I'm trying to understand how the calculus of variations works in the setting of smooth manifolds. The texts I'm reading tend to switch from euclidean space to manifolds as if there is no difference, but I'm someone who (at least once) needs to see things done more formally.
The Case of Euclidean Space: $ \ $ Let $I = [a, b] \subset \mathbb{R}$ and let $q_0: I \to \mathbb{R}^n$ be a curve. We define a deformation of $q_0$ with fixed endpoints as another curve $q: I \times (-\epsilon, \epsilon) \to \mathbb{R}^n$ satisfying:
- $q(t, 0) = q_0(t)$ for all $t \in I$.
- $q(a, s) = q_0(a)$ and $q(b, s) = q_0(b)$ for all $s \in (-\epsilon, \epsilon)$.
The variation of $q_0$ with respect to the deformation $q$, denoted $\delta q_0$, is defined by:
$$ \left. \delta q_0(t) \right. = \left. \frac{\partial}{\partial s}q(t, s) \right|_{s=0} $$
Smooth Manifolds: The definitions of curves and thus deformations work essentially the same, just replacing $\mathbb{R}^n$ by some manifold $Q$. To define the variation, one option is to move to local charts and then—carefully—glue the pieces together.
Partition $[a, b]$ as $\{a = t_0, t_1, \cdots, t_{n-1}, t_n = b\}$ so that the restriction of $q$ to the subinterval $[t_i, t_{i+1}]$ is contained in the local chart $(U_i, \phi_i)$. Within each chart, the previous definition of $\delta q_0$ is well-defined:
$$ \left. \delta (\phi \circ q_0)(t) \right. = \left. \frac{\partial}{\partial s} (\phi \circ q)(t, s) \right|_{s=0} $$
It can also be shown that these "local" deformations agree with each other on the intersection of charts, so conceptually speaking it shouldn't be an issue to glue the pieces back together. However, I'm not completely sure where this object lives or why, so I don't feel comfortable proceeding.
Should the deformation be another curve on $Q$? A curve on $TQ$? Does anybody know of a more clear way of doing this?
(I'm just recently learning this stuff myself, so take whatever I say with some caution, check the details of it, and take this more as an extended comment rather than a full answer.)
Let $Q$ be a smooth manifold, and let $q_0 : I \to Q$ be a given smooth curve. I think a smooth deformation of $q_0$ with fixed endpoints should be a smooth map $q: I \times (-\epsilon , \epsilon) \to Q$ such that:
However, I think the variation of $q_0$ at the point $t$, denoted by $\delta q_0(t)$, should be an element of the tangent space $T_{q_0(t)}Q$. Ideally, we want this to mimic the euclidean case of it being $\dfrac{\partial}{\partial s} \bigg|_{s=0} q(t,s)$. So, to formalise this idea, fix a point $t \in I$, and define $q^t:(-\epsilon, \epsilon) \to Q$ by $q^t(s) = q(t,s)$. Now, define \begin{align} \delta q_0(t) &= (q^t)_{*,0}(1) \in T_{q^t(0)}Q = T_{q_0(t)}Q \end{align}
Just to clarify the notation, if $f:M \to N$ is a smooth map between smooth manifolds, and $p\in M$, then I use $f_{*,p}: T_pM \to T_{f(p)}N$ to mean the push-forward mapping between tangent spaces. (The $1$ in $(q^t)_{*,0}(1)$ is the unit tangent vector in the tangent space $T_0(-\epsilon, \epsilon)$)
So, just to recap, the deformation is defined in the same way as the Euclidean case; just replace $\Bbb{R}^n$ with $Q$ everywhere. Now, the variation of $q_0$ at $t$ is a tangent vector $\delta q_0(t) \in T_{q_0(t)}Q$, which in essence (after using a chart) is pretty much the same partial derivative at $s=0$ which you mentioned.
The variation $\delta q_0: I \to TQ$, $t \mapsto \delta q_0(t)$ is now a curve in the tangent bundle.