2

Why is the closed Poincare dual of a point in $R^n$ trivial but the compact dual is a "bump"?

Please provide very detailed answer.

alireza
  • 1,241

1 Answers1

9

Let M be an oriented manifold. Recall that the closed Poincare dual of a $k$-dimensional submanifold of $M$ is an element in $H^{n-k}(M)$. In our case, as a point is $0$-dimensional and $H^n(\mathbb{R}^n)=0$, there is nothing to compute.

The compact dual should be some $\omega\in H_c^n(\mathbb{R}^n)$, such that for every closed $0$-form $\tau$ one has $$\int_{\mathbb{R}^n}\omega\wedge\tau=\int_p\tau,$$where $p$ denotes our point. Since a closed $0$-form is in fact a constant function, this equality holds for $$\omega=fdx_1\wedge\ldots\wedge dx_n,$$where $f$ is a function with compact support whose integral is equal to $1$.

Amitai Yuval
  • 19,308
  • thanks but this means the integral is the constant, but is the integral a constant function over a single point not zero? The integral of constant function over a single point is zero that is. (the integral over a single point is zero unless we have a "delta" function) – alireza Oct 22 '14 at 16:17
  • The integral of a function on a point is equal to $\pm$ the value of the function at this point. If this function is a closed $0$-form, it is constant. If in addition it has a compact support, it is necessarily $\equiv0$. – Amitai Yuval Oct 22 '14 at 17:00
  • thanks I think I now get it; the integral of a zero form which is always just a function is by definition the value of the function? For higher forms the integral over a point is always zero. – alireza Oct 22 '14 at 17:10
  • Note - you can't integrate other forms on points, because points are $0$-dimensional. The rank of the integrated form is always equal to the dimension of the manifold on which it is integrated. – Amitai Yuval Oct 22 '14 at 17:12
  • right I see thanks – alireza Oct 22 '14 at 17:15