Skip to main content
3 events
when toggle format what by license comment
May 20, 2023 at 18:58 comment added Sergiy Maksymenko Thank you Ryan. This might work. Then the next step is to properly glue all those Riemannian metrics. - Suppose we can create such metric for the case when f is of index 0 i.e. it is a sum of squares. - Then we can choose such metric independently on positive and negative bundles, - Then together they might give a global Riemannian metric near C, and f will be still squared distance to C of coordinated on positive subbundle minus squared distance to C of coordinated on negative subbundle.
May 20, 2023 at 18:02 comment added Ryan Budney You can state (2) as saying that the normal bundle to $C$ is a direct sum of two sub-bundles, call them the positive and negative sub-bundles. There is a Riemann metric in the tubular neighbourhood of $C$ such that $f$ is the distance squared (from $C$) of the projection to the positive sub-bundle minus the distance-squared (from $C$) of the projection to the negative sub-bundle.
May 20, 2023 at 17:50 history asked Sergiy Maksymenko CC BY-SA 4.0