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Is $\delta \Delta^{-1} d$ the identity operator?

Let $d, \delta, \Delta = (d\delta + \delta d)$ be the exterior derivative, codifferential and Laplace-de-Rham operator. Let $\omega$ be a closed $k$-form, one can then say $\Delta \omega = d \delta \...
Theo Diamantakis's user avatar
2 votes
0 answers
145 views

Closed $n$ form from Harvard Qualifying Exam

In the most recent Harvard Qualifying exam, one is asked to prove that if $M$ is a compact oriented manifold, and there are two nonvanishing orientation forms $\xi$ and $\eta$ who's integrals over $M$ ...
Chris's user avatar
  • 4,163
0 votes
1 answer
139 views

Exterior derivative of a 2-form defined by composition of metric with an endomorphism

Im interested in calculating the exterior derivative of a 2-form defined by $\omega(x,y) = g(x,Ay)$ where $A \in \Gamma(End(T) )$ is skew and $g$ is some metric. I hope to reach some coordinate-...
Tom Ariel's user avatar
  • 681
2 votes
0 answers
90 views

Geometric Interpretation of the Exterior Derivative of a 1-form

I was reading this link where says that the geometric interpretation of the exterior derivative of a 1-form $\varphi$ is “the sum of $\varphi$ on the boundary of the surface defined by its arguments” ...
user1234567890's user avatar
3 votes
0 answers
1k views

The wedge of an exact form with a closed form is exact.

I'm trying to prove that the wedge of a closed form $\xi$ with an exact form $\omega$ is exact. We already have that half of it is exact. Maybe we can use the equation of $\xi$ being closed to rewrite ...
edenstar's user avatar
  • 974
1 vote
1 answer
319 views

Confusion about Tangent vs Cotangent differentials

In Lee's Introduction to Smooth manifolds, let $M$ be a smooth manifolds and given a smooth map $f \in C^\infty(M)$, we have two definitions of the "differential" at $f$: (1): for each $p \in M$, ...
Anthony Peter's user avatar