# One-parameter family of metrics

1. Nov 29, 2008

### mach4

I have a manifold M=S^4 which is endowed with a physical metric g.
I can define another metric on this manifold h (a pullback metric).

Does it make sense to define a one-parameter family of metrics G(u) on the manifold M in the form

G(u) = (1-u)*g + u*h , where u is a parameter in [0,1] ?

Are there any compatibility conditions?
Any help would be appreciated - Thx!

2. Nov 29, 2008

### StatusX

If you're talking about euclidean metrics, then that should work, since the sum of two symmetric positive definite matrices is symmetric and positive definite, and so qualifies as a metric. It won't work for minkoswkian metrics though, since, eg, diag(1,1,1,-1) and diag(1,1,-1,1) are both valid metrics, but their sum is not.

3. Nov 30, 2008

### mach4

Both metrics are symmetric positive definite but non-Euclidean.

When I check G for
-symmetry
-bilinearity
-non-degeneracy
all criteria of a metric seemed to be satisfied.
I was just bothered by the fact that g and h are associated with different curvature tensors, but it seems that they simply add to define the new curvature tensor of G.

Did I understand correctly? In the case of the Minkowskian-metrics the 'non-degeneracy' is not satisfied and thus it does not define a metric.

4. Nov 30, 2008

### StatusX

The curvature is not linear in the metric, so will not simply add. But it's true, you can get a continuous family of metrics with different curvatures (obviously the curvature will then vary continuously over this family). And yes, the problem is that the sum of Minkowski metrics is not necessarily non-degenerate.

By the way, by "Euclidean" I mean a positive definite metric, not a flat one. It's just to distinguish from "Minkowskian".

5. Dec 1, 2008

### mach4

uups - you are right. The Riemannian is clearly not linear in the metric. Bad mistake :(.
Thus, the operation of adding two positive definite metric is possible and does not lead to any inconsistencies. Great!