BO(-) example in Weiss Calculus

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I'nm reading Orthogonal Calculus by Michael Weiss, and trying to understand example 2.7, concerning the derivatives of the functor $BO$, which sends a (finite dimensional) inner product space to the classifying space of its orthogonal group.

My question concerns the homotopy equivalence between the homotopy fiber of the map
$$
BO(V) to BO(Voplus BbbR)
$$
and $O(BbbR oplus V)/ O(V)$, and the subsequent homotopy equivalence of this, and the one-point compactificiation of $V$.

By guess for the first one is to use the fact that $BO(V) = Gr(dim V, BbbR^infty)$, where $Gr$ denotes the Grassmannian manifold, and that there is fibre sequence
$$
O(V) to Vr(dim V, BbbR^infty) to Gr(dim V+1, BbbR^infty)
$$
where $Vr$ denotes the Steifel manifold (which is contractible).

Although, I'm not overly sure how this helps.

I'm clueless on the second on currently.

Any references or suggestions would be greatly appreciated.

group-theory algebraic-topology grassmannian classifying-spaces stiefel-manifolds

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edited Jul 26 at 21:48

asked Jul 26 at 19:57

AlgTop

1113

add a comment | 

up vote
2
down vote

favorite

I'nm reading Orthogonal Calculus by Michael Weiss, and trying to understand example 2.7, concerning the derivatives of the functor $BO$, which sends a (finite dimensional) inner product space to the classifying space of its orthogonal group.

My question concerns the homotopy equivalence between the homotopy fiber of the map
$$
BO(V) to BO(Voplus BbbR)
$$
and $O(BbbR oplus V)/ O(V)$, and the subsequent homotopy equivalence of this, and the one-point compactificiation of $V$.

By guess for the first one is to use the fact that $BO(V) = Gr(dim V, BbbR^infty)$, where $Gr$ denotes the Grassmannian manifold, and that there is fibre sequence
$$
O(V) to Vr(dim V, BbbR^infty) to Gr(dim V+1, BbbR^infty)
$$
where $Vr$ denotes the Steifel manifold (which is contractible).

Although, I'm not overly sure how this helps.

I'm clueless on the second on currently.

Any references or suggestions would be greatly appreciated.

group-theory algebraic-topology grassmannian classifying-spaces stiefel-manifolds

share|cite|improve this question

edited Jul 26 at 21:48

asked Jul 26 at 19:57

AlgTop

1113

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I'nm reading Orthogonal Calculus by Michael Weiss, and trying to understand example 2.7, concerning the derivatives of the functor $BO$, which sends a (finite dimensional) inner product space to the classifying space of its orthogonal group.

My question concerns the homotopy equivalence between the homotopy fiber of the map
$$
BO(V) to BO(Voplus BbbR)
$$
and $O(BbbR oplus V)/ O(V)$, and the subsequent homotopy equivalence of this, and the one-point compactificiation of $V$.

By guess for the first one is to use the fact that $BO(V) = Gr(dim V, BbbR^infty)$, where $Gr$ denotes the Grassmannian manifold, and that there is fibre sequence
$$
O(V) to Vr(dim V, BbbR^infty) to Gr(dim V+1, BbbR^infty)
$$
where $Vr$ denotes the Steifel manifold (which is contractible).

Although, I'm not overly sure how this helps.

I'm clueless on the second on currently.

Any references or suggestions would be greatly appreciated.

group-theory algebraic-topology grassmannian classifying-spaces stiefel-manifolds

share|cite|improve this question

edited Jul 26 at 21:48

asked Jul 26 at 19:57

AlgTop

1113

I'nm reading Orthogonal Calculus by Michael Weiss, and trying to understand example 2.7, concerning the derivatives of the functor $BO$, which sends a (finite dimensional) inner product space to the classifying space of its orthogonal group.

My question concerns the homotopy equivalence between the homotopy fiber of the map
$$
BO(V) to BO(Voplus BbbR)
$$
and $O(BbbR oplus V)/ O(V)$, and the subsequent homotopy equivalence of this, and the one-point compactificiation of $V$.

By guess for the first one is to use the fact that $BO(V) = Gr(dim V, BbbR^infty)$, where $Gr$ denotes the Grassmannian manifold, and that there is fibre sequence
$$
O(V) to Vr(dim V, BbbR^infty) to Gr(dim V+1, BbbR^infty)
$$
where $Vr$ denotes the Steifel manifold (which is contractible).

Although, I'm not overly sure how this helps.

I'm clueless on the second on currently.

Any references or suggestions would be greatly appreciated.

group-theory algebraic-topology grassmannian classifying-spaces stiefel-manifolds

share|cite|improve this question

edited Jul 26 at 21:48

asked Jul 26 at 19:57

AlgTop

1113

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 21:48

edited Jul 26 at 21:48

edited Jul 26 at 21:48

asked Jul 26 at 19:57

AlgTop

1113

asked Jul 26 at 19:57

AlgTop

1113

asked Jul 26 at 19:57

AlgTop

1113

1113

add a comment | 

add a comment | 

1 Answer
1

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I haven't looked at the paper in a while, but I think you want

$$F^(1)(V)simeq hofibleft(BO(V)rightarrow BO(VoplusmathbbR)right)$$

without the loop functor, rather than with it, since we want $m=0$ in Proposition 2.2. (Here I'm using Weiss's notation of $F(V)=BO(V)$ from the Example 2.7.)

Then the map in the definition of $F^(1)$ is just the map induced by the incusion $O(V)hookrightarrow O(VoplusmathbbR)$ so clearly

$$F^(1)(V)simeq O(VoplusmathbbR)/O(V).$$

Now by definition $O(VoplusmathbbR)$ acts transitively on the unit sphere $S(VoplusmathbbR)$ of length one vectors in $VoplusmathbbR$, and the stabiliser of the line $mathbbR$ is exactly $O(V)$. Hence $O(VoplusmathbbR)/O(V)cong S(VoplusmathbbR)$. Finally applying stereographic projection we get $S(VoplusmathbbR)cong S^V$ (see, for instance, the nlab page https://ncatlab.org/nlab/show/representation+sphere). Therefore

$$F^(1)(V)simeq S^V$$

as claimed.

share|cite|improve this answer

answered Jul 26 at 20:51

Tyrone

3,23611025

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
  – AlgTop
  Jul 26 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
  – Tyrone
  Jul 27 at 9:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
  – AlgTop
  Jul 27 at 18:49
  

  
  
  
  

add a comment | 

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1 Answer
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active

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votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

I haven't looked at the paper in a while, but I think you want

$$F^(1)(V)simeq hofibleft(BO(V)rightarrow BO(VoplusmathbbR)right)$$

without the loop functor, rather than with it, since we want $m=0$ in Proposition 2.2. (Here I'm using Weiss's notation of $F(V)=BO(V)$ from the Example 2.7.)

Then the map in the definition of $F^(1)$ is just the map induced by the incusion $O(V)hookrightarrow O(VoplusmathbbR)$ so clearly

$$F^(1)(V)simeq O(VoplusmathbbR)/O(V).$$

Now by definition $O(VoplusmathbbR)$ acts transitively on the unit sphere $S(VoplusmathbbR)$ of length one vectors in $VoplusmathbbR$, and the stabiliser of the line $mathbbR$ is exactly $O(V)$. Hence $O(VoplusmathbbR)/O(V)cong S(VoplusmathbbR)$. Finally applying stereographic projection we get $S(VoplusmathbbR)cong S^V$ (see, for instance, the nlab page https://ncatlab.org/nlab/show/representation+sphere). Therefore

$$F^(1)(V)simeq S^V$$

as claimed.

share|cite|improve this answer

answered Jul 26 at 20:51

Tyrone

3,23611025

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
  – AlgTop
  Jul 26 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
  – Tyrone
  Jul 27 at 9:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
  – AlgTop
  Jul 27 at 18:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

I haven't looked at the paper in a while, but I think you want

$$F^(1)(V)simeq hofibleft(BO(V)rightarrow BO(VoplusmathbbR)right)$$

without the loop functor, rather than with it, since we want $m=0$ in Proposition 2.2. (Here I'm using Weiss's notation of $F(V)=BO(V)$ from the Example 2.7.)

Then the map in the definition of $F^(1)$ is just the map induced by the incusion $O(V)hookrightarrow O(VoplusmathbbR)$ so clearly

$$F^(1)(V)simeq O(VoplusmathbbR)/O(V).$$

Now by definition $O(VoplusmathbbR)$ acts transitively on the unit sphere $S(VoplusmathbbR)$ of length one vectors in $VoplusmathbbR$, and the stabiliser of the line $mathbbR$ is exactly $O(V)$. Hence $O(VoplusmathbbR)/O(V)cong S(VoplusmathbbR)$. Finally applying stereographic projection we get $S(VoplusmathbbR)cong S^V$ (see, for instance, the nlab page https://ncatlab.org/nlab/show/representation+sphere). Therefore

$$F^(1)(V)simeq S^V$$

as claimed.

share|cite|improve this answer

answered Jul 26 at 20:51

Tyrone

3,23611025

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
  – AlgTop
  Jul 26 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
  – Tyrone
  Jul 27 at 9:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
  – AlgTop
  Jul 27 at 18:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

I haven't looked at the paper in a while, but I think you want

$$F^(1)(V)simeq hofibleft(BO(V)rightarrow BO(VoplusmathbbR)right)$$

without the loop functor, rather than with it, since we want $m=0$ in Proposition 2.2. (Here I'm using Weiss's notation of $F(V)=BO(V)$ from the Example 2.7.)

Then the map in the definition of $F^(1)$ is just the map induced by the incusion $O(V)hookrightarrow O(VoplusmathbbR)$ so clearly

$$F^(1)(V)simeq O(VoplusmathbbR)/O(V).$$

Now by definition $O(VoplusmathbbR)$ acts transitively on the unit sphere $S(VoplusmathbbR)$ of length one vectors in $VoplusmathbbR$, and the stabiliser of the line $mathbbR$ is exactly $O(V)$. Hence $O(VoplusmathbbR)/O(V)cong S(VoplusmathbbR)$. Finally applying stereographic projection we get $S(VoplusmathbbR)cong S^V$ (see, for instance, the nlab page https://ncatlab.org/nlab/show/representation+sphere). Therefore

$$F^(1)(V)simeq S^V$$

as claimed.

share|cite|improve this answer

answered Jul 26 at 20:51

Tyrone

3,23611025

I haven't looked at the paper in a while, but I think you want

$$F^(1)(V)simeq hofibleft(BO(V)rightarrow BO(VoplusmathbbR)right)$$

without the loop functor, rather than with it, since we want $m=0$ in Proposition 2.2. (Here I'm using Weiss's notation of $F(V)=BO(V)$ from the Example 2.7.)

Then the map in the definition of $F^(1)$ is just the map induced by the incusion $O(V)hookrightarrow O(VoplusmathbbR)$ so clearly

$$F^(1)(V)simeq O(VoplusmathbbR)/O(V).$$

Now by definition $O(VoplusmathbbR)$ acts transitively on the unit sphere $S(VoplusmathbbR)$ of length one vectors in $VoplusmathbbR$, and the stabiliser of the line $mathbbR$ is exactly $O(V)$. Hence $O(VoplusmathbbR)/O(V)cong S(VoplusmathbbR)$. Finally applying stereographic projection we get $S(VoplusmathbbR)cong S^V$ (see, for instance, the nlab page https://ncatlab.org/nlab/show/representation+sphere). Therefore

$$F^(1)(V)simeq S^V$$

as claimed.

share|cite|improve this answer

answered Jul 26 at 20:51

Tyrone

3,23611025

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 20:51

Tyrone

3,23611025

answered Jul 26 at 20:51

Tyrone

3,23611025

answered Jul 26 at 20:51

Tyrone

3,23611025

3,23611025

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
  – AlgTop
  Jul 26 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
  – Tyrone
  Jul 27 at 9:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
  – AlgTop
  Jul 27 at 18:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
  – AlgTop
  Jul 26 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
  – Tyrone
  Jul 27 at 9:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
  – AlgTop
  Jul 27 at 18:49
  

  
  
  
  

Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
– AlgTop
Jul 26 at 21:48

Why is it clear that the homotopy fiber of the map induced by inclusion is $O(V oplus BbbR)/O(V)$?
– AlgTop
Jul 26 at 21:48

1

1

Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
– Tyrone
Jul 27 at 9:27

Since we have a fibration sequence $O(V)hookrightarrow O(VoplusmathbbR)rightarrow O(Voplus mathbbR)/O(V)$, and the map $BO(V)rightarrow BO(VoplusmathbbR)$, of which $F^(1)$ is the homotopy fibre, is exactly a delooping of the first map in this sequence.
– Tyrone
Jul 27 at 9:27

Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
– AlgTop
Jul 27 at 18:49

Thanks for the help! Do you happen to know of any places where I can read about homotopy fibers of this kind?
– AlgTop
Jul 27 at 18:49

add a comment | 

 

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