# Lagrange multipliers on manifolds

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

To enhance understanding, proofs and intuitive explanations of the Lagrange multipler method will 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 $f\colon N\to\mathbb{R}$, and $g_{i}\colon N\to\mathbb{R}$, for $i=1,\ldots,k$, be continuously differentiable. Set $M=\bigcap_{i=1}^{k}g_{i}^{-1}(\{0\})$.

### 1.1 Formulation with differential forms

###### Theorem 1.

Suppose $dg_{i}$ are linearly independent  at each point of $M$. If $p\in M$ is a local minimum  or maximum point of $f$ restricted to $M$, then there exist Lagrange multipliers $\lambda_{1},\ldots,\lambda_{k}\in\mathbb{R}$, depending on $p$, such that

 $df(p)=\lambda_{1}\,dg_{1}(p)+\cdots+\lambda_{k}\,dg_{k}(p)\,.$

Here, $d$ denotes the exterior derivative  .

Of course, as in one-dimensional calculus, the condition $df(p)=\sum_{i}\lambda_{i}\,dg_{i}(p)$ by itself does not guarantee $p$ is a minimum or maximum point, even locally.

### 1.2 Formulation with gradients

###### Theorem 2.

Suppose $\nabla g_{i}$ are linearly independent at each point of $M$. If $p\in M$ is a local minimum or maximum point of $f$ restricted to $M$, then there exist Lagrange multipliers $\lambda_{1},\ldots,\lambda_{k}\in\mathbb{R}$, depending on $p$, such that

 $\nabla f(p)=\lambda_{1}\,\nabla g_{1}(p)+\cdots+\lambda_{k}\,\nabla g_{k}(p)\,.$

### 1.3 Formulation with tangent maps

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

###### Theorem 3.

Let $g=(g_{1},\ldots,g_{k})\colon N\to\mathbb{R}^{k}$. Suppose the tangent map $\operatorname{D}g$ is surjective  at each point of $M$. If $p\in M$ is a local minimum or maximum point of $f$ restricted to $M$, then there exists a Lagrange multiplier vector $\lambda\in(\mathbb{R}^{k})^{*}$, depending on $p$, such that

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

Here, $\operatorname{D}g(p)\colon(\mathbb{R}^{k})^{*}\to(\mathrm{T}_{p}N)^{*}$ denotes the pullback of the linear transformation (http://planetmath.org/DualHomomorphism) $\operatorname{D}g(p)\colon\mathrm{T}_{p}N\to\mathbb{R}^{k}$.

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

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

If the expression $\operatorname{D}g(p)^{*}\lambda$ is written out in coordinates   , then it is apparent that the components $\lambda_{i}$ of the vector $\lambda$ are exactly those Lagrange multipliers from Theorems 1 and 2.

## 2 Proofs

### 2.1 Beautiful abstract proof

###### Proof.

Since $dg_{i}$ 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 $\alpha\colon U\to M$, with $U$ open in $\mathbb{R}^{m}$, be a 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   $\left(\mathrm{T}_{p}M\right)^{*}\to\left(\mathrm{T}_{0}\mathbb{R}^{m}\right)^{*}$, 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 $0=d(\alpha^{*}g_{i})=\alpha^{*}dg_{i}$. So like $df$, $dg_{i}$ vanishes on $\mathrm{T}_{p}M$.

In other words, $dg_{i}(p)$ 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 $dg_{i}(p)\in\left(\mathrm{T}_{p}M\right)^{0}$ 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 combination  of the $dg_{i}(p)$:

 $df(p)=\lambda_{1}\,dg_{1}(p)+\cdots+\lambda_{k}\,dg_{k}(p)\,.\qed$

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 vector lives in the dual space $(\mathbb{R}^{k})^{*}$.

###### Alternative argument..

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

Yet another proof could be devised by observing that the result is obvious if $N=\mathbb{R}^{n}$ and the constraint functions are just coordinate projections on $\mathbb{R}^{n}$:

 $g_{i}(y_{1},\ldots,y_{n})=y_{i}\,,\quad i=1,\ldots,k\,.$

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

###### Alternate argument..

Since $dg_{i}$ are linearly independent, we can find a coordinate chart for $N$ about the point $p$, with coordinate functions $y_{1},\ldots,y_{n}\colon N\to\mathbb{R}$ such that $y_{i}=g_{i}$ for $i=1,\ldots,k$. Then

 $\displaystyle df$ $\displaystyle=\frac{\partial f}{\partial y_{1}}dy_{1}+\cdots+\frac{\partial f}% {\partial y_{n}}dy_{n}$ $\displaystyle=\frac{\partial f}{\partial g_{1}}dg_{1}+\cdots+\frac{\partial f}% {\partial g_{k}}dg_{k}+\frac{\partial f}{\partial y_{k+1}}dy_{k+1}+\cdots+% \frac{\partial f}{\partial y_{n}}dy_{n}\,,$

but $\partial f/\partial y_{k+1}=\cdots=\partial f/\partial y_{n}=0$ 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 $N=\mathbb{R}^{n}$. Consider the list vector $g=(g_{1},\ldots,g_{k})$ discussed earlier, and its Jacobian matrix $\operatorname{D}g$ in Euclidean  coordinates. The $i$th row of this matrix is

 $\begin{bmatrix}\dfrac{\partial g_{i}}{\partial x_{1}}&\ldots&\dfrac{\partial g% _{i}}{\partial x_{n}}\end{bmatrix}=\left(\nabla g_{i}\right)^{\mathrm{T}}\,.$

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

Consider each solution $q\in M$ of $g(q)=0$. Since $\operatorname{D}g$ has full rank, we can apply the implicit function theorem, which states that there exist smooth solution parameterizations $\alpha\colon U\to M$ around each point $q\in M$. ($U$ is an open set in $\mathbb{R}^{m}$, $m=n-k$.) 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 $0=\operatorname{D}(f\circ\alpha)(0)=\operatorname{D}f(p)\cdot\operatorname{D}% \alpha(0)$. In other words, $\ker\operatorname{D}f(p)\supseteq\textrm{range of }\operatorname{D}\alpha(0)=% \mathrm{T}_{p}M$. Intuitively, this says that the directional derivatives  at $p$ 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 $0=\operatorname{D}g(p)\cdot\operatorname{D}\alpha(0)$.

Let the columns of $\operatorname{D}\alpha(0)$ be the column vectors $v_{1},\ldots,v_{m}$, which span the $m$-dimensional space $\mathrm{T}_{p}M$, and look at the matrix equation $0=\operatorname{D}f(p)\cdot\operatorname{D}\alpha(0)$ again. The equation for each entry of this matrix, which consists of only one row, is:

 $\nabla f(p)\cdot v_{j}=0\,,\quad j=1,\ldots,m\,.$

In other words, $\nabla f(p)$ is orthogonal    to $v_{1},\ldots,v_{m}$, and hence it is orthogonal to the entire tangent space $\mathrm{T}_{p}M$.

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

 $\nabla g_{i}(p)\cdot v_{j}=0\,,\quad i=1,\ldots,k,\;j=1,\ldots,m\,.$

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 for the 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 $\nabla g_{i}(p)$:

 $\nabla f(p)=\lambda_{1}\nabla g_{1}(p)+\cdots+\lambda_{k}\nabla g_{k}(p)\,.\qed$

## 3 Intuitive interpretations

### 3.1 Normals to tangent hyperplanes Each equation $g_{i}=0$ defines a hypersurface $M_{i}$ in $\mathbb{R}^{n}$, 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 direction of the greatest change in $f$ at $p$, should be perpendicular to $\mathrm{T}_{p}M$. 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 the inner product  structure of $\mathbb{R}^{n}$, which is not present in an abstract manifold (without a Riemannian metric  ).

On the other hand, this explains the mysterious appearance of annihilators in the last paragraph of the abstract proof. Annihilators and dual space theory serve as the proper tools to formalize the manipulations we made with the matrix equations $0=\operatorname{D}f(p)\cdot\operatorname{D}\alpha(0)$ and $0=\operatorname{D}g(p)\cdot\operatorname{D}\alpha(0)$, 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 $dg_{i}$ 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 $\Delta p$ about this point. The infinitesimal movement $\Delta p$ 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 $\Delta p$.

Furthermore, the change $df$ may be decomposed as 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 be any change of $f$ along $M$; thus $f$ only changes when 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 $dg_{i}$. As $df$ is linear in the change $\Delta p$, we ought to be able to write it as a weighted sum of the changes $dg_{i}$. The weights are, of course, the Lagrange multipliers $\lambda_{i}$.

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

### 3.3 As rates of substitution

Observe that the formula for Lagrange multipliers is formally very similar   to the standard formula for expressing a differential form in terms of a basis:

 $df(p)=\frac{\partial f}{\partial y_{1}}dy_{1}+\cdots+\frac{\partial f}{% \partial y_{k}}dy_{k}\,.$

In fact, if $dg_{i}(p)$ 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 differential  of $f$ with respect to changes in $g_{i}$.

In applications of Lagrange multipliers to economic problems, 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 in finding stationary points $p$ of $f$ — defined as points $p$ such that $df$ vanishes on $\mathrm{T}_{p}M$, or equivalently, that the Taylor expansion of $f$ at $p$, under any system of coordinates for $M$, has no terms of first order. Then the Lagrange multiplier method works for this situation too.

The following theorem incorporates the more general notion of stationary points.

###### Theorem 4.

Let $N$ be a $n$-dimensional differentiable manifold (without boundary), and $f\colon N\to\mathbb{R}$, $g_{i}\colon N\to\mathbb{R}$, for $i=1,\ldots,k$, be continuously differentiable. Suppose $p\in M=\bigcap_{i=1}^{k}g_{i}^{-1}(\{0\})$, and $dg_{i}(p)$ 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 $\lambda_{1},\ldots,\lambda_{k}\in\mathbb{R}$ such that

 $df(p)=\lambda_{1}\,dg_{1}(p)+\cdots+\lambda_{k}\,dg_{k}(p)\,.$

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 the annihilated subspace of $\operatorname{span}\{dg_{i}(p)\}$.

It is also enough that $dg_{i}$ be linearly independent only at the point $p$. For $dg_{i}$ are continuous   , so they will be linearly independent for points near $p$ anyway, and we may restrict our viewpoint to a sufficiently small neighborhood around $p$, and the proofs carry through.

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

## References

• 1 Friedberg, Insel, Spence. Linear Algebra. Prentice-Hall, 1997.
• 2 David Luenberger. Optimization by Vector Space Methods. John Wiley & Sons, 1969.
• 3 James R. Munkres. Analysis on Manifolds. Westview Press, 1991.
• 4 R. Tyrrell Rockafellar. “Lagrange Multipliers and Optimality”. SIAM Review. Vol. 35, No. 2, June 1993.
• 5 Michael Spivak. Calculus on Manifolds. Perseus Books, 1998.
Title Lagrange multipliers on manifolds LagrangeMultipliersOnManifolds 2013-03-22 15:25:45 2013-03-22 15:25:45 stevecheng (10074) stevecheng (10074) 24 stevecheng (10074) Topic msc 58C05 msc 49-00 Manifold