Editing Lagrange multiplier (section)

===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>