0
$\begingroup$

The following is an exercise that I have attempted to work through but whose statement I find difficult to parse through:

For an open set $U$ in $\mathbb{R}^n$, let $R$ be the ring of $C^\infty$ real-valued funcions on $U$. Let $\Omega^1(U)$ be the vector space of differential ($C^\infty)$ forms in $U$, which is the free $R$-module with basis $dx_1$, $\ldots, dx_n$. Let $\Omega^k(U)$ be the space of $k$-forms on $U$, which is the $k$-th exterior power of $\Omega^1(U)$ over $R$; it has basis over $R$ of expressions $dx_{i_1} \wedge \cdots \wedge dx_{i_k}$ for $1 \le i_1 < \cdots < i_k \le n$.

Define linear mappings $d^k: \Omega^k(U) \rightarrow \Omega^{k+1}(U)$, which, for $k=0$, takes a function $f$ to its differential $df = \sum_{i=1}^n (\partial f/ \partial x_i) d x_i$. Show that:

(a) $d^{k+1} \circ d^k = 0$;

(b) For $U$ the complement of the origin in $\mathbb{R}^2$, $Ker(d^1)/Im(d^0)$ is not zero. In this case, can you compute it.

This exercise looks interesting but I am not confident on how to begin working through it, so I post it here to see if it draws anyone's interest and if anyone visiting could suggest a strategy for one or both parts.

  • 0
    de Rham cohomology of punctured Euclidean space is done in detail in both James R. Munkres, Analysis on Manifolds, and Michael Spivak, Calculus on Manifolds.2012-02-20

2 Answers 2

2

This answer addresses part (a) of your question:

For ease of notation, for $f \in C^{\infty}(U) := \Omega^{0}(U)$ let $\partial_if := \frac{\partial f}{\partial x^i}$ and interpret repeated upper/lower indices as a summation over the index. The first thing that needs to be verified is that the statement holds for $k=1$. As Joe indicated, this follows from the the algebra of differentials and equality of mixed partials of a continuous function. So, by definition $ df = \partial_if \; dx^i $ and therefore $d(df) = d(\partial_if \; dx^i) = \partial_j \partial_i f dx^j \wedge dx^i$

Now, from this resulting double sum, we can immediately eliminate all terms along the "diagonal" $i=j$ because they evaluate to $0$. The sum then reduces to $ d(df) = \sum_{i < j} \partial_j \partial_i f dx^j \wedge dx^i + \sum_{i > j} \partial_j \partial_i f dx^j \wedge dx^i. $ If you stare at this long enough you will realize that each term in the first sum is paired asymmetrically with exactly one term in the second sum. For example, assuming the dimension of your space is at least $2$, $\partial_{21}f dx^2 \wedge dx^1$ is a term in the first sum and $ \partial_{12}fdx^1 \wedge dx^2 = -\partial_{12}dx^2 \wedge dx^1 $ is a paired term in the second sum where the further abbreviation $\partial_{ji} := \partial_j \partial_i$ has been applied. So in this example,

$ \partial_{21}f dx^2 \wedge dx^1 + \partial_{12}fdx^1 \wedge dx^2 = (\partial_{21}f - \partial_{12}f)dx^2 \wedge dx^1 = 0, $ the last result following from equality of mixed partials. Using this example as a guide, it is not hard to see that all such terms in the first and second sum are paired analogously.

Finally, an inductive step will prove the desired result.

1

For (a) you will need that $dx_idx_i=0$ and that $dx_idx_j=-dx_jdx_i$. The rest is a straightforward computation. For (b) you can find an explicit generator. Try to pick something that has a discontinuity at the origin.