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Is it true to say that the total space of the tautological line bundle over $\mathbb{R}P^{n}$ is a non orientable manifold?

Perhaps the question can be indirectly related to the following question:

The blow up of of the plane and the Moebius band

As another question which is indirectly related to our question we ask:

What is an example of a non orientable manifold $M$ such that $TM$ is an orientable manifold?

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Ali Taghavi
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  • What do you think? – Amitai Yuval Oct 18 '15 at 17:03
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    @AmitaiYuval The first motivation is the case $n=1$. – Ali Taghavi Oct 18 '15 at 17:04
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    The tangent bundle is always orientable. See http://math.stackexchange.com/questions/129514/why-is-the-tangent-bundle-orientable for various answers. – Saal Hardali Oct 18 '15 at 17:37
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    Note that the tautological bundle is not the tangent bundle of $\Bbb{RP}^n$. The total space of the tautological bundle is diffeomorphic to a punctured $\Bbb{RP}^{2n}$, so it's always non-orientable. –  Oct 18 '15 at 17:39
  • @MikeMiller I think that that some thing is missing in your comment. their dimension are not equal.right? – Ali Taghavi Oct 18 '15 at 17:45
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    Oh, I'm so sorry, had a thinko. You're right. The tautological bundle is diffeomorphic to $\Bbb{RP}^{n+1}$ minus a point, so is orientable iff $\Bbb{RP}^n$ is not! –  Oct 18 '15 at 17:46
  • @MikeMiller thank you for very interesting comment. I guess that the diffeomorphism send $([x],v)$ to $([x,x.v])$. – Ali Taghavi Oct 18 '15 at 18:03
  • So an alternative proof of nontriviality of tautological line bundle either in real or complex case? – Ali Taghavi Oct 18 '15 at 18:07
  • @MikeMiller is there a natural description of tautological 2-plane bundle over real Grassmanian $G(2,n)$? A natural embedding into $G(2,n+1)$ or some thing like this? – Ali Taghavi Oct 18 '15 at 18:15
  • I don't know a similar description for the total space of tautological bundles over Grassmannians off the top of my head. –  Oct 18 '15 at 19:14
  • @MikeMiller simillar to projective space, there is a dimension compatibility:the total space of 2 plane bundle over $G(2,n)$ has the same dimension as $G(2, n+1)$. – Ali Taghavi Oct 19 '15 at 05:49

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The Whitney product formula for Stiefel-Whitney classes specializes to $w_1(V\oplus W)=w_1(V)+w_1(W)$, where $w_1$ is the first Stiefel-Whitney class.

If $V$ is a vector bundle over $M$ (a CW-complex), $w_1(V)=0$ if and only if $V$ is orientable as a vector bundle.

If $E$ is the total space of a smooth vector bundle over a manifold $M$, then $T(E)$ always splits as $p^*(T(M))\oplus p^*(V)$ where $p:E \to M$ is the projection.

In particular, for any manifold $M$, the vector bundle $T(TM)$ splits as $p^*(TM)\oplus p^*(TM)$ where $p$ is the projection $TM \to M$. By the Whitney product formula(as mentioned in an answer to a linked question), $w_1(T(TM))=2w_1(p^*(TM))=0$.

Let $\gamma_n$ be the total space of the canonical bundle over $\Bbb RP^n$. Now $T(\gamma_n) \cong p^*(\gamma_n)\oplus p^*(T(\Bbb RP^n)). $ A standard computation shows that $w_1(\gamma_n)= \alpha$ the unique nonzero class in $H^1(\Bbb RP^n,\Bbb Z_2$). It is also standard to check that $\Bbb RP^n$ is orientable if and only if $n$ is odd. Equivalently $w_1(T(\Bbb RP^n))=\alpha$ if and only if $n$ is even. By naturality of $w_1$, $w_1(T(\gamma_n))=p^*(\alpha) +0\ne 0$ when $n$ is odd and $w_1(T(\gamma_n))=p^*(\alpha) +p^*(\alpha)=0$ when $n$ is even (which gives the total space orientable iff $n$ is even as mentioned in the comments).

One can generalize this idea quite easily to show the total space of a smooth vector bundle $V$ over a smooth manifold $M$ is orientable if and only if $w_1(TM)=w_1(V)$.

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