Dtjytyk

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 $F(N) = int_NLambda$. 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:

$0 = frac{d}{dt}(F(N_t))|_{t=0} = frac{d}{dt}(int_{N_t}Lambda)|_{t=0} = frac{d}{dt}(int_{N}phi_t^*Lambda)|_{t=0} = int_{N}frac{d}{dt}(phi_t^*Lambda)|_{t=0} = int_{N}mathcal{L}_XLambda = int_N(di_XLambda + i_XdLambda ) = int_{partial N}i_XLambda + int_N i_XdLambda$

where $X = frac{dphi_t}{dt}|_{t=0}$

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 $i_XdLambda = 0 mod I$ for every $X$. But I think that's not correct because if I apply to my example I get

$i_X(frac{partial L}{partial q}dqwedge dt + frac{partial L}{partial dot{q}}ddot{q}wedge dt) = 0$

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}ddot q = 0
$

that doesn't led me to anything...

For what I have understood

1. 
   

   If the ideal is trivial, we have the equation $i_XdLambda = 0$ 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 $i_X(dp wedge dq - frac{partial H}{partial q}dqwedge dt - frac{partial H}{partial p}dpwedge dt) = 0$. 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 $Omega = Lambda + lambda_iphi^i$, where $lambda_i$ are the new extra variables. Now we can do the same of point 1. So the Euler-Lagrange equation is $i_Xd(Lambda + lambda_iphi^i) = 0$. 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 $i_X(frac{partial L}{partial q}dq wedge dt + frac{partial L}{partial dot{q}}ddot{q} wedge dt -lambda ddot{q}wedge dt + dlambda wedge (dq - dot{q}dt)) = 0$. We get

   
   
   
   
   
   • $frac{partial L}{partial q} dt - dlambda = 0$
   
   
   • $frac{partial L}{partial dot{q}}dt - lambda dt = 0$
   
   
   • (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} $