# Does killing vector change with change in metric signature or with change in coordinate system?

For the Schwarzschild metric with (-,+,+,+) metric signature, the killing vector is (1,0,0,0). If we use another metric signature, i.e (+,-,-,-), will the killing vector change?

Also, if we change the coordinate system i.e instead of Schwarzschild coordinates if we use Eddington Finkelstein coordinate system, will the killing vector change?

• "Killing" is a proper noun, thankfully.
– JEB
Dec 21, 2022 at 20:13

Changing the metric signature doesn't affect whether a vector is a Killing vector. In particular, because it corresponds to just multiplying the metric by $$-1$$, which keeps the Killing equation unchanged.
Vectors are geometrical objects, so they do not change under coordinate transformations. The Killing vector is a Killing vector in any coordinates, because it is defined in a coordinate free manner. However, it's components do change, as with any other vector. For example, $$(1,0,0,0)$$ are the components for a Killing vector in the Schwarzschild spacetime, as you pointed out. However, these coordinates won't descobre a Killing vector if you perform the coordinate transformations where I simply change the order of the coordinates and write $$(r,t,\theta,\phi)$$ instead of $$(t,r,\theta,\phi)$$. Hence, using the same components in different coordinate systems might not lead to a Killing vector. However, if you do the correct component transformations for the coordinate transformations, the Killing vector will always be a Killing vector, regardless of the coordinates. This is because "being a Killing vector" is a geometrical property defined without making any references to a coordinate system.