Isometry & Inverse: Does Every Isometry Have an Inverse?

In summary: C. However, I cannot seem to get past the first step.In summary, the author is trying to solve a problem in Modern Geometry with Applications from chapter 1. The problem is asking for a proof that an isometry (defined as a distance-preserving function) maps a Euclidean circle in ExE (all those points at a fixed distance from the center point P) into a circle as its image. The author states that all points in the image are at the proper distance/radius from the image of P, but it is not clear what prohibits the image from having other points at the correct distance and which do not have preimages under the isometry.
  • #1
Pearce_09
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Does every isometry have an inverse?
 
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  • #2
Have you tried proving it? How far did you get?

Have you tried coming up with a counterexample? What things have you tried?
 
  • #3
Do you know that in order to have an inverse, a function must be a bijection? Do you know that in order to be a bijection, a function must be an injection and a surjection? You should be able to easily prove that every isometry is an injection. Suppose x is some point that is in the codomain of an isometry but not in its range. Now choose a line passing through x which contains at least two points in the range of the isometry (if no line passing through x contains two points in the range of the isometry, you should be able to show that there is something severly wrong with this "isometry"). Since the isometry is injective, those two points have unique preimages. Can you see where to go from here? As an additional hint, think about when the triangle inequality gives an equality.
 
  • #4
I just noticed this old thread earlier today, and I would appreciate some clarification on one or two of the points above.

A function need only be injective to have an inverse; bijectivity isn't required.

Some authors do not define an isometry as a bijection, and state only that it is a function F which is distance-preserving: d(F(a),F(b)) = d(a,b). It is then easy to show that F is one-to-one, but the issue of surjection doesn't seem to be explicit or implicit -- at least to me. Others (say, Roe, Silvester) define an isometry as a bijection to begin with.

I have wondered for some time about one of the early problems in George Jennings' text, Modern Geometry with Applications. In chapter 1 he asks for a proof that an isometry (defined as above) maps a Euclidean circle in ExE (all those points at a fixed distance from the center point P) into a circle as its image.

Here too it is clear that all points in the image are at the proper distance/radius from the image of P -- d(F(p),F(P)) = d(p,P). But it is NOT clear (to me ..) what prohibits the image from having other points at the correct distance and which do not have preimages under the isometry.

Short of invoking the (topological) completeness property of bounded and closed subsets of the plane -- and thus inferring surjection for the isometry -- is there another, easier method to prove the simple assertion above?

B.
 
  • #5
You can certainly have maps with the property that d(F(a),F(b))=d(a,b) which are not surjective. Beyond that I'm not sure what your question is. What is it?
 
  • #6
Clarification (from Jennings):

"A function F from ExE to ExE" (he uses E^n for this definition) "is an isometry if for all points p, q in ExE, d(F(p),F(q)) = d(p,q).

Let C (in ExE) be the circle with center P and radius r.

If F is an isometry from ExE to ExE, prove that F(C) is the circle with center F(P) and radius r."

As in my earlier post, it is trivial to show that all points in F(C) have distance r from F(P). So F(C) is contained in the circle with center F(P) at distance r. But short of assuming or otherwise demonstrating that F is surjective, my question is,

What is the proof that there are no other points at distance r from F(P) except those in F's image?

B.
 
  • #7
Take the circle in ExE to be centered on (0,0) with radius r. Then the distance from (r,0) and the distance from (0,r) determine a unique point on the circle. Since this point is determined completely by distance relations, it also is in the image ExE. I was hoping for a really more elegant proof of this, but it doesn't seem to be coming. It's late. Sorry.
 
  • #8
I appreciate your response. At the same time, Modern Geometry with Applications distinguishes RxR (with a distinguished origin and coordinates), from ExE which has neither. In this sense "distance" in ExE appears to be a primitive term, not otherwise defined for the exercise at hand.

So barring consideration of the compactness of the Euclidean plane over RxR, I am having a small problem finding the logic that demonstrates

F^-1(F(C)) = C (true by the definition of isometry of F, since F is injective)

F^-1(C') = C (where C' is the circle centered on F(P))

Without surjection of F, the rationale for stmt. 2 is .. vague.

I am trying to not reproduce too much of the text here, but the "problem hint" suggests proving that F(C) is contained in the circle C' centered on F(P) (easy task); and then prove that F^-1(C') is contained in C (which is where the path stops for me w/o surjection).

I concur as well that it's getting late. Perhaps this will become a bit more clear in the daylight.

B.
 
  • #9
Hmm. I guess the general point (aside from subtlties of definitions distinguishing ExE from RxR) is that I can pick three points in the plane a, b and c so that any other point x is uniquely identified by d(a,x), d(b,x) and d(c,x). Is there a problem with that program?
 
  • #10
I agree that point uniqueness in the manner you describe is a successful approach to proving that F^-1(C') = C, but I'd consider it an "analytic" vs. "synthetic" solution. It is entirely possible that I am guilty of over-analyzing this particular, simple problem. Perhaps the author expects readers to use the compactness of the reals, or to just plain "know" that there can't be holes in a circle in ExE, for the pre-image and the image of the isometry.

B.
 

What is an isometry?

An isometry is a type of transformation in geometry that preserves the size, shape, and distance of objects. This means that after an object is transformed, it remains congruent to its original form.

What is an inverse isometry?

An inverse isometry is the transformation that “undoes” the original isometry. It brings the object back to its original form, preserving its size, shape, and distance.

Does every isometry have an inverse?

Yes, every isometry has an inverse. This is because isometries are bijective, meaning they are both one-to-one and onto. This allows for the inverse transformation to exist and bring the object back to its original form.

How is the inverse of an isometry calculated?

The inverse of an isometry can be calculated by using the properties of the original isometry. For example, if the original isometry is a translation, the inverse would be the opposite translation in the same direction. If the original isometry is a reflection, the inverse would be the same reflection. For rotations, the inverse would be the rotation in the opposite direction.

Can a non-isometric transformation have an inverse?

No, a non-isometric transformation cannot have an inverse. This is because non-isometric transformations do not preserve the size, shape, and distance of objects. Therefore, there is no way to “undo” the transformation and bring the object back to its original form.

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