I'm reading some lecture notes on Lagrangian mechanics. The author defines the variational derivative of a function of curves roughly as follows:

Let $Q$ be a manifold and $\Gamma_{a,b}$ be the space of all smooth paths $q: [0,1] \rightarrow Q$ with $ q(0)=a$ and $q(1)=b$ with $a,b \in Q$. Then the variational derivative of a function $S: \Gamma_{a,b} \rightarrow \mathbb{R}$ at $q \in \Gamma_{a,b}$ is defined as $\delta f(q):=\frac{d}{ds}f(q_s)$ at $s=0$, where $q_s : \mathbb{R} \rightarrow \Gamma_{a,b}$ is a "smooth 1-parameter family of paths" with $q_0 = q$.

I don't quite understand how this definition is rigorous. Why is the variational derivative independent of the family of paths chosen? If one wants to talk about the family $(q_s)_{s \in \mathbb{R}}$ being smooth, doesn't $\Gamma_{a,b}$ need some kind of differentiable structure? Otherwise, why would $f(q_s)$ be nessecarily be differentiable with respect to $s$?


I find this an interesting topic which is often ignored in certain Riemannian texts when introducing variational methods when dealing with energy and length minimization. Some of the rigorous specifics which are needed here deals with Hilbert Manifolds, and I believe Klingenberg's text has a chapter on it, but I think we can get an understanding of what's happening here without going into too much of those details.

I'm going to switch notation on you here, as it's just more natural for me type, so apologies for that, but I'll be explicit: Let $M$ be a smooth manifold and fix two points $p,q\in M$. Let $\mathcal{C}(p,q)$ denote the space of all smooth curves $\gamma:[a,b]\to M$ such that $\gamma(a)=p$ and $\gamma(b)=q$. Given $\gamma\in\mathcal{C}(p,q)$, (let $I_\epsilon=[-\epsilon,\epsilon]$) and define a variation of $\gamma$ to be the map $\Gamma:I_\epsilon\times[a,b]\to M$ such that (i.) $\Gamma$ is smooth; (ii.) the map $t\mapsto\Gamma(s,t)$ is in $\mathcal{C}(p,q)$, i.e., $\Gamma(s,\cdot)\in\mathcal{C}(p,q)$ for all $s\in I_\epsilon$; and (iii.) $\Gamma(0,\cdot)=\gamma$.

Let's now think of $\mathcal{C}(p,q)$ as a smooth manifold itself (I know we should be explicit here, but let's just assume we have a smooth structure, as I don't think the technicalities are too beneficial to the question at hand). Fix $\gamma\in\mathcal{C}(p,q)$ and let $\Gamma$ be a variation of $\gamma$. Since $\mathcal{C}(p,q)$ is a manifold of curves, we have the longitudinal curves $t\mapsto\Gamma(s,t)=\Gamma_s(t)$ are points in the manifold $\mathcal{C}(p,q)$ (this is by definition of our variation). This means that our transverse curves $s\mapsto\Gamma(s,t)=\Gamma_t(s)$ are curves in the manifold $\mathcal{C}(p,q)$ (since for each $s$, we get a new point in $\mathcal{C}(p,q)$).

Consider the vector field $V(t)$ defined along $\gamma$ given by $$V(t)=\partial_s\Gamma(0,t)=\Gamma_*\left(\frac{\partial}{\partial s}|_{s=0}\right).$$ Let's call this the variational field of $\Gamma$. Now consider for comparison, a curve on a usual manifold, $\alpha:I_\epsilon\to M$, and we have initial velocity of the curve is given by $\alpha'(0)=\alpha_*(\frac{d}{dt}|_0)\in T_{\alpha(0)}M$. Thus this variational field $V$ is the initial velocity vector of the curve $\Gamma(s,\cdot)$.

Since tangent vectors to a manifold at a point are exactly all initial velocities to curves starting at a point, we've actually characterized all tangent vectors at a point on the manifold $\mathcal{C}(p,q)$. That is, we see that $$V(t)\in T_\gamma\mathcal{C}(p,q).$$

Can we be more explicit with this description of the tangent space? Since, we require that $\Gamma(s,a)=p$ for all $s\in I_\epsilon$, this means that $V(a)=0$ (since it's constant in the $s$-direction). Similarly, since $\Gamma(s,b)=q$ for all $s\in I_\epsilon$, we have that $V(b)=0$. Thus $V\in T_\gamma\mathcal{C}(p,q)$ are precisely the smooth vector fields along $\gamma$ which vanish at the end points. Moreover, if we suppose $M$ has a Riemannian metric $g$, then given any $V(t)\in T_\gamma\mathcal{C}(p,q)$, we can define the variation $\Gamma$ by $$\Gamma(s,t)=\exp_{\gamma(t)}(sV(t)).$$ This shows we actually have a complete characterization $T_\gamma\mathcal{C}(p,q)$ (when $M$ is Riemannian). Moreover, by the usual treatment of characterizing tangent vectors by curve, we know they're independent of choice of curve (this will answer your question of independence of variation).

Now, suppose we have some smooth function $A:\mathcal{C}(p,q)\to\mathbb{R}$ (some might call this an action in Lagrangian mechanics). With our setting, the variational derivative of $A$ at a point $\gamma$ is just the usual notion of exterior differentiation of smooth manifolds. That is, the variational derivative of $A$ at a point $\gamma\in\mathcal{C}(p,q)$ is the exterior derivative $dA_\gamma:T_\gamma\mathcal{C}(p,q)\to\mathbb{R}$.

How do we compute this map acting on tangent vectors in our usual setting of smooth manifolds? Well, we use our curve characterization of tangent vectors. That is, for $V\in T_\gamma\mathcal{C}(p,q)$, we wish to consider $dA_\gamma(V)$. To this end, let $\Gamma$ be variation of $\gamma$ with variational field $V(t)$ (we know this exists by preceding paragraphs), and we have that $$dA_\gamma(V)=\frac{d}{ds}|_{s=0}(A(\Gamma(s,\cdot)),$$ which is exactly how your variational derivative is defined.

Some closing remarks: The above constructions typically use $\mathcal{C}(p,q)$ to consist of piecewise smooth curves, and have piecewise smooth variations and variational vector fields as well (this allows for the concatenation of curves in certain topics, e.g., showing the triangle formula for the distance function on Riemannian manifolds).

Also, in this setting, it can generalize nicely to when we don't necessarily have fixed endpoints of $p,q\in M$. That is, we could actually consider a submanifold $B\subseteq M\times M$ and consider the space of curves $\mathcal{C}(B)$ containing curves $\gamma:[a,b]\to M$ such that $(\gamma(a),\gamma(b))\in B$ and then the tangent space $T_\gamma\mathcal{C}(B)$ consisting of vector fields $V$ along $\gamma$ such that $(V(a),V(b))\in T_{(\gamma(a),\gamma(b))}B$. When $B=\{p\}\times\{q\}$ we recover our initial setting since $T_{(p,q)}(\{p\}\times\{q\})\cong\{0\}\times\{0\}$.


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