Euclidean Killing Field Question

  • Thread starter Kreizhn
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  • #1
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Main Question or Discussion Point

Hey,

This may seem like a simple question, but hopefully someone can answer it quickly.

Consider the Euclidean 2-metric [itex] ds^2 = dx^2 + dy^2 [/itex]. There are three killing fields, two translations
[tex] K_1 = \frac{\partial}{\partial x}, \qquad K_2 = \frac{\partial}{\partial y} [/tex]
and a rotation. Now my issue is this, if [itex] K_3 [/itex] is the rotational Killing field, with coordinate decomposition
[tex] K_3 = K_3^x \frac{\partial}{\partial x} + K_3^y \frac{\partial}{\partial y} [/tex]
We can show that [itex] K_3[/itex]'s components must satisfy the differential question
[tex] \frac{\partial K^y}{\partial x} + \frac{\partial K^x}{\partial y} = 0 [/tex]
Hence we can have two solutions [itex] (K^x,K^y) = (-y,x) [/itex] and [itex] (K^x,K^y) = (y, -x) [/itex]. Clearly these are just a multiple of -1 different. But if one were to plot these, they would give rotations in opposite directions. One would move clockwise, the other counter-clockwise.

Do we care about the orientation of the field? Or just it's general flow? Do these both represent the same field all the same?
 

Answers and Replies

  • #2
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Is it because the flow (as a one parameter semi-group) can move in both directions? Hence the flow generated by each is equivalent, and just corresponds to taking a negative "time?"
 

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