Line bundle on $S^2$

Question

How do you prove that a line bundle (vector bundle of rank 1) on $S^2$ is isomorphic to the trivial line bundle? Can you give a reference?

Thanks.

Answer 1

Since $GL(1,\mathbf R)$ and the symmetric group $S_2$ are homotopy equivalent as topological groups, there is a bijective correspondence between isomorphism classes of real line bundles and double covers. It follows that a simply connected space has no nontrivial real line bundles.

Answer 2

In general, there is a bijection $$\Phi \colon [S^{k-1}, \textrm{GL}_n^+(\mathbb{R})] \to \textrm{Vect}_+^n(S^k),$$ where $[X,Y ]$ denotes the set of homotopy classes of continous maps from $X$ to $Y$ and $\textrm{Vect}_+^n(Z)$ denotes the set of isomorphism classes of real (oriented) vector bundles of rank $n$ on $Z$.

Using the retraction of $\textrm{GL}_n^+(\mathbb{R})$ onto $\textrm{SO}(n)$ one obtains another bijection $$\Psi \colon [S^{k-1}, \textrm{SO}(n)] \to \textrm{Vect}_+^n(S^k).$$

In the case of line bundles we have $n=1$ and $\textrm{SO}(1)$ is just a point, hence the bijection $\Psi$ shows that any orientable real line bundle on $S^k$ is trivial. Furthermore, it is not difficult to prove that when $k \geq 2$ any real vector bundle on $S^k$ is orientable, so any real line bundle is trivial.

Notice that this is not true for $S^1$, where there is an orientable bundle (the trivial one) and a nonorientable one (the Moebius band).

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Similar methods of classification can be applied to the case of complex vector bundles in order to show that $S^2$ admits nontrivial complex line bundles.

In fact, one finds that there is a bijection $$\Phi_{\mathbb{C}} \colon [S^{k-1}, \textrm{GL}_n(\mathbb{C})] \to \textrm{Vect}_{\mathbb{C}}^n(S^k).$$ For $n=1$, moreover, there is a bijection between $[S^{k-1}, \textrm{GL}_1(\mathbb{C})]$ and $H^2(S^k, \mathbb{Z})$. In particular, if $k \neq 2$ the only complex line bundle on $S^k$ is the trivial one, whereas if $k =2$ there is a discrete family $\{L_t \}$ of non-isomorphic complex line bundles, parametrized by $t \in \mathbb{Z}$. In fact, viewing $S^2$ as the Riemann sphere with complex coordinate $z$, the transiction function of $L_t$ is $z^t$.

For further details you can look at Hatcher's book Vector bundles and K-theory, Chapter 1.

Answer 3

Another solution is to consider clutching functions. The sphere $S^2$ is obtained by gluing two disks along their boundaries, it is not hard to prove that the information is encoded by a continuous map: $$f:S^1\rightarrow GL_1(\mathbb{R})$$ and that we have a map: $$[S^1,GL_1(\mathbb{R})]\rightarrow Vect^1(S^2)$$ where on the left hand side you have homotopy classes of clutching functions and on the right hand side you have isomorphism classes of real line bundles over $S^2$. As our continuous map $f$ is homotopic to a constant map the associated vector bundle is trivial. In general for $n$-dimensional real vector bundles over $S^k$ you will have a map: $$[S^{k-1},GL_n(\mathbb{R})]\rightarrow Vect^n(S^k).$$ This map becomes a bijection when you consider oriented vector bundles.

edit: I just realize that I write my post at the same time as Francesco.

Answer 4

Along the same lines. Rank $1$ real vector bundles over a compact CW complex $X$ are all pullbacks of the tautological line bundle over $\mathbb{RP}^\infty$. The space of isomorphism classes of such bundles can be identified with the space $[X,\mathbb{RP}^\infty]$ of homotopy classes of continuous maps $X\to \mathbb{RP}^\infty$. Since $\mathbb{RP}^\infty$ is the Eilenberg-MacLane space $K(\mathbb{Z}/2, 1)$ (the universal cover $S^\infty$ is contractible) we deduce that $[X,\mathbb{RP}^\infty]$ can be identified with the cohomology group $H^1(X,\mathbb{Z}/2)$. When $X=S^2$ this group is trivial.

Answer 5

copy of answer on question on MSE:

Here is a nice (in my opinion) elementary proof. It only assumes you know transistion functions for vector bundles, the standard charts for $S^2$, and that a nowhere-zero global section demonstrates triviality of a line bundle.

Since $\mathbb{R}^2\cong S^2\setminus \{\infty\}$ is contractible, we have that $\mathcal{L}|_{S^2\setminus \{\infty\}}$ must be trivial, and similarly for $\mathcal{L}|_{S^2\setminus \{0\}}$. A global section of the bundle consists of two smooth functions $f_1,f_2:\mathbb{R}^2\to \mathbb{R}$ such that on $\mathbb{R}^2\setminus\{0\}$: $$f_2(x)=g(x)f_1(x^{-1}) \quad(*)$$ where $g(x)$ is some nowhere vanishing smooth function on $\mathbb{R}^2\setminus\{0\}$, and $x^{-1}$ is $1/x$ the complex sense. So after specifying $f_1$, we know all values of $f_2$, except $f_2(0)$. We want to show that we can find some $f_1$ and $f_2$ both nowhere $0$ such that $f_2$ satisfies $(*)$.

We construct $f_1$: denote by $U_1$ the ball $B(0,1)$, $A$ the annulus A[1,2], and $U_2$ the set $\mathbb{R}^2\setminus B(0,2)$. Let $b_1,b_2$ be positive bump functions such that: $b_1(x)=1$ on $U_1$, $b_1(x)=0$ on $U_2$, $b_2=1$ on $U_2$, $b_2=0$ on $U_1$ and $b_1+b_2=1$ on $A$. Then define $$f_1(x)=b_1(x)+g(x^{-1})^{-1}b_2(x)$$ Note that $f_1$ is nowhere 0 and smooth. Then on $\mathbb{R}^2\setminus\{0\}$: $$f_2(x)=g(x)(b_1(x^{-1})+g(x)^{-1}b_2(x))=g(x)b_1(x^{-1})+b_2(x^{-1})$$ Note that this is also smooth on $\mathbb{R}^2\setminus\{0\}$. Also, as $|x|\to 0$, $f_2(x)=b_2(x^{-1})=1$, so $f_2$ extends to a nowhere zero smooth function on $\mathbb{R}^2$.

The reason for the bump functions is that the naive definition $f_1(x)=g(x^{-1})^{-1}$ does not work: this might not extend to a smooth function on $\mathbb{R}^2$ if $g$ is annoying. The bump functions get rid of this problem. The need for bump functions is no coincidence: every proof of this fact has to fail in the analytic category. This proof fails there because bump functions are not analytic.