Decomposition of odd-dimensional oriented vector bundles

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As it is well known, every oriented vector bundle $E$ of rank $2k+1$ over a manifold $M$ of dimension $2k+1$ has a nowhere vanishing section. Thus we may decompose $E=Foplusvarepsilon^1$.



Is $F$ uniquely determined (up to isomorphism)? This question can be rephrased to homotopy theory: if $M to BSO(2k+1)$ is the classifying map of $E$, how many lifts of this map to $M to BSO(2k)$ exists?




algebraic-topology homotopy-theory




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    As it is well known, every oriented vector bundle $E$ of rank $2k+1$ over a manifold $M$ of dimension $2k+1$ has a nowhere vanishing section. Thus we may decompose $E=Foplusvarepsilon^1$.



    Is $F$ uniquely determined (up to isomorphism)? This question can be rephrased to homotopy theory: if $M to BSO(2k+1)$ is the classifying map of $E$, how many lifts of this map to $M to BSO(2k)$ exists?




    algebraic-topology homotopy-theory




    share|cite|improve this question





















      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      As it is well known, every oriented vector bundle $E$ of rank $2k+1$ over a manifold $M$ of dimension $2k+1$ has a nowhere vanishing section. Thus we may decompose $E=Foplusvarepsilon^1$.



      Is $F$ uniquely determined (up to isomorphism)? This question can be rephrased to homotopy theory: if $M to BSO(2k+1)$ is the classifying map of $E$, how many lifts of this map to $M to BSO(2k)$ exists?




      algebraic-topology homotopy-theory




      share|cite|improve this question











      As it is well known, every oriented vector bundle $E$ of rank $2k+1$ over a manifold $M$ of dimension $2k+1$ has a nowhere vanishing section. Thus we may decompose $E=Foplusvarepsilon^1$.



      Is $F$ uniquely determined (up to isomorphism)? This question can be rephrased to homotopy theory: if $M to BSO(2k+1)$ is the classifying map of $E$, how many lifts of this map to $M to BSO(2k)$ exists?





      algebraic-topology homotopy-theory





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      share|cite|improve this question




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      asked Aug 1 at 19:58









      Wilhelm L.

      1655




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          Not necessarily. For example, consider the space $S^2times S^1$ and the vector bundle $T(S^2times S^1)$. We have $T(S^2times S^1) cong pi_1^*TS^2opluspi_2^*TS^1 cong pi_1^*TS^2oplusvarepsilon^1$ where $pi_i$ are the natural projections. On the other hand, we have $T(S^2times S^1) cong varepsilon^3 cong varepsilon^2oplusvarepsilon^1$. Note that $pi_1^*TS^2 notcong varepsilon^2$ as the former has non-zero Euler class.






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            Not necessarily. For example, consider the space $S^2times S^1$ and the vector bundle $T(S^2times S^1)$. We have $T(S^2times S^1) cong pi_1^*TS^2opluspi_2^*TS^1 cong pi_1^*TS^2oplusvarepsilon^1$ where $pi_i$ are the natural projections. On the other hand, we have $T(S^2times S^1) cong varepsilon^3 cong varepsilon^2oplusvarepsilon^1$. Note that $pi_1^*TS^2 notcong varepsilon^2$ as the former has non-zero Euler class.






            share|cite|improve this answer

























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              Not necessarily. For example, consider the space $S^2times S^1$ and the vector bundle $T(S^2times S^1)$. We have $T(S^2times S^1) cong pi_1^*TS^2opluspi_2^*TS^1 cong pi_1^*TS^2oplusvarepsilon^1$ where $pi_i$ are the natural projections. On the other hand, we have $T(S^2times S^1) cong varepsilon^3 cong varepsilon^2oplusvarepsilon^1$. Note that $pi_1^*TS^2 notcong varepsilon^2$ as the former has non-zero Euler class.






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                up vote
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                up vote
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                Not necessarily. For example, consider the space $S^2times S^1$ and the vector bundle $T(S^2times S^1)$. We have $T(S^2times S^1) cong pi_1^*TS^2opluspi_2^*TS^1 cong pi_1^*TS^2oplusvarepsilon^1$ where $pi_i$ are the natural projections. On the other hand, we have $T(S^2times S^1) cong varepsilon^3 cong varepsilon^2oplusvarepsilon^1$. Note that $pi_1^*TS^2 notcong varepsilon^2$ as the former has non-zero Euler class.






                share|cite|improve this answer













                Not necessarily. For example, consider the space $S^2times S^1$ and the vector bundle $T(S^2times S^1)$. We have $T(S^2times S^1) cong pi_1^*TS^2opluspi_2^*TS^1 cong pi_1^*TS^2oplusvarepsilon^1$ where $pi_i$ are the natural projections. On the other hand, we have $T(S^2times S^1) cong varepsilon^3 cong varepsilon^2oplusvarepsilon^1$. Note that $pi_1^*TS^2 notcong varepsilon^2$ as the former has non-zero Euler class.







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                answered Aug 1 at 22:22









                Michael Albanese

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