# 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?

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