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 ($n\geq5$) closed orientable manifold and some $S^2$ is embedded in $M$.
Does $S^2$ always have a trivial normal bundle in $M$?
If the answer 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.
Best Answer
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 $V$.