Return to Answer

I like to play around with projective spaces for these type of questions. The tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle would split, let's say $T\mathbb{RP}^4=E\oplus F$, we would have for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side has terms maximally of degree $a^4$. The Stiefel Whitney $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]/a^5$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

Upd

I like to play around with projective spaces for these type of questions. The tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle splitted, let's say $T\mathbb{RP}^4=E\oplus F$, we would find for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side maximally has terms of the form $a^4$. The Stiefel Whitney class $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]/a^5$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

I like to play around with projective spaces for these type of questions. The the tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle would split, let's say $T\mathbb{RP}^4=E\oplus F$, we would have for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side has terms maximally of degree $a^4$. The Stiefel Whitney $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]/a^5$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

Upd

I like to play around with projective spaces for these type of questions. The tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle would split, let's say $T\mathbb{RP}^4=E\oplus F$, we would have for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side has terms maximally of degree $a^4$. The Stiefel Whitney $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]/a^5$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

Upd

I like to play around with projective spaces for these type of questions. The the tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle would split, let's say $T\mathbb{RP}^4=E\oplus F$, we would have for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side has terms maximally of degree $a^4$. The Stiefel Whitney $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

I like to play around with projective spaces for these type of questions. The the tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have stiefel whitney class $1+a+a^4$. This polynomial is irreducible hence the tangent bundle does not split.

EDIT:

The comment below made me realise I was not clear. If the tangent bundle would split, let's say $T\mathbb{RP}^4=E\oplus F$, we would have for the stiefel Whitney class that $$ sw(T\mathbb{RP}^4)=sw(E)sw(F) $$ But $\dim E+ \dim F=4$ and the right hand side has terms maximally of degree $a^4$. The Stiefel Whitney $1+a+a^4$ would factor then also in $\mathbb{Z}_2[a]$ (not only in $\mathbb{Z}_2[a]/a^5$). It is true that $1+a+a^4$ is reducible in $\mathbb{Z}_2[a]/a^5$, and in fact stably the tangent bundle does split as $T\mathbb{RP}^4\oplus \mathbb{R}\cong\tau^5$, where $\tau$ is the dual of the tautological bundle.

Upd