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  A Theorem Due to Hodge: Hawking/Ellis

+ 7 like - 0 dislike

This is probably quite an obscure question but hopefully somebody has a simple answer. I'm studying the proof of the topology theorem on black holes due to Hawking and Ellis (Proposition 9.3.2, p. 335 of their famous book, see also Heusler ``black hole uniqueness theorems" p. 99 Theorem 6.17).

Their proof relies critically on a `theorem due to Hodge' which I have had no success in locating. I own Hodge's book, to which they refer, ``The theory and applications of harmonic integrals", but cannot find the actual theorem they are using.

Specifically, the important expression is (eq. (9.6), p. 336 of Hawking Ellis):

$$p_{b ; d} \hat{h}^{bd} + y_{; bd} \hat{h}^{bd} - R_{ac} Y^{a}_{1} Y^{c}_{2} + R_{adcb} Y^{d}_{1} Y^{c}_{2} Y^{a}_{2} Y^{b}_{1} + p'^{a} p'_{a} \tag{1}$$

They claim one can choose $y$ such that $(1)$ is constant with sign depending on the integral: $$\int_{\partial \mathscr{B}(\tau)} (- R_{ac} Y^{a}_{1} Y^{c}_{2} + R_{adcb} Y^{d}_{1} Y^{c}_{2} Y^{a}_{2} Y^{b}_{1})$$

In the above we have: $\partial \mathscr{B}$ is the horizon surface, $Y^{j}_{1}, Y^{\ell}_{2}$ are future directed null vectors orthogonal to $\partial \mathscr{B}$, $\hat{h}^{ij}$ is the induced metric on $\partial \mathscr{B}$ from the space-time, $p^{a} = - \hat{h}^{ba} Y_{2 c ; b} Y^{c}_{1}$, $y$ is the transformation $\boldsymbol{Y}'_{1} = e^{y} \boldsymbol{Y}_{1}$, $\boldsymbol{Y}'_{2} = e^{-y} \boldsymbol{Y}_{2}$ and fnally $p'^{a} = p^{a} + \hat{h}^{a b} y_{; b}$. So $(1) = \text{cst}$ becomes a differential equation in $y$.

Any ideas on which theorem is invoked?

This post imported from StackExchange Physics at 2014-10-16 11:14 (UTC), posted by SE-user Arthur Suvorov
asked Oct 8, 2014 in Theoretical Physics by Arthur Suvorov (180 points) [ no revision ]
LaTeX tip: use \tag{1} to number your equations. In addition, if you want to add extra spacing, instead of resorting to \, \, \, and so on, you can use the larger \quad.

This post imported from StackExchange Physics at 2014-10-16 11:14 (UTC), posted by SE-user JamalS

1 Answer

+ 4 like - 0 dislike

The claim is not that there exists a $y$ such that (1) is constant but that there exists $y$ such that the sum of the first four terms of (1) is constant. In these four terms, the only dependence in $y$ is the second term, which is exactly the Laplacian of $y$ that I denote $\Delta y$. So we have to solve an equation of the form $\Delta y = f + c$ where $f$ is a given function and where $c$ is a function.

The theorem due to Hodge is the following: Let $X$ be a compact Riemannian manifold and let $g$ be a smooth function on $X$. Then the equation $\Delta y = g$ of unknown $y$ has a solution if and only if $\int_X g vol =0$ (where $vol$ is the natural Riemannian volume form on $X$).

In this theorem, the necessity of the integral condition is clear by Stokes theorem. The non-trivial part is the existence of a solution given this integral condition. This is essentially the "existence fundamental theorem" of p119 of Hodge's book (in fact, Hodge proved much more: what is generally called the Hodge decomposition theorem can be found in many textbooks: see http://en.wikipedia.org/wiki/De_Rham_cohomology for example). In fact, Hawking and Ellis apply this result for $X$ the horizon which is a surface, in this case, the result was known long before Hodge (in his book, Hodge recalls that the case of dimension 2 is classical in the theory of Riemann surfaces).

Applying the theorem shows that given $f$, there exists a constant $c$ such that there exists $y$ with $\Delta y = f+c$, simply $c = - \int_X f$. In their case, Hawking and Ellis compute this integral in the passage following (1).  

answered Oct 16, 2014 by 40227 (5,140 points) [ revision history ]

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