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wonderich
wonderich
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$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible for $n>1$?

Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $m\alpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $\beta$ in $\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

p.s. I am asking $n>1$ so if you vote down for $n=1$, you may reconsider your vote... Lol

$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $m\alpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $\beta$ in $\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible for $n>1$?

Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $m\alpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $\beta$ in $\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

p.s. I am asking $n>1$ so if you vote down for $n=1$, you may reconsider your vote... Lol

Not torsion and indivisible -> indivisible and not torsion (to clarify scope of 'not'); link to referenced book
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LSpice
LSpice
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$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

Question: Is the element $\alpha$ in $\pi_{2n-1}(SO(2n))$$\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion and indivisible?

My understanding so far ---—

An $SO(2n)$$\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}BSO(2n) =\pi_{2n-1}SO(2n)$$\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$.

Not torsion: It There does not exist any integer $m$$m > 0$ such that $m \cdot \alpha$$m\alpha$ is a trivial element. The $\alpha$ is the element in $\pi_{2n-1}SO(2n)$ representing the tangent bundle $TS^{2n}$.

Indivisible: It means that itThere does not exist any integer $k$$k > 1$ and any element $\beta$ in $\pi_{2n-1}SO(2n)$$\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groupsTopology of Lie groups.

Chapter Chapter IV Corollary 6.14.

$\pi_{2n-1}(SO(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

Question: Is the element in $\pi_{2n-1}(SO(2n))$ representing tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ not torsion and indivisible?

My understanding so far ---

An $SO(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}BSO(2n) =\pi_{2n-1}SO(2n)$.

Not torsion: It does not exist any integer $m$ such that $m \cdot \alpha$ is a trivial element. The $\alpha$ is the element in $\pi_{2n-1}SO(2n)$ representing the tangent bundle $TS^{2n}$.

Indivisible: It means that it does not exist any integer $k$ and any element $\beta$ in $\pi_{2n-1}SO(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups.

Chapter IV Corollary 6.14.

$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $m\alpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $\beta$ in $\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

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wonderich
wonderich
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$\pi_{2n-1}(SO(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

Question: Is the element in $\pi_{2n-1}(SO(2n))$ representing tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ not torsion and indivisible?

My understanding so far ---

An $SO(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}BSO(2n) =\pi_{2n-1}SO(2n)$.

Not torsion: It does not exist any integer $m$ such that $m \cdot \alpha$ is a trivial element. The $\alpha$ is the element in $\pi_{2n-1}SO(2n)$ representing the tangent bundle $TS^{2n}$.

Indivisible: It means that it does not exist any integer $k$ and any element $\beta$ in $\pi_{2n-1}SO(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups.

Chapter IV Corollary 6.14.