Editing Lagrange multiplier (section)

==Modern formulation via differentiable manifolds==

The problem of finding the local maxima and minima subject to constraints can be generalized to finding local maxima and minima on a [[differentiable manifold]] <math>\ M ~.</math><ref>{{cite book |last=Lafontaine |first=Jacques |year=2015 |title=An Introduction to Differential Manifolds |publisher=Springer |isbn=978-3-319-20735-3 |page=70 |url=https://books.google.com/books?id=KNhJCgAAQBAJ&pg=PA70}}</ref> In what follows, it is not necessary that <math>M</math> be a Euclidean space, or even a [[Riemannian manifold]]. All appearances of the gradient <math>\ \nabla\ </math> (which depends on a choice of Riemannian metric) can be replaced with the [[exterior derivative]] <math>\ \operatorname{d} ~.</math>

===Single constraint===
Let <math>\ M\ </math> be a [[smooth manifold]] of dimension <math>\ m ~.</math> Suppose that we wish to find the stationary points <math>\ x\ </math> of a smooth function <math>\ f:M \to \mathbb{R}\ </math> when restricted to the submanifold <math>\ N\ </math> defined by <math>\ g(x) = 0\ ,</math> where <math>\ g:M \to \mathbb{R}\ </math> is a smooth function for which {{math|0}} is a [[regular value]].

Let <math>\ \operatorname{d}f\ </math> and <math>\ \operatorname{d}g\ </math> be the [[exterior derivative]]s of <math>\ f\ </math> and <math>\ g\ </math>. Stationarity for the restriction <math>\ f|_{N}\ </math> at <math>\ x\in N\ </math> means <math>\ \operatorname{d}(f|_N)_x=0 ~.</math> Equivalently, the kernel <math>\ \ker(\operatorname{d}f_x)\ </math> contains <math>\ T_x N = \ker(\operatorname{d}g_x) ~.</math> In other words, <math>\ \operatorname{d}f_x\ </math> and <math>\ \operatorname{d}g_x\ </math> are proportional 1-forms. For this it is necessary and sufficient that the following system of <math>\ \tfrac{1}{2}m(m-1)\ </math> equations holds:
<math display="block"> \operatorname{d}f_x \wedge \operatorname{d}g_x = 0 \in \Lambda^2(T^{\ast}_x M) </math>
where <math>\ \wedge\ </math> denotes the [[exterior algebra|exterior product]]. The stationary points <math>\ x\ </math> are the solutions of the above system of equations plus the constraint <math>\ g(x) = 0 ~.</math> Note that the <math>\ \tfrac{1}{2} m(m-1)\ </math> equations are not independent, since the left-hand side of the equation belongs to the subvariety of <math>\ \Lambda^{2}(T^{\ast}_x M)\ </math> consisting of [[exterior algebra|decomposable elements]].

In this formulation, it is not necessary to explicitly find the Lagrange multiplier, a number <math>\ \lambda\ </math> such that <math>\ \operatorname{d}f_x = \lambda \cdot \operatorname{d}g_x ~.</math>

===Multiple constraints===
Let <math>\ M\ </math> and <math>\ f\ </math> be as in the above section regarding the case of a single constraint. Rather than the function <math>g</math> described there, now consider a smooth function <math>\ G:M\to \R^p (p>1)\ ,</math> with component functions <math>\ g_i: M \to \R\ ,</math> for which <math>0\in\R^p</math> is a [[regular value]]. Let <math>N</math> be the submanifold of <math>\ M\ </math> defined by <math>\ G(x)=0 ~.</math>

<math>\ x\ </math> is a stationary point of <math>f|_{N}</math> if and only if <math>\ \ker( \operatorname{d}f_x )\ </math> contains <math>\ \ker( \operatorname{d}G_x ) ~.</math> For convenience let <math>\ L_x = \operatorname{d}f_x\ </math> and <math>\ K_x = \operatorname{d}G_x\ ,</math> where <math>\ \operatorname{d}G</math> denotes the tangent map or Jacobian <math>\ TM \to T\R^p ~</math> (<math>\ T_x\R^p</math> can be canonically identified with <math>\ \R^p</math>). The subspace <math>\ker(K_x)</math> has dimension smaller than that of <math>\ker(L_x)</math>, namely <math>\ \dim(\ker(L_x)) = n-1\ </math> and <math>\ \dim(\ker(K_x)) = n-p ~.</math> <math>\ker(K_x)</math> belongs to <math>\ \ker(L_x)\ </math> if and only if <math>L_x \in T^{\ast}_x M</math> belongs to the image of <math>\ K^{\ast}_x: \R^{p\ast}\to T^{\ast}_x M ~.</math> Computationally speaking, the condition is that <math>L_x</math> belongs to the row space of the matrix of <math>\ K_x\ ,</math> or equivalently the column space of the matrix of <math>K^{\ast}_x</math> (the transpose). If <math>\ \omega_x \in \Lambda^{p}(T^{\ast}_x M)\ </math> denotes the exterior product of the columns of the matrix of <math>\ K^{\ast}_x\ ,</math> the stationary condition for <math>\ f|_{N}\ </math> at <math>\ x\ </math> becomes
<math display="block"> L_x \wedge \omega_x = 0 \in \Lambda^{p+1} \left (T^{\ast}_x M \right ) </math>
Once again, in this formulation it is not necessary to explicitly find the Lagrange multipliers, the numbers <math>\ \lambda_1, \ldots, \lambda_p\ </math> such that
<math display="block">\ \operatorname{d}f_x = \sum_{i=1}^p \lambda_i \operatorname{d}(g_i)_x ~.</math>