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I'm a fourth-year graduate student in Princeton.


Jul
2
awarded  Curious
Dec
26
awarded  Yearling
Oct
15
awarded  Caucus
Oct
15
awarded  Constituent
Jul
22
awarded  Informed
Jun
25
awarded  Citizen Patrol
Jun
19
answered Aubin's book - construction of Green's function on compact manifold
Mar
7
awarded  Popular Question
Feb
27
asked “Mathai-Quillen-type” form on $M\times M$?
Dec
26
awarded  Yearling
Dec
8
awarded  Organizer
Dec
7
revised Is the closure of the orbits of the mean curvature flow compact for a finite time?
edited tags
Oct
18
comment Invariance group of Morse charts
(2.) Yes, we're interested in the diffeomorphisms satisfying $\varphi \circ f = \varphi$. Such a diffeo $f$ has the property that, on each level set $\{x:|x|^2=r\}$, $f$ restricts to a diffeo of the level set. Moreover the possible $f$ are basically characterized by this property. Think of $f$ as a (germ of a) 1-parameter family of diffeomorphisms of the $(n-1)$-sphere, modulo some boundary conditions (tending to the identity near 0) to ensure smoothness at $p$.
Oct
18
comment Invariance group of Morse charts
Hi Will (and Kofi). (1.) It seems to me that one can work with the set of germs of charts near p, and the group of germs of diffeomorphisms fixing p. Then the action is well-defined and free and transitive.
Sep
23
comment Dolbeault cohomology of Hopf manifolds
Very interesting and relevant. Thanks!
Aug
28
comment Kahler manifolds with constant bisectional curvature
Regarding Walker's comment on Hawley's paper: The Bochner paper which is cited by Hawley is "Curvature in Hermitian metric" (1947). In this paper Bochner proves the local version of the result: that the metric of constant holomorphic bisectional curvature $b$ is unique up to local isometry. Maybe Walker felt that passing to the global version (as done by Hawley/Igusa) was straighforward.
Jun
22
answered Constant scalar curvature metrics in a conformal class
Jun
2
awarded  Nice Question
May
9
comment Finite groups admitting free isometric actions on round spheres
$\mathbb{Z}$ does act freely on $S^{2k+1}$ in the same way. This doesn't contradict Wolf's theorem -- the point is that this action is not properly discontinuous, so the quotient by this action is not a manifold.
May
9
comment Finite groups admitting free isometric actions on round spheres
In your new answer, you're missing one group in the even-$n$ case: the group $\mathbb{Z}/2$, generated by the antipodal map $x\mapsto -x$. (Its eigenvalues are all $-1$, which is real.)