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Principal bundle: Why does the fundamental vector field live in the vertical subspace?

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Let us consider a principal bundle \(P \overset{\pi }{\to} M\), let \(u \in P\) and let \(G_p\) be the fibre at \(p = \pi(u)\). According to my book, the vertical subspace \(V_uP\) is defined as the subspace of \(T_uP\)  which is tangent to \(G_p\) at \(u\). Subsequently, let \(A \in \mathfrak{g}\), where \(\mathfrak{g}\) denotes the Lie algebra, then the fundamental vector \(A^\sharp \in T_uP\) is defined by:

\(A^\sharp f(u) = \frac{\mathrm{d}}{\mathrm{d} t} f (u \exp (t A))|_{t=0}\)

where \(f: P \to \mathbb{R}\) is an arbitrary smooth function. Now the book states:

The vector \(A^\sharp\) is tangent to \(P\) at \(u\), hence \(A^\# \in V_uP\).

Why does \(A^\sharp \in V_uP\)?

P.S. I do understand that since we are considering a principal bundle, we have:

\(\pi(u) = \pi(u \exp (t A)) = p\)

and so \(u \exp (t A) \in G_p\), but I'm not sure how/if this is helpful.

asked May 21, 2014 in Mathematics by Hunter (510 points) [ revision history ]
edited May 21, 2014 by Hunter

1 Answer

+ 2 like - 0 dislike

As a general (quite obvious) result of differential geometry, if $t\mapsto g_t$ is a one-parameter group of diffeomorphisms on a manifold $M$, and $x\in M$, it holds

$$\frac{d}{dt}|_{t=0} f(g_t(x)) = X_x(f)$$

where $X$ is the vector field whose flow is $\{g_t\}_{t\in \mathbb R}$. In other words, for each fixed $x\in M$ the vector $X_x$ is the initial (i.e. at $t=0$) tangent vector to the curve $t \mapsto g_t(x)$  which just starts at $x$ for $t=0$.

In case of a principal bundle, $P$, the structure group $G$ moves points $u\in P$ along each fibre $G_u$ separately (this is exactly you wrote in your final comment). If $u\in P$, the curve $t \mapsto u\exp\{tA\}$ is the action of the one-parameter subgroup $\{\exp\{tA\}\}_{t\in \mathbb R} \subset G$ on the point $u$. This curve is completely included in $G_u$ since the action of $G$ moves $u$ along the fibre $G_u$. Hence the  vector field $X= A^\sharp$, which is by definition tangent to the curve $t \mapsto u\exp\{tA\} \subset G_u$,  must be tangent to the fibre by construction.  In other words it is vertical. 

answered May 21, 2014 by Valter Moretti (2,025 points) [ revision history ]
edited May 21, 2014 by Valter Moretti

@V.Moretti thanks for your answer. I'm still a bit confused. If I understand your notation correctly, then you say that are actually considering the general result:

\(\frac{\mathrm{d}}{\mathrm{d}t} (g \exp(tA)) |_{t=0} = X(g)\)

for \(g \in G\).

If we are to convert the above equation to the principal bundle, then shouldn't we actually consider the equation:

\(A^\sharp (u) = \frac{\mathrm{d}}{\mathrm{d}t} (u \exp(tA))|_{t=0}\)

instead of what my book says? Why are we interested in \(f: P \to \mathbb{R}\)\(f\) cannot be the one-parameter group, right (because then it should be the curve \(f : \mathbb{R} \to G\))? Note that the book I'm reading has a ridiculous high amount of errors (even though it is the second edition), and so I can never trust it.

The general result (valid for $f:M \to \mathbb R$ with $M$ generic manifold) is $$d/dt|_{t=0}f(g_t(x)) = X_x(f)$$

where $X_x = dg_t(x)/dt|_{t=0}$.

In your case $x=u$ and the action of the one parameter group generated by $A$ is indicated by  $g_t(u) = u \exp\{tA\}$, since the action of the group $G$ on $u\in P$ is denoted by $ug$. Is it OK?

Actually there is no  reason for defining $A^\sharp$ using smooth functions, even if the identity you quoted is correct. On can equivalently and more simply define $$A^\sharp_u :=\left. \frac{d}{dt} u\exp\{t A\}\right|_{t=0}\:.$$

But then should we not consider:

\(A^\sharp (f(u)) = \frac{\mathrm{d}}{\mathrm{d} t} f (u \exp (t A))|_{t=0}\)

instead of:

\(A^\sharp f(u) = \frac{\mathrm{d}}{\mathrm{d} t} f (u \exp (t A))|_{t=0}\)?

Well, perhaps the most correct notation should be $(A^\sharp f)_u$: First the vector field/differential operator $A^\sharp$ acts on the smooth function $f$ producing another smooth function $A^\sharp f$, next the obtained function is evaluated at $u$. Independently from the used notation, that is the meaning of the said identity.

$A^\sharp_u \in T_uP$ is the differential operator localized at $u$ defined by

$$A^\sharp_u f := (A^\sharp f)_u\:,$$

for every $f\in C^\infty(P)$.

Ok, thanks a lot for your help! I think it makes sense now.

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