# Vector space structure on the fiber of a vector bundle

While trying to get accustomed to the very definitions of some usual objects in geometry/topology, a lot of questions I could find complete answers to came to my mind. I will try to be clear :

Def (Vector Bundle over a base B) : Let $$E$$ be a manifold and $$\pi: E \rightarrow B$$ a submersion, $$E$$ is a locally trivial vector bundle of rank $$r$$ over B if we can cover B with open subsets : $$B = \cup_{i} U_i$$ such that we have local trivialisations (diffeomorphisms) : $$\Phi_i : \pi^{-1}(U_i) \rightarrow U_i \times \mathbb{K}^{r}$$ such that the classical diagram commutes, i.e : $$pr_1 \circ \Phi_i = \pi$$.

Now we ask for a compatibility condition : on any $$U_i \cap U_j$$, we know from the above condition that the diffeomorphism $$\Phi_i \circ \Phi_j^{-1} : U_i \cap U_j \times \mathbb{K}^r \rightarrow U_i \cap U_j \times \mathbb{K}^r$$ is of the form : $$\Phi_i \circ \Phi_j^{-1}(p,v) = (p, \varphi_{ij}(p).v)$$

And we ask that $$\varphi_{ij}(p)$$ is a linear isomorphism instead of just a diffeo, and that it depends smoothly on p, in other words : $$\varphi_{ij}: U_i \cap U_j \rightarrow GL_r(\mathbb{K})$$ smooth.

Now, we say that as a consequence, the vector space structure defined on a given fiber does not depend on the choice of the trivialisation (say the choice of $$i$$ or $$j$$ on the interesection).

I understand it is natural, but there is something that bugs me.

Here is how I see things : Pick a point $$b$$ in $$U_i \cap U_j$$, "at the beginning" the fiber $$E_b$$ is only a topological space/submanifold of E. but by identifying $$E_b \rightarrow \{b\}\times \mathbb{K}^s \approx \mathbb{K}^s$$ through $$\Phi_i$$ or $$\Phi_j$$ we give it a vector space structure in the following way :

For instance if $$v_1,v_2 \in E_b$$ we say $$v_1 + v_2 = \Phi_i^{-1}(\Phi_i(v_1) + \Phi_i(v_2))$$ or $$v_1 + v_2 = \Phi_j^{-1}(\Phi_j(v_1) + \Phi_j(v_2))$$ right ?

I agree this gives two different vector space structure, and we have : $$\Phi_i(v_1) + \Phi_i(v_2) = (\Phi_i \circ \Phi_j^{-1})(\Phi_j(v_1) + \Phi_j(v_2))$$

But : as we compose by $$\Phi_i^{-1}$$ or $$\Phi_j^{-1}$$ to go back in $$E_b$$ I don't see we relate the two structures on $$E_b$$ ? How to quantize that this structure is "independent of the choice of $$i$$ or $$j$$" ?

Given a "blank" space $$V$$, "how many" different vector space structures on a given field can we give to it ? Are they not isomorphic ? Does it relate to the theorem that says that if you have an homomorphism between two subsets of $$R^n$$ they have the same dimension (as manifolds) (so that would mean given a "blank" space, at least the dimension it will have as a vector space is unique ?

Any clarification welcome !

I think you are homing in on the right ideas, but have made some slip-ups. You say that $$v_1$$ is in the fiber $$E_b = \pi^{-1}(b) \subseteq E$$, but then you start talking about $$\Phi_i(b,v_1)$$ which does not make sense. You wrote that the domain of the chart $$\Phi_i$$ is $$\pi^{-1}(U_i) \subseteq E$$, but here you are applying $$\Phi_i$$ to $$(b,v_1) \in B \times E$$.
Instead, what you should be doing is saying that $$\Phi_i(v_1)$$ is in $$\{b\} \times \mathbb{K}^r$$ so that $$\Phi_i(v_1) = (b,w_1)$$ and, similarly, $$\Phi_i(v_2)=(b,w_2)$$ where $$w_1,w_2 \in \mathbb{K}^r$$. Then the addition of $$v_1$$ and $$v_2$$ which you define wil be based on the the operation $$(b,w_1) + (b,w_2) = (b,w_1+w_2)$$ in $$B \times \mathbb{K}^r$$.
I think after making these fixes you will have no trouble establishing that the resulting addition operation on $$E_b$$ is independent of which chart you used to define it.
• Thanks for your answer ! That's right, it was just a typing error though, I'm facing the same issues after correction $(b,w_1) + (b,w_2) = (b,w_1 + w_2)$ was what I meant by $\{b\}\times \mathbb{K}^s \approx \mathbb{K}^s$ Additionnally, would you have some kind of answer to the question : "how many" vector structure could we have put on $E_b$ ? Nov 17, 2018 at 16:53
• @Gericault: So do you now understand why the vector space structure (i.e. addition and scalar multiplication) on $E_b$ coming from $\Phi_i$ is equal to the vector space structure coming from $\Phi_j$? Nov 17, 2018 at 17:53
• @Gericault: Regarding your question about the number of possible vector space structures on a "blank" space $V$, it depends what $V$ is to begin with. Is it just a set? Does it come with a smooth manifold structure? Do we have a distinguished point in $V$ which we must use as the zero vector? Nov 17, 2018 at 17:56
• Yes, it became clear indeed ! As for my question, if we fix the field to be $\mathbb{R}$ for instance. I'm guessing: - if V is only a set, then if 2<card(V) < card(R), then V cannot be equipped with a vector space structure, and if card(V) = card(R) for instance, we can get any vectorial space of finite dimension or something like this ? - On the contrary, if V is a differential manifold, the dimension is fixed, and we can get as many structures as diffeomorphisms : $V \rightarrow \mathbb{R}^n$ mod the relation $\phi \approx \psi$ if $\phi\psi^{-1}$ is a linear isomorphism of $R^n$ ? Nov 17, 2018 at 18:39