Mixing
Homotopy fiber of $BSO(n) to BSO$
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By definition there is an inclusion of $BSO(n)$ into $BSO$. I am asking for the homotopy fiber of this inclusion. Actually I would like to know what is the first non trivial homotopy group of that fiber.
algebraic-topology homotopy-theory
asked Jul 23 at 7:26
Wilhelm L.
30515
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up vote
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By definition there is an inclusion of $BSO(n)$ into $BSO$. I am asking for the homotopy fiber of this inclusion. Actually I would like to know what is the first non trivial homotopy group of that fiber.
algebraic-topology homotopy-theory
asked Jul 23 at 7:26
Wilhelm L.
30515
mathoverflow.net/a/296093/40804
– Mike Miller
Jul 23 at 8:09
sadly, i don't understand the connection to my question.
– Wilhelm L.
Jul 23 at 8:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
By definition there is an inclusion of $BSO(n)$ into $BSO$. I am asking for the homotopy fiber of this inclusion. Actually I would like to know what is the first non trivial homotopy group of that fiber.
algebraic-topology homotopy-theory
asked Jul 23 at 7:26
Wilhelm L.
30515
By definition there is an inclusion of $BSO(n)$ into $BSO$. I am asking for the homotopy fiber of this inclusion. Actually I would like to know what is the first non trivial homotopy group of that fiber.
algebraic-topology homotopy-theory
asked Jul 23 at 7:26
Wilhelm L.
30515
asked Jul 23 at 7:26
Wilhelm L.
30515
asked Jul 23 at 7:26
Wilhelm L.
30515
asked Jul 23 at 7:26
Wilhelm L.
30515
30515
mathoverflow.net/a/296093/40804
– Mike Miller
Jul 23 at 8:09
sadly, i don't understand the connection to my question.
– Wilhelm L.
Jul 23 at 8:44
add a comment |Â
mathoverflow.net/a/296093/40804
– Mike Miller
Jul 23 at 8:09
sadly, i don't understand the connection to my question.
– Wilhelm L.
Jul 23 at 8:44
mathoverflow.net/a/296093/40804
– Mike Miller
Jul 23 at 8:09
mathoverflow.net/a/296093/40804
– Mike Miller
Jul 23 at 8:09
sadly, i don't understand the connection to my question.
– Wilhelm L.
Jul 23 at 8:44
sadly, i don't understand the connection to my question.
– Wilhelm L.
Jul 23 at 8:44
add a comment |Â
1 Answer
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2
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It is the infinite Stiefel manifold
$$SO/SO(n)=colim_k(SO(n+k)/SO(n))$$
and I'm afraid there isn't a whole lot more to say.
Now the good news is that for each $kgeq 0$ the canonical projection $p:SO(n+k+1)rightarrow S^n+k$ induces a map $SO(n+k+1)/SO(n)rightarrow S^n+k$ with fibre $SO(n+k)/SO(n)$ so that we have a fibration sequence
$$SO(n+k)/SO(n)rightarrow SO(n+k+1)/SO(n)rightarrow S^n+k.$$
Since $SO(n+1)/SO(n)cong S^n$ we infer that the homotopy groups of $SO/SO(n)$ are trivial below degree $n$, and in degree $n$ we have, from the above fibration, a surjection
$$rho:pi_n S^nrightarrow pi_n(SO(n+2)/SO(n))$$
where the right-hand group is already the stable group,
$$pi_n(SO(n+2)/SO(n))cong pi_n(SO/SO(n))$$
and these groups were calculated by G. Paechter, here. In fact this group is never trivial, so is in fact the first non-zero homotopy group of $SO/SO(n)$.
The kernel of $rho$ is generated by the image of the composite $pi_n+1S^n+1xrightarrowDelta pi_nSO(n+1)xrightarrowp pi_nS^n$ where the first map is the connecting homomorphism in the long exact sequence of $SO(n+1)rightarrow SO(n+1)$, and you might like to play around with this if you want to make an explicit calculation yourself.
Now the bigger picture is detailed in the link that Mike Miller has kindly provided in the comments details, where the mod 8 periodicity of the homotopy groups of $SO$ is discussed. The group you are interested in will depend on the mod $8$ value of $n$ (there are some hiccups for low values of $n$, caused by the exceptional isomorphisms).
answered Jul 23 at 11:11
Tyrone
3,23611025
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It is the infinite Stiefel manifold
$$SO/SO(n)=colim_k(SO(n+k)/SO(n))$$
and I'm afraid there isn't a whole lot more to say.
Now the good news is that for each $kgeq 0$ the canonical projection $p:SO(n+k+1)rightarrow S^n+k$ induces a map $SO(n+k+1)/SO(n)rightarrow S^n+k$ with fibre $SO(n+k)/SO(n)$ so that we have a fibration sequence
$$SO(n+k)/SO(n)rightarrow SO(n+k+1)/SO(n)rightarrow S^n+k.$$
Since $SO(n+1)/SO(n)cong S^n$ we infer that the homotopy groups of $SO/SO(n)$ are trivial below degree $n$, and in degree $n$ we have, from the above fibration, a surjection
$$rho:pi_n S^nrightarrow pi_n(SO(n+2)/SO(n))$$
where the right-hand group is already the stable group,
$$pi_n(SO(n+2)/SO(n))cong pi_n(SO/SO(n))$$
and these groups were calculated by G. Paechter, here. In fact this group is never trivial, so is in fact the first non-zero homotopy group of $SO/SO(n)$.
The kernel of $rho$ is generated by the image of the composite $pi_n+1S^n+1xrightarrowDelta pi_nSO(n+1)xrightarrowp pi_nS^n$ where the first map is the connecting homomorphism in the long exact sequence of $SO(n+1)rightarrow SO(n+1)$, and you might like to play around with this if you want to make an explicit calculation yourself.
Now the bigger picture is detailed in the link that Mike Miller has kindly provided in the comments details, where the mod 8 periodicity of the homotopy groups of $SO$ is discussed. The group you are interested in will depend on the mod $8$ value of $n$ (there are some hiccups for low values of $n$, caused by the exceptional isomorphisms).
answered Jul 23 at 11:11
Tyrone
3,23611025
add a comment |Â
up vote
2
down vote
accepted
It is the infinite Stiefel manifold
$$SO/SO(n)=colim_k(SO(n+k)/SO(n))$$
and I'm afraid there isn't a whole lot more to say.
Now the good news is that for each $kgeq 0$ the canonical projection $p:SO(n+k+1)rightarrow S^n+k$ induces a map $SO(n+k+1)/SO(n)rightarrow S^n+k$ with fibre $SO(n+k)/SO(n)$ so that we have a fibration sequence
$$SO(n+k)/SO(n)rightarrow SO(n+k+1)/SO(n)rightarrow S^n+k.$$
Since $SO(n+1)/SO(n)cong S^n$ we infer that the homotopy groups of $SO/SO(n)$ are trivial below degree $n$, and in degree $n$ we have, from the above fibration, a surjection
$$rho:pi_n S^nrightarrow pi_n(SO(n+2)/SO(n))$$
where the right-hand group is already the stable group,
$$pi_n(SO(n+2)/SO(n))cong pi_n(SO/SO(n))$$
and these groups were calculated by G. Paechter, here. In fact this group is never trivial, so is in fact the first non-zero homotopy group of $SO/SO(n)$.
The kernel of $rho$ is generated by the image of the composite $pi_n+1S^n+1xrightarrowDelta pi_nSO(n+1)xrightarrowp pi_nS^n$ where the first map is the connecting homomorphism in the long exact sequence of $SO(n+1)rightarrow SO(n+1)$, and you might like to play around with this if you want to make an explicit calculation yourself.
Now the bigger picture is detailed in the link that Mike Miller has kindly provided in the comments details, where the mod 8 periodicity of the homotopy groups of $SO$ is discussed. The group you are interested in will depend on the mod $8$ value of $n$ (there are some hiccups for low values of $n$, caused by the exceptional isomorphisms).
answered Jul 23 at 11:11
Tyrone
3,23611025
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It is the infinite Stiefel manifold
$$SO/SO(n)=colim_k(SO(n+k)/SO(n))$$
and I'm afraid there isn't a whole lot more to say.
Now the good news is that for each $kgeq 0$ the canonical projection $p:SO(n+k+1)rightarrow S^n+k$ induces a map $SO(n+k+1)/SO(n)rightarrow S^n+k$ with fibre $SO(n+k)/SO(n)$ so that we have a fibration sequence
$$SO(n+k)/SO(n)rightarrow SO(n+k+1)/SO(n)rightarrow S^n+k.$$
Since $SO(n+1)/SO(n)cong S^n$ we infer that the homotopy groups of $SO/SO(n)$ are trivial below degree $n$, and in degree $n$ we have, from the above fibration, a surjection
$$rho:pi_n S^nrightarrow pi_n(SO(n+2)/SO(n))$$
where the right-hand group is already the stable group,
$$pi_n(SO(n+2)/SO(n))cong pi_n(SO/SO(n))$$
and these groups were calculated by G. Paechter, here. In fact this group is never trivial, so is in fact the first non-zero homotopy group of $SO/SO(n)$.
The kernel of $rho$ is generated by the image of the composite $pi_n+1S^n+1xrightarrowDelta pi_nSO(n+1)xrightarrowp pi_nS^n$ where the first map is the connecting homomorphism in the long exact sequence of $SO(n+1)rightarrow SO(n+1)$, and you might like to play around with this if you want to make an explicit calculation yourself.
Now the bigger picture is detailed in the link that Mike Miller has kindly provided in the comments details, where the mod 8 periodicity of the homotopy groups of $SO$ is discussed. The group you are interested in will depend on the mod $8$ value of $n$ (there are some hiccups for low values of $n$, caused by the exceptional isomorphisms).
answered Jul 23 at 11:11
Tyrone
3,23611025
It is the infinite Stiefel manifold
$$SO/SO(n)=colim_k(SO(n+k)/SO(n))$$
and I'm afraid there isn't a whole lot more to say.
Now the good news is that for each $kgeq 0$ the canonical projection $p:SO(n+k+1)rightarrow S^n+k$ induces a map $SO(n+k+1)/SO(n)rightarrow S^n+k$ with fibre $SO(n+k)/SO(n)$ so that we have a fibration sequence
$$SO(n+k)/SO(n)rightarrow SO(n+k+1)/SO(n)rightarrow S^n+k.$$
Since $SO(n+1)/SO(n)cong S^n$ we infer that the homotopy groups of $SO/SO(n)$ are trivial below degree $n$, and in degree $n$ we have, from the above fibration, a surjection
$$rho:pi_n S^nrightarrow pi_n(SO(n+2)/SO(n))$$
where the right-hand group is already the stable group,
$$pi_n(SO(n+2)/SO(n))cong pi_n(SO/SO(n))$$
and these groups were calculated by G. Paechter, here. In fact this group is never trivial, so is in fact the first non-zero homotopy group of $SO/SO(n)$.
The kernel of $rho$ is generated by the image of the composite $pi_n+1S^n+1xrightarrowDelta pi_nSO(n+1)xrightarrowp pi_nS^n$ where the first map is the connecting homomorphism in the long exact sequence of $SO(n+1)rightarrow SO(n+1)$, and you might like to play around with this if you want to make an explicit calculation yourself.
Now the bigger picture is detailed in the link that Mike Miller has kindly provided in the comments details, where the mod 8 periodicity of the homotopy groups of $SO$ is discussed. The group you are interested in will depend on the mod $8$ value of $n$ (there are some hiccups for low values of $n$, caused by the exceptional isomorphisms).
answered Jul 23 at 11:11
Tyrone
3,23611025
answered Jul 23 at 11:11
Tyrone
3,23611025
answered Jul 23 at 11:11
Tyrone
3,23611025
answered Jul 23 at 11:11
Tyrone
3,23611025
3,23611025
add a comment |Â
add a comment |Â
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mathoverflow.net/a/296093/40804
– Mike Miller
Jul 23 at 8:09
sadly, i don't understand the connection to my question.
– Wilhelm L.
Jul 23 at 8:44