# Integration on Manifolds: Compact Support

Compact support for a metric of a manifold is often required to preform integration on the manifold. However, why is it required? Can one preform integration on a manifold without compact support? We will explore those questions here!

To start, let’s slowly define support and build the definition to the version that we require.

Support: Suppose that $f : X \to\mathbb{R}$ is a real-valued function whose domain is an arbitrary set $X$. The set theoretic support of $f$, written $\textbf{supp}(f)$, is the set of points in $X$ where $f$ is non-zero, i.e.

$$\textbf{supp}(f)=\left\{x\in X\bigg|f(x)\ne0\right\}.$$

Topological Support: If $X$ is a topological space and $f:X\to\mathbb{R}$ is a continuous real (or complex)–valued function, the support of $f$ is defined as the closure of the subset of $X$ where $f$ is non-zero, i.e. $$\textbf{supp}(f)=\overline{\left\{x\in X\bigg| f(x)\ne0\right\}},$$ meaning that all the limit points of sequences in $\textbf{supp}(f)$ are contained inside $\textbf{supp}(f).$

Compact Support: Lastly, functions with compact support on a topological space $X$ are those whose support is a compact subset of $X$, i.e. a closed and bounded subset of $X$.

To see why it is essential for the metric to be provided with compact support,  we’ll consider the Einstein$-$Hilbert action as our example. The action is given by:

$$\mathcal{S}[g] =\int_\mathcal{M} \text{d}^4x\sqrt{-|g|}\mathcal{R}$$

where $g$ is an arbitrary $4D$ metric, $|g|$ is the determinant of the metric, and $\mathcal{R}$ is the Ricci scalar. Upon variation, we have the identities \begin{align} {\delta\sqrt{-|g|}}&=\frac{1}{2} \sqrt{-|g|}g^{ab}\delta g_{ab}\\ \delta g^{ab}&=-g^{ac}g^{bd}\delta g_{cd}\\ {\delta\mathcal{R}} = \delta(g^{ab}\mathcal{R}_{ab})&= {-\mathcal{R}^{ab}\delta g_{ab}+\nabla_a X^a} \end{align}

with $X^a = g^{bc} \delta \Gamma^{a}_{bc} – g^{ab} \delta \Gamma_{bc}^c$ being a rank 1 tensor field. The action is therefore

$$\delta\mathcal{S}[g] = \int_\mathcal{M}\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}\delta g_{ab}-\mathcal{R}^{ab}\delta g_{ab}+\nabla_a X^a\right).$$

We see that there is an extra term, that being the last term in the above equation. Usually, such a term would disappear by applying the divergence theorem and providing the usual boundary conditions, i.e. setting it to zero on the boundary. For a generalized manifold, the divergence theorem is given as follows:

Divergence Theorem: If $\partial\mathcal{M}$ is timelike or spacelike then $$\int_\mathcal{M}\text{d}^nx\sqrt{-|g|}\nabla_aX^a = \int_{\partial\mathcal{M}}\text{d}^{n-1}x\sqrt{-|h|}n_aX^a$$ where $X^a$ is a vector field on $\mathcal{M},$ $\nabla$ is the Levi-Civita connection, and $h$ denotes the determinant of the metric on $\partial\mathcal{M}$ induced by pulling back the metric on $\mathcal{M}.$

Using the following theorem, the prior action becomes \begin{align} \delta\mathcal{S}[g]&=\int_\mathcal{M}\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}\delta g_{ab}-\mathcal{R}^{ab}\delta g_{ab}+\nabla_a X^a\right)\\ &= \int_\mathcal{M}\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}-\mathcal{R}^{ab}\right)\delta g_{ab} + \int_{\partial\mathcal{M}} \text{d}^{3}x\sqrt{-|h|}n_aX^a. \end{align} We would like to eliminate the third boundary term naturally, and this is where the metric having compact support comes in handy.

Supposition: The variation of the metric, $\delta g_{ab}$, has support in a compact region \underline{interior to} the boundary. This supposition is pretty strong as $\delta g_{ab}\ne0$ only on a set $U\subset \mathcal{M}\backslash\partial\mathcal{M}$.

Given this supposition, the first term in the sum may be decomposed into the following, \begin{align*} \delta\mathcal{S}[g]= \int_U\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}-\mathcal{R}^{ab}\right)\delta g_{ab} &+ \int_{U^c}\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}-\mathcal{R}^{ab}\right)\delta g_{ab}\\  &+ \int_{\partial\mathcal{M}} \text{d}^{3}x\sqrt{-|h|}n_aX^a, \end{align*} where $U^{c}$ denotes the complement. Then breaking the action in the following manner  $$\delta\mathcal{S}[g] = \delta\mathcal{S}_U[g] + \delta\mathcal{S}_{U^c}[g] = 0$$ reveals the set of equations \begin{align} \delta\mathcal{S}_{U}[g] &=\int_U\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}-\mathcal{R}^{ab}\right)\delta g_{ab} = 0,\text{ and}\\ \delta\mathcal{S}_{U^c}[g] &= \int_{U^c}\text{d}^4x\sqrt{-|g|}\left(\frac{1}{2}\mathcal{R} g^{ab}-\mathcal{R}^{ab}\right)\delta g_{ab} + \int_{\partial\mathcal{M}} \text{d}^{3}x\sqrt{-|h|}n_aX^a = 0. \end{align} In $\delta\mathcal{S}_{U^c}[g]$, the first term vanishes as $\delta g_{ab}$ does not have support in this set, thus  $$\int_{\partial\mathcal{M}} \text{d}^{3}x\sqrt{-|h|}n_aX^a = 0,$$ which is out desired result and what we expect. Now that $n_aX^a$ is a scalar on the boundary, and, as the metric $h$ is arbitrary, we must assert that the scalar $n_aX^a=0$. The cases are either the inner product is 0, or the unit normal, $n_a=0$, or $X^a=0.$ If the boundary, $\partial\mathcal{M}$, is either spacelike or timelike, $X^a$ must vanish. Otherwise, if $\partial\mathcal{M}$ is a hypersurface that is null, it’s normal will be null everywhere and thus $n_a=0.$ In all cases, $n_aX^a$ vanishes.