# Topology of maps between fibers of vector bundles

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

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Are E and F over the same space? – Ketil Tveiten Jan 14 '11 at 15:19
I don't know about a topology (actually yes) but it certainly exists a natural diffeology, and by the way a D-topology. I use a similar construction to define fiber bundles in diffeology (considering the groupoid of diffeomorphisms between fibers of a projection). Just comment if you are interested, I'll elaborate. – Patrick I-Z Jan 14 '11 at 15:37
You will have to clean up the question and clarify a bit: if the fibers are not finite dimensional, what structure do they have? Secondly, the question as stated makes it sound like you are interested in the topology of the space of linear maps from one v.s. to another; there are several different answers to this question depending on your clarification. – Sheikraisinrollbank Jan 14 '11 at 15:51
Ketil, E and F are not necessarily the same space. – Ferraiol Jan 14 '11 at 15:53
Patrick, I know nothing diffeology, but maybe the construction you use can clarify something to me. If you could say something about it i'll appreciate – Ferraiol Jan 14 '11 at 15:57

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$(pr_1^*(p_2,E_2,X_2),pr_2^*(p_1,E_1,X_1))$.

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Apparently he is interested infinite dimensional fibres. – Sheikraisinrollbank Jan 14 '11 at 15:53
Henrik... thanks for the answer. I'll try develop something in this way and post the answer soon – Ferraiol Jan 14 '11 at 16:09

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 $G_i$.

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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.

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