On the definition of fundamental vector field

Question

I've encountered with following question while reading Morita's book "Geometry of Differential Forms" (pp.263)

Let $(P,\pi,M,G)$ be a principal G-bundle, $\mathfrak{g}$ be the Lie algebra of the Lie group $G$,

He is trying to define fundamental vector field by showing that there is an isomorphism $$V_u:=T_u(\pi^{-1}(p))\cong\mathfrak{g}\tag{1}$$

Here he uses two claims:

Claim.1:for any $p\in M$,there exist a diffeomorphism $i_p\colon G\to \pi^{-1}(p)$ such that $$i_p(hg)=i_p(h)g\tag{2}$$where $h,g\in G$.

In fact, $i_p=\varphi^{-1}(p,\cdot):G\to \pi^{-1}(p)$ where $\varphi \colon \pi^{-1}(U)\to U\times G$ is any local trivialization on $U$ around $p$. It's quite straight forward to see that $i_p$ is a diffeomorphism and satisfies $(2)$

Claim.2:if $\psi :\pi^{-1}(V)\to V\times G$ is another local trivialization, $j_p=\psi^{-1}(p,\cdot):G\to \pi^{-1}(p)$, then $$j_p=i_p\circ L_g$$ for some $g\in G$

Then he concludes:for any $u\in \pi^{-1}(p)$, there exists an isomorphism $(1)$.

I have several questions here:

1.Can we write down the isomorphism $(1)$ explicitly? Claim.2 seems to be used by letting $(i_p)_\ast$ act on a left invariant vector field on $G$, so is the isomorphism $(1)$ given by $$\mathfrak{g}\to V_u,X\mapsto (i_p)_\ast(X)\arrowvert_u?$$

2.Do we really need property $(2)$, which looks like an equivariant condition?

3.Does the isomorphism $(1)$ depend on $u\in \pi^{-1}(p)$ or does it only depend on $p=\pi(u)\in M$?

So much effort, thanks for any hints or answers, as well as any references!

Answer 1

The isomorphism depends on $u$ because it occurs at the point $u \in \pi^{-1}(p)$. You could say that you simultaneously identify all tangent spaces of each fiber by identifying the fiber with $G$ and then right translating the tangent spaces of $G$ to trivialize the tangent bundle of $G$. The isomorphism indeed does not depend on property (2), since property (1) uniquely determines the diffeomorphism to $G$, up to left translation. We can write down the isomorphism explicitly as you did.

Answer 2

To understand the isomorphism (1) it may help to look at the principal bundle of frames $P\to M$ of a real vector bundle $E\to M$ of rank $r$. The fiber of this bundle over $x\in M$ consists of all the frames (bases) of the $r$-dimensional vector spaces $E_x$. This space cannot be canonically identified with the Lie group $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\GL}{GL}$ $\GL_r(\bR)$.

However, once you fix a basis $\beta_0$ of $E_x$, any other basis $\beta$ can be identified with an invertible $r\times r$-matrix $T(\beta_0,\beta)$ whose columns represent the coordinates of the vectors in $\beta$ with respect to the fixed basis $\beta_0$.

The diffeomorphism $P_x\to \GL_r(\bR)$ then takes the form

$$ P_x\ni \beta\mapsto T(\beta_0,\beta)\in \GL_r(\bR). $$

In particular, the diffeomorphism (1) it depends on the choice of $\beta_0$. This isomorphism is also equivariant because

$$T(\beta_0,\beta g)= T(\beta_0,\beta)g,\;\;\forall g\in\GL_r(\bR). $$

(The natural action of $\GL_r(\bR)$ on the space of frames of $E_x$ is a right action.)

You may want to have a look at Sec. 2.3.3. of these lectures.

Answer 3

The action of $G$ on $P$ defines, for each $u \in P$, a smooth embedding $\Phi_u: G \rightarrow P$, where $\Phi_u(1) = u$. If $p = \pi(u)$, then $\Phi_u(G) = \pi^{-1}(p).$ Therefore, the differential of $\Phi_u$ at $1 \in G$, $$(\Phi_u)_*: \mathfrak{g} = T_1G \rightarrow T_u(\pi^{-1}(p)) \subset T_uP,$$ is an isomorphism, proving (1). Moreover, there is a natural map from $\mathfrak{g}$ to vector fields on $P$, where if $X \in \mathfrak{g}$, the corresponding vector field $V_X$ is given by $$ V_X(u) = (\Phi_u)_*X. $$

Answer 4

Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a right invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also define: For every $p\in P$, you can define ${d\over{dt}}exp(tu).p\in T_pP$.

Answer 5

In the book Foundations of Differential Geometry by Kobayashi & Nomizu, the fundamental vector field is defined in section 4, pg.42:

If $G$ acts on $M$ on the right, we assign to each element $X\in G'$ a vector field $X^*$ [called the fundamental vector field] on $M$ as follows. The action of the 1-parameter subgroup $a_t:=exp(tX)$ on $M$ induces a vector field $X^*$ on $M$.

Where we have denoted $G'$ as the Lie algebra of $G$. Also the following proposition 4.2 states using the same setup

The mapping $\sigma: G' \rightarrow X(M) $ which sends $X$ into $X^*$ is a Lie algebra morphism. If $G$ acts effectively on $M$, then $\sigma$ is an isomorphism of $G'$ into $X(M)$. If $G$ acts freely on $M$, then, for each non-zero $X\in G'$, $\sigma(X)$ never vanishes on $M$.

Then in section 5, where $P\rightarrow M$ is a principal bundle with structure group $G$ they write:

Since the action of $G$ sends each fibre into itself, $(X^*)_p$ is tangent to the fibre at each $p\in P$. As $G$ acts freely on $P$ [by definition of a principal bundle], $X^*$ never vanishes on $P$ (if $X!=0$) by proposition 4.1. The dimension of each fibre being equal to that of $G'$, the mapping $X\rightarrow (X^*)_p$ of $G'$ into $T_pP$ is a linear isomorphism of $G'$ onto the tangent space at $p$ of the fibre through $p$.

Also, on the book Natural Operations in Differential Geometry by Michor, Kolar & Slovak which is freely available on the net they define the fundamental field and establish some of its properties in lemma 5.12 and then in theorem 10.18 they write:

The vertical bundle $VP\rightarrow P$ of the principal bundle $P\rightarrow M$ is trivial as a vector bundle over $P$; hence $VP =~ P \times G'$

Then in the following section they show how to obtain a connection form $\omega$ from any connection on $P$; the form $\omega$ is a 1-form on $P$ valued in $G'$ and they relate this in a lemma to the fundamental field:

$\omega$ reproduces the generators of the fundamental fields: $\omega(X^*_p) = X$

Finally, Nlab defines the dual notion of a fundamental form:

Given a Lie group $G$ with a right action on a manifold $M$ then a fundamental 1-form is a 1-form valued in $G'$, that is $\omega \in \Gamma(T^*M \otimes G')$, such that $\omega_p(X^*_p)=X$

In this language, a connection 1-form is simply a fundamental 1-form as Michor Lemma 11.1 points out; and it's a principal connection iff it is $G$ equivariant, that is:

$((r^g)\omega)(X_p) = Ad(g^{-1}.\omega(X_p)$