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9 votes

$\Lambda^n(M)$ is not isomorphic to $C^{\infty}(M)$ if M is not orientable

Assume that $\Omega^n(M)$ and $\mathcal{C}^\infty(M)$ are isomorphic as $\mathcal{C}^\infty(M)$-modules. Let $\varphi : \mathcal{C}^\infty(M) \rightarrow \Omega^n(M)$ be an isomorphism and $\omega = \...
Cactus's user avatar
  • 7,021
2 votes

Can this closed form not exist on the sphere?

I suggest you just write \begin{align} 0=\int_{B^3}d\alpha=\int_{\partial B^3}\alpha=\int_{S^2}\alpha=\int_{S^2}\omega=-\frac{4\pi}{3}. \end{align} Or if you want to be fully precise, you’re going to ...
peek-a-boo's user avatar
  • 58.4k
1 vote

Čech cohomolog of a good cover of the real projective plane

In the below, I use the convention of strictly ordered simplices i.e. I don't consider $\omega_{10}$ or whatever. One option is to fix bases and just write down the linear maps. You could perform row ...
FShrike's user avatar
  • 42.7k
1 vote

Showing a form is closed using integrals

Let's show $d\omega=0$ pointwise. Take a point $x\in M$ and using charts just assume $M=\mathbb{R}^n, \omega \in \Omega^r(\mathbb{R}^n)$ such that $\int_N \omega = 0$ for all submanifolds $N$ of ...
Csaba Daniel Farkaš's user avatar
1 vote

Relation between curvature form and the Riemann tensor

Given vector fields $X,Y,Z$, denote $$\nabla^2_{XY}Z = \nabla_X(\nabla_YZ) - \nabla_{\nabla_XY}Z. $$ The Riemann curvature tensor is defined by $$ R(X,Y)Z = \nabla^2_{XY}Z - \nabla^2_{YX}Z. $$ Let $(...
Deane's user avatar
  • 8,517
1 vote

Why are Alternating Forms Zero when Evaluated for a Set of Linear Dependent Vectors?

A $k$-form is not necessarily simple/decomposable, but is a sum of such $k$-forms, so it is sufficient to verify the condition for a simple $k$-form. In this case the condition is equivalent to a ...
Mikhail Katz's user avatar
  • 43.8k

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