to a zero mode (although the converse does not hold: not every zero mode can be “exponentiated" to produce a continuous family of solutions). The Euclidean Schwarzschild metric has a single negative mode, concentrated near the horizon [@GPY], as does the “small black hole" solution for gravity in a cavity [@Gregory:2001bd].

We now turn to the spectrum of $\Delta_H$. Note that this operator is not self-adjoint, and may have complex eigenvalues. It maps the pure-gauge fluctuation $\nabla_{(\mu}w_{\nu)}$ to $\nabla_{(\mu}\Delta_Vw_{\nu)}$. It therefore has the same spectrum as $\Delta_V$ acting on the space of vector fields (modulo Killing fields), which may include a finite number of eigenvalues with negative or zero real part. From we see that the action of $\Delta_H$ equals $\Delta_R$ when acting on the quotient $\mathcal{H}/\mathcal{H}_l$, so they have the same spectrum.

In the presence of a boundary, there is one technical subtlety we should address. For concreteness we assume that we are imposing Dirichlet boundary conditions on the metric. This puts a boundary condition $h_{ij}|_{\partial M} = 0$ on the metric fluctuations in $\mathcal{H}$. Furthermore, the vector fields $w^\mu$ generating elements of $\mathcal{H}_l$ should vanish on the boundary. Now, when we study the operator $\Delta_H$, we impose an 