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On page $10$ of Hatcher's Vector Bundles and K Theory, he gives a proof that the Whitney sum of the trivial line bundle over $\mathbb{R}P^n$ and the tangent bundle is equal to the Whitney sum of copies of the tautological line bundle.

A summary of the proof is

1) The image of the tangent bundle of the sphere under the quotient is the tangent bundle of real projective space, and the image of the normal bundle is the trivial line bundle over real projective space.

2) He says the sum of the tangent bundle and normal bundle of the sphere is trivial.

3) He then proves that the trivial line bundle over the sphere is isomorphic to the normal bundle and proves that the normal bundle under the quotient is mapped to the tautological line bundle over real projective space.

But this doesn't make any sense because he claimed earlier that the image of the normal bundle is trivial and the tautological bundle is NOT trivial. Could someone please clarify what has happened? Is the image of the normal bundle of the sphere under the quotient the tautological bundle or the trivial bundle?

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Your point $1$ is correct. If you meant the direct sum of the tangent bundle and the normal bundle of the sphere is trivial, then point $2$ is also correct. However, point $3$ is wrong.

Hatcher shows that under the quotient map, the trivial bundle $S^n\times\mathbb{R}^{n+1}$ on $S^n$ is sent to $E^{\oplus(n+1)}$ on $\mathbb{RP}^n$ where $E$ is the line bundle on $\mathbb{RP}^n$ obtained by making the identification $(x, t) \sim (-x, -t)$ in the trivial bundle $S^n\times\mathbb{R}$. As has already been shown, there is an isomorphism $S^n\times\mathbb{R}\cong NS^n$, which is given by $(x, t) = (x, tx)$. Under this isomorphism, the identification $(x, t) \sim (-x, -t)$ becomes $(x, tx) \sim (-x, tx)$. This is not the identification used when showing the trivial bundle is sent to the trivial bundle, that identification is $(x, tx) \sim (-x, -tx)$.


You may not know why yet, but every line bundle on $S^n$ is trivial. This is not true on $\mathbb{RP}^n$ though; there is the trivial bundle $\varepsilon^1$ and the tautological bundle $\gamma^1$ which is non-trivial. If $f : S^n \to \mathbb{RP}^n$ is the quotient map, $f^*\varepsilon^1$ and $f^*\gamma^1$ are line bundles on $S^n$ and are therefore trivial. That is, both $\varepsilon^1$ and $\gamma^1$ on $\mathbb{RP}^n$ arise from the trivial bundle on $S^n$, even though $\varepsilon^1$ and $\gamma^1$ are not isomorphic. That's what's going on here.

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  • $\begingroup$ Thank you for your answer but I am afraid I still don't quite understand. It seems to me we have the quotient of the trivial bundle over the sphere, which gets sent to the tautological line bundle and then say this is isomorphic to the normal bundle over the sphere, which gets sent to the trivial bundle over projective space and use this to conclude. But why can I not follow this exact argument to conclude that the tautological bundle over projective space is isomorphic to the trivial bundle over projective space, which is false? $\endgroup$ – R Mary Jul 6 '16 at 12:58
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    $\begingroup$ As I pointed out, the identifications in the normal bundle $NS^n$ are different in the two cases. If we use $(x, tx) \sim (-x, tx)$, we get the tautological line bundle on $\mathbb{RP}^n$. If we use $(x, tx) \sim (-x, -tx)$, we get the trivial line bundle on $\mathbb{RP}^n$. You need to know which identification you are using. In particular, "the quotient of the trivial bundle" doesn't make sense as there are two quotients, not just one. Which quotient you obtain depends on the identification being made. $\endgroup$ – Michael Albanese Jul 6 '16 at 22:03

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