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Does there exist a manifold M which all iterated tangent bundles are non parallelizable manifolds? That is$ M, TM , T^2(M), \ldots ,T^n(M)\ldots$ are non parallelizable manifold?

What is an example of a manifold which is not parallelizable, but $T^{n}(M)$ is parallelizable for some n?

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    $\begingroup$ This is equivalent to the question "Does parallelizability of $TM$ implies parallelizability of $M$?" applied to $T^{n-1}M$. $\endgroup$
    – Michael
    Commented Dec 12, 2013 at 19:26

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If $\xi$ is a vector bundle over a manifold $B$ with total space $E(\xi)$, then the restriction of $TE(\xi)$ to the the zero section of $\xi$ is isomorphic to $\xi\oplus TB$. It follows that $T^k M$ restricted to $T^{k-1}M$ is isomorphic to $T^{k-1}M\oplus T^{k-1}M$, and iterating we conclude that $T^k M$ restricted to $M$ is the sum of $2^{k-1}$ copies of $TM$. Thus $TM$ is non-torsion in the group $(KO(M), \oplus)$ if and only if $T^kM$ is nontrivial for all $k$. The simplest example of a manifold with non-torsion tangent bundle is $M=CP^2$.

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  • $\begingroup$ Igor, could you explain something in your answer? First, why is it valid to restrict $TE(\xi)$ to the 0-section of $\xi$ for the purpose of answering the question, i.e. how one would show that $TE(\xi)$ parallelizable iff the 0-section is? The reason I am asking is that it's possible apriori that parallelizability is not stable property relative to direct sum with a trivial bundle. $\endgroup$
    – Michael
    Commented Dec 12, 2013 at 22:51
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    $\begingroup$ @Michael, I never claimed that $TE(\xi)$ is parallelizable iff its $0$-section is (this is false, think of $S^2$). The point is that a rank $k$ vector bundle $\xi$ over $B$ is a homotopy class of maps $B\to BO(k)$, so if $B$ deformation retracts to $S$, then $\xi$ is completely determined (up to isomorphism) by its restriction to $S$. $\endgroup$
    – Igor Belegradek
    Commented Dec 12, 2013 at 23:01
  • $\begingroup$ I see. In particular, Stiefel-Whitney of $TTM$ would be 0, therefore $TTM$ is always orientable. Could you remind me the criterion of parallelizability? $\endgroup$
    – Michael
    Commented Dec 12, 2013 at 23:21
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    $\begingroup$ @Ali: yes, $TS^2$ is parallelizable. In general, an open manifold is parallelizable iff it is stably parallelizable. As I explained $T(TS^2)$ is $2TS^2$ in $K$-theory, but $TS^2$ is stably trivial, and hence so is $2TS^2$. Hence $S^2$ answers your second question. $\endgroup$
    – Igor Belegradek
    Commented Dec 12, 2013 at 23:39
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    $\begingroup$ Ali: because an open manifold is homotopy equivalent to a complex of lower dimension. Michael: there is no criterion of parallelizability in terms of characteristic classes. Ali: $CP^2$ has nonzero first Pontryagin class. Actually, what you ask are good exercises. You should do them yourself. $\endgroup$
    – Igor Belegradek
    Commented Dec 13, 2013 at 0:03

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