Register to reply 
How the inner product changes under nonlinear transformation 
Share this thread: 
#1
Nov2412, 06:30 PM

P: 625

Hi,
if we suppose x and y are two elements of some vector space V (say ℝ^{n}), 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 nonlinear, for example a diffeomorphism ? 


#2
Nov2512, 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.



#3
Nov2712, 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  
Linear algebra  inner product and linear transformation question  Calculus & Beyond Homework  0 