Mixing
Do the octonions contain infinitely many copies of the quaternions?
$begingroup$
Note: by "infinitely many", I'm confident I always mean $beth_1$ many herein.
We can easily show the quaternions contain infinitely many copies of $Bbb C$ because, given any unit vector $inBbb R^3$ of components $b,,c,,d$, $Bbb R[h]$ is isomorphic to $Bbb C$ with $h:=bi+cj+dk$. Sure, these aren't "independent" copies of $Bbb C$ in the same way $Bbb R[i],,Bbb R[j],,Bbb R[k]$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $Bbb C$ provide an easy definition of a determinant, even though general matrices of quaternions prohibit this. For example, if $A_1,,cdots,,A_n$ are matrices and $O$ denotes a contextually appropriate zero matrix, the block matrix $$left(begin{array}{cccc}
A_{1} & O & cdots & O\
O & A_{2} & cdots & O\
vdots & vdots & ddots & O\
O & O & cdots & A_{n}
end{array}right)$$can be said to have determinant $prod_ldet A_l$. The order is important, but one in particular is natural.
I may be overlooking some details of the above benefit to arbitrarily many copies of $Bbb C$ in $Bbb H$, wherein a very rich commuting family of matrices is constructed, but my question isn't about that. I'm wondering how we'd prove there are infinitely many copies of $Bbb H$ in $Bbb O$. (Again, block matrices provide a benefit, in this case inheriting the associativity of the $A_l$.) I suspect a proof exists that admits the following sketch:
- In $Bbb O$, create some $h_1,,h_2$ each analogous to $iinBbb C$, generating an associative algebra and satisfying $h_1h_2=-h_2h_1$;
- Note that any such pair of square roots of $-1$ can model the quaternions viz. $i=h_1,,j=h_2,,k=h_1h_2$;
- Show the above can be done in infinitely many different ways.
Of course there are infinitely many ways to choose the pair $(h_1,,h_2)$, but $Bbb R[h_1,,h_2]$ won't always be a different set for such pairs. That's why I suspect the proof requires a few clever i-dotting t-crosses.
group-isomorphism quaternions octonions
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
Note: by "infinitely many", I'm confident I always mean $beth_1$ many herein.
We can easily show the quaternions contain infinitely many copies of $Bbb C$ because, given any unit vector $inBbb R^3$ of components $b,,c,,d$, $Bbb R[h]$ is isomorphic to $Bbb C$ with $h:=bi+cj+dk$. Sure, these aren't "independent" copies of $Bbb C$ in the same way $Bbb R[i],,Bbb R[j],,Bbb R[k]$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $Bbb C$ provide an easy definition of a determinant, even though general matrices of quaternions prohibit this. For example, if $A_1,,cdots,,A_n$ are matrices and $O$ denotes a contextually appropriate zero matrix, the block matrix $$left(begin{array}{cccc}
A_{1} & O & cdots & O\
O & A_{2} & cdots & O\
vdots & vdots & ddots & O\
O & O & cdots & A_{n}
end{array}right)$$can be said to have determinant $prod_ldet A_l$. The order is important, but one in particular is natural.
I may be overlooking some details of the above benefit to arbitrarily many copies of $Bbb C$ in $Bbb H$, wherein a very rich commuting family of matrices is constructed, but my question isn't about that. I'm wondering how we'd prove there are infinitely many copies of $Bbb H$ in $Bbb O$. (Again, block matrices provide a benefit, in this case inheriting the associativity of the $A_l$.) I suspect a proof exists that admits the following sketch:
- In $Bbb O$, create some $h_1,,h_2$ each analogous to $iinBbb C$, generating an associative algebra and satisfying $h_1h_2=-h_2h_1$;
- Note that any such pair of square roots of $-1$ can model the quaternions viz. $i=h_1,,j=h_2,,k=h_1h_2$;
- Show the above can be done in infinitely many different ways.
Of course there are infinitely many ways to choose the pair $(h_1,,h_2)$, but $Bbb R[h_1,,h_2]$ won't always be a different set for such pairs. That's why I suspect the proof requires a few clever i-dotting t-crosses.
group-isomorphism quaternions octonions
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
$endgroup$
$begingroup$
In fact, we can identify the space of copies of $Bbb H$ in $Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$.
$endgroup$
– Travis
Feb 2 at 5:29
add a comment |
$begingroup$
Note: by "infinitely many", I'm confident I always mean $beth_1$ many herein.
We can easily show the quaternions contain infinitely many copies of $Bbb C$ because, given any unit vector $inBbb R^3$ of components $b,,c,,d$, $Bbb R[h]$ is isomorphic to $Bbb C$ with $h:=bi+cj+dk$. Sure, these aren't "independent" copies of $Bbb C$ in the same way $Bbb R[i],,Bbb R[j],,Bbb R[k]$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $Bbb C$ provide an easy definition of a determinant, even though general matrices of quaternions prohibit this. For example, if $A_1,,cdots,,A_n$ are matrices and $O$ denotes a contextually appropriate zero matrix, the block matrix $$left(begin{array}{cccc}
A_{1} & O & cdots & O\
O & A_{2} & cdots & O\
vdots & vdots & ddots & O\
O & O & cdots & A_{n}
end{array}right)$$can be said to have determinant $prod_ldet A_l$. The order is important, but one in particular is natural.
I may be overlooking some details of the above benefit to arbitrarily many copies of $Bbb C$ in $Bbb H$, wherein a very rich commuting family of matrices is constructed, but my question isn't about that. I'm wondering how we'd prove there are infinitely many copies of $Bbb H$ in $Bbb O$. (Again, block matrices provide a benefit, in this case inheriting the associativity of the $A_l$.) I suspect a proof exists that admits the following sketch:
- In $Bbb O$, create some $h_1,,h_2$ each analogous to $iinBbb C$, generating an associative algebra and satisfying $h_1h_2=-h_2h_1$;
- Note that any such pair of square roots of $-1$ can model the quaternions viz. $i=h_1,,j=h_2,,k=h_1h_2$;
- Show the above can be done in infinitely many different ways.
Of course there are infinitely many ways to choose the pair $(h_1,,h_2)$, but $Bbb R[h_1,,h_2]$ won't always be a different set for such pairs. That's why I suspect the proof requires a few clever i-dotting t-crosses.
group-isomorphism quaternions octonions
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
$endgroup$
Note: by "infinitely many", I'm confident I always mean $beth_1$ many herein.
We can easily show the quaternions contain infinitely many copies of $Bbb C$ because, given any unit vector $inBbb R^3$ of components $b,,c,,d$, $Bbb R[h]$ is isomorphic to $Bbb C$ with $h:=bi+cj+dk$. Sure, these aren't "independent" copies of $Bbb C$ in the same way $Bbb R[i],,Bbb R[j],,Bbb R[k]$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $Bbb C$ provide an easy definition of a determinant, even though general matrices of quaternions prohibit this. For example, if $A_1,,cdots,,A_n$ are matrices and $O$ denotes a contextually appropriate zero matrix, the block matrix $$left(begin{array}{cccc}
A_{1} & O & cdots & O\
O & A_{2} & cdots & O\
vdots & vdots & ddots & O\
O & O & cdots & A_{n}
end{array}right)$$can be said to have determinant $prod_ldet A_l$. The order is important, but one in particular is natural.
I may be overlooking some details of the above benefit to arbitrarily many copies of $Bbb C$ in $Bbb H$, wherein a very rich commuting family of matrices is constructed, but my question isn't about that. I'm wondering how we'd prove there are infinitely many copies of $Bbb H$ in $Bbb O$. (Again, block matrices provide a benefit, in this case inheriting the associativity of the $A_l$.) I suspect a proof exists that admits the following sketch:
- In $Bbb O$, create some $h_1,,h_2$ each analogous to $iinBbb C$, generating an associative algebra and satisfying $h_1h_2=-h_2h_1$;
- Note that any such pair of square roots of $-1$ can model the quaternions viz. $i=h_1,,j=h_2,,k=h_1h_2$;
- Show the above can be done in infinitely many different ways.
Of course there are infinitely many ways to choose the pair $(h_1,,h_2)$, but $Bbb R[h_1,,h_2]$ won't always be a different set for such pairs. That's why I suspect the proof requires a few clever i-dotting t-crosses.
group-isomorphism quaternions octonions
group-isomorphism quaternions octonions
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
edited Feb 1 at 21:40
J.G.
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
asked Feb 1 at 21:23
J.G.J.G.
33.2k23252
33.2k23252
$begingroup$
In fact, we can identify the space of copies of $Bbb H$ in $Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$.
$endgroup$
– Travis
Feb 2 at 5:29
add a comment |
$begingroup$
In fact, we can identify the space of copies of $Bbb H$ in $Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$.
$endgroup$
– Travis
Feb 2 at 5:29
$begingroup$
In fact, we can identify the space of copies of $Bbb H$ in $Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$.
$endgroup$
– Travis
Feb 2 at 5:29
$begingroup$
In fact, we can identify the space of copies of $Bbb H$ in $Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$.
$endgroup$
– Travis
Feb 2 at 5:29
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
Yes. One can show the following 2 things:
Any octonion $x$ not in $mathbb R$ generates a subalgebra $A$ isomorphic to $mathbb C$.
For any octonion $y$, the subalgebra generated $B$ by $x$ and $y$ is associative.
This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.
Hence if we take $y$ outside of $A$ in 1, by Frobenius' classification of $mathbb R$-division algebras, $B$ must be isomorphic to $mathbb H$ (a priori it may not be obvious $B$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $B$, which necessarily lie in $mathbb O$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $1, x, y, xy$. Since no finite collection of $mathbb R^4$ subspaces of $mathbb R^8$ cover $mathbb R^8$, you get infinitely many copies of $mathbb H$.
To say that you get (at least) the cardinality of the continuum copies of $mathbb H$, it suffices to show
- For any nonreal octonion $z$, there is a subalgebra $B simeq mathbb H$ as above not containing $z$.
To see this, take $x$ as above not in $mathbb R z$. Then $x$ and $z$ generate a space $B' simeq mathbb H$. Simply take $y not in B'$. Then the algebra $B$ generated by $x$ and $y$ cannot contain $B'$ since $B ne B'$ but $dim B = dim B'$.
answered Feb 2 at 3:14
KimballKimball
2,0451029
$endgroup$
1
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
add a comment |
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$begingroup$
Yes. One can show the following 2 things:
Any octonion $x$ not in $mathbb R$ generates a subalgebra $A$ isomorphic to $mathbb C$.
For any octonion $y$, the subalgebra generated $B$ by $x$ and $y$ is associative.
This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.
Hence if we take $y$ outside of $A$ in 1, by Frobenius' classification of $mathbb R$-division algebras, $B$ must be isomorphic to $mathbb H$ (a priori it may not be obvious $B$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $B$, which necessarily lie in $mathbb O$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $1, x, y, xy$. Since no finite collection of $mathbb R^4$ subspaces of $mathbb R^8$ cover $mathbb R^8$, you get infinitely many copies of $mathbb H$.
To say that you get (at least) the cardinality of the continuum copies of $mathbb H$, it suffices to show
- For any nonreal octonion $z$, there is a subalgebra $B simeq mathbb H$ as above not containing $z$.
To see this, take $x$ as above not in $mathbb R z$. Then $x$ and $z$ generate a space $B' simeq mathbb H$. Simply take $y not in B'$. Then the algebra $B$ generated by $x$ and $y$ cannot contain $B'$ since $B ne B'$ but $dim B = dim B'$.
answered Feb 2 at 3:14
KimballKimball
2,0451029
$endgroup$
1
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
add a comment |
$begingroup$
Yes. One can show the following 2 things:
Any octonion $x$ not in $mathbb R$ generates a subalgebra $A$ isomorphic to $mathbb C$.
For any octonion $y$, the subalgebra generated $B$ by $x$ and $y$ is associative.
This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.
Hence if we take $y$ outside of $A$ in 1, by Frobenius' classification of $mathbb R$-division algebras, $B$ must be isomorphic to $mathbb H$ (a priori it may not be obvious $B$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $B$, which necessarily lie in $mathbb O$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $1, x, y, xy$. Since no finite collection of $mathbb R^4$ subspaces of $mathbb R^8$ cover $mathbb R^8$, you get infinitely many copies of $mathbb H$.
To say that you get (at least) the cardinality of the continuum copies of $mathbb H$, it suffices to show
- For any nonreal octonion $z$, there is a subalgebra $B simeq mathbb H$ as above not containing $z$.
To see this, take $x$ as above not in $mathbb R z$. Then $x$ and $z$ generate a space $B' simeq mathbb H$. Simply take $y not in B'$. Then the algebra $B$ generated by $x$ and $y$ cannot contain $B'$ since $B ne B'$ but $dim B = dim B'$.
answered Feb 2 at 3:14
KimballKimball
2,0451029
$endgroup$
1
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
add a comment |
$begingroup$
Yes. One can show the following 2 things:
Any octonion $x$ not in $mathbb R$ generates a subalgebra $A$ isomorphic to $mathbb C$.
For any octonion $y$, the subalgebra generated $B$ by $x$ and $y$ is associative.
This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.
Hence if we take $y$ outside of $A$ in 1, by Frobenius' classification of $mathbb R$-division algebras, $B$ must be isomorphic to $mathbb H$ (a priori it may not be obvious $B$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $B$, which necessarily lie in $mathbb O$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $1, x, y, xy$. Since no finite collection of $mathbb R^4$ subspaces of $mathbb R^8$ cover $mathbb R^8$, you get infinitely many copies of $mathbb H$.
To say that you get (at least) the cardinality of the continuum copies of $mathbb H$, it suffices to show
- For any nonreal octonion $z$, there is a subalgebra $B simeq mathbb H$ as above not containing $z$.
To see this, take $x$ as above not in $mathbb R z$. Then $x$ and $z$ generate a space $B' simeq mathbb H$. Simply take $y not in B'$. Then the algebra $B$ generated by $x$ and $y$ cannot contain $B'$ since $B ne B'$ but $dim B = dim B'$.
answered Feb 2 at 3:14
KimballKimball
2,0451029
$endgroup$
Yes. One can show the following 2 things:
Any octonion $x$ not in $mathbb R$ generates a subalgebra $A$ isomorphic to $mathbb C$.
For any octonion $y$, the subalgebra generated $B$ by $x$ and $y$ is associative.
This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.
Hence if we take $y$ outside of $A$ in 1, by Frobenius' classification of $mathbb R$-division algebras, $B$ must be isomorphic to $mathbb H$ (a priori it may not be obvious $B$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $B$, which necessarily lie in $mathbb O$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $1, x, y, xy$. Since no finite collection of $mathbb R^4$ subspaces of $mathbb R^8$ cover $mathbb R^8$, you get infinitely many copies of $mathbb H$.
To say that you get (at least) the cardinality of the continuum copies of $mathbb H$, it suffices to show
- For any nonreal octonion $z$, there is a subalgebra $B simeq mathbb H$ as above not containing $z$.
To see this, take $x$ as above not in $mathbb R z$. Then $x$ and $z$ generate a space $B' simeq mathbb H$. Simply take $y not in B'$. Then the algebra $B$ generated by $x$ and $y$ cannot contain $B'$ since $B ne B'$ but $dim B = dim B'$.
answered Feb 2 at 3:14
KimballKimball
2,0451029
edited Feb 2 at 5:20
answered Feb 2 at 3:14
KimballKimball
2,0451029
answered Feb 2 at 3:14
KimballKimball
2,0451029
answered Feb 2 at 3:14
KimballKimball
2,0451029
2,0451029
1
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
add a comment |
1
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
1
1
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
For what it's worth, property (2) is called alternativity.
$endgroup$
– Travis
Feb 2 at 5:30
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
$begingroup$
I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry
$endgroup$
– Vincent
Mar 19 at 22:55
add a comment |
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In fact, we can identify the space of copies of $Bbb H$ in $Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$.
$endgroup$
– Travis
Feb 2 at 5:29