# Trivializations of Tangent Bundle over 1-skeleton that do not extend to 2-skeleton, etc. Examples

I am reading up on Stiefel-Whitney classes $w_k(T_M)$ on $m$-manifolds; $k=1,2,\dots,m$, which are described as obstacles to extending a trivialization of a bundle, from the $k$-skeleton (assume manifolds are nice-enough to admit a cell-decomposition); for $k=1$, it is just the obstruction to orientability (we just define $w_1$ over embedded circles $C$, and define $w_1(C)=0$ if $C$ preserves orientation, and $w_1(C)=1$ otherwise. A nowhere-zero vector field is equivalent to orientability)

Question: anyone know of an example of a manifold $M$, say for $\dim \leq4$ that admits a trivialization over its $1$-skeleton (i.e., $M$ is orientable), but so that the trivialization does not extend to the $2$-skeleton? (and, is this trivialization over the $1$-skeleton a line bundle as the restriction to the $1$-skeleton of the $\mathbb R^4$-bundle?)

Even better if there are examples of trivialization over $1$- and $2$-skeleton, but not $3$-skeleton, etc.

Jeff.

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Most compact orientable surfaces work, no? –  Mariano Suárez-Alvarez May 21 '11 at 0:24
In dimensions larger then $2$, the answer is any orientable manifold which does not admit a spin structure . Any trivialization over the $2$-skeleton automatically extends to the $3$-skeleton, since $\pi_2 SO_n$ is trivial. –  Ryan Budney Dec 18 '11 at 8:23
Well this has been open for a while, but you want a manifold with non-trivial second homology class right? What about $\mathbb{C}P^2$?
Sorry, but I think he wanted an example of a non-spinable manifold, of which $\mathbb{C}P^2$ is. Did I miss the point? –  levitopher Nov 23 '11 at 22:46
Dear cduston, Actually, maybe I missed the point. I thought that he wanted a manifold whose tangent bundle could be trivialized over the $1$-skeleton, but not over the $2$-skeleton, hence any compact orientable surface of genus different from $1$ seems to do the job. But your interpretation, that he wants a manifold with $w_2 \neq 0$, also makes sense. (I presume that when you say "non-trivial second homology class", you mean non-trivial second Stiefel--Whitney class.) Regards, –  Matt E Nov 24 '11 at 3:15