Typeships17

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Your reference desk question

Hello there. I thought I'd try and answer your question about orthogonals. Your exact question wasn't very clear, but I think I know what you meant. If this isn't what you meant then I'm sorry. I'll give a bit of theory first, if you don't understand any of it then try to read the articles, and/or send me a message (or the reference desk a message). Then I'll carry on and try to answer your question.

Let f : R^m → Rⁿ be a smooth function. For a fixed point x₀ in R^m the differential of f at x₀ , denoted by $df|_{{\mathbf {x} }_{0}}$ is a linear map from the tangent space of R^m at x₀ to the tangent space of Rⁿ at f(x₀); i.e. $df|_{{\mathbf {x} }_{0}}:T_{{\mathbf {x} }_{0}}\mathbb {R} ^{m}\to T_{f({\mathbf {x} }_{0})}\mathbb {R} ^{n}$. As with any linear map, if we take a basis then we can find a matrix representation of the linear map. In this case, the matrix if the Jacobian matrix. The differential df (not evaluating at a point anymore) gives a map from the tangent bundle of R^m to the tangent bundle of Rⁿ, i.e. $df:T\mathbb {R} ^{m}\to T\mathbb {R} ^{n}.$

Now, in you question you were talking about the case n = 1, and about the level sets of f, i.e. for some real number y in R

$L_{y}=\{{\mathbf {x} }\in \mathbb {R} ^{m}:f({\mathbf {x} })=y\}.$

In this case the Jacobian matrix of f is a 1 × m matrix, i.e. a vector. The Jacobian matrix, in the n = 1 case, is just the gradient vector of f. And you say that you understand why this vector if orthongonal to the level sets.

Now, in the case of general n, things become more complicated. The Jacobian matrix will be an n × m matrix, and it doesn't make sense to say that someting is orthogonal to a matrix. (Orthogonal is a word used with vectors). If we think of a vector as a 1 × m matrix, then another vector is orthogonal to it if and only if it lies in the kernel of the 1 × m matrix. And in the case of the Jacobian matrix the kernel also plays a key role. But we need to impose some conditions on f.

First of all we need that n ≤ m and that the Jacobian matrix has maximal rank, i.e. n. This just means that the differential at each point is a surjective map between tangent spaces. If n < m then f is called a submersion. If n = m then f is called a diffeomorphism. For the level set idea it's normal to have n < m.

We can prove (using the implicit function theorem) that if the Jacobian matrix of f evaluated at a point x₀ of R^m has rank n and f(x₀) = y₀ then the level set L_y0 is a parametrisable p-dimensional manifold in a neighbourhood of x₀, where p = m – n. (In this case we say that x₀ is a regular point of f, and that y₀ is a regular value of f.) Now, here's the key point:

If f is a submersion, x₀ is a regular point of f, and y₀ is a regular value of f then the kernel of the Jacobian matrix evaluated at x₀ spans the tangent space of Ly₀ at x₀.

(Note that in the case where n = 1, the submersion condition if just that the gradient vector is not the zero vector. In this case the gradient vector is perpendicular to the level sets, so the orthogonal will be tangent to the level sets.)

To see why this is true, we make calculations along the lines of the n = 1 case. Assume that f is a submersion. I'll drop any refernce to point, but just keep in mind that everything needs to be evaluated at a point. We know that the level set is a parametrisable p-dimensional manifold sitting inside R^m, say M. So let's parametrise it with $\gamma :U\to \mathbb {R} ^{m},$ where $U\subset \mathbb {R} ^{p}$ is an open neighbourhood and

$\gamma (u_{1},\ldots ,u_{p})=(x_{1}(u_{1},\ldots ,u_{p}),\ldots ,x_{m}(u_{1},\ldots ,u_{p}))$

for smooth functions $x_{1},\ldots ,x_{m}.$ To keep things simple lets assume that M is the zero level set of f, i.e.

$M=\{{\mathbf {x} }\in \mathbb {R} ^{m}:f({\mathbf {x} })={\mathbf {0} }\}\ .$

This just makes the calculations look simpler. We would just need to add constants to all of the $x_{i}$ if you wanted a different level set.

Since $\gamma$ parametrises M we have $f\circ \gamma \equiv {\mathbf {0} },$ i.e.

$f(x_{1}(u_{1},\ldots ,u_{p}),\ldots ,x_{m}(u_{1},\ldots ,u_{p}))\equiv {\mathbf {0} }.$

Notice that f is a mapping from m-dimensions to n-dimensions, so we can break down the components of f as

$f(x_{1},\dots ,x_{m})=(f_{1}({\mathbf {x} }),\ldots ,f_{n}({\mathbf {x} }))\ ,$

where for 1 ≤ i ≤ n we have $f_{i}:\mathbb {R} ^{m}\to \mathbb {R} .$

Given that $f(x_{1}(u_{1},\ldots ,u_{p}),\ldots ,x_{m}(u_{1},\ldots ,u_{p}))\equiv {\mathbf {0} },$ we calculate the partial derivatives of both sides of this equality. We get that for 1 ≤ i ≤ n and 1 ≤ j ≤ p

${\frac {\partial f_{i}}{\partial u_{j}}}={\frac {\partial f_{i}}{\partial x_{1}}}{\frac {\partial x_{1}}{\partial u_{j}}}+\cdots +{\frac {\partial f_{i}}{\partial x_{m}}}{\frac {\partial x_{m}}{\partial u_{j}}}=0\ .$

Letting i vary from 1 to n we can put these equations into matrix form:

$\left({\begin{array}{ccc}\partial f_{1}/\partial x_{1}&\cdots &\partial f_{1}/\partial x_{m}\\\vdots &&\vdots \\\partial f_{n}/\partial x_{1}&\cdots &\partial f_{n}/\partial x_{m}\end{array}}\right)\left({\begin{array}{c}\partial x_{1}/\partial u_{j}\\\vdots \\\partial x_{m}/\partial u_{j}\end{array}}\right)={\mathbf {0} }\ .$

Notice that the matrix on the left is exactly the Jacobian matrix of f. This last expressions tells us that each of the partial derivatives of the parametrising functions lie in the kernel of the Jacobian matrix. But these partial derivatives of the parametrisation of M are exactly what span the tangent space to M. It now follows that the kernel of the Jacobian matrix coincides with the tangent space to M. ~~ Dr Dec (Talk) ~~ 12:08, 10 September 2009 (UTC)