Mathematics Asked by probably123 on February 27, 2021
Suppose $Eto S^2$ is an (smooth) oriented disk bundle. Let $D_1$ and $D_2$ denote the upper and lower hemispheres of $S^2$, respectively. Then $E|_{D_1}$ and $E|_{D_2}$ are trivial, so they are both diffeomorphic to $D_itimes D^2$. So $E$ may be seen to be obtained from $D_1times D^2$ and $D_2times D^2$ by attaching their boundaries, for some diffeomorphism $f:partial D_1 times D^2 to partial D_2times D^2$ which is identity on the first factor, and an element of $SO(2)$ on the second factor. (Thus $f(x,y)=(x,g(x)y)$ for some $g:partial D_1to SO(2)$. Thus $g$ gives an element $pi_1(SO(2))cong Bbb Z$, say $n$. We may assume $g$ is the $n$-fold product of the standard generator of $pi_1(SO(2))$.
Note that $E$ is homotopy equivalent to the zero section $cong S^2$. I want to show that the euler number of $E$ is $n$, by showing that the self intersection number of the zero section is $n$. To this end, it suffices to show that there is a section $h:D_1cup D_2to E=D_1times D^2 cup_f D_2times D^2$ which intersects the zero section $|n|$ times positively if $n>0$ (negatively if $n<0$). First, fix an arbitrary point $p(neq 0)in D^2$ and define $h(x)=(x,p)in D_1times D^2$ for $xin D_1$. Then we must have $h(x)=(x,g(x)p)in D_2times D^2$ for $xin partial D_2$. Now extend $h$ over $D_2$ by defining $h(rx)=(rx,r^2g(x)p+(1-r^2)q)$ where $q=(1,0)in D^2$ and $0leq rleq 1$. This gives a section of $Eto S^2$ that intersects the zero section exactly $n$ times, but how can we know whether the intersection is positive or negative?
Or is there another method for computing the euler number of $E$?
P.S. Does every oriented $D^2$-bundle over $S^2$ is the unit disk bundle of some complex line bundle over $S^2$ with a hermitian metric?
The Euler class depends homomorphically on $n$ so it suffices to calculate it for $n=1$. In this case, the disk bundle we get is fiberwise diffeomorphic to the disk bundle of the tautological bundle over $mathbb{C}P(1)$. The Thom space of this bundle is $mathbb{C}P(2)$. The self intersection number can be calculated via the cup square. The zero section corresponds to the inclusion of $mathbb{C}P^1$ into $mathbb{C}P^2$ and so the cup square of the dual of this class is the fundamental class by the cohomology structure. This means the Euler number is 1.
Hence, for general $n$ the Euler number will be $n$.
To address your other question: I highly recommend figuring this issue with orientations out yourself. It will be very instructive.
And yes this follows from the fact that $pi_2(BO(2))$ is generated by the tautological bundle over $mathbb{C}P^1$. If one wanted to do this calculation via complex means, we could use the fact that the Chern number of the bundle is 1 and that this is equal to its Euler number.
Correct answer by Connor Malin on February 27, 2021
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