Is there an algebraic or geometric explanation for why $Spin_7simeq_frac13S^7times G_2$?
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The octonions $mathbbO$ are the $8$-dimension real (non-associative) normed division algebra. Forgetting the algebra structure leads to an identification of $mathbbO$ with $mathbbR^8$ as real normed spaces, and so generates a tautological representation of $SO(8)$ on $mathbbO$.
Now one may define the compact exceptional group $G_2$ as the $mathbbR$-linear algebra automorphism group of $mathbbO$. That is,
$$G_2=alphain SO(8)mid alpha(xy)=alpha(x)alpha(y),,forall, x,yinmathbbOleq SO(8).$$
On the other hand we can appeal to Cartan's principle of triality in $SO(8)$ and define a subgroup of $S0(8)$ which identifies with a double cover of $SO(7)$. In short, we have
$$Spin_7congalphain SO(8)midexists tildealphain SO(7),textsuch that,alpha(xy)=tildealpha(x)alpha(y),forall,x,yinmathbbOleq SO(8).$$
Then clearly $G_2leq Spin_7$. Moreover this $G_2$ subgroup preserves the algebra unit $1inmathbbO$, and if we let $Spin_7$ act on the unit sphere $S^7subseteq mathbbO$ through its inclusion in $SO(8)$ we are led to a fibre sequence
$$G_2rightarrow Spin_7rightarrow S^7.$$
Now this fibration is principal, classified by a map $gamma:S^7rightarrow BG_2$ where $BG_2$ is the classifying space of $G_2$. If we check the homotopy groups of $G_2$ we find that $pi_7BG_2congpi_6G_2congmathbbZ_3$ and it turns out that $gamma$ is a generator for the first group. On the other hand, if we localise these spaces away from $3$, then clearly $(pi_7BG_2)_(frac13)=0$ and (the localisation of) $gamma$ is null-homotopic. In this case the fibration above splits, and we get a $frac13$-local homotopy equivalence
$$Spin_7simeq_frac13S^7times G_2,$$
although one should be quick to remark that this homotopy equivalence is one of spaces, not $H$-spaces.
This brings me to my actual question.
Is there an algebraic or geometric explanation for why the $frac13$-local homotopy equivalence $Spin_7simeq_frac13S^7times G_2$ should exist?
The splitting is clearly a homotopy-theoretic artifact, rather than directly a geometric or algebraic consequence. So perhaps the question should rather be, is there an algebraic or geometric reason why the homotopy-theoretic data aligns itself in this way?
As an example of what I'm looking for, the exceptional isomorphism $Spin_4cong S^3times S^3$ can be very explicitly seen by writing down a simple representation of $Spin_4$ and determining the consequences. It seems that the explanation for the homotopy equivalence I am considering, however, would be much more subtle.
algebraic-topology lie-groups homotopy-theory topological-groups
asked Sep 10 at 11:10
Tyrone
3,68111125
add a comment |Â
up vote
3
down vote
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The octonions $mathbbO$ are the $8$-dimension real (non-associative) normed division algebra. Forgetting the algebra structure leads to an identification of $mathbbO$ with $mathbbR^8$ as real normed spaces, and so generates a tautological representation of $SO(8)$ on $mathbbO$.
Now one may define the compact exceptional group $G_2$ as the $mathbbR$-linear algebra automorphism group of $mathbbO$. That is,
$$G_2=alphain SO(8)mid alpha(xy)=alpha(x)alpha(y),,forall, x,yinmathbbOleq SO(8).$$
On the other hand we can appeal to Cartan's principle of triality in $SO(8)$ and define a subgroup of $S0(8)$ which identifies with a double cover of $SO(7)$. In short, we have
$$Spin_7congalphain SO(8)midexists tildealphain SO(7),textsuch that,alpha(xy)=tildealpha(x)alpha(y),forall,x,yinmathbbOleq SO(8).$$
Then clearly $G_2leq Spin_7$. Moreover this $G_2$ subgroup preserves the algebra unit $1inmathbbO$, and if we let $Spin_7$ act on the unit sphere $S^7subseteq mathbbO$ through its inclusion in $SO(8)$ we are led to a fibre sequence
$$G_2rightarrow Spin_7rightarrow S^7.$$
Now this fibration is principal, classified by a map $gamma:S^7rightarrow BG_2$ where $BG_2$ is the classifying space of $G_2$. If we check the homotopy groups of $G_2$ we find that $pi_7BG_2congpi_6G_2congmathbbZ_3$ and it turns out that $gamma$ is a generator for the first group. On the other hand, if we localise these spaces away from $3$, then clearly $(pi_7BG_2)_(frac13)=0$ and (the localisation of) $gamma$ is null-homotopic. In this case the fibration above splits, and we get a $frac13$-local homotopy equivalence
$$Spin_7simeq_frac13S^7times G_2,$$
although one should be quick to remark that this homotopy equivalence is one of spaces, not $H$-spaces.
This brings me to my actual question.
Is there an algebraic or geometric explanation for why the $frac13$-local homotopy equivalence $Spin_7simeq_frac13S^7times G_2$ should exist?
The splitting is clearly a homotopy-theoretic artifact, rather than directly a geometric or algebraic consequence. So perhaps the question should rather be, is there an algebraic or geometric reason why the homotopy-theoretic data aligns itself in this way?
As an example of what I'm looking for, the exceptional isomorphism $Spin_4cong S^3times S^3$ can be very explicitly seen by writing down a simple representation of $Spin_4$ and determining the consequences. It seems that the explanation for the homotopy equivalence I am considering, however, would be much more subtle.
algebraic-topology lie-groups homotopy-theory topological-groups
asked Sep 10 at 11:10
Tyrone
3,68111125
I am confused: you just gave an explanation, which is both algebraic and geometric.
– Pedro Tamaroff♦
Sep 10 at 11:57
@PedroTamaroff, perhaps you are correct, but as I see it, the algebra and geometry flows together beautifully until you ask that the prime $3$ be inverted. The exceptional isomorphisms lead to the $G_2leq Spin_7$ subgroup and geometric observations lead to the fibration sequence. But why has fate conspired to make $pi_6G_2congmathbbZ_3$? For example it is also true that $Spin_8simeq S^7times Spin_7$ (integrally), since $pi_6Spin_7=0$, however this homotopy equivalence - derived by the same reasoning - is actually a homeomorphism.
– Tyrone
Sep 10 at 13:45
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The octonions $mathbbO$ are the $8$-dimension real (non-associative) normed division algebra. Forgetting the algebra structure leads to an identification of $mathbbO$ with $mathbbR^8$ as real normed spaces, and so generates a tautological representation of $SO(8)$ on $mathbbO$.
Now one may define the compact exceptional group $G_2$ as the $mathbbR$-linear algebra automorphism group of $mathbbO$. That is,
$$G_2=alphain SO(8)mid alpha(xy)=alpha(x)alpha(y),,forall, x,yinmathbbOleq SO(8).$$
On the other hand we can appeal to Cartan's principle of triality in $SO(8)$ and define a subgroup of $S0(8)$ which identifies with a double cover of $SO(7)$. In short, we have
$$Spin_7congalphain SO(8)midexists tildealphain SO(7),textsuch that,alpha(xy)=tildealpha(x)alpha(y),forall,x,yinmathbbOleq SO(8).$$
Then clearly $G_2leq Spin_7$. Moreover this $G_2$ subgroup preserves the algebra unit $1inmathbbO$, and if we let $Spin_7$ act on the unit sphere $S^7subseteq mathbbO$ through its inclusion in $SO(8)$ we are led to a fibre sequence
$$G_2rightarrow Spin_7rightarrow S^7.$$
Now this fibration is principal, classified by a map $gamma:S^7rightarrow BG_2$ where $BG_2$ is the classifying space of $G_2$. If we check the homotopy groups of $G_2$ we find that $pi_7BG_2congpi_6G_2congmathbbZ_3$ and it turns out that $gamma$ is a generator for the first group. On the other hand, if we localise these spaces away from $3$, then clearly $(pi_7BG_2)_(frac13)=0$ and (the localisation of) $gamma$ is null-homotopic. In this case the fibration above splits, and we get a $frac13$-local homotopy equivalence
$$Spin_7simeq_frac13S^7times G_2,$$
although one should be quick to remark that this homotopy equivalence is one of spaces, not $H$-spaces.
This brings me to my actual question.
Is there an algebraic or geometric explanation for why the $frac13$-local homotopy equivalence $Spin_7simeq_frac13S^7times G_2$ should exist?
The splitting is clearly a homotopy-theoretic artifact, rather than directly a geometric or algebraic consequence. So perhaps the question should rather be, is there an algebraic or geometric reason why the homotopy-theoretic data aligns itself in this way?
As an example of what I'm looking for, the exceptional isomorphism $Spin_4cong S^3times S^3$ can be very explicitly seen by writing down a simple representation of $Spin_4$ and determining the consequences. It seems that the explanation for the homotopy equivalence I am considering, however, would be much more subtle.
algebraic-topology lie-groups homotopy-theory topological-groups
asked Sep 10 at 11:10
Tyrone
3,68111125
The octonions $mathbbO$ are the $8$-dimension real (non-associative) normed division algebra. Forgetting the algebra structure leads to an identification of $mathbbO$ with $mathbbR^8$ as real normed spaces, and so generates a tautological representation of $SO(8)$ on $mathbbO$.
Now one may define the compact exceptional group $G_2$ as the $mathbbR$-linear algebra automorphism group of $mathbbO$. That is,
$$G_2=alphain SO(8)mid alpha(xy)=alpha(x)alpha(y),,forall, x,yinmathbbOleq SO(8).$$
On the other hand we can appeal to Cartan's principle of triality in $SO(8)$ and define a subgroup of $S0(8)$ which identifies with a double cover of $SO(7)$. In short, we have
$$Spin_7congalphain SO(8)midexists tildealphain SO(7),textsuch that,alpha(xy)=tildealpha(x)alpha(y),forall,x,yinmathbbOleq SO(8).$$
Then clearly $G_2leq Spin_7$. Moreover this $G_2$ subgroup preserves the algebra unit $1inmathbbO$, and if we let $Spin_7$ act on the unit sphere $S^7subseteq mathbbO$ through its inclusion in $SO(8)$ we are led to a fibre sequence
$$G_2rightarrow Spin_7rightarrow S^7.$$
Now this fibration is principal, classified by a map $gamma:S^7rightarrow BG_2$ where $BG_2$ is the classifying space of $G_2$. If we check the homotopy groups of $G_2$ we find that $pi_7BG_2congpi_6G_2congmathbbZ_3$ and it turns out that $gamma$ is a generator for the first group. On the other hand, if we localise these spaces away from $3$, then clearly $(pi_7BG_2)_(frac13)=0$ and (the localisation of) $gamma$ is null-homotopic. In this case the fibration above splits, and we get a $frac13$-local homotopy equivalence
$$Spin_7simeq_frac13S^7times G_2,$$
although one should be quick to remark that this homotopy equivalence is one of spaces, not $H$-spaces.
This brings me to my actual question.
Is there an algebraic or geometric explanation for why the $frac13$-local homotopy equivalence $Spin_7simeq_frac13S^7times G_2$ should exist?
The splitting is clearly a homotopy-theoretic artifact, rather than directly a geometric or algebraic consequence. So perhaps the question should rather be, is there an algebraic or geometric reason why the homotopy-theoretic data aligns itself in this way?
As an example of what I'm looking for, the exceptional isomorphism $Spin_4cong S^3times S^3$ can be very explicitly seen by writing down a simple representation of $Spin_4$ and determining the consequences. It seems that the explanation for the homotopy equivalence I am considering, however, would be much more subtle.
algebraic-topology lie-groups homotopy-theory topological-groups
algebraic-topology lie-groups homotopy-theory topological-groups
asked Sep 10 at 11:10
Tyrone
3,68111125
asked Sep 10 at 11:10
Tyrone
3,68111125
asked Sep 10 at 11:10
Tyrone
3,68111125
asked Sep 10 at 11:10
Tyrone
3,68111125
asked Sep 10 at 11:10
Tyrone
3,68111125
3,68111125
I am confused: you just gave an explanation, which is both algebraic and geometric.
– Pedro Tamaroff♦
Sep 10 at 11:57
@PedroTamaroff, perhaps you are correct, but as I see it, the algebra and geometry flows together beautifully until you ask that the prime $3$ be inverted. The exceptional isomorphisms lead to the $G_2leq Spin_7$ subgroup and geometric observations lead to the fibration sequence. But why has fate conspired to make $pi_6G_2congmathbbZ_3$? For example it is also true that $Spin_8simeq S^7times Spin_7$ (integrally), since $pi_6Spin_7=0$, however this homotopy equivalence - derived by the same reasoning - is actually a homeomorphism.
– Tyrone
Sep 10 at 13:45
add a comment |Â
I am confused: you just gave an explanation, which is both algebraic and geometric.
– Pedro Tamaroff♦
Sep 10 at 11:57
@PedroTamaroff, perhaps you are correct, but as I see it, the algebra and geometry flows together beautifully until you ask that the prime $3$ be inverted. The exceptional isomorphisms lead to the $G_2leq Spin_7$ subgroup and geometric observations lead to the fibration sequence. But why has fate conspired to make $pi_6G_2congmathbbZ_3$? For example it is also true that $Spin_8simeq S^7times Spin_7$ (integrally), since $pi_6Spin_7=0$, however this homotopy equivalence - derived by the same reasoning - is actually a homeomorphism.
– Tyrone
Sep 10 at 13:45
I am confused: you just gave an explanation, which is both algebraic and geometric.
– Pedro Tamaroff♦
Sep 10 at 11:57
I am confused: you just gave an explanation, which is both algebraic and geometric.
– Pedro Tamaroff♦
Sep 10 at 11:57
@PedroTamaroff, perhaps you are correct, but as I see it, the algebra and geometry flows together beautifully until you ask that the prime $3$ be inverted. The exceptional isomorphisms lead to the $G_2leq Spin_7$ subgroup and geometric observations lead to the fibration sequence. But why has fate conspired to make $pi_6G_2congmathbbZ_3$? For example it is also true that $Spin_8simeq S^7times Spin_7$ (integrally), since $pi_6Spin_7=0$, however this homotopy equivalence - derived by the same reasoning - is actually a homeomorphism.
– Tyrone
Sep 10 at 13:45
@PedroTamaroff, perhaps you are correct, but as I see it, the algebra and geometry flows together beautifully until you ask that the prime $3$ be inverted. The exceptional isomorphisms lead to the $G_2leq Spin_7$ subgroup and geometric observations lead to the fibration sequence. But why has fate conspired to make $pi_6G_2congmathbbZ_3$? For example it is also true that $Spin_8simeq S^7times Spin_7$ (integrally), since $pi_6Spin_7=0$, however this homotopy equivalence - derived by the same reasoning - is actually a homeomorphism.
– Tyrone
Sep 10 at 13:45
add a comment |Â
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I am confused: you just gave an explanation, which is both algebraic and geometric.
– Pedro Tamaroff♦
Sep 10 at 11:57
@PedroTamaroff, perhaps you are correct, but as I see it, the algebra and geometry flows together beautifully until you ask that the prime $3$ be inverted. The exceptional isomorphisms lead to the $G_2leq Spin_7$ subgroup and geometric observations lead to the fibration sequence. But why has fate conspired to make $pi_6G_2congmathbbZ_3$? For example it is also true that $Spin_8simeq S^7times Spin_7$ (integrally), since $pi_6Spin_7=0$, however this homotopy equivalence - derived by the same reasoning - is actually a homeomorphism.
– Tyrone
Sep 10 at 13:45