Normal vectors on tangent spaces.

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Homework Help Overview

The discussion revolves around finding the normal vector to a hyperplane at a point on the tangent space of a surface parameterized by a function f(x). The original poster attempts to relate the normal vector to the gradient of f at a given point p on the surface.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the gradient of the function and the normal vector to the tangent hyperplane. There are attempts to manipulate equations involving the gradient and points on the tangent space to derive the normal vector.

Discussion Status

Some participants have offered insights regarding the properties of normal vectors and their relationship to the tangent space. There is an ongoing exploration of how to project points back onto the manifold and the implications of using orthogonal vectors for defining search directions.

Contextual Notes

There is a mention of a multivariate, matrix-valued function and the complexity of defining tangent manifolds in higher dimensions. The discussion includes considerations of numerical methods and constraints related to the problem setup.

Kreizhn
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Homework Statement


Given a surface parameterized by the function [itex]f(x)[/itex] and a point p on that surface, assume that [itex]P[/itex] is a point on the tangent space of f at p. Find the normal vector to the hyperplane at [itex]P[/itex].

The Attempt at a Solution



The tangent hyperplane to f at p is given by the equation
[tex]\nabla f(p) \cdot ( x- p) = 0[/tex]
Since we know that [itex]P[/itex] is on the tangent space, we must have that [itex]\nabla f(p) \cdot ( P - p ) = 0[/itex]. However, here is where I am stuck. I'm not sure how to use this to calculate the normal vector. Any ideas?
 
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Hmm. The normal vector to a plane is the vector n such that n.(a-b)=0 for all a and b in the plane, right?
 
If I'm not mistaken, in the case of n.(a-b) = 0 the normal is very specifically the point passing through b. The set of all a for which this is satisfied help to define the plane.

In this case, the planes should be the same, so it seems that I should be able to do something like equation the equations

[tex]\nabla f(p)\cdot(x-p) = n\cdot (x-P) [/itex]<br /> <br /> In this case, solving for n will give the desired result. I'm playing around with this right now, but haven't gotten anywhere.[/tex]
 
So if n.(p-a)=0 and n.(p-b)=0 isn't n.(a-b)=0? Just subtract the first two equations. But let me rephrase the hint. Isn't grad(f)(p) a normal vector?
 
Yes, [itex]\nabla f(p)[/itex] is normal, and I see what you're saying as far determining it being invariant under choice of internal coordinates.

This yields some important insight into my problem and answers my posed question, so thank you. Now perhaps you can help me with my real question.

Suppose that [itex]f:\mathbb R^n \to \mathbb R^{m\times m}[/itex] is a multivariate, matrix-valued function so that [itex]\nabla f(p)[/itex] is a rank-3 tensor defining the tangent manifold to the orientable surface defined by by [itex]f(x) = 0[/itex]. Namely, [itex]\nabla f(p)[/itex] generates [itex]T_p M[/itex] where [itex]M = \{ x \in \mathbb R^n : f(x) = 0 \}[/itex].

Now through some magic I get a point P (solving a linear programming problem under the constraint that P lies on the tangent manifold) and I want to project P back onto M. Numerically, this is best done by considering vectors orthogonal to [itex]T_p M[/itex] at P. This being said, what is the ``best direction'' to consider? Analogously, I would take the normal vector to P. Should I just use orthogonal vectors at p to define a search direction? Or is there a frame in which the unique ``vector'' passing through P defines the best search direction?
 

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