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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 !

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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.

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  • $\begingroup$ 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 $E_b$ ? $\endgroup$
    – Gericault
    Nov 17, 2018 at 16:53
  • $\begingroup$ @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 $\Phi_j$? $\endgroup$
    – Mike F
    Nov 17, 2018 at 17:53
  • $\begingroup$ @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? $\endgroup$
    – Mike F
    Nov 17, 2018 at 17:56
  • $\begingroup$ 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 $R^n$ ? $\endgroup$
    – Gericault
    Nov 17, 2018 at 18:39
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    $\begingroup$ @Gericault: I agree with all that $\endgroup$
    – Mike F
    Nov 17, 2018 at 18:46

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