There has been a previous discussion about concrete constructions using the Hodge theorem , Construction of Hodge decomposition

Let me try to ask here about a specific case where one is trying to write a metric fluctuation $h$ (about a background metric $g$) on a product of two 2-manifolds. Let the indiced $\mu, \nu$ be on the first manifold and $\alpha, \beta$ on the second manifold. (possibly what is said below works only for $\mathbb{H}^2 \times S^2$)

Then the $10$ components of this metric on this $4-manifold$ can apparently be parametrized in terms of $10$ scalars $A_1,..,A_{10}$ and a harmonic scalar $f$ as follows,

we first define two vector fields as,

$X_a = A_1 \partial_a f + A_2 \epsilon_{a b} \partial ^b f$

$Y_a = A_3 \partial_a f + A_4 \epsilon_{a b} \partial ^b f$

and then in terms of that the metric components are defined as,

$h_{\mu \nu} = A_5 g_{\mu \nu} f + (D_\mu X_\nu + D_\nu X_\mu - g_{\mu \nu} D^\lambda X_\lambda )$

$h_{\alpha \beta} = A_6 g_{\alpha \beta} f + (D_\alpha Y_\beta + D_\beta Y_\alpha - g_{\alpha \beta} D^\gamma Y_\gamma )$

$h_{\alpha \mu} = A_7 \partial_\alpha \partial_\mu f + A_8 \epsilon_{\mu \nu} \partial_\alpha \partial ^\nu f + A_9\epsilon_{\alpha \beta} \partial_\mu \partial ^\beta f + A_{10}\epsilon_{\alpha \beta} \epsilon_{\mu \nu}\partial^\nu \partial ^\beta f $

where the derivative w.r.t the metric $g$ is defined as, $D_a X_b = \partial_a X_b + \Gamma ^{c}_{ab} X_c$

  • I would like to understand the above construction and how it might be generalizable to higher dimensions.

Roughly the following ideas have possibly gone into this,

  • In two dimensions the eigenfunctions for the vector and tensor can be constructed out of the eigenfunctions of the scalar field.(why?) So possibly the "h" constructed above is an harmonic symmetric transverse traceless tensor of rank 2

  • Hodge decomposition says that any k form on (any?) manifold can uniquely be written as a sum of a harmonic form, an exact form and a co-exact form. Apply this to a vector field on $S^2$ (the $X_a$ and the $Y_a$ above?). The vector field is a one-form and and so we can decompose it into a harmonic form, an exact form and a co-exact form. On S^2 there is no harmonic one form (why?). So it is just a sum of an exact form which is just this $\partial f$ and a co-exact form which is just this $\epsilon\partial f$.

  • In getting the first ``gf" term on $h_{\mu \nu}$ and $h_{\alpha \beta}$ one has possibly used the fact that the cohomology of 2 forms is the same as the cohomology of 0-forms for 2-manifolds?

But I still don't see the origin of the $( )$ terms in $h_{\mu \nu}$ and $h_{\alpha \beta}$ and the structure of $h_{\alpha \mu}$!

  • $\begingroup$ I don't think I understand what you are asking generally. But specifically, in what sense is $h$ a "metric fluctuation"? Is $h$ a metric itself, or is it a family of metrics, or is $g + h$ a metric? Are $X$ and $Y$ vector fields on the two factor manifolds in the product, respectively? What do you mean by eigenfunctions of the vector, tensor, and scalar field? $\endgroup$ – Phillip Andreae Jun 10 '14 at 0:08
  • $\begingroup$ To answer a couple of your questions regarding cohomology: the Hodge decomposition does not necessarily apply to non-compact manifolds. Note that $S^2$ is compact and $\mathbb{H}^2$ is not compact, so their product is not compact. A vector field is not a one-form, as you say; rather, I would say that a vector field is dual to a one-form, and to each vector field one can associate a unique one-form via the metric (this is the "musical isomorphism"). On compact manifolds, harmonic $k$-forms are in one-to-one correspondence with degree $k$ cohomology classes... $\endgroup$ – Phillip Andreae Jun 10 '14 at 1:37
  • $\begingroup$ (continued) Since $H^1(S^2) = 0$, there are no harmonic one-forms on $S^2$. Finally, for compact $n$-manifolds, Poincare duality says that $H^k \equiv H^{n-k}$ for all $k$. So, yes, as you say, on $S^2$ for example, degree-0 cohomology is isomorphic to degree-2 cohomology. $\endgroup$ – Phillip Andreae Jun 10 '14 at 1:44

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