how the inner product changes under non-linear transformation

by mnb96
Tags: nonlinear, product, transformation
mnb96 is online now
Nov24-12, 06:30 PM
P: 625

if we suppose x and y are two elements of some vector space V (say ℝn), and if we consider a linear function f:V→V', we know that the inner product of the transformed vectors is given by: [tex]\left\langle f\mathbf{x} , f\mathbf{y} \right\rangle = \left\langle \mathbf{x} , \overline{f}f\mathbf{y} \right\rangle = \left\langle \overline{f}f\mathbf{x} , \mathbf{y} \right\rangle[/tex] where [itex]\overline{f}[/itex] is the adjoint operator of [itex]f[/itex].

What can we say about [itex]\left\langle f\mathbf{x} , f\mathbf{y} \right\rangle[/itex] when f is non-linear, for example a diffeomorphism ?
Phys.Org News Partner Science news on
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
Muphrid is offline
Nov25-12, 03:21 PM
P: 834
I admit I'm not well versed in stuff with manifolds and such, but isn't a diffeomorphism basically a way of mapping positions on one manifold to positions on another? If so, then ##f## is just the Jacobian of the mapping, and it is inherently dependent on position within the manifold. It's important to note that positions won't obey this transformation law, only the full, nonlinear transformation.
mnb96 is online now
Nov27-12, 12:01 PM
P: 625
Hi Muphrid,
thanks for the answer. That's exactly what I wanted to know.

Register to reply

Related Discussions
Decompose any linear transformation as a product of INJECTIVE and SURJECTIVE ones Calculus & Beyond Homework 0
Inner Product and Linear Transformation Calculus & Beyond Homework 3
Matrix vector product and linear transformation proof Precalculus Mathematics Homework 2
Inner Product of a Linear Transformation Calculus & Beyond Homework 3
[SOLVED] linear algebra - inner product and linear transformation question Calculus & Beyond Homework 0