Let $S^{1}$ bundle over $S^{2}$ is homeomorphic (diffeomorphic) to $S^{3}$. Is the Chern class of the fibration 1, i.e. is the Hopf fibration up to isomorphism the only fibre bundle with total space $S^{3}$?


1 Answer 1


Yes, this is correct. You first use long exact sequence of homotopy groups of the fibration to show that the base of a fibration $p: S^3\to B$ has to be simply-connected. Therefore, $B=S^2$. Now, split $B$ in two disks $B_1, B_2$ meeting along a circle $C$. The fibration $p$ has to be trivial over $B_1, B_2$. Therefore, this gives us a representation of $S^3$ as a union of two solid tori $B_i\times S^1$. Suppose that the absolute value $n$ of the Euler number $e(p)$ of $p$ is not $1$. Orient the fiber of the bundle and the circle $C$. The oriented loops $b=x\times S^1$ (where $x$ is a point in $C$) and $a_i=\partial B_i\times y$ (where $y$ is a point on $S^1$) serve as a bases of $H_1(\partial B_i\times S^1; Z_n)$. The gluing map between the boundaries of the solid tori sends (on the $Z_n$-homology level) $b$ to itself and sends $a_1$ to $a_2 + n b=a_2$ (the first equality holds on the level of integer homology, just by definition of the Euler number). Thus, the natural map $$ H_1(T^2; Z_n)\to H_1(B_1\times S^1; Z_n)\oplus H_1(B_2\times S^1; Z_n) $$ has image isomorphic to $Z_n$. Now, feed this information into the Mayer-Vietoris sequence for the homology with $Z_n$-coefficients, to conclude that $H_1(S^3, Z_n)\ne 0$, which is, of course, nonsense. Therefore, $e(p)=\pm 1$, from which you see that the fibration is unique. (Changing the sign of $e(p)$ is achieved by changing orientation on the fibers.)

  • $\begingroup$ Thanks for your answer. Would you mind to explain me a little bit more how to get the contradiction? I don't understand how to relate the euler class with the M-V. $\endgroup$
    – yess
    Apr 23, 2014 at 14:52
  • $\begingroup$ I will when I have a bit more time. $\endgroup$ Apr 23, 2014 at 15:48

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