5
$\begingroup$

Let $M$ be a manifold with boundary $\partial M$ and interior $M_0$. Let $E\rightarrow M_0$ be a fixed vector bundle. How many extensions of $E$ to a vector bundle $E'\rightarrow M$ are there, up to isomorphism? In terms of the bundles monoid: the restriction of $E'$ to $M_0$ gives a monoid morphism $\mathrm{Vec}_k(M)\rightarrow \mathrm{Vec}_k(M_0)$. Is it surjective/injective?

Intuitively, the bundle $E'|_{\partial M}$ is "the limit" of $E$ at $\partial M$, and therefore should be fixed up to isomorphism.

And perhaps in the same vein, the inclusion $\iota : M_0 \rightarrow M$ induces $\iota_*:\pi_1(M_0)\rightarrow \pi_1(M)$. Is this map surjective/injective? Can the bijectivity be deduced from a tubular neighborhood of $\partial M$ in $M$?

Counter-examples are appreciated.

$\endgroup$
3
  • 9
    $\begingroup$ The inclusion $M_0 \hookrightarrow M$ is a homotopy equivalence, so a vector bundle on $M$ is determined by its restriction to $M_0$ $\endgroup$ Commented Jun 17, 2020 at 18:54
  • $\begingroup$ @MichaelAlbanese: I don't get this one, maybe you can point out my mistake, please? If $X$ is a smooth manifold, and if $D$ is a relatively compact open subset with smooth boundary $\partial D$, then we have two bundles over $\overline D$: the intrinsic tangent bundle $T \overline D$ and the restriction of $TX$ to $\overline D$. These bundles coincide on $D$, but have different ranks on $\partial D$, apparently contradicting your comment above. Where is my mistake? Maybe the naive restriction of $TX$ to $\overline D$ is not a bundle over $\overline D$? $\endgroup$
    – Alex M.
    Commented May 17, 2021 at 9:27
  • 1
    $\begingroup$ @AlexM. They have the same rank. Maybe you're thinking of $T(\partial\overline{D})$ which has rank one less than that of $T\overline{D}$ and $T\overline{X}$. The relationship between these bundles is that $T\overline{D} \cong TX|_{\overline{D}}$ and $T\overline{D}|_{\partial\overline{D}} \cong TX|_{\partial\overline{D}} \cong T(\partial\overline{D})\oplus\varepsilon^1$. $\endgroup$ Commented Jun 17, 2021 at 16:53

1 Answer 1

9
$\begingroup$

As I indicated in my comment, the inclusion $\iota : M_0 \to M$ is a homotopy equivalence. This can be shown using the fact that the boundary $\partial M$ has a collar neighbourhood; it then boils down to showing the inclusion $(0, 1) \hookrightarrow [0, 1)$ is a homotopy equivalence. Actually, one needs to show that there is a homotopy inverse $j : [0, 1) \to (0, 1)$ to $i$ such that $i\circ j$ and $j\circ i$ are homotopic to identity maps relative to $[\frac{1}{2}, 1)$. This is not difficult, see this answer for some details.

On any paracompact space $X$, there is a natural bijection between isomorphism classes of real vector bundles on $X$ of rank $r$ and $[X, BO(r)]$, the set of homotopy classes of maps $X \to BO(r)$; see section $1.2$ of Hatcher's Vector Bundles and K-Theory for example. In particular, given a map $f : X \to Y$, we get an induced map $f^* : [Y, BO(r)] \to [X, BO(r)]$ which corresponds to pulling back a vector bundle by $f$. The analogous statement is true for complex vector bundles too, one just replaces $BO(r)$ with $BU(r)$.

In the case that $f$ is a homotopy equivalence, then $f^*$ is a bijection: if $g$ is the homotopy inverse of $f$, then $g^*$ is the inverse of $f^*$. In particular, for the homotopy equivalence $\iota : M_0 \to M$, we see that there is a bijection between isomorphism classes of real/complex rank $r$ bundles on $M$ and $M_0$ given by $E \mapsto \iota^*E = E|_{M_0}$.

Finally, as $\iota : M_0 \to M$ is a homotopy equivalence, the induced map $\iota_* : \pi_1(M_0) \to \pi_1(M)$ is an isomorphism.


As Ben McKay indicates in the comment below, the above does not deal with smooth bundles but topological bundles. The statement for smooth bundles is also true, but requires a bit more work. The key is that every real rank $r$ vector bundle on a smooth manifold $M$ has a classifying map $M \to \operatorname{Gr}_r(\mathbb{R}^N)$ which is unique up to homotopy where $N = r + \dim M + 1$; this is Theorem 3.3.4 of Hirsch's Differential Topology. It follows that isomorphism classes of topological real rank $r$ vector bundles on $M$ are in bijection with $[M, \operatorname{Gr}_r(\mathbb{R}^N)]$; that is, the inclusion $\operatorname{Gr}_r(\mathbb{R}^N) \hookrightarrow \operatorname{Gr}_r(\mathbb{R}^{\infty})$ induces a bijection $[M, \operatorname{Gr}_r(\mathbb{R}^N)] \to [M, \operatorname{Gr}_r(\mathbb{R}^{\infty})] = [M, BO(r)]$.

If the classifying map of a bundle is smooth, then the bundle itself is smooth (the pullback of a smooth bundle by a smooth map is smooth). As every continuous map between smooth manifolds is homotopic to a smooth one, every topological vector bundle on $M$ is isomorphic to a smooth one. Moreover, two smooth maps are homotopic if and only if they are smoothly homotopic which implies that every topological vector bundle is isomorphic to a unique smooth vector bundle up to smooth isomorphism. It follows that isomorphism classes of smooth real rank $r$ vector bundles on $M$ are in bijection with $[M, \operatorname{Gr}_r(\mathbb{R}^N)]$.

Now we can argue as before to deduce that $\iota^*$ induces a bijection between the set of isomorphism classes of smooth real rank $r$ bundles on $M$ and $M_0$. Again, the statement is also true for smooth complex bundles.

$\endgroup$
8
  • $\begingroup$ Thanks. Can you explain what the space $BO(k)$ (and $BU(k)$) is? $\endgroup$ Commented Jun 18, 2020 at 0:24
  • 3
    $\begingroup$ For a topological group $G$, there is a space $BG$ called its classifying space; here the two groups of interest are the orthogonal group and the unitary group. In this case, we have explicit models for $BO(r)$ and $BU(r)$, they are the grassmannians $\operatorname{Gr}_r(\mathbb{R}^{\infty})$ and $\operatorname{Gr}_r(\mathbb{C}^{\infty})$ respectively. $\endgroup$ Commented Jun 18, 2020 at 0:35
  • $\begingroup$ There is still the problem of smooth structure of the vector bundle, once the topological structure is determined. $\endgroup$
    – Ben McKay
    Commented Jun 18, 2020 at 8:54
  • $\begingroup$ @BenMcKay: I believe this can also be dealt with. I will edit my answer later today. $\endgroup$ Commented Jun 18, 2020 at 11:43
  • 1
    $\begingroup$ @QGravity: Yes, by the same argument: principal $G$-bundles have a classifying space $BG$, and a manifold with boundary is homotopy equivalent to its interior. $\endgroup$ Commented May 30 at 18:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .