Calculus of variations with differential forms

Question

I want to generalize calculus of variations with differential forms. Or better, I saw it somewhere some time ago, but now I cannot re-build it. Here is what I remember.

Let be $(M, I, \Lambda)$ a triplet where $M$ is a manifold, $I$ an ideal of forms and $\Lambda \in \Omega^n(M)$ a differential n-form on M. Let's denote $S^nM$ the set of all $n$-dimensional sub-manifold of $M$ which annihilate $I$. We can define $F: S^nM \to \mathbb{R}$ such that $F(N) = \int_N\Lambda$. We want to minimize (or maximize) $F$.

Suppose that exists $N \in S^nM$ such that minimizes $F.$ We can define a variation with fixed boundary as a map $\phi: (-1, 1) \to M$ such that:

• $\phi_t: N \to N_t$
• $\phi_0 = id_N$
• $\phi_t | _{\partial N} = id_{\partial N}$
• $\phi_t^*\omega = 0$ for every $\omega \in I$

Now we want to get Euler-Lagrange equation:

$0 = \frac{d}{dt}(F(N_t))|_{t=0} = \frac{d}{dt}(\int_{N_t}\Lambda)|_{t=0} = \frac{d}{dt}(\int_{N}\phi_t^*\Lambda)|_{t=0} = \int_{N}\frac{d}{dt}(\phi_t^*\Lambda)|_{t=0} = \int_{N}\mathcal{L}_X\Lambda = \int_N(di_X\Lambda + i_Xd\Lambda ) = \int_{\partial N}i_X\Lambda + \int_N i_Xd\Lambda$

where $X = \frac{d\phi_t}{dt}|_{t=0}$

Now I suppose the first term is zero because $\phi_t | _{\partial N}$ doesn't depend on $t$.

And the second term? How can I manipulate it to delete the variation as in the normal Euler-Lagrange equation? How can I use the ideal $I$?

Next, I would also like recover from this the classical E-L equations with $M = TQ \times \mathbb{R}$, $I = (dq -\dot{q}dt)$ and $\Lambda = L(q, \dot{q}, t)dt$

Note: I am undergraduate student and I am studying these things by myself

EDIT

My trouble is that I would "easily" conclude that $i_Xd\Lambda = 0 \mod I$ for every $X$. But I think that's not correct because if I apply to my example I get

$i_X(\frac{\partial L}{\partial q}dq\wedge dt + \frac{\partial L}{\partial \dot{q}}d\dot{q}\wedge dt) = 0$

Then I would try with the basis $X = \frac{\partial}{\partial q},\frac{\partial}{\partial \dot q}, \frac{\partial}{\partial t}$ and I get

$ \frac{\partial L}{\partial q}dt = 0 \\ \frac{\partial L}{\partial \dot q}dt = 0 \\ -\frac{\partial L}{\partial q}dq - \frac{\partial L}{\partial \dot q}d\dot q = 0 $

that doesn't led me to anything...

Answer 1

For what I have understood

1. If the ideal is trivial, we have the equation $i_Xd\Lambda = 0$ for every $X$ vector field. For example, this is the case when $M = T^*Q$ with coordinates $(q, p)$ and $\Lambda = pdq - H(q, p)dt$ is the Poincaré-Cartan 1-form. The Euler-Lagrange equation is $i_X(dp \wedge dq - \frac{\partial H}{\partial q}dq\wedge dt - \frac{\partial H}{\partial p}dp\wedge dt) = 0$. Taking $X = \frac{\partial}{\partial q},\frac{\partial}{\partial \dot q}, \frac{\partial}{\partial t}$ I got

   • $-dp - \frac{\partial H}{\partial q}dt = 0$
   • $dq - \frac{\partial H}{\partial p}dt = 0$
   • $\frac{\partial H}{\partial q}dq + \frac{\partial H}{\partial p}dp = 0$

   that become

   • $\dot{q} = \frac{\partial H}{\partial p}$
   • $\dot{p} = -\frac{\partial H}{\partial q}$
   • $\dot{H} = \frac{\partial H}{\partial t}$

2. If the ideal is non-trivial, we can use (don't know why) the method of Lagrange multipliers, i.e if $I = (\phi^i)$ we can move to another space with trivial ideal with the form $\Omega = \Lambda + \lambda_i\phi^i$, where $\lambda_i$ are the new extra variables. Now we can do the same of point 1. So the Euler-Lagrange equation is $i_Xd(\Lambda + \lambda_i\phi^i) = 0$. Evaluating it on $X = \frac{\partial}{\partial \lambda_i}$ give $\phi^i = 0$, so the condition of the ideal. For example, take $M = TQ$ with coordinates $(q, \dot{q})$, $I = (dq - \dot{q}dt)$ and $\Lambda = L(q, \dot{q})dt$.The Euler-Lagrange equation is $i_X(\frac{\partial L}{\partial q}dq \wedge dt + \frac{\partial L}{\partial \dot{q}}d\dot{q} \wedge dt -\lambda d\dot{q}\wedge dt + d\lambda \wedge (dq - \dot{q}dt)) = 0$. We get

   • $\frac{\partial L}{\partial q} dt - d\lambda = 0$
   • $\frac{\partial L}{\partial \dot{q}}dt - \lambda dt = 0$
   • (The last is linear combination of the first two)

   that become the classical

   • $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q} $