# Non trivial vector bundle over non-paracompact contractible space

The proof that the set of classes of vector bundles is homotopy invariant relies on the paracompactness and the Hausdorff property of the base space. Are there any known examples of:

Non trivial vector bundles over a paracompact non-Hausdorff contractible space

Non trivial vector bundles over a Hausdorff non-paracompact contractible space

Non trivial vector bundles over a non-Hausdorff non-paracompact contractible space

• You can always use a nontrivial bundle over a sphere one point join with a trivial bundle over a space satisfying those properties to construct such example. – Xiaolei Wu Sep 6 '12 at 18:16
• @Xiaolei Wu: I'd like to hear more about this example. Suppose I want a non-trivial bundle over the Sierpinski space. What's the base space your comment would give? – David White Sep 6 '12 at 19:58
• @Xiaolei Wu: I guess by one-point join you mean the wedge of the two spaces, i.e. $B\vee B'$. This won't be contractible unless both pieces are, so by sphere I guess you mean $S^\infty$? Another problem is that there is no wedge sum of vector spaces, so why should the resulting thing be a vector bundle? – David White Sep 6 '12 at 20:50
• I tried to find an example with a finite base space and have concluded that I can't do it. My hope was to use non-Hausdorffness to make a bundle which was $\mathbb{R}$ on some component and $\mathbb{R}^2$ on another. Unfortunately, it seems that in order for a finite space to be contractible you need to know that there is one point whose only neighborhood is the whole space. That's bad, because then locally trivial implies trivial. I'll think it over some more later tonight. – David White Sep 6 '12 at 21:50
• David White, You are write. I was wrong. – Xiaolei Wu Sep 6 '12 at 22:41

Let $U$ and $V$ be two copies of the real line and make a space $X$ by gluing them by the identity along the strictly positive half-line: $x\in U$ equals $x\in V$ for $x>0$. Now make a rank one vector bundle over this space by taking a trivial bundle over each of the lines: glue $U\times\mathbb R$ to $V\times \mathbb R$ by identifying $(x,y)\in U\times\mathbb R$ with $(x,f(x)y)\in V\times\mathbb R$ for $x>0$. Let's choose the clutching function to be $f(x)=x$, or any other nowhere zero continuous function of positive real $x$ that cannot be extended to a nowhere zero function of all real $x$. Surely $X$ is contractible, but the bundle can't be trivial since the function $f:U\cap V\to GL_1(\mathbb R)$ cannot be expressed as the quotient of a function extendible to $U$ and a function extendible to $V$, since that would make it itself extendible to all of $\mathbb R$.

• Did you mean to end in mid-sentence? – Todd Trimble Sep 7 '12 at 5:09
• Thanks for your answer. This also deals with the non-paracompact non-Hausdorff case, using a suitable wedge sum with the trivial bundle over a non-paracompact contractible space, so only the Hausdorff non-paracompact case remains without an example. – Ramón Barral Sep 7 '12 at 13:15
• It's been a long time since I actively studied point-set topology, but isn't $\mathbb{R}^J$ for some uncountable $J$ an example of a space which is not paracompact? It seems like it should be Hausdorff, since it's a product of Hausdorff spaces. It also feels contractible, but maybe it's not. Anyway, it's the best I could come up with for a potential base-space to handle this last example. Another I thought of was the order topology $S_\Omega$ (see Munkres), and I also feel like it's contractible because it's not too long yet to be reeled in. – David White Sep 7 '12 at 15:05
• Well, $\mathbb{R}^J$ is certainly contractible: the map $(t; x_1, x_2, \ldots) \mapsto (t x_1, t x_2, \ldots)$ is continuous. The long line however is not contractible; see here: mathoverflow.net/questions/35087/… – Todd Trimble Sep 7 '12 at 16:50

To complement the answer providing a non-Hausdorff example, here is an example on a non-paracompact Hausdorff space as given in Schröer's Pathologies in cohomology of non-paracompact Hausdorff spaces arxiv:1309.2524.

I will slightly simplify the space in question, though. Let $(X,0)$ be the wedge sum of countably many copies of intervals $([0,1],0)$. We denote the $1$ of the $n$-th copy by $1_n\in X$, $n\geq 1$. Of course, we have to alter the topology a little to get a non-paracompact space. Namely, the open subsets $U\subset X$ are those which are open in the CW-topology, and which either do not contain $0$, or if they do, they have to contain all but finitely many of the half-open intervals $[0,1_n)$.

This space is Hausdorff, as is easily seen, and contractible: each interval $[0,1_n]\subset X$ carries the usual topology and so $\{0\}\subset [0,1_n]$ is a strong deformation retract, which implies that each arm of $X$ can be contracted onto $0$ simultaneously.

Instead of an explicit example of a (complex) line bundle, Schröer shows that $H^1(X,U(1))\not = 0$, as follows. Using that even for non-paracompact spaces, $H^1$-sheaf cohomology is computed by Čech-cohomology, it is first shown that $H^1(X,C^0(X))\not = 0$; then $H^1(X,U(1))\not = 0$ follows from the real Euler sequence $0\to\underline{\mathbb Z}_X\to C^0(X)\xrightarrow{exp(\cdot 2\pi i)}\underline{U(1)}_X\to0$, and $H^1(X,\underline{\mathbb Z}_X) = 0$.

Instead of giving all the details, let me directly construct a non-trivial real line bundle, but using the same idea. Consider the open cover of $X$ defined by $U_0 = \bigcup_{n\geq 1}[0,1_n)$ and $U_n = (0,1_n]\subset X$, $n\geq 1$. Then clearly $U_n\cap U_m=\emptyset$ unless $n = m$ or $n = 0$ or $m=0$. Hence, we have to specify transition functions $g_n\colon U_0\cap U_n = (0,1_n)\to\mathbb{R}^\times$, $n\geq 1$. Let $f_n\colon U_n\to\mathbb{R}$, $n\geq 1$, be continuous functions with $f_n(1) = 0$, but $f_n(x)\not=0$ for $x\not=1$, and let $g_n = \tfrac{1}{f_n}$. (E.g., $g_n(x) = (1-x)^{-1}$.) Let $L$ be the line bundle defined by those transition functions. I claim that every section $s\colon X\to L$ is trivial at all but finitely many $1_n$. In particular, $L$ itself is non-trivial.

Let $s\colon X\to L$ be any section. It defines (and is defined by) continuous functions $s_n\colon U_n\to \mathbb{R}$, $n\geq 0$, satisfying $s_n(x) = f_n(x)s_0(x)$ for all $x\in (0,1_n)$, $n\geq 1$. From the existence of the limit $\lim_{x\to 1}f_n(x) = f_n(1) = 0$, we conclude that $s_n(1_n) = 0$ as soon as the limit $\lim_{x\to 1_n}s_0(x)$ exists as well. To see that this happens for all but finitely many $n$, it suffices to observe that $s_0$ is bounded away from finitely many $U_n$, which is a consequence of our choice of topology.

Now that Tom Goodwillie has basically answered the question, I feel I can undelete my non-answer. I wrote this when I had misunderstood what the OP wanted, but I feel like it's worth putting out there for people to see. The point is to show how far you can go in developing the theory without the hypotheses of Hausdorff and paracompact.

You can define vector bundles in a great level of generality. Here is the most general approach I know, which I learned from Mark Hovey. Let $B$ be a topological space. A space over $B$ is a space $E$ and a continuous map $p:E\to B$, i.e. an element in the category $Top \downarrow B$. A vector space over $B$ is an object $V$ of this category together with a field $F$ and maps $+:V\times V\to V, \ast: F\times V\to V, (-1):V\to V$, and $0:pt \to V$ such that a bunch of diagrams commute. These are the diagrams for associativity of $+$ and $\ast$, commutativity for $+$, distributivity, identity, inverse, etc. As an example, here is the diagram which says $x+0=x$:

$\begin{array}{ccc} V & \rightarrow & V\times pt\\\ \\\ || & & \downarrow (id,0) \\\ \\\ || & & V\times V\\\ \\\ || & & \downarrow +\\\ \\\ V & = & V \end{array}$

The product of two vector spaces $p:E\to B$ and $p':E'\to B$ is the pullback $E\times_B E'\to B$. It is an easy exercise to deduce from the axioms above that $p$ must be surjective. For all $b\in B$ define $F_b = p^{-1}(b)$. Note that for topological vector space $V$ you can form one of these vector spaces over $B$ by setting $E = V\times B$ (so all $F_b = V$) and $p$ to be projection. This is the trivial vector space over $B$. More generally, a vector space over $B$ is called trivial if there is an isomorphism in the category of spaces over $B$ between $E$ and $V\times B$ for some topological vector space $V$.

Given any $U\subset B$ with $i:U\hookrightarrow B$, one has a vector space over $U$, denoted $i^*(p)$, defined by $p:p^{-1}(U)\to U$. A vector bundle over $V$ is a locally trivial vector space over $B$, i.e. for all $b$ there is a neighborhood $U$ s.t. $i^*{p}$ is trivial.

Note that none of this development required any hypotheses on $B$. As the OP mentions, the proof of homotopy invariance of the functor $X\mapsto Vect(X)$ requires paracompactness, though it's not clear to me that it requires the Hausdorff property. Even without paracompactness you can still prove $Vect(X)$ is a contravariant functor and that for any bundle $p$ over $X\times I$ there are open sets $U\times I$ covering $X\times I$ over which $p$ is trivial (by the Lebesgue covering lemma, since $I$ is a compact metric space, and ordered). However, without partitions of unity you can't compare $i_0^*p$ with $i_1^*p$ and conclude that they are the same if there's a homotopy $H$ between $f=Hi_0$ and $g=Hi_1$. It seems to me that you really can't get away with less than partitions of unity, and this is the same as paracompactness. You might be able to push the theory a little bit further without the Hausdorff hypothesis, but you'll probably need the hypothesis eventually anyway in order to say nice things.

• Paracompactness is necessary in order that every locally trivial vector bundle is numerable = trivialisable over a numerable cover (i.e. a cover with a subordinate partition of unity). Numerable vector bundles on any space are classified by the usual classifying spaces. Since the universal vector bundle is numerable, and pulled back numerable bundles are trivial, this is necessary condition to have classification. However there should be locally trivial bundles that don't trivialise over a numerable cover that aren't classifiable. – David Roberts Sep 8 '12 at 4:36
• Interesting. This is an even further level of generality than I was aware of. Thanks for sharing. – David White Sep 8 '12 at 16:25