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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
24
votes
7
answers
5k
views
Difference between parallel transport and derivative of the exponential map
This is a crosspost from math.stackexchange
Given a Riemannian manifold $M$, let $c(t) = \exp_p(tX)$ be the geodesic emanating from $p \in M$ with initial value $X$. Let $t_0$ be small enough, then w …
23
votes
Why differential forms are important?
In Chern-Weil Theory, characteristic classes appear as certain closed differential forms associated to vector bundles with connections (constructed via the curvature form of the connection).
When yo …
16
votes
1
answer
3k
views
Heat Kernel Asymptotics on Manifold with Boundary
This is crosspost from math.stackexchange https://math.stackexchange.com/questions/311213/heat-kernel-asymptotics-on-manifold-with-boundary where the question did not yield any answer
On a closed Rie …
16
votes
Accepted
Formal adjoint of the covariant derivative
Ad 1: Yes, there is. The formula is
$$\nabla^*(X^\flat \otimes u) = - \nabla_X u -\mathrm{div}(X) \cdot u,$$
as can easily seen by local computation. Here, $X$ is a vector field and $X^\flat$ is the d …
11
votes
How to prove Liouville measure is invariant under geodesic flow?
A hands down proof not using the theory of Hamiltonian systems can be done by just proving that the Jacobian determinant of the transformation is zero.
We have
$$ TSM \cong \pi^* TM \oplus VSM,$$
wh …
9
votes
0
answers
439
views
Invariant polynomials in curvature tensor vs. characteristic classes
Let $M$ be an $4m$-dimensional Riemannian manifold. We can then form the Pontryagin classes $p_k(TM)$ of the tangent bundle using Chern-Weil theory. For any sequence of numbers $k_1, \dots, k_l$ such …
9
votes
1
answer
600
views
Long-time decay of heat kernel on compact manifolds
Let $M$ be a compact Riemannian manifold and let $V \in C^\infty(M)$. Consider the operator $\Delta + V$ and let $p_t(x, y)$ be the corresponding heat kernel. If $\Delta + V$ is a positive operator su …
8
votes
What are "good" examples of spin manifolds?
If $M$ is a spin manifold, then any submanifold of codimension 1 is also a spin manifold. This yields a lot of examples, for example, that $S^n$ is spin etc.
(I may not have understood your point com …
8
votes
0
answers
207
views
(Un)bounded Geometry and Sobolev Spaces
This post is related to this and this post.
It is known that on a complete Riemannian manifold, the space $C^\infty_c(M)$ is generally not dense in the Sobolev spaces $W^{k, p}(M)$ ($1 \leq p < \inft …
8
votes
0
answers
1k
views
Can you make the cotangent bundle to a complex manifold?
The cotangent bundle of a manifold has a canonical symplectic form and if we choose a riemannian metric on $M$, we can give it an almost complex structure.
Is this structure integrable, and if it is …
8
votes
1
answer
2k
views
Solutions to the eikonal equation
Theorem. Let V be a $C^\infty$ function on a riemannian manifold $M$ and $p$ be a nondegenerate local minimum with $V(p)=0$. Then there is a unique positive function $\varphi \in C^\infty(U)$ such tha …
7
votes
2
answers
799
views
Asymptotic expansion of the Schrödinger kernel?
My stackexchange post was somewhat unsatisfactory (also because I may not have stated clear enough what my interest was). So here it goes!
Let $M$ be a compact Riemannian manifold and $\Delta$ be the …
7
votes
1
answer
894
views
Sharp Gaussian upper bounds on Heat Kernel
I am looking for references (with proof) for the following statement:
Let $(M, g)$ be a Riemannian manifold with bounded curvature and let $p_t(x , y)$ be the heat kernel of $M$. Let $K$ be compac …
6
votes
1
answer
184
views
Zeta-Determinant for shifted Laplacians on the circle
Consider on the circle $S^1$ the operator
$$L := - \frac{\partial^2}{\partial \theta^2} + c$$
for some constant $c \in \mathbb{R}$.
What is its $\zeta$-regularized determinant?
This should be well- …
6
votes
Accepted
Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
Given a connection on the tangent bundle, you can define the second covariant derivative $\nabla^2f$ by
$$ \nabla^2f[X, Y] := \partial_X\partial_Y f - \partial_{\nabla_X Y} f.$$
Then $\nabla^2f$ is a …