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Added a remark about nonassociativity.

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edited Mar 18, 2021 at 18:08

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The quaternions are generated by any two imaginary elements $x$ and $y$ that are orthonormal, i.e., $\bigl(1,\, x,\, y,\, xy\bigr)$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions.

Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, $x$, $y$, and $z$ that are orthonormal and $z$ is perpendicular to $xy$. In fact, $\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $x'$, $y'$, $z'$ are another orthonormal triple of imaginary octonions such that $z'$ is perpendicular to $x'y'$, there is a unique automorphism of the octonions that carries $(x,y,z)$ to $(x',y',z')$. (Note

Note that the set of orthonormal triples $(x,y,z)$ in the imaginary octonions is the Stiefel manifold $V_{7,3}$, which has dimension $6+ 5+ 4 = 15$, and the single extra relation $z\cdot xy = 0$ cuts out a submanifold of dimension $14$. Hence the automorphism group of the octonions is a Lie group of dimension $14$.)

Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $\mathrm{SO}(7)$ (which has dimension $21$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions.

(Sadly, when $x$, $y$, $xy$, and $z$ are imaginary orthogonal octonions, the defining property $uv\cdot uv = (u\cdot u)(v\cdot v)$ turns out to imply that $(xy)z = -x(yz)$, so the octonions are not associative. However, this is not really the reason that $\mathrm{SO}(7)$ is not the automorphisms of the octonions.)

The quaternions are generated by any two imaginary elements $x$ and $y$ that are orthonormal, i.e., $\bigl(1,\, x,\, y,\, xy\bigr)$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions.

Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, $x$, $y$, and $z$ that are orthonormal and $z$ is perpendicular to $xy$. In fact, $\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $x'$, $y'$, $z'$ are another orthonormal triple of imaginary octonions such that $z'$ is perpendicular to $x'y'$, there is a unique automorphism of the octonions that carries $(x,y,z)$ to $(x',y',z')$. (Note that the set of orthonormal triples $(x,y,z)$ in the imaginary octonions is the Stiefel manifold $V_{7,3}$, which has dimension $6+ 5+ 4 = 15$, and the single extra relation $z\cdot xy = 0$ cuts out a submanifold of dimension $14$. Hence the automorphism group of the octonions is a Lie group of dimension $14$.)

Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $\mathrm{SO}(7)$ (which has dimension $21$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions.

The quaternions are generated by any two imaginary elements $x$ and $y$ that are orthonormal, i.e., $\bigl(1,\, x,\, y,\, xy\bigr)$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions.

Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, $x$, $y$, and $z$ that are orthonormal and $z$ is perpendicular to $xy$. In fact, $\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $x'$, $y'$, $z'$ are another orthonormal triple of imaginary octonions such that $z'$ is perpendicular to $x'y'$, there is a unique automorphism of the octonions that carries $(x,y,z)$ to $(x',y',z')$.

Note that the set of orthonormal triples $(x,y,z)$ in the imaginary octonions is the Stiefel manifold $V_{7,3}$, which has dimension $6+ 5+ 4 = 15$, and the single extra relation $z\cdot xy = 0$ cuts out a submanifold of dimension $14$. Hence the automorphism group of the octonions is a Lie group of dimension $14$.

Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $\mathrm{SO}(7)$ (which has dimension $21$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions.

(Sadly, when $x$, $y$, $xy$, and $z$ are imaginary orthogonal octonions, the defining property $uv\cdot uv = (u\cdot u)(v\cdot v)$ turns out to imply that $(xy)z = -x(yz)$, so the octonions are not associative. However, this is not really the reason that $\mathrm{SO}(7)$ is not the automorphisms of the octonions.)

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created Mar 18, 2021 at 18:02

108.4k
8
342
453

The quaternions are generated by any two imaginary elements $x$ and $y$ that are orthonormal, i.e., $\bigl(1,\, x,\, y,\, xy\bigr)$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions.

Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, $x$, $y$, and $z$ that are orthonormal and $z$ is perpendicular to $xy$. In fact, $\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $x'$, $y'$, $z'$ are another orthonormal triple of imaginary octonions such that $z'$ is perpendicular to $x'y'$, there is a unique automorphism of the octonions that carries $(x,y,z)$ to $(x',y',z')$. (Note that the set of orthonormal triples $(x,y,z)$ in the imaginary octonions is the Stiefel manifold $V_{7,3}$, which has dimension $6+ 5+ 4 = 15$, and the single extra relation $z\cdot xy = 0$ cuts out a submanifold of dimension $14$. Hence the automorphism group of the octonions is a Lie group of dimension $14$.)

Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $\mathrm{SO}(7)$ (which has dimension $21$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions.