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)) &amp;= \phi&#039;(HX(t)) H<br /> \end{align*}<br />
and between not knowing what \phi&#039; 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.