20
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

17
+50
$\begingroup$

My favorite reference for these sorts of things that straddle physics and geometry is Frankel's "The geometry of physics". In the chapter on harmonic forms, you will find what he refers to simply as "Hodge's Theorem". It's a little more general than you need, because it applies to general $p$-forms, and you only need functions ($0$-forms). So I'll specialize it to functions.


Hodge's Theorem (for functions): Let $M^n$ be a closed Riemannian manifold. Then Poisson's equation \begin{equation} \Delta \alpha = \rho \tag{A} \end{equation} (where $\alpha$ and $\rho$ are real-valued functions, and $\Delta$ is the Laplacian) has a solution $\alpha$ if and only if $\rho$ has mean value $0$ on $M^n$: \begin{equation} \int_M \rho\ \mathrm{vol}^n = 0. \tag{B} \end{equation}


Now, to translate between Frankel's notation and Hawking & Ellis's, we should substitute \begin{align} M &\leftrightarrow \partial \mathscr{B}, \\ \alpha &\leftrightarrow y, \\ \rho &\leftrightarrow \mathrm{const} - \left( p_{b ; d} \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} \right). \\ \end{align} Also note that Hawking & Ellis use a more explicit notation for the Laplacian, so that \begin{equation} \Delta y \leftrightarrow y_{;bd} \hat{h}^{bd}. \end{equation} Now, plugging in these translations, we can rewrite Poisson's equation [Eq. (A)] as \begin{equation} y_{;bd} \hat{h}^{bd} = \mathrm{const} - \left( p_{b ; d} \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} \right). \tag{A'} \end{equation} So Hodge's theorem tells us that it is possible to find a function $y$ that satisfies this equation if and only if the integral of the right-hand side of Eq. (A') over $\partial \mathscr{B}$ is zero.

Alternatively, we could rewrite Eq. (B) and say that a function $y$ exists to solve Eq. (A') if and only if \begin{equation} \int_{\partial \mathscr{B}} \mathrm{const}\ \mathrm{vol}^n = \int_{\partial \mathscr{B}} \left( p_{b ; d} \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} \right)\ \mathrm{vol}^n. \tag{B'} \end{equation} But we get to adjust the value of "$\mathrm{const}$", so we can just set it to whatever we need to make this equation true.

Hawking & Ellis point out that $p_{b ; d} \hat{h}^{bd}$ is a pure divergence. So you can use Stokes' theorem to transform its integral over $\partial \mathscr{B}$ into an integral over the boundary of $\partial \mathscr{B}$. But the boundary of a boundary is always empty,†† so that integral has value $0$. Therefore, this term disappears when you do the integral on the right-hand side of Eq. (B'). So the theorem now states that a solution for $y$ exists if and only if \begin{equation} \mathrm{const}\ \int_{\partial \mathscr{B}} \mathrm{vol}^n = \int_{\partial \mathscr{B}} \left( - R_{ac} Y^{a}_{1} Y^{c}_{2} + R_{adcb} Y^{d}_{1} Y^{c}_{2} Y^{a}_{2} Y^{b}_{1} \right)\ \mathrm{vol}^n. \end{equation} The integral on the left-hand side is just the area of $\partial \mathscr{B}$, and Hawking & Ellis also leave the volume form implicit, so we can rewrite this as \begin{equation} \mathrm{const} = \frac{\int_{\partial \mathscr{B}} \left( - R_{ac} Y^{a}_{1} Y^{c}_{2} + R_{adcb} Y^{d}_{1} Y^{c}_{2} Y^{a}_{2} Y^{b}_{1} \right)} {\mathrm{Area}}. \tag{C} \end{equation} The area is assumed to be finite and nonzero, and is necessarily non-negative, so — as Hawking & Ellis claimed — the sign of the constant is indeed determined by this integral.

Now, a math-classy way of stating the conclusion would be that given $p_a$, $Y_1^b$, $Y_2^c$, $R_{ijkl}$, and $\hat{h}^{mn}$, one can choose a constant [given by Eq. (C)] such that there exists a function $y$ that solves Eq. (A'). Hawking & Ellis change the emphasis to suit their goals, but the statement is also true: there exists a $y$ such that the first four terms in the original question's Eq. (1) add up to a constant, the sign of which is determined by the integral on the right-hand side of Eq. (C).


Note that I have left out the $p'^a p'_a$ term in Eq. (1) from the original question; that term involves derivatives of $y$ other than the Laplacian, so Hodge's theorem doesn't apply to them. But also note that Hawking & Ellis don't actually claim that it should be included in what is equal to a constant. So it's not actually relevant here.

†† Just to clarify, the boundary of a boundary is always empty when dealing with manifolds. This is not true of more general topological spaces, because in those settings the word "boundary" means something different.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.