Differentiating a Linear Functional

[Kreizhn]

Hey All,

Here's a stupid and probably ridiculously easy question, but I want to make sure that I have it right.

Let $G$ be a Lie group with Lie algebra $\mathfrak g$. Assume that $H \in \mathfrak g$ and $\phi \in \mathfrak g^*$ the algebraic dual. Assume that $X(t)$ is an integral curve satisfying
$$\dot X(t) = HX(t)$$
and we have a function defined as $\mathcal H(X,\phi) = \phi(HX(t))$.

(For anyone familiar with geometric control theory, this is essentially Pontryagin's principle only greatly simplified for non-control theorists)

Now I want to calculate $\frac{d\mathcal H}{dX}$ so my question is as follows: Can we pull the $X(t)$ out of the linear functional since it's only a functional on H?

See, because it's a linear function, there's something that is telling me that differentiating it with respect to X should just give $\phi$. Something that I can partly corroborate by the fact that we should get
$$\frac{d}{dX} \langle \phi, X \rangle = \phi$$
However, if I cannot pull the X outside of the functional, then I end up getting something along the lines of
$$\begin{align*}<br /> \frac d{dX} \phi(HX(t)) &= \phi'(HX(t)) H<br /> \end{align*}$$
and between not knowing what $\phi'$ is and that last statement looking pretty useless, I'm not sure if I've done something wrong.

Edit: Messed up last equation. Fixed it.

 

[holy_toaster]

I find your notation a bit confusing, especially because I do not know how to make sense of a derivative by an integral curve $$\frac d{dX}$$.

If you mean by this the derivative along the integral curve, which should be denoted $$\frac d{dt}$$ the second of your possibilities seems right to me. Because $$\phi(HX(t))$$ is a function along the integral curve in this case. Its derivative or its change along the curve does of course also depend on how $$\phi$$ is changing along the curve.

Things simplify to a chain rule in this case.

 

[Kreizhn]

Yes, I can see what you're saying and unfortunately, I'm honestly not sure how to respond. I will briefly try to introduce the theory and hopefully it'll clear something up (for either of us).

Let M be a smooth manifold and H be a smooth, complete vector field on M. We know that the differential equation
$$\frac{d}{dt} x(t) = H(x(t))$$
then defines an integral curve on for any point on M. We can bump the vector field H to a smooth Hamiltonian function on the cotangent bundle $$\mathcal H \in C^\infty(T^*M)$$ by demanding that if $(x,\phi) \in T^*M$ then
$$\mathcal H(x,\phi) = \phi(H(x))$$
Then $\mathcal H$ can be lifted to a Hamiltonian vector field $\mathfrak h$ on $T^*M$ by demanding that is satisfy
$$\iota_{\mathfrak h} \omega = -dH$$
Then the integral curves of the new differential equation
$$\dot\xi(t) = \mathfrak h (\xi(t))$$
are given by (if $\xi(t) = (x(t),\phi(t))$)
$$\frac{ dx}{dt} = \frac{\partial \mathcal H}{\partial \phi}, \qquad \frac{ d\phi}{dt} = -\frac{\partial \mathcal H}{\partial x}$$
These are the Hamiltonian equations of motion.

The issue comes in that I'm not certain how to calculate
$$\frac{ d\phi}{dt} = -\frac{\partial \mathcal H}{\partial x}$$

 

[Kreizhn]

Now, perhaps the problem comes in the in my problem, we're working on a Lie group/algebra so that the vector field H is right invariant and we can write $H(x) = Hx$.