\mathbb{R})$ Clifford algebra? Can you give a reference why it is a direct sum of two central simple algebras isomorphic to $M_n(D)$? How do you get $2^{d/2}$ in the end? – Māris Ozols Aug 19 '15 at 14:03
• It seems that there must be a gap in your argument, since $d=8$ is actually possible (see the answer by Adam Przeździecki). Could you elaborate a bit more on step 5? – Māris Ozols Aug 19 '15 at 15:43
• You right, I was not careful enough. I edited my answer. – Ehud Meir Aug 19 '15 at 16:17

One more example: for $d=8$ based on octonions. And this are all possible dimensions. $R_1$ takes vectors of length one to vectors of length one hence it is an isometry. By multiplying from the left by $R_1^{-1}$ we may assume that $R_1=\mathop{\rm Id}$. Then for every $v\in S^{d-1}$, the unit sphere in $\mathbb{R}^d$, the map $v\mapsto(R_2v,R_3v,\ldots,R_dv)$ gives a parallelization of the tangent bundle of $S^{d-1}$. Then by a celebrated Bott-Kervaire-Milnor theorem such a parallelization exists only when $d\in\{1,2,4,8\}$.

Edit: More explicitly, the solution for $d = 8$ is given by $$\sum_{i=1}^8 a_i R_i = \begin{pmatrix} a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8 \\ -a_2 & a_1 & -a_4 & a_3 & -a_6 & a_5 & a_8 & -a_7 \\ -a_3 & a_4 & a_1 & -a_2 & -a_7 & -a_8 & a_5 & a_6 \\ -a_4 & -a_3 & a_2 & a_1 & -a_8 & a_7 & -a_6 & a_5 \\ -a_5 & a_6 & a_7 & a_8 & a_1 & -a_2 & -a_3 & -a_4 \\ -a_6 & -a_5 & a_8 & -a_7 & a_2 & a_1 & a_4 & -a_3 \\ -a_7 & -a_8 & -a_5 & a_6 & a_3 & -a_4 & a_1 & a_2 \\ -a_8 & a_7 & -a_6 & -a_5 & a_4 & a_3 & -a_2 & a_1 \\ \end{pmatrix}$$

• Could you elaborate more on how this works (e.g., what exactly are the matrices $R_i$)? Since octonions are not associative, I don't think you can represent them by matrices as matrix multiplication is associative. – Māris Ozols Aug 19 '15 at 13:34
• You may view the octonions $\bf{O}$ as a vector space $\mathbb{R}^8$ plus a bilinear map $\bf{O}\times\bf{O}\to\bf{O}$. You also have the standard basis $\{e_i\mid i=0,1,\ldots,7\}$ of $\bf{O}$. The bilinear map is not associative, but all you need is that each $e_i$ acts linearly from the left on $\bf{O}$ and that $\forall_{v\in\bf{O}}\{e_0v,e_1v,\ldots,e_7v\}$ is an orthonormal basis. Thus you may put $R_i=e_{i-1}$. – Adam Przeździecki Aug 19 '15 at 14:02
• I checked that this indeed works, so I added an explicit expression to your answer. Do you have any good reference for why there is no solution other than $d \in \{1,2,4,8\}$? – Māris Ozols Aug 19 '15 at 14:46
• @AdamPrzeździecki Cześć Adam ! wikipedia article en.wikipedia.org/wiki/Parallelizable_manifold says that it was proved in 1958 by Kervaire, Bott and Milnor that the only paralelizable spheres are $S^1$, $S^3$ and $S^7$. – user21230 May 27 '16 at 15:23

This is an alternative way to see this question. By argument given by the current answer given by Ehud Meir we may assume that $R_1$ is the identity matrix. Then the hypotheses imply that for $i=2,\cdots,d$ the maps $v\rightarrow R_i v$ are $d-1$ independent vector fields on the unit sphere $S^{d-1}$. But it is known that the maximal number of independent vector fields on the unit sphere $S^{n-1}$ is $n-1$ if and only if $n=1, 2, 4$ or $8$. This follows from a much stronger result due to J. F. Adams (Annals of Mathematics, 1962)