8
$\begingroup$

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.

$\endgroup$
4
  • 6
    $\begingroup$ Take a metric on your line bundle and a metric connection. This connection cannot have curvature, and as the 2-sphere is simply connected, there is no monodromy, so parallel transport trivializes the bundle. Of course, this is only the differential geometry version of the answers of Francesco Polizzi and Dan Petersen. $\endgroup$
    – Sebastian
    Nov 20, 2012 at 12:02
  • 1
    $\begingroup$ Huh. And yet there are nontrivial vector bundles on $S^$. $\endgroup$ Nov 20, 2012 at 14:27
  • 6
    $\begingroup$ I'm assuming we mean Real line bundles? There are non-trivial complex line bundles on S^2. $\endgroup$
    – cheyne
    Nov 20, 2012 at 15:03
  • 3
    $\begingroup$ See here $\endgroup$ Oct 1, 2017 at 20:27

5 Answers 5

19
$\begingroup$

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.

$\endgroup$
10
$\begingroup$

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).


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.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ This can also be seen via the fact that $[S^k,BG] = [\Sigma S^{k-1},BG] = [S^{k-1},\Omega BG] = [S^{k-1},G]$. $\endgroup$ Nov 20, 2012 at 12:16
7
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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