In this survey by Lubotzky, he has the following:
Proposition 2.1 (Hodge decomposition): The following are true:
- $C^i=B^i\oplus\mathcal{H}^i\oplus\mathcal{B}_i$,
- $\mathcal{H}^i\cong H^i(X;\mathbb{R})$, and
- $\Delta_i^\wedge(B^i\oplus\mathcal{H}^i)=0$.
I'll make all the definitions clear momentarily, but to avoid getting bogged down, first I'll ask my questions:
- Why is this true? In particular I've thought a bit about the third one and don't see how it could possibly be true on $\mathcal{H}^i$ (will elaborate below). I don't really see how any part is as elementary as is claimed in the survey though.
- He traces the result back to this paper of Eckmann, which I can't seem to find in English. Does there exist a translation or a later document recapping its proofs?
- More generally, the survey doesn't give many proofs (which makes sense, as it's a survey). Is there a treatment of the topic which does provide more proofs?
Thanks in advance for any thoughts.
Now, here are all the definitions. $X$ is a finite $d$-dimensional simplicial complex. We study real (co)homology, so the $i$th cochain group $C^i$ is the $\mathbb{R}$-space of $\mathbb{R}$-valued functions on $X^i$ (the set of pure $i$-faces). The coboundary map is defined on pure faces $F\in X^{i+1}$ and extends linearly: $$(\delta_if)(F):=\sum\limits_{G\in X^i}[F:G]f(G)$$ where $[F:G]$ is $-1$ if $G$ comes from $F$ by removing an odd-indexed entry, $1$ if it arises by removing an even-indexed entry, and $0$ otherwise. Put $B^i:=\mathrm{im\,}\delta_{i-1}$. Cohomology is defined in the standard way here (kernel mod image). We define $\deg F:=\#\{G\in X^d:F\subseteq G\}$, so that we have the adjoint map generated by $$(\delta_i^*f)(G)=\frac{1}{\deg G}\sum\limits_{F\in X^{i+1}}[F:G]\deg F \cdot f(F)$$ and the inner product $$(f,g):=\sum\limits_{F\in X^i}\deg F\cdot f(F)g(F).$$ Put \begin{align*} \Delta_i^\wedge&:=\delta_i^*\circ\delta_i, \\ \Delta_i^\vee&:=\delta_{i-1}\circ\delta_{i-1}^*, \\ \Delta_i&:=\Delta_i^\wedge+\Delta_i^\vee. \end{align*} We put $\mathcal{H}^i:=\ker\Delta_i$ and $\mathcal{B}_i:=\mathrm{im\,}\delta_i^*$.
Thinking just about the third claim in the Proposition, it's clear to me from the definitions why $\Delta_i^\wedge(B^i)=0$. However, $\Delta_i^\wedge(\mathcal{H}^i)=0$ doesn't make much sense: $\mathcal{H}^i$ is precisely those $f$ for which $\Delta_i^\vee f=-\Delta_i^\wedge f$. But if the RHS here is $0$ (as the Proposition seems to claim) then this would mean that individually, $\Delta_i^\vee$ and $\Delta_i^\wedge$ vanish on their sum's kernel. This seems like way too strong a statement.
EDIT: This paper by Parzanchevski and Rosenthal also mentions the decomposition (page 10) and chalks it up to just linear algebra (and also clarifies that $\mathcal{H}^i$ is precisely the intersection of the two kernels). I guess what I'm looking for then is how to get started on thinking about the linear algebraic computations needed to show the results.
EDIT: This paper by Parzanchevski, Rosenthal, and Tessler seems to have most of the details so I think my questions should be mainly resolved.