Topology of maps between fibers of vector bundles - MathOverflow

Topology of maps between fibers of vector bundles

2011-01-14T15:07:36Z

First of all sorry for the (possible) incorrect english. I don't know english very well.

I'm with a doubt about topology of maps between fibres of vector bundles.

Consider $E$ and $F$ vector bundles and the set of all linear maps from a fibre of $E$ to a fibre of $F$, ie, the set of all linear maps $T:E_x \rightarrow F_y$, where $E_x$ is the fiber over $x$ and $F_y$ is the fiber over $y$.

I want to know a way to define the topology of this set!

I need this topology for this question: Consider $f: E \rightarrow F$, a map that preserves each fiber and its restriction to each fibre, $f_x : E_x \rightarrow F_y$, is differentiable. The differential of $f_x$ calculated in a vector $v \in E_x$ is the linear maps $df_x(v):E_x \rightarrow F_y$. I want to say that $f$ is a $C^1$ map if the function $v \in E \rightarrow df_{\pi(v)} (v)$ is continuous. And for this I need a topology for the set defined above.

Anybody know how to define the topology? What means two of these maps are close to each other

Note that $df_x (v)$ is not necessarily a homomorphism, because this is only defined over the fiber that contains $v$.

Thanks

Answer by HenrikRüping for Topology of maps between fibers of vector bundles

2011-01-14T15:51:11Z

If both bundles were trivial, say $X\times \mathbb{R}^n\rightarrow X$ and $Y\times \mathbb{R}^n\rightarrow Y$ one could just take $X\times Y\times M(m,n,\mathbb{R})$. Different choices of trivializations should give homeomorphic spaces, so this topology seems to be right.

In general the bundles are just locally trivial. However this is enough to write down a basis for the topology.

I don't see, what nice properties this topology might have but it seems to be the canonical choice.

EDIT: I think one can reduce this to classical vector bundle constructions in the following way. Given two vector bundles $p_i:E_i\rightarrow X_i$ (for $i=1,2$). Let $pr_i:X_1\times X_2\rightarrow X_i$ denote the projections. Then the desired bundle is just Hom$(pr_1^*(p_2,E_2,X_2),pr_2^*(p_1,E_1,X_1))$.

Answer by Patrick I-Z for Topology of maps between fibers of vector bundles

2011-01-14T16:53:23Z

I elaborate a little bit on what your question inspires me. I will treat a more general question, but you can reduce it to finite linear dimensional fiber bundles over manifolds. Let $\pi : E \to X$ and $\pi' : E' \to X$ be two projections (smooth in some sense, and you can assume that you deal with manifolds). You consider the set $$ M = \{ f \in C^\infty(E_x,E'_y) \mid \mbox{for some } x, y \in X \} $$ You have two natural projections ${\it src} : f \mapsto x$ and ${\it trg} : f \mapsto y$, with $ \in C^\infty(E_x,E'_y)$. Let $r \mapsto f_r$ be a parametrization, that means a map defined on some open subset $U$ of some vector space ${\bf R}^n$, for some $n \in {\bf N}$. We will say that $r \mapsto f_r$ is smooth if

1) The map $r \mapsto x_r = {\it src}(f_r)$ is smooth.

2) The evaluation map $(r,e) \mapsto f_r(e)$ defined $$ \mbox{from } \ \{ (r,e) \in U \times E \mid e \in E_{x_r} \} \ \mbox{ to }\ E' $$ is smooth. The source of the evaluation map has a nice induced smooth structure.

These parametrizations define on $M$ what is called a diffeology. Now, every time you have a diffeology you get a topology by considering the finest topology such that the smooth parametrizations are continuous. You may change everywhere $C^\infty$ by $C^0$ I don't think it will change much, if you prefer continuous maps. So, a subset $\Omega \subset M$ is open iff its preimage by any smooth parametrization is open.

Writing this stuff down, I realize that it is maybe overkilling for the question you ask. Since the bundle are locally trivial, by using a trivialization and using the fact that you consider just linear maps (changing $f \in C^\infty(E_x,E'_y)$ to $f \in {\rm L}(E_x,E'_y)$, you may simplify this construction. But it will get you a natural topology.

I don't know if it may help you but it can inspire you to build what you are looking for.

Actually, in your case, this construction gives you a fiber bundle, as notices Johannes Ebert in next answer. So, for finite dimension fiber bundle, it is really overkill. Sorry.

Answer by Johannes Ebert for Topology of maps between fibers of vector bundles

2011-01-14T17:06:54Z

Let $G_i$, $i=0,1$, be topological groups and $P_i \to X_i$ be $G_i$-principal bundles. Then $P_0 \times P_1 \to X_0 \times X_1$ is a $G_0 \times G_1$-principal bundle. Let $V_i$ be topological vector spaces with continuous $G_i$-actions. The group $G_0 \times G_^$ acts on $Hom(V_0,V_1)$. If the topology on the Hom-space is appropriately chosen, the action is continuous and you can form the bundle $(P_0 \times P_1) \times_{G_0 \times G_1 } Hom (V_0,V_1)$ on $X_0 \times X_1$. A point in the total space is precisely a linear map between two fibres. In the finite-dimensional case, you can take $P_i$ to be the frame bundle of $E_i$. All the topologies on linear groups and Hom-spaces are unique and you get a unique answer. In the infinite-dimensional case, you need to be very careful about the topology on the groups $G_i$.