Lagrange multipliers on manifolds

We discuss in this article the theoretical aspects ofthe Lagrange multiplier method.

To enhance understanding,proofs and intuitive explanations of the Lagrange multipler methodwill be given from several different viewpoints,both elementary and advanced.

1 Statements of theorem

Let $N$ be a $n$-dimensional differentiable manifold (without boundary), and $f:N\to ℝ$, and$g_i:N\to ℝ$, for $i=1,\mathrm{\dots },k$, be continuously differentiable.Set $M={\bigcap }_{i=1}^k{g}_i^{-1}(\{0\})$.

1.1 Formulation with differential forms

Theorem 1.

Suppose $d\mathit{}{g}_i$ are linearly independentat each point of $M$.If $p\mathrm{\in }M$ is a local minimum or maximum point of $f$ restricted to $M$,then there exist Lagrange multipliers${\lambda }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\lambda }_k\mathrm{\in }\mathrm{R}$, depending on $p$,such that

$$df(p)={\lambda }_{1}d{g}_1(p)+\mathrm{\cdots }+{\lambda }_{k}d{g}_k(p).$$

Here, $d$ denotes the exterior derivative.

Of course, as in one-dimensional calculus, the condition $df(p)={\sum }_i{\lambda }_{i}d{g}_i(p)$ by itself does not guarantee $p$ is aminimum or maximum point, even locally.

1.2 Formulation with gradients

The version of Lagrange multipliers typicallyused in calculus is the special case$N={ℝ}^n$ in Theorem 1.In this case,the conclusion of thetheorem can also be writtenin terms of gradients instead of differential forms:

Theorem 2.

Suppose $\mathrm{\nabla }\mathit{}{g}_i$ are linearly independentat each point of $M$.If $p\mathrm{\in }M$ is a local minimum or maximum point of $f$ restricted to $M$,then there exist Lagrange multipliers${\lambda }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\lambda }_k\mathrm{\in }\mathrm{R}$, depending on $p$,such that

$$\nabla f(p)={\lambda }_1\nabla g_1(p)+\mathrm{\cdots }+{\lambda }_k\nabla g_k(p).$$

This formulation and the first oneare equivalent sincethe 1-form $df$ can be identified with the gradient$\nabla f$, via the formula $\nabla f(p)\cdot v=df(p;v)=d{f}_p(v)$.

1.3 Formulation with tangent maps

The functions $g_i$ can also be coalesced into a vector-valued function$g:N\to {ℝ}^k$. Then we have:

Theorem 3.

Let $g\mathrm{=}\mathrm{(}{g}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{g}_k\mathrm{)}\mathrm{:}N\mathrm{\to }{\mathrm{R}}^k$.Suppose the tangent map $\mathrm{D}\mathit{}g$ is surjectiveat each point of $M$.If $p\mathrm{\in }M$ is a local minimum or maximum point of $f$ restricted to $M$,thenthere exists a Lagrange multiplier vector $\lambda \mathrm{\in }{\mathrm{(}{\mathrm{R}}^k\mathrm{)}}^{\mathrm{*}}$,depending on $p$, such that

$$\mathrm{D}f(p)=\mathrm{D}g{(p)}^{*}\lambda .$$

Here, $\mathrm{D}\mathit{}g\mathit{}\mathrm{(}p\mathrm{)}\mathrm{:}{\mathrm{(}{\mathrm{R}}^k\mathrm{)}}^{\mathrm{*}}\mathrm{\to }{\mathrm{(}{\mathrm{T}}_p\mathit{}N\mathrm{)}}^{\mathrm{*}}$denotes the pullback of the lineartransformation (http://planetmath.org/DualHomomorphism) $\mathrm{D}\mathit{}g\mathit{}\mathrm{(}p\mathrm{)}\mathrm{:}{\mathrm{T}}_p\mathit{}N\mathrm{\to }{\mathrm{R}}^k$.

If $\mathrm{D}g$ is represented by its Jacobian matrix,then the condition that it be surjective is equivalent toits Jacobian matrix having full rank.

Note the deliberate use of the space ${({ℝ}^k)}^{*}$ instead of ${ℝ}^k$— to which the former is isomorphic to —for the Lagrange multiplier vector. It turns out that theLagrange multiplier vector naturallylives in the dual space and not the original vector space ${ℝ}^k$.This distinction is particularly important in the infinite-dimensionalgeneralizations of Lagrange multipliers.But even in the finite-dimensional setting,we do see hints that the dual spacehas to be involved, because a transpose is involvedin the matrix expression for Lagrange multipliers.

If the expression $\mathrm{D}g{(p)}^{*}\lambda$ is writtenout in coordinates, then it is apparent that the components ${\lambda }_i$of the vector $\lambda$ are exactly thoseLagrange multipliers from Theorems 1 and 2.

2 Proofs

The proof of the Lagrange multiplier theorem is surprisingly short and elegant,when properly phrased in the language of abstract manifolds anddifferential forms.

However, for the benefit of the readers not versed in these topics,we provide, in addition to the abstract proof, a concrete translation of the argumentsin the more familiar setting $N={ℝ}^n$.

2.1 Beautiful abstract proof

Proof.

Since $d{g}_i$ are linearly independent at each point of $M={\bigcap }_{i=1}^k{g}_i^{-1}(\{0\})$,$M$ is an embedded submanifold of $N$,of dimension $m=n-k$. Let $\alpha :U\to M$, with $U$ open in ${ℝ}^m$, bea coordinate chart for $M$ such that $\alpha (0)=p$.Then ${\alpha }^{*}f$ has a local minimum or maximum at $0$,and therefore $0=d({\alpha }^{*}f)={\alpha }^{*}df$ at $0$.But ${\alpha }^{*}$ at $p$ is an isomorphism ${\left({\mathrm{T}}_{p}M\right)}^{*}\to {\left({\mathrm{T}}_0{ℝ}^m\right)}^{*}$,so the preceding equation says that $df$ vanishes on ${\mathrm{T}}_{p}M$.

Now, by the definition of $g_i$, we have ${\alpha }^{*}{g}_i=0$,so $0=d({\alpha }^{*}{g}_i)={\alpha }^{*}d{g}_i$. So like $df$, $d{g}_i$ vanishes on ${\mathrm{T}}_{p}M$.

In other words, $d{g}_i(p)$ is in the annihilator (http://planetmath.org/AnnihilatorOfVectorSubspace)${\left({\mathrm{T}}_{p}M\right)}^0$ of the subspace ${\mathrm{T}}_{p}M\subseteq {\mathrm{T}}_{p}N$.Since ${\mathrm{T}}_{p}M$ has dimension $m=n-k$, and ${\mathrm{T}}_{p}N$ has dimension $n$,the annihilator ${\left({\mathrm{T}}_{p}M\right)}^0$ has dimension $k$.Now $d{g}_i(p)\in {\left({\mathrm{T}}_{p}M\right)}^0$ are linearly independent,so they must in fact be a basis for ${\left({\mathrm{T}}_{p}M\right)}^0$.But we had argued that $df(p)\in {\left({\mathrm{T}}_{p}M\right)}^0$.Therefore $df(p)$ may be written as a unique linear combinationof the $d{g}_i(p)$:

$$df(p)={\lambda }_{1}d{g}_1(p)+\mathrm{\cdots }+{\lambda }_{k}d{g}_k(p).\mathit{∎}$$

The last paragraph of the previous proof can also be rephrased,based on the same underlying ideas,to make evident the fact that the Lagrange multiplier vectorlives in the dual space ${({ℝ}^k)}^{*}$.

Alternative argument..

A general theorem in linear algebra states that for any lineartransformation $L$,the image of the pullback $L^{*}$is the annihilator of the kernel of $L$.Since $\ker \mathrm{D}g(p)={\mathrm{T}}_{p}M$and $df(p)\in {({\mathrm{T}}_{p}M)}^0$,it immediately follows that $\lambda \in {({ℝ}^k)}^{*}$exists such that $df(p)=\mathrm{D}g{(p)}^{*}\lambda$.∎

Yet another proofcould be devised by observingthat the result is obvious if $N={ℝ}^n$ and the constraint functionsare just coordinate projections on ${ℝ}^n$:

$$g_i(y_1,\mathrm{\dots },y_n)=y_i,i=1,\mathrm{\dots },k.$$

We clearly must have$\partial f/\partial y_{k+1}=\mathrm{\cdots }=\partial f/\partial y_n=0$at a point $p$ that minimizes $f(y)$ over $y_1=\mathrm{\cdots }=y_k=0$.The general case can be deduced to this by a coordinate change:

Alternate argument..

Since $d{g}_i$ are linearly independent,we can find a coordinate chart for $N$ about the point $p$,with coordinate functions $y_1,\mathrm{\dots },y_n:N\to ℝ$such that $y_i=g_i$ for $i=1,\mathrm{\dots },k$.Then

$df$$={\displaystyle \frac{\partial f}{\partial y_1}}d{y}_1+\mathrm{\cdots }+{\displaystyle \frac{\partial f}{\partial y_n}}d{y}_n$

$={\displaystyle \frac{\partial f}{\partial g_1}}d{g}_1+\mathrm{\cdots }+{\displaystyle \frac{\partial f}{\partial g_k}}d{g}_k+{\displaystyle \frac{\partial f}{\partial y_{k+1}}}d{y}_{k+1}+\mathrm{\cdots }+{\displaystyle \frac{\partial f}{\partial y_n}}d{y}_n,$

but $\partial f/\partial y_{k+1}=\mathrm{\cdots }=\partial f/\partial y_n=0$at the point $p$. Set ${\lambda }_i=\partial f/\partial g_i$ at $p$.∎

2.2 Clumsy, but down-to-earth proof

Proof.

We assume that $N={ℝ}^n$.Consider the list vector $g=(g_1,\mathrm{\dots },g_k)$ discussed earlier,and its Jacobian matrix $\mathrm{D}g$ in Euclidean coordinates.The $i$th row of this matrixis

So the matrix $\mathrm{D}g$ has full rank (i.e. $\mathrm{rank}\mathrm{D}g=k$) if and onlyif the $k$ gradients $\nabla g_i$ are linearly independent.

Consider each solution $q\in M$ of $g(q)=0$.Since $\mathrm{D}g$ has full rank, we can apply the implicit function theorem,which states that there exist smooth solution parameterizations$\alpha :U\to M$ around each point $q\in M$. ($U$ is an open set in ${ℝ}^m$, $m=n-k$.)These $\alpha$ are the coordinate charts which give to $M=g^{-1}(\{0\})$ a manifold structure.

We now consider specially the point $q=p$; without loss of generality, assume $\alpha (0)=p$.Then $f\circ \alpha$ is a function on Euclidean space having a local minimum or maximum at $0$,so its derivative vanishes at $0$.Calculating by the chain rule, we have$0=\mathrm{D}(f\circ \alpha )(0)=\mathrm{D}f(p)\cdot \mathrm{D}\alpha (0)$.In other words, $\ker \mathrm{D}f(p)\supseteq \text{range of }\mathrm{D}\alpha (0)={\mathrm{T}}_{p}M$.Intuitively, this says that the directional derivativesat $p$ of $f$ lying in the tangent space ${\mathrm{T}}_{p}M$ of the manifold $M$ vanish.

By the definition of $g$ and $\alpha$, we have $g\circ \alpha =0$.By the chain rule again, we derive $0=\mathrm{D}g(p)\cdot \mathrm{D}\alpha (0)$.

Let the columns of $\mathrm{D}\alpha (0)$ be the column vectors $v_1,\mathrm{\dots },v_m$, which span the$m$-dimensional space ${\mathrm{T}}_{p}M$,and look at the matrix equation $0=\mathrm{D}f(p)\cdot \mathrm{D}\alpha (0)$ again.The equation for each entry of this matrix, which consists of only one row, is:

$$\nabla f(p)\cdot v_j=0,j=1,\mathrm{\dots },m.$$

In other words, $\nabla f(p)$ is orthogonal to $v_1,\mathrm{\dots },v_m$,and hence it is orthogonal to the entire tangent space ${\mathrm{T}}_{p}M$.

Similarly, the matrix equation $0=\mathrm{D}g(p)\cdot \mathrm{D}\alpha (0)$ can be split into individualscalar equations:

$$\nabla g_i(p)\cdot v_j=0,i=1,\mathrm{\dots },k,j=1,\mathrm{\dots },m.$$

Thus $\nabla g_i(p)$ is orthogonal to ${\mathrm{T}}_{p}M$.But $\nabla g_i(p)$ are, by hypothesis, linearly independent,and there are $k$ of these gradients, so they must form a basis forthe orthogonal complement of ${\mathrm{T}}_{p}M$, of $n-m=k$ dimensions.Hence $\nabla f(p)$ can be written as a unique linear combination of $\nabla g_i(p)$:

$$\nabla f(p)={\lambda }_1\nabla g_1(p)+\mathrm{\cdots }+{\lambda }_k\nabla g_k(p).\mathit{∎}$$

3 Intuitive interpretations

We now discuss the intuitive and geometricinterpretations of Lagrange multipliers.

3.1 Normals to tangent hyperplanes

Each equation $g_i=0$ defines a hypersurface $M_i$ in ${ℝ}^n$, a manifold of dimension $n-1$.If we consider the tangent hyperplane at $p$ of these hypersurfaces, ${\mathrm{T}}_p{M}_i$, the gradient $\nabla g_i(p)$gives the normal vector to these hyperplanes.

The manifold $M$ is the intersection of the hypersurfaces $M_i$.Presumably, the tangent space ${\mathrm{T}}_{p}M$ is the intersection of the ${\mathrm{T}}_p{M}_i$, and the subspace perpendicular to ${\mathrm{T}}_{p}M$ would be spanned by the normals $\nabla g_i(p)$.Now, the direction derivatives at $p$ of $f$ with respect to each vector in ${\mathrm{T}}_{p}M$, as we have proved,vanish. So the direction of $\nabla f(p)$, the directionof the greatest change in $f$ at $p$, should be perpendicularto ${\mathrm{T}}_{p}M$. Hence $\nabla f(p)$ can be written as a linear combination of the $\nabla g_i(p)$.

Note, however, that this geometric picture, and the manipulations with the gradients $\nabla f(p)$and $\nabla g_i(p)$, do not carry over to abstract manifolds.The notions of gradients and normals to surfaces depend on theinner product structure of ${ℝ}^n$, which isnot present in an abstract manifold (without a Riemannian metric).

On the other hand, this explains the mysterious appearance of annihilators inthe last paragraph of the abstract proof.Annihilators and dual space theory serve as the proper toolsto formalize the manipulations we made with the matrix equations $0=\mathrm{D}f(p)\cdot \mathrm{D}\alpha (0)$ and $0=\mathrm{D}g(p)\cdot \mathrm{D}\alpha (0)$, without resorting to Euclidean coordinates, which, of course, are not even defined on an abstract manifold.

3.2 With infinitesimals

If we are willing to interpret the quantities $df$ and $d{g}_i$ as infinitesimals,even the abstract version of the result has an intuitive explanation.Suppose we are at the point $p$ of the manifold $M$,and consider an infinitesimal movement $\mathrm{\Delta }p$ about this point.The infinitesimal movement $\mathrm{\Delta }p$ is a vector in the tangent space${\mathrm{T}}_{p}M$, because, near $p$, $M$ looks like the linear space ${\mathrm{T}}_{p}M$.And as $p$ moves, the function $f$ changes by a corresponding infinitesimal amount $df$that is approximately linear in $\mathrm{\Delta }p$.

Furthermore, the change $df$ may be decomposedas the sum of a change as $p$ moves along the manifold $M$,and a change as $p$ moves out of the manifold $M$.But if $f$ has a local minimum at $p$, then there cannot beany change of $f$ along $M$; thus $f$ only changeswhen moving out of $M$.Now $M$ is described by the equations $g_i=0$,so a movement out of $M$ is described by the infinitesimal changes$d{g}_i$.As $df$ is linear in the change $\mathrm{\Delta }p$,we ought to be able to write it as a weighted sum of the changes $d{g}_i$.The weights are, of course, the Lagrange multipliers ${\lambda }_i$.

The linear algebra performed in the abstract proof can be regarded as the precise, rigoroustranslation of the preceding argument.

3.3 As rates of substitution

Observe that the formula for Lagrange multipliers is formallyvery similar to the standard formula for expressinga differential form in terms of a basis:

$$df(p)=\frac{\partial f}{\partial y_1}d{y}_1+\mathrm{\cdots }+\frac{\partial f}{\partial y_k}d{y}_k.$$

In fact, if $d{g}_i(p)$ are linearly independent,then they do form a basis for ${({\mathrm{T}}_{p}M)}^0$,that can be extended to a basis for ${({\mathrm{T}}_{p}N)}^{*}$.By the uniqueness of the basis representation,we must have

$${\lambda }_i=\frac{\partial f}{\partial g_i}.$$

That is, ${\lambda }_i$ is the differentialof $f$ with respect to changes in $g_i$.

In applications of Lagrange multipliers to economicproblems, the multipliers ${\lambda }_i$ are rates of substitution —they give the rate of improvement in the objective function $f$as the constraints $g_i$ are relaxed.

4 Stationary points

In applications,sometimes we are interested infinding stationary points $p$ of $f$ — defined aspoints $p$ such that $df$ vanishes on ${\mathrm{T}}_{p}M$, or equivalently,that the Taylor expansion of $f$ at $p$, under any system of coordinatesfor $M$, has no terms of first order. Then the Lagrange multiplier methodworks for this situation too.

The following theorem incorporatesthe more general notion of stationary points.

Theorem 4.

Let $N$ be a $n$-dimensional differentiable manifold (without boundary), and $f\mathrm{:}N\mathrm{\to }\mathrm{R}$,$g_i\mathrm{:}N\mathrm{\to }\mathrm{R}$, for $i\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}k$,be continuously differentiable. Suppose $p\mathrm{\in }M\mathrm{=}{\mathrm{\bigcap }}_{i\mathrm{=}\mathrm{1}}^k{g}_i^{\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\{}\mathrm{0}\mathrm{\}}\mathrm{)}$, and$d\mathit{}{g}_i\mathit{}\mathrm{(}p\mathrm{)}$ are linearly independent.

Then $p$ is a stationary point (e.g. a local extremum point) of $f$ restricted to $M$,if and only if there exist ${\lambda }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\lambda }_k\mathrm{\in }\mathrm{R}$such that

$$df(p)={\lambda }_{1}d{g}_1(p)+\mathrm{\cdots }+{\lambda }_{k}d{g}_k(p).$$

The Lagrange multipliers ${\lambda }_i$, which depend on $p$, are unique when they exist.

In this formulation, $M$ is not necessarily a manifold,but it is one when intersected with a sufficiently small neighborhood about $p$.So it makes sense to talk about ${\mathrm{T}}_{p}M$, although we are abusing notation here.The subspace in question can be more accurately described as theannihilated subspace of $\mathrm{span}\{d{g}_i(p)\}$.

It is also enough that $d{g}_i$ be linearly independentonly at the point $p$.For $d{g}_i$ are continuous, so they will belinearly independent for points near$p$ anyway,and we may restrict our viewpoint to a sufficiently small neighborhoodaround $p$, and the proofs carry through.

The proof involves only simple modifications tothat of Theorem 1 — for instance,the converse implication followsbecause we have already proved that the $d{g}_i(p)$ form a basis for the annihilator of ${\mathrm{T}}_{p}M$,independently of whether or not $p$ is a stationary point of $f$ on $M$.