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Is reflection in the $x$-axis or in the line $y=x$ in a $2$-dimensional normed space isometric? How about rotation through a right angle? If so, what is the proof?

LoveMath
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  • Do you understand what it means for a map to be isometric? – Zev Chonoles Mar 08 '13 at 05:38
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    Any open convex set that is symmetric about zero is the unit ball for some norm. If this set is not preserved by the transformation you consider, then the transformation is not an isometry. Use this to construct counterexamples to all statements above. –  Mar 08 '13 at 05:39
  • Yes,the map to be isometric is that it preserve the norm. And my example is that I am taking a unit circle and therefore the function norm{e1+x}=norm{e1-x}, where (e1 is a unit vector) for some value of x. So my question is in this unit circle the reflection in x-axis and the line y=x, and the rotation through a right angle will be isometric?? I thing it is but how to prove? – LoveMath Mar 08 '13 at 05:56
  • @5PM -- the transformation $(x,y) \mapsto (x+3, y)$ is certainly an isometry. But it doesn't preserve the unit ball --in fact it shifts it to the right by a distance of 3. I think the OP will have a lot of trouble constructing counterexamples, because all three mappings are in fact isometries. – bubba Mar 08 '13 at 13:07
  • I'm assuming that he's using the usual Euclidean norm, which may or may not be a correct assumption. – bubba Mar 08 '13 at 13:35
  • @bubba I meant the maps preserving the origin, as only such maps are mentioned in the question. –  Mar 08 '13 at 15:03

1 Answers1

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You don't say what norm/metric you're using, or whether it is allowed to vary. So, I'll assume for now that you're using the conventional Euclidean metric: $d\left((x_1, y_1), (x_2,y_2)\right) = \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 }$.

Then, since we're dealing with the space that we all live in, we can trust our intuition, and the answers should be obvious, geometrically -- reflecting things or rotating them is not going to change distances. Maybe this doesn't qualify as a "proof" (though, personally, it's good enough for me).

If you want a formal algebraic proof, proceed as follows: take two arbitrary points $P_1 = (x_1,y_1)$, and $P_2 = (x_2,y_2)$. Their reflections in the $x$-axis are $Q_1 = (x_1,-y_1)$, and $Q_2 = (x_2,-y_2)$. You have to show that the distance $Q_1Q_2$ is the same as the distance $P_1P_2$. That shouldn't be too hard.

The other two examples can be handled with the same sort of approach -- take two points, calculate their images under the mapping, and then show that the distance between is unchanged.

However, if we're allowed to use any norm we like, then things get more interesting. For each of the three given mappings, you can find some norm on $\mathbb R^2$ for which it's not an isometry. But, my guess is that this is not the question you're asking.

bubba
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