Topology of maps between fibers of vector bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:18:04Z http://mathoverflow.net/feeds/question/52079 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52079/topology-of-maps-between-fibers-of-vector-bundles Topology of maps between fibers of vector bundles Ferraiol 2011-01-14T15:07:36Z 2011-01-14T19:03:37Z <p>First of all sorry for the (possible) incorrect english. I don't know english very well.</p> <p>I'm with a doubt about topology of maps between fibres of vector bundles.</p> <p>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 $y$.</p> <p>I want to know a way to define the topology of this set!</p> <p>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.</p> <p>Anybody know how to define the topology? What means two of these maps are close to each other</p> <p>Note that $df_x (v)$ is not necessarily a homomorphism, because this is only defined over the fiber that contains $v$.</p> <p>Thanks</p> http://mathoverflow.net/questions/52079/topology-of-maps-between-fibers-of-vector-bundles/52084#52084 Answer by HenrikRüping for Topology of maps between fibers of vector bundles HenrikRüping 2011-01-14T15:51:11Z 2011-01-14T19:03:37Z <p>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.</p> <p>In general the bundles are just locally trivial. However this is enough to write down a basis for the topology. </p> <p>I don't see, what nice properties this topology might have but it seems to be the canonical choice.</p> <p>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$(pr_1^*(p_2,E_2,X_2),pr_2^*(p_1,E_1,X_1))$.</p> http://mathoverflow.net/questions/52079/topology-of-maps-between-fibers-of-vector-bundles/52087#52087 Answer by Patrick I-Z for Topology of maps between fibers of vector bundles Patrick I-Z 2011-01-14T16:53:23Z 2011-01-14T17:22:17Z <p>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 <code>$$M = \{ f \in C^\infty(E_x,E'_y) \mid \mbox{for some } x, y \in X \}$$</code> 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 </p> <p>1) The map $r \mapsto x_r = {\it src}(f_r)$ is smooth.</p> <p>2) The evaluation map $(r,e) \mapsto f_r(e)$ defined <code>$$\mbox{from } \ \{ (r,e) \in U \times E \mid e \in E_{x_r} \} \ \mbox{ to }\ E'$$</code> is smooth. The source of the evaluation map has a nice induced smooth structure.</p> <p>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. </p> <p>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. </p> <p>I don't know if it may help you but it can inspire you to build what you are looking for.</p> <hr> <p>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.</p> http://mathoverflow.net/questions/52079/topology-of-maps-between-fibers-of-vector-bundles/52093#52093 Answer by Johannes Ebert for Topology of maps between fibers of vector bundles Johannes Ebert 2011-01-14T17:06:54Z 2011-01-14T17:06:54Z <p>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 $G_i$.</p>