MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
Every bundle is a normal bundle. – Tom Goodwillie Jul 7 '11 at 14:49
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
@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

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 $V$.

share|cite|improve this answer
Thanks for your answer. Are some obstructions to get a trivial normal bundle? – yeshengkui Jul 7 '11 at 16:09
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.