Definition of projective bundle

Question

What is the definition of projective bundle?

In this website, projective bundle is defined as

and in this article, the definition is

I don't know whether they are equivalent, could you please give me a formal definition? Thanks!

Answer 1

The definitions are equivalent. The short answer is that the sources you cite are describing vector bundles in two different ways.

The first one considers a vector bundle to be a map $\pi:E\to B$ in which the fibers $\pi^{-1}(b)$ are vector spaces and such that every point $b\in B$ has a neighborhood $U$ such that $\pi^{-1}(U)\cong U\times V$ where $V$ is a vector space.

The second one says that a vector bundle is a map $\pi:E\to B$ such that $\pi^{-1}(U_i)\cong U_i\times V$ for suitable $U_i$ along with the data of the "transition maps", i.e. the maps that tell you how to go from one "chart" $\pi^{-1}(U_i)$ to another $\pi^{-1}(U_j)$ when $U_i\cap U_j$ is nonempty. It's a routine albeit tedious exercise to show that these definitions are equivalent, meaning that any $\pi:E\to B$ which is a vector bundle in one sense is a vector bundle in the other sense. If you remain unconvinced then I suggest you try proving this yourself.

Let's assume you take the above equivalence for granted. Notice that any linear transformation of $T:V\to V$ induces a map $\bar T:\mathbb P(V)\to \mathbb P(V)$ by the formula $\bar T([v])=[Tv]$. Indeed this is well-defined precisely because $T$ preserves scalar multiplication. Well then morally speaking the first definition you give is certainly the correct one, yes? You simply take the space you get when you identify vectors in $E$ with their scalar multiples ($\pi(v_1)=\pi(v_2)$ ensures that they are in the same fiber, so that they are comperable at all, and $v_2=cv_1$ says $v_2$ is a scalar multiple of $v_1$). The second definition is slightly harder to parse, but what this is saying is that the transition maps are just the projectivizations of the transition maps of the parent vector bundle, which should be expected.

The above is not a proof however, so how about this? Consider a vector bundle $\pi:E\to B$. This bundle has transition maps $\phi_{ij}:(U_i\cap U_j)\times V$. Now let's form the projectivization $\bar \pi:\mathbb{P}(E)\to B$ using the first definition. We can show relatively easily that the second definition is also satisfied. Indeed consider two open sets $U_i$ and $U_j$ as above. What does the equivalence relation in the first definition look like on $\pi^{-1}(U_i\cap U_j)\cong (U_i\cap U_j)\times V$? Well, any two equivalent elements have to be in the same fiber, so their first coordinates (in $U_i\cap U_j$) have to be the same. Also, their second coordinates have to be scalar multiples of each other. So this relation turns $(U_i\cap U_j)\times V$ into $(U_i\cap U_j)\times \mathbb P(V)$ ($(b,v)\mapsto (b,[v])$). With this, we see that the map $\phi_{ij}:(U_i\cap U_j)\times \mathbb P(V)\to (U_i\cap U_j)\times V$ induces a map $\bar\phi_{ij}:(U_i\cap U_j)\times \mathbb P(V)\to (U_i\cap U_j)\times \mathbb P(V)$ given by $$(b,[v])\mapsto(b,\bar\phi_{ij}([v]))=(b,[\phi_{ij}(v)]).$$ So our transition maps are the maps induced by the maps from the parent vector bundle. This is exactly what the second definition says, so the first definition implies the second. I'll leave it to you to do the converse.

EDIT: I guess I should note here that this only works in the topological and smooth settings. When moving to the holomorphic setting this is no longer true. After talking with a friend about this, it seems that not every projective bundle lifts to a vector bundle, even in the topological setting; thus in fact the second definition does not imply the first unless you assume this. The details are hairy and beyond the scope of this answer, but I thought I should relay this information here. This answer is incorrect.