Vectors tangent to the fiber of a Principal Fiber bundle

Question

I am trying to understand the definition of a connection on a principal fiber bundle and I don't understand the following. I have taken this from the book "Foundation of differential geometry vol. 1":

Let $P(M,G)$ be a principal fibre bundle over M. Let $ T_pP$ be the tangent space of $P$ at point $p \in P$ and $G_p$ the subspace of $T_pP$ consisting of vectors tangent to the fibre through p.

Questions: (1):What is the definition of a vector tangent to the fibre through p?

(2)Is there a way to visualize the subspace $G_p$ in a simple example? Any help would be appreciated. Thank you and have a nice day.

Answer 1

Let's say $\pi \colon P \rightarrow M$ is a fiber bundle. The fiber $\pi^{-1}(q)$ through $q \in M$ is a submanifold of $P$ (diffeomorphic to $G$ in your case, but this is not really relevant for what follows). Choose a point in the fiber $p \in \pi^{-1}(q)$. A tangent vector at $p \in P$ which is tangent to the fiber is just a tangent vector which belongs to the tangent space of the fiber $\pi^{-1}(q)$ (identified as a subspace of the tangent space to the total space $T_pP$). Such a tangent vector is represented by an equivalence class $[\alpha]$ of smooth curves $\alpha \colon I \rightarrow \pi^{-1}(q)$ with $\alpha(0) = p$ (that is, curves in $P$ that pass through $p$ and stay on the fiber). Since $\pi(\alpha(t)) \equiv q$ is a constant curve on $M$, by differentiating we get that

$$ d\pi|_{p}(\dot{\alpha}(0)) = 0 $$

so the tangent vector $\dot{\alpha}(0)$ lies in $\ker(d\pi|_p)$. By using the local model of a fiber bundle, you can see that the converse also holds so the tangent space to the fiber $\pi^{-1}(q)$ at $p$ is just the kernel $\ker(d\pi|_p)$.

Answer 2

Just as in levap's answer, the best way to think of the vertical tangent space $G_{p}$ at any point $p \in P$ is as the kernel $\text{ker}(d\pi_{p})$ of the pushforward of the projection. The projection $\pi$ is part of the data defining the principal bundle, so intuitively, you should think of the vertical vectors as those projecting to zero when pushedforward by $\pi$.

It's also not hard to establish an isomorphism $G_{p} \cong \mathfrak{g}$ where $\mathfrak{g}$ is the Lie algebra of $G$. Associated to any $X \in \mathfrak{g}$, you have the canonical vector field $\sigma(X)$ on $G$, whose vector at $g \in G$

$$\sigma_{g}(X) = \frac{d}{dt}\bigg(g \cdot e^{tX}\bigg)\bigg|_{t=0}.$$

Here, $g \cdot e^{tX}$ just denotes $e^{tX}$ acting on the right on another group element. So this is really a canonical vector field on the Lie group but since in a principal bundle, your fibers are diffeomorphic to $G$, you can think of this as a canonical vertical vector field on $P$ associated to $X \in \mathfrak{g}$. Replace the expression above by

$$\sigma_{p}(X) = \frac{d}{dt}\bigg(p \cdot e^{tX}\bigg)\bigg|_{t=0},$$

for all $p \in P$ and not that this is a vertical vector. Why? Because the right G-action on $P$ is fiberwise! The map $\mathfrak{g} \to G_{p}$ sending $X$ to $\sigma_{p}(X)$ is an isomorphism, which identifies the vertical tangent space as the Lie algebra.

From this, note that when you hear a connection described as a Lie algebra-valued form this may also be described as a form valued in a vertical tangent space in other sources/contexts.