Fourth obstruction, Pontryagin and Euler class

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Assume the first three obstruction classes of a rank 4 vector bundle vanish and look at the fourth obstruction class. This fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (since $pi_3(SO_4) simeq mathbbZ oplus mathbbZ$). Is there a geometric description of a system of generators in $pi_3(SO_4)$ which is associated to these classes?

Recall that $SO_4$ is double covered by $SU_2 times SU_2$ and since
$SU_2 cong S_3$, $π_3(SO_4)=pi_3(S_3) oplus pi_3(S_3)= mathbbZ oplus mathbbZ$. The question is: how do the Euler and Pontryagin classes relate to this double cover? In other words, what is the system of generator $langle alpha, beta rangle$ of $mathbbZ oplus mathbbZ$ so that given an element, if one writes it down as $aalpha+bbeta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class

dg.differential-geometry at.algebraic-topology vector-bundles characteristic-classes

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asked Aug 9 at 14:42

ARG

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up vote
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Assume the first three obstruction classes of a rank 4 vector bundle vanish and look at the fourth obstruction class. This fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (since $pi_3(SO_4) simeq mathbbZ oplus mathbbZ$). Is there a geometric description of a system of generators in $pi_3(SO_4)$ which is associated to these classes?

Recall that $SO_4$ is double covered by $SU_2 times SU_2$ and since
$SU_2 cong S_3$, $π_3(SO_4)=pi_3(S_3) oplus pi_3(S_3)= mathbbZ oplus mathbbZ$. The question is: how do the Euler and Pontryagin classes relate to this double cover? In other words, what is the system of generator $langle alpha, beta rangle$ of $mathbbZ oplus mathbbZ$ so that given an element, if one writes it down as $aalpha+bbeta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class

dg.differential-geometry at.algebraic-topology vector-bundles characteristic-classes

share|cite|improve this question

asked Aug 9 at 14:42

ARG

1,4561030

add a comment | 

up vote
9
down vote

favorite

1

up vote
9
down vote

favorite

1

1

Assume the first three obstruction classes of a rank 4 vector bundle vanish and look at the fourth obstruction class. This fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (since $pi_3(SO_4) simeq mathbbZ oplus mathbbZ$). Is there a geometric description of a system of generators in $pi_3(SO_4)$ which is associated to these classes?

Recall that $SO_4$ is double covered by $SU_2 times SU_2$ and since
$SU_2 cong S_3$, $π_3(SO_4)=pi_3(S_3) oplus pi_3(S_3)= mathbbZ oplus mathbbZ$. The question is: how do the Euler and Pontryagin classes relate to this double cover? In other words, what is the system of generator $langle alpha, beta rangle$ of $mathbbZ oplus mathbbZ$ so that given an element, if one writes it down as $aalpha+bbeta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class

dg.differential-geometry at.algebraic-topology vector-bundles characteristic-classes

share|cite|improve this question

asked Aug 9 at 14:42

ARG

1,4561030

Assume the first three obstruction classes of a rank 4 vector bundle vanish and look at the fourth obstruction class. This fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (since $pi_3(SO_4) simeq mathbbZ oplus mathbbZ$). Is there a geometric description of a system of generators in $pi_3(SO_4)$ which is associated to these classes?

Recall that $SO_4$ is double covered by $SU_2 times SU_2$ and since
$SU_2 cong S_3$, $π_3(SO_4)=pi_3(S_3) oplus pi_3(S_3)= mathbbZ oplus mathbbZ$. The question is: how do the Euler and Pontryagin classes relate to this double cover? In other words, what is the system of generator $langle alpha, beta rangle$ of $mathbbZ oplus mathbbZ$ so that given an element, if one writes it down as $aalpha+bbeta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class

dg.differential-geometry at.algebraic-topology vector-bundles characteristic-classes

share|cite|improve this question

asked Aug 9 at 14:42

ARG

1,4561030

share|cite|improve this question

share|cite|improve this question

asked Aug 9 at 14:42

ARG

1,4561030

asked Aug 9 at 14:42

ARG

1,4561030

asked Aug 9 at 14:42

ARG

1,4561030

1,4561030

add a comment | 

add a comment | 

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Geometric generators for $pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $pi_3(SO(4))$ denoted by $alpha_3$ and left multiplication induces an element denoted by $beta_3$. These generate $pi_3(SO(4))congmathbbZoplusmathbbZ$.

The relation between obstruction classes and characteristic classes is discussed in

• A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677.

I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $beta_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d_1-2d_2$ where $d_1$ is the part of the obstruction class corresponding to $alpha_3$ and $d_2$ is the part of the obstruction class corresponding to $beta_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $pi_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2).

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answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
  – ARG
  Aug 9 at 18:37
  
  

  
  
  
  

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up vote
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accepted

Geometric generators for $pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $pi_3(SO(4))$ denoted by $alpha_3$ and left multiplication induces an element denoted by $beta_3$. These generate $pi_3(SO(4))congmathbbZoplusmathbbZ$.

The relation between obstruction classes and characteristic classes is discussed in

• A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677.

I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $beta_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d_1-2d_2$ where $d_1$ is the part of the obstruction class corresponding to $alpha_3$ and $d_2$ is the part of the obstruction class corresponding to $beta_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $pi_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2).

share|cite|improve this answer

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
  – ARG
  Aug 9 at 18:37
  
  

  
  
  
  

add a comment | 

up vote
12
down vote



accepted

Geometric generators for $pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $pi_3(SO(4))$ denoted by $alpha_3$ and left multiplication induces an element denoted by $beta_3$. These generate $pi_3(SO(4))congmathbbZoplusmathbbZ$.

The relation between obstruction classes and characteristic classes is discussed in

• A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677.

I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $beta_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d_1-2d_2$ where $d_1$ is the part of the obstruction class corresponding to $alpha_3$ and $d_2$ is the part of the obstruction class corresponding to $beta_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $pi_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2).

share|cite|improve this answer

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
  – ARG
  Aug 9 at 18:37
  
  

  
  
  
  

add a comment | 

up vote
12
down vote



accepted

up vote
12
down vote



accepted

Geometric generators for $pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $pi_3(SO(4))$ denoted by $alpha_3$ and left multiplication induces an element denoted by $beta_3$. These generate $pi_3(SO(4))congmathbbZoplusmathbbZ$.

The relation between obstruction classes and characteristic classes is discussed in

• A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677.

I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $beta_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d_1-2d_2$ where $d_1$ is the part of the obstruction class corresponding to $alpha_3$ and $d_2$ is the part of the obstruction class corresponding to $beta_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $pi_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2).

share|cite|improve this answer

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

Geometric generators for $pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $pi_3(SO(4))$ denoted by $alpha_3$ and left multiplication induces an element denoted by $beta_3$. These generate $pi_3(SO(4))congmathbbZoplusmathbbZ$.

The relation between obstruction classes and characteristic classes is discussed in

• A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677.

I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $beta_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d_1-2d_2$ where $d_1$ is the part of the obstruction class corresponding to $alpha_3$ and $d_2$ is the part of the obstruction class corresponding to $beta_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $pi_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2).

share|cite|improve this answer

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

share|cite|improve this answer

share|cite|improve this answer

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

answered Aug 9 at 15:46

Matthias Wendt

12.9k24387

12.9k24387

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
  – ARG
  Aug 9 at 18:37
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
  – ARG
  Aug 9 at 18:37
  
  

  
  
  
  

2

2

How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
– ARG
Aug 9 at 18:37

How nice! and thanks for correcting my misunderstanding of the obstruction class... It was based on the idea the values of both the Euler and Pontryagin classes can be identifies with $mathbbZ$ and that there is no other class in this dimension. So I (wrongly, hastily and naively) assumed there could be no other way to express the obstruction class. Should I edit the question?
– ARG
Aug 9 at 18:37

add a comment | 

 

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