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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 rearranging Eq. (A')] 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.

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.

Hodge's Theorem (for functions): Let $M^n$ be a closed Riemannian manifold. Then Poisson's equation \begin{equation} \Delta \alpha = \rho \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{A} \end{equation}

So, to apply this theorem to your question, we have^† \begin{align} M &= \partial \mathscr{B}, \\ \alpha &= y, \\ \rho &= \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{B} \\ \end{align} To put that another way, there is a solution for $y$ such that \begin{equation} y_{;bd} \hat{h}^{bd} + 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} = \mathrm{const} \tag{C} \end{equation} if and only if the integral of $\rho$ over $\partial \mathscr{B}$ is zero. But we get to adjust the value of $\mathrm{const}$, so we can just set it to whatever we need to make that integral zero.

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 integrate $\rho$. Now, combining this fact with (A) and (B), the theorem states that $y$ has a solution 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. \tag{D} \end{equation} The volume (actually, the area of the surface) is assumed to be finite and nonzero, and any volume (area) is non-negative, so the left-hand side is just the constant in question times some positive number.

Now, a math-class 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 rearranging Eq. (D)] such that there exists a function $y$ that solves Eq. (C). 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. (D).

^† Note that I have left out the $p'^a p'_a$ term in (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.

^†† As an interesting side note, 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.

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 rearranging Eq. (A')] 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.

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 \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{A} \end{equation}

So, to apply this theorem to your question, we have^† \begin{align} M &= \partial \mathscr{B}, \\ \alpha &= y, \\ \rho &= \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{B} \\ \end{align} To put that another way, there is a solution for $y$ such that \begin{equation} y + 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} = \mathrm{const} \end{equation} if and only if the integral of $\rho$ over $\partial \mathscr{B}$ is zero.

As Hawking & Ellis point out, $p_{b ; d} \hat{h}^{bd}$ 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 (at least for manifolds), so that integral has value $0$. Therefore, this term disappears when you integrate $\rho$. Now, combining this fact with (A) and (B), the theorem states that $y$ has a solution 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 volume (actually, the area of the surface) is assumed to be nonzero, and any volume (area) is non-negative, so the left-hand side is just the constant in question times some positive number.

Thus, as Hawking & Ellis claim, there exists a $y$ such that the first four terms in the original question's (1) add up to a constant, the sign of which is determined by the integral on the right-hand side above.

^† Note that I have left out the $p'^a p'_a$ term in (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.

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 \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{A} \end{equation}

So, to apply this theorem to your question, we have^† \begin{align} M &= \partial \mathscr{B}, \\ \alpha &= y, \\ \rho &= \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{B} \\ \end{align} To put that another way, there is a solution for $y$ such that \begin{equation} y_{;bd} \hat{h}^{bd} + 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} = \mathrm{const} \tag{C} \end{equation} if and only if the integral of $\rho$ over $\partial \mathscr{B}$ is zero. But we get to adjust the value of $\mathrm{const}$, so we can just set it to whatever we need to make that integral zero.

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 integrate $\rho$. Now, combining this fact with (A) and (B), the theorem states that $y$ has a solution 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. \tag{D} \end{equation} The volume (actually, the area of the surface) is assumed to be finite and nonzero, and any volume (area) is non-negative, so the left-hand side is just the constant in question times some positive number.

Now, a math-class 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 rearranging Eq. (D)] such that there exists a function $y$ that solves Eq. (C). 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. (D).

^† Note that I have left out the $p'^a p'_a$ term in (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.

^†† As an interesting side note, 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.