Why Does the Missing Lagrange Multiplier Matter in Differential Forms?

In summary: I'm sure you're right and it's necessary, but I can't imagine why.In summary, the conversation discusses the use of Lagrange multipliers in mathematical contexts, specifically in the case of sub-Riemannian and symplectic manifolds. It is mentioned that the multiplier lambda_0 often arises in these cases, leading to the need for additional proofs and considerations. However, the exact reason for its necessity is unclear and it is suggested that further study of differential geometry may provide insight.
  • #1
Kreizhn
743
1
I think that many of us have had to endure working with Lagrange multipliers in the past, but it seems to me that it has always been taught incorrectly.

So the statement (if you will allow me to use differential forms) is

Let M be a smooth manifold. Suppose [itex] dg_i [/itex] are linearly independent at each point [itex] p \in M [/itex]. If p is a local extremum of f restricted to M then [itex] \exists \lambda_1,\ldots,\lambda_k \in \mathbb R [/itex] such that
[tex] df(p) = \lambda_1 dg_1(p) + \cdots + \lambda_k dg_k(p) [/tex]

Now my issue is that it's well-known that this should be

[tex] \lambda_0 df(p) = \lambda_1 dg_1(p) + \cdots + \lambda_k dg_k(p) [/tex]

Now normally of course, this wouldn't matter. We could just "normalize by [itex]\lambda_0[/itex]" to get rid of it, unless of course [itex] \lambda_0 = 0 [/itex]. I've seen most of the proofs of this theorem and the [itex] \lambda_0 [/itex] never arises. I can't convince myself why it should indeed arise, yet I can give a great deal of evidence that it is necessary.

Am I confusing this with another LM theorem? Does the [itex]\lambda_0 [/itex] only arise when we consider problems on sub-Riemannian and symplectic manifolds? In those cases, the [itex] \lambda_i [/itex] are not elements of [itex] \mathbb R [/itex] but instead are covectors in [itex] T^*_p M [/itex].
 
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  • #2
Why should there exist any such missing multiplier??

We have a function [tex]f(x_{1},...,x_{n}[/tex], subject to m constraints [tex]g_{i}(x_{1},...,x_{n})=0[/tex]

Now, we create the uxiliary function:
[tex]F(x_{i},...,x_{n},\lambda_{1},...,\lambda_{m})=f+\sum_{i=1}^{m}\lambda_{i}g_{i}[/tex]

We minimize F, i.e, we solve the n+m equations [tex]\nabla_{n+m}F=0[/tex]

This minimum will necessarily meet the constraints upon f, without any additional Lagrange multiplier.
 
  • #3
Hey,

Thanks for the reply. Indeed, this seems to be how most of the proofs proceed. However, in higher level settings (especially in differential geometry) the [itex]\lambda_0 [/itex] appears in the literature. For example, in sub-Riemannian geometry there have been a great deal of false published proofs that the only length minimizing horizontal geodesics are normal trajectories (satisfy Hamilton's equations when considering the cotangent bundle with the induced cometric). However, Montgomery has given a relative simple counter example of a length minimizing singular trajectory that arises as a critical point of the endpoint mapping. Such critical points are directly analogous to the case where [itex] \lambda_0 = 0 [/itex].

Alternatively, the theory of Lagrange multipliers is used in the proof of Pontryagin's principle, arguably the seminal theorem of optimal control theory. While singular controls do not arise often in linear control theory, non-linear theory on manifolds sees singular controls (corresponding to [itex] \lambda_0 = 0 [/itex]) become an integral part of all solutions.

In both of these situations the multipliers themselves are adjoint orbits on a lifted trajectory. In the case of Pontryagin, the multipliers are covectors on the cotangent bundle induced with its natural symplectic structure. This is why I had alluded to symplectic and sub-Riemannian geometries in my earlier post.
 
  • #4
Interesting.
Since I haven't studied any differential geometry, I fully accede that for non-nice manifolds, the "simple" increase in dimensions the sketched proof relies upon (i.e, going from the study of f to F) might not be so simple after all.

But as I said, it is quite a bit outside my area of competence to go into this. :smile:
 
  • #5
Don't worry at all! I don't see why it's necessary, which is why I'm asking.
 

Related to Why Does the Missing Lagrange Multiplier Matter in Differential Forms?

What is a "Missing Lagrange Multiplier"?

A Missing Lagrange Multiplier is a mathematical concept used in optimization problems. It refers to the situation where a Lagrange multiplier, which is a constant used to find the maximum or minimum value of a function subject to a set of constraints, is not present in the final solution.

Why is a Missing Lagrange Multiplier important?

A Missing Lagrange Multiplier can indicate that the optimization problem has been solved incorrectly or that there is an error in the constraints. It is important to identify and address any Missing Lagrange Multipliers in order to ensure an accurate solution to the optimization problem.

How is a Missing Lagrange Multiplier identified?

A Missing Lagrange Multiplier can be identified by checking the final solution to the optimization problem. If the Lagrange multiplier is not present in the solution, it is considered to be missing.

What causes a Missing Lagrange Multiplier?

A Missing Lagrange Multiplier can be caused by a variety of factors, such as a mistake in the formulation of the optimization problem, incorrect constraints, or errors in the calculations used to solve the problem. It is important to carefully check all steps in the optimization process to identify and correct the cause of the Missing Lagrange Multiplier.

How can a Missing Lagrange Multiplier be fixed?

If a Missing Lagrange Multiplier is identified, it is important to go back and check all steps in the optimization process. This may involve double-checking the constraints, recalculating the solution, or adjusting the formulation of the problem. Once the issue is identified and corrected, the Missing Lagrange Multiplier should no longer be present in the final solution.

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