# Can De Rham cohomology be defined distributionally?

Classically, when we have a smooth manifold $M^n$, we define the smooth differential $p$-forms $\Omega^p(M)$ to be the smooth sections of $\Lambda^p M$, and we define $$H_{\mathrm{dR}}^p(M;\Bbb R)=H^p(\Omega^\bullet(M),d).$$

However, it is possible (and sometimes useful) to expand the class of $p$-forms significantly to include those with distributional coefficients, $\mathcal{D}'^p(M)$. I detail the construction below:

First, for any coordinate chart $(\Omega,(x^i))$ on $M$, any compact subset $L\Subset\Omega$, and any $s\in\Bbb N$, we define the family of seminorms $$p_L^s(u) = \max_{x\in L}\max_{|\alpha|\le s,I}|\partial_x^\alpha u_I(x)|$$ for $u =\sum_Iu_I\mathrm{d}x^I\subset\Omega^p(M)$ which induces a Fréchet topology on $\Omega^p(M)$. We then set $\mathcal{D}^p(M)\subset\Omega^p(M)$ to be the forms with compact support, and we define $$\mathcal{D}'^{n-p}(M)=\mathcal{D}_p'(M)=\mathcal{D}^p(M)'$$ to be its topological dual. For $T\in\mathcal{D}'^p(M)$ and $u\in\Omega^p(M)$, we define $$\int_M T\wedge u := \langle T,u\rangle.$$ As this notation suggests, we have an identification $T:\Omega^p(M)\hookrightarrow\mathcal{D}'^p(M)$ by setting, for $f\in\Omega^p(M)$ and $u\in\mathcal{D}^{n-p}(M)$, $$\int_M T_f\wedge u = \int_M f\wedge u.$$

We can therefore define $\mathrm{d}T$ by $$\int_M \mathrm{d}T\wedge u = (-1)^{p+1}\int_M T\wedge\mathrm{d}u.$$

But does this give us the same cohomology theory?

• The answer is Yes. Distributional forms are called currents. They were defined and their cohomology was studied already in de Rham's original book Variétés Différentiables (1955). – Igor Khavkine Jul 10 '17 at 23:36
• Is there an easy way to see why that's true? – Monstrous Moonshine Jul 10 '17 at 23:45
• It is true because the Poincaré lemma also holds for currents. The identity map from the distributional de Rham complex on $\mathbb{R}^n$ into itself is homotopic to one that smooths currents to forms, which reduces everything to the standard Poincaré lemma. This is all covered in de Rham's book. You should have a look, it's a great read! – Igor Khavkine Jul 10 '17 at 23:59