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We know that the middle circle $S^1$ in Mobius band has a nontrivial normal bundle. Now consider the higher dimensional case. Let $M$ be a $n$-dimensional ($n\geq5$ or even larger) closed orientable manifold and some $S^2$ is embeded in $M$.

Does $S^2$ always have a trivial normal bundle in $M$?

If the anwer is 'No', what conditions on $M$ can make sure the answer is 'yes'? Spin manifolds? In surgery theory, we always want an embedding of $S^k \times D^{n-k}$ in $M$ to do surgery and the trivial normal bundle is necessary. Any references or comments are welcomed.

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Isn't this equivalent to asking whether every vector bundle over $S^2$ of rank $r\ge 3$ is trivial? Since the set of isomorphism classes of such bundles is naturally identified with $\pi_1(SO(r))\simeq \mathbb{Z}_2$, it would appear that the answer is 'no'. Explicitly, if you take the 2-plane bundle over $S^2$ with Euler class equal to $1$ and add a trivial bundle of rank $r{-}2$, this should give you a nontrivial rank $r$ bundle over $S^2$. –  Robert Bryant Jul 7 '11 at 12:03
    
To Robert: thanks for your comments. Is the bundle you constructed a normal bundle? Actually, I know little about this field. Would you like to give a reference for your argument? –  yeshengkui Jul 7 '11 at 12:17
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Every bundle is a normal bundle. –  Tom Goodwillie Jul 7 '11 at 14:49
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As Tom wrote, every bundle over a compact manifold is a normal bundle of an embedding into a compact manifold: If $E$ is a vector bundle over $N$, then $E$ is the normal bundle of the obvious section of the unit sphere bundle in $E\oplus R$ where $R$ is the trivial bundle. As for the other, look in Steenrod, where you'll see that the isomorphism classes of orientable rank $r$ bundles over $S^n$ are given by the elements of $\pi_{n-1}\bigl(SO(r)\bigr)$. As for the spin assumption, yes, that's enough to establish triviality because then $w_1$ and $w_2$ of the normal bundle will vanish. –  Robert Bryant Jul 7 '11 at 15:05
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@yeshengkui: Yes, the point is that a bundle of rank $r\ge3$ over $S^2$ is trivial if and only if its second Stiefel-Whitney class vanishes. By the Whitney sum formula, the second Stiefel-Whitney class of the normal bundle is the same as the second Stiefel-Whitney class of the pullback of the ambient bundle, which vanishes exactly when you assume that the ambient manifold is spin. –  Robert Bryant Jul 8 '11 at 2:37

1 Answer 1

The normal bundle to $\mathbb{C}P^1\simeq S^2$ in $\mathbb{C}P^2$ is the dual of the tautological bundle. This is nontrivial (even as a real bundle rather than a complex bundle); indeed, we have $H^*(\mathbb{C}P^1;\mathbb{Z}/2)=\mathbb{Z}/2[x]/x^2$, where $|x|=2$ and $x$ is the second Stiefel-Whitney class of the bundle in question.

Also, to expand on Tom Goodwillie's comment, every vector bundle is a normal bundle. In more detail, if $V$ is a vector bundle over a manifold $M$, then we can identify $M$ with the zero section in the total space $EV$, and then the normal bundle to $M$ in $EV$ is just $V$. If you prefer to work with compact manifolds, you can choose an inner product on $V$, and put $$ P=\{(x,t,v) : x\in M, t\in\mathbb{R}, v\in V_x, t^2+\|v\|^2=1\} $$ (the unit sphere bundle in $\mathbb{R}\oplus V$). We have an embedding $i:M\to P$ given by $i(x)=(x,1,0)$, and it is not hard to see that the normal bundle is $V$. Equivalently, $P$ is the fibrewise one-point compactification of $V$.

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Thanks for your answer. Are some obstructions to get a trivial normal bundle? –  yeshengkui Jul 7 '11 at 16:09
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On $S^2$ there is one oriented vector bundle of rank $2$ for every integer. This follows from the fact that $H^2(S^2)\cong \mathbb Z$. The integer for a bundle corresponds to the number of zeroes (properly counted with signs) of a generic section. The tangent bundle corresponds to $2$. Changing the orientation changes the integer by a minus sign. Every bundle of rank $3$ is a trivial line bundle plus a bundle of rank $2$, but there are only two of them: all even integers give trivial bundle and all odds give the other (which is not spin). Classification for rank $r>2$ is independent of $r$. –  Tom Goodwillie Jul 7 '11 at 17:18
    
That's so nice. Thanks a lot. –  yeshengkui Jul 7 '11 at 19:17

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