What does it mean for a fiber bundle to be local? equivariant theory

Question

Let $\pi:P \rightarrow B$ be a fiber bundle with fiber $F$. What does it mean to say that locally, $\pi$ is of the form $U\times F\rightarrow U$?

Moreover, suppose, say, product fiber bundles with fiber $F$. Say, that the local trivializations of $U\times F\rightarrow U$ satisfy certain local conditions. How does one the obtain local trivializations on $\pi:E\rightarrow B$ that satisfy the same local conditions?

This idea seems to be a reocorring theme in the course im taking. Though, I don't really understand this.

Answer 1

What does it mean to say that locally, $\pi$ is of the form $U\times F\rightarrow U$?

By definition of a (smooth) fiber bundle, for each point $b\in B$, there is an open neighborhood $U$ of $b$ in $B$, and a (smooth) map $\phi:\pi^{-1}(U)\to U\times F$ such that $\text{pr}_1\circ \phi=\pi|_U$, i.e the map $\phi$ has the format $\phi(x)=(\pi(x),f(x))$ for some (smooth) $f:\pi^{-1}(U)\to F$.

You can consider $(\pi^{-1}(U),\pi,U)$ as a (smooth) fiber bundle in its own right (why? just unwind the definitions). The mapping $\phi$ above provides an isomorphism, in the sense of (smooth) fiber-bundles, from $(\pi^{-1}(U),\pi,U)$ onto the trivial (smooth) fiber bundle $(U\times F,\text{pr}_1,U)$. So, the adjective 'locally' is referring to locality with respect to the base manifold, i.e the clause 'for each $b\in B$, there exists an open neighborhood $U$ of $b$ in $B$'. The phrase 'is of the form' means there exists an isomorphism (in the fiber bundle sense).

Next, you ask

Say, that the local trivializations of $U\times F\rightarrow U$ satisfy certain local conditions. How does one the obtain local trivializations on $\pi:E\rightarrow B$ that satisfy the same local conditions? This idea seems to be a reocorring theme in the course im taking. Though, I don't really understand this.

This is not just a theme in fiber bundles, but all over mathematics. Say I have two objects of the same type $X_1$ and $X_2$, and I have an isomorphism $\phi:X_1\to X_2$. Suppose you already know how to solve a problem on the space $X_2$. Then, you can definitely solve the problem on the space $X_1$ simply by invoking the isomorphism $\phi$ (usually, this involves composing with $\phi$ or $\phi^{-1}$ or maybe in the manifold setting, it may involve the tangent mapping $T\phi$ etc).

Imagine you're in school and you spoke two languages, say English and Russian. Suppose one week your English teacher asked you to write an essay on 'Why math is fun', and the next week, your Russian teacher asked you for the same essay. Then, you could in essence take your English essay, and translate it into Russian and you'd get an essay for free. Same thing above: an isomorphism allows you to translate everything about one space/object to another. So, everything (solving an equation or defining functions/vector fields/differential forms/connections etc) you can do on one, you can do on the other. To get familiar with this, you just have to write out a few proofs in detail till you see the pattern, and then it will become obvious.