Maximal trivialising subspace for a vector bundle

Let $X$ be a locally compact Hausdorff space. Given a vector bundle $p: E\to X$, a subspace $Y$ of $X$ is called trivialising (for this bundle), if after restricting this bundle to $Y$, it is a trivial bundle. In other words, $p: p^{-1}(Y)\to Y$ is trivial. $Y$ is called maximally trivial, if it is trivial and there is no trivial subspace of $X$ which contains $Y$ as a proper subset.

Given a point $x$ in $X$, a maximal trivialising subspace containing $x$ may not be unique. Does any maximal trivialising subspace must be open? Can we say something others about maximal trivialising subspaces?

There is no reference about this topic in the existing literatures.

A similar question has been posted on stackexchange: https://math.stackexchange.com/questions/1581861/maximal-trivial-subspace-in-vector-bundles

• Is every trivial subspace contained in a maximal one? – Julian Rosen Dec 24 '15 at 20:05
• @ Julian Rosen I think yes. Given any trivial subspace, keeping it trivial, we can enlarge it as much as possible, at last we get a maximal one. – Strongart Dec 26 '15 at 4:53
• Unfortunately, the question is not so meaningful. – jhgfd Dec 26 '15 at 11:45
• It isn't obvious to me that it works to enlarge a trivial subspace as much as possible. The union of a nested sequence of trivial subspaces need not be trivial. – Julian Rosen Dec 26 '15 at 20:05
• The example I had in mind was $X=S^1=\mathbb{R}/\mathbb{Z}$, $E$ the Mobius bundle. Then $[0,1-1/n]$ (for $n=2,3,4,\ldots$) is an increasing sequence of trivial subspaces, but there is no trivial subspace containing them all because their union is $X$. Of course, the complement of any point is a maximal trivial subspace, this example is just to illustrates that it isn't obvious how to produce a maximal trivial from an increasing sequence of trivials. – Julian Rosen Dec 27 '15 at 18:09

In addition to Mark's argument, a maximal trivialising $Y\subset X$ is also open, if we assume that $E$ is a $\Bbbk$-vector bundle with $\Bbbk=\mathbb R$ or $\mathbb C$. So every maximal trivialising subset is open and dense, but it is not (yet) clear that every trivialising subset is contained in a maximal one.

Assume $Y$ is trivialising, but not open. Then there exists $x\in Y$ such that no neighbourhood of $x$ is contained in $Y$. We choose a trivialising neighbourhood $U$ for $E$. We have maps $\varphi\colon E|_Y\to\Bbbk^r$ and $\psi\colon E|_U\to\Bbbk^r$ such that $$p\times\varphi\colon E|_Y\to Y\times\Bbbk^r\quad\text{and}\quad p\times\psi\colon E|_U\to U\times\Bbbk^r$$ are vector bundle isomorphisms. Hence there exists $g\colon Y\cap U\to GL_r(\Bbbk)$ such that $\psi(e)=g(p(e))\cdot\varphi(e)$ on $E|_{Y\cap U}$.

Because $g$ is continuous and $Y\cap U$ carries the subspace topology, there exists a compact neighbourhood $K\subset U$ of $x$ and a map $\xi\colon Y\cap K\to\mathfrak{gl}_r(\Bbbk)$ such that $g|_{K\cap Y}=\exp\circ\xi$. By Urysohn's lemma, there also exists a cutoff function $\rho\colon X\to[0,1]$ with $\mathrm{supp}(\rho)\subset K$ and such that $W=\rho^{-1}(1)$ is a neighbourhood of $x$. We replace the chosen trivialisation of $E|_Y$ by $$\varphi'(e)=\exp((\rho\cdot\xi)(p(e)))\cdot\varphi(e)\;.$$ Then $\varphi'$ agrees with $\varphi$ on $Y\setminus\mathring K$, and with $\psi$ on $W\cap Y$. Hence, we can extend $\varphi'$ by $\psi$ on $W$, contradicting the maximality of $Y$.

• Does the Uryshon lemma can be used in a Tychonoff space ( not a normal space)? In the Tychonoff space, if U is a open set and x in U, there is a continuous function f such that supp(f)$\subseteq$U and f(x)=1, but maybe f$^{-1}$(1)={x}. – Strongart Dec 30 '15 at 14:09
• @Strongart I only need Urysohn on the compact set $K$, which is normal in its subspace topology inherited from $X$. I wrote $\rho\colon X\to [0,1]\to\mathbb R$, but you construct $\rho$ on $K$, then extend by zero. – Sebastian Goette Dec 30 '15 at 15:10
• I see, choose a closed neighbourhood L of x such that L in K, then Urysohn' lemma can be uesd to get a continuous function f such that f(L)=1 and f($\partial$K)=0, then extend f by zero. – Strongart Jan 1 '16 at 4:49

I claim that any maximal trivial subspace $Y$ is dense in $X$.

Otherwise, take $x\in X\setminus\operatorname{Cl}( Y) \neq\emptyset$. Since $X$ is locally compact Hausdorff, and hence regular, there are non-intersecting open neighbourhoods of $x$ and $\operatorname{Cl}(Y)$. It follows that the subspace topology on $Y\cup\{x\}$ is the disjoint union topology. Hence $E$ is trivial when restricted to $Y\cup\{x\}$ (since its trivial when restricted to each of $Y$ and $\{x\}$). Hence $Y\cup \{x\}$ is a trivial subspace, contradicting the maximality of $Y$.

• Please do not change the topology of X,, we can choose a small open set U of x, which is trivial subspace and do not intersect Y, then we can get your claim. – Strongart Dec 26 '15 at 4:49
• @Strongart: I don't think we can say that there is a open neighbourhood of $x$ which does not intersect $Y$ and is a trivial subspace, without assuming $X$ is locally contractible or something similar. However, any vector bundle over a point is trivial, hence my argument involving the disjoint union topology. – Mark Grant Dec 26 '15 at 11:37
• I am careless, I do this on the manifold unconscious, there are some topological questions, it need has enough small open set locally. – Strongart Dec 27 '15 at 4:55
• @Sebastian Goette I go back to the basic topology, local compact Hausdorff space maybe be not normal, it is a Tychonoff space. If it also has second countable axiom, it will be normal. Here is a countexample:mathoverflow.net/questions/53300/… – Strongart Dec 29 '15 at 5:19
• @Strongart: I added some clarifying details to my answer. In particular, I think the separation axiom I need is regularity, which follows from locally compact Hausdorff. – Mark Grant Dec 29 '15 at 13:27