# Satisfies Killing equation, but not a Killing field?

1. Aug 22, 2010

### bcrowell

Staff Emeritus
A cone has all the same local geometrical properties as a plane, so if you take a piece of graph paper and form it into a cone, $\partial_x$ and $\partial_y$ still satisfy the Killing equation. On the other hand, the cone has intrinsic geometrical properties that are different from those of the plane, e.g., parallel transport around a loop enclosing the tip will cause a vector to rotate. This singles out the tip and gives it a special geometrical role, which is clearly not consistent with translational symmetry. Does this mean that we can have a field that satisfies the Killing equation without being a Killing vector, or is the Killing equation violated at the tip of the cone? Does it matter if you extend the cone to make a double cone, so that orbits of a Killing vector can pass smoothly through the tip?

2. Aug 22, 2010

### petergreat

I think due to topological reasons, $$\partial_x$$ is no longer a global Killing field, i.e. it doesn't join up at the cut of the cone.

The mapping between the plane and the cone fails to be a diffeomorphism at the tip because it's not smooth. The tip must be excluded from the cone for the cone to form a differentiable manifold.

Feel free to correct me as I'm not entirely familiar with these issues.

3. Aug 22, 2010

### bcrowell

Staff Emeritus
Due to topological reasons, not only is $$\partial_x$$ not a global Killing field, $$\partial_x$$ is not a smooth global vector field.