- #1

docnet

- 587

- 242

- Homework Statement:
- psb

- Relevant Equations:
- psb

Solution attempt:

$$F(F(x,y))=(z,w)$$

is the map given by

$$x=z$$

$$y=w$$

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- Thread starter docnet
- Start date

- #1

docnet

- 587

- 242

- Homework Statement:
- psb

- Relevant Equations:
- psb

Solution attempt:

$$F(F(x,y))=(z,w)$$

is the map given by

$$x=z$$

$$y=w$$

- #2

mitochan

- 294

- 133

I observe for x=-1,y=0, u=0/0,v=0/0 not defined so this point should be excluded.

- #3

- 17,817

- 19,037

No. The map is given if you express ##z=z(u,v)=z(u(x,y),v(x,y))=z(x,y)##. You have to substitute the ##u,v## with their definitions in terms of ##x,y##.Homework Statement::psb

Relevant Equations::psb

View attachment 278552

Solution attempt:

$$F(F(x,y))=(z,w)$$

is the map given by

$$x=z$$

$$y=w$$

- #4

mitochan

- 294

- 133

I assume after the substitution he got this resultYou have to substitute the with their definitions in terms of .

##F^2=E##

##F=F^{-1}##

- #5

- 17,817

- 19,037

Maybe, but I guess there is no way but calculation to be sure. It looks a bit like a local Lie group.I assume after the substitution he got this result

##F^2=E##

##F=F^{-1}##

- #6

mitochan

- 294

- 133

- #7

- 17,817

- 19,037

It is called reflection, or involution.

- #8

mitochan

- 294

- 133

Thanks. And for this special idempotent transformation, is it used in Physics ?

- #9

- 17,817

- 19,037

I corrected it. Idempotent functions are ##F^2=F##. We have theoretically ##F^3=F## and any element with ##F^n=F## qualifies to be idempotent, but Wiki said idempotent functions are those with ##n=2##.Thanks. And for this special idempotent transformation, is it used in Physics ?

So let's call it reflection or involution. These terms already suggest their usage. Any reflection satisfies ##F^2=1:## reflected twice is the original. This is a geometric property but as such also used in physics whenever geometry plays a role (crystallography, particle physics, mechanics). Other examples for involutions are matrix transposition, complex conjugation or the Legendre transformation.

- #10

docnet

- 587

- 242

This problem was assigned last semester in differential geometry by the person who invented the Ricci flow and was partially credited for solving the Poincaré conjecture. This was probably too difficult to solve for most undergraduate students in differential geometry.

- #11

- 24,042

- 15,745

$$u' = \frac{2x'}{x'^2 + y^2}, \ \ v = \frac{2y}{x'^2 + y^2}$$ With the same inverse transformation: $$x' = \frac{2u'}{u'^2 + v^2}, \ \ y =\frac{v}{u'^2 + v^2}$$ Then express that in terms of ##x, u## we have:

$$u +1 = \frac{2(x+1)}{(x+1)^2 + y^2}, \ \ v = \frac{2y}{(x+1)^2 + y^2}$$ With, again, the same inverse transformation. Finally, we re-express ##u##:

$$u = \frac{1 - x^2 - y^2}{(x+1)^2 + y^2}$$ And we see that it's the same transformation as above.

- #12

mitochan

- 294

- 133

The transformation for (x,y)=(-1,0) ## u(-1,0)=-1, v(-1,0)=0##, otherwise as above mentioned would be defined for all xy, uv plane.Although my answer is probably wrong, because is not defined everywhere on the plane to start with. I wonder if the composite map would be different if I computed everything out?

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