# Invertibility of discrete Laplacian

In QFT and Statistical Mechanics the discrete Laplacian usually plays a key role when we want to discretize the theory. However, few books (at least to my knowledge) really work the properties of this operator in details, so I'm trying to figure out some of these properties myself.

Let $$\Lambda := \epsilon Z^{d}/L\mathbb{Z}^{d}$$ be a finite lattice where $$\epsilon> 0$$ and $$L > 1$$ are integers such that $$L/\epsilon \in \mathbb{N}$$ is even. An scalar field over the lattice $$\Lambda$$ is simply a function $$\phi : \Lambda \to \mathbb{C}$$, so that the space of all fields is $$\mathbb{C}^{\Lambda}$$. Because the lattice is a quotient space, we're dealing with periodic boundary conditions. Thus, we can introduce the discrete Laplacian as the linear operator $$-\Delta: \mathbb{C}^{\Lambda} \to \mathbb{C}^{\Lambda}$$ defined by: $$(-\Delta \phi)(x) := \frac{1}{\epsilon^{2}}\sum_{k=1}^{d}[2\phi(x)-\phi(x+\epsilon e_{k})-\phi(x-\epsilon e_{k})]$$ with $$\{e_{1},...,e_{d}\}$$ being the canonical basis for $$\mathbb{R}^{d}$$. Now, let $$\langle \phi, \varphi \rangle_{\Lambda} := \epsilon^{d}\sum_{x\in \Lambda}\overline{\phi(x)}\varphi(x)$$ be an inner product on $$\mathbb{C}^{\Lambda}$$. If I'm not mistaken, the follwing identity holds: $$\begin{eqnarray} \langle \phi, -\Delta \phi\rangle_{\Lambda} = \sum_{x\in \Lambda}\sum_{y\sim x}|\phi(x)-\phi(y)|^{2} = \sum_{x\in \Lambda}\sum_{y\sim x}(\overline{\phi(x)}-\overline{\phi(y)})(\phi(x)-\phi(y)) \tag{1}\label{1} \end{eqnarray}$$ where $$y\sim x$$ denotes that $$|x-y| = 1$$, where $$|\cdot|$$ is the maximum 'norm' on $$\mathbb{Z}^{d}$$.

My point is the following. We could have assumed $$\phi = 0$$ outise $$\Lambda$$ as a boundary condition, instead of our periodic one. In this case, I know that the discrete Laplacian is positive in the sense that: $$\langle \phi, -\Delta \phi \rangle_{\Lambda} > 0 \quad \mbox{if} \quad \langle \phi, \phi \rangle_{\Lambda} > 0$$ and I'd expect the same property with periodic bondary conditions. However, because of the first equality in relation (\ref{1}), it seems that if we take $$\phi$$ to be constant everywhere, say $$\phi(x) = 1$$ for every $$x \in \Lambda$$, it'd follow that $$\langle \phi, -\Delta \phi \rangle_{\Lambda} = 0$$ even with $$\langle \phi, \phi\rangle_{\Lambda} > 0$$. This would lead to a non-invertibility of this operator. I don't know if this is a known fact that I just didn't know yet or if my reasoning is not correct, but I'd appreciate any help here.

• Yes, this is well known. The derivative operator with periodic boundary conditions has a zero mode and is not invertible. This remains true for your discretized version. A wealth of information about much more than this is in any textbook on lattice gauge theory. – Michael Engelhardt May 29 at 17:53
• It shouldn't be too surprising that the injectivity, and the spectral properties in general, are affected by the choice of boundary conditions. You see this in the continuous situation too: the Dirichlet Laplacian on a nice domain is injective, the Neumann Laplacian isn't. – Nate Eldredge May 29 at 19:18

Your reasoning is correct in that the discrete Laplacian for periodic boundary conditions has a zero mode. On the space of fields satisfying $$\sum_x\phi(x)=0$$, its spectrum is, however, strictly positive, and it can be inverted on that space. This is all well known.
• Thanks for the comment! Does this fact depend on the dimension $d$? – IamWill May 29 at 18:54