The Hessians ${\bf H}_L$, ${\bf H}_g$

The Hessian ${\bf H}_L$ of $-E_L$ is defined as the matrix or operator of second derivatives

$${\bf H}_L (L) (x,y; x^\prime y^\prime) = \frac{\delta^2 (-E_L)} {\delta L (x,y)\delta L(x^\prime ,y^\prime )}\Bigg\vert _{L}.$$ (154)

For functional (109) and fixed $\Lambda_X$ we find the Hessian by taking the derivative of the gradient in (127) with respect to $L$ again. This gives

$${\bf H}_L(L) (x,y; x^\prime y^\prime ) =-{{\bf K}}(x,y;x^\pr... ...ta (x-x^\prime )\delta (y-y^\prime ) \Lambda_X (x )e^ {L(x,y)}$$ (155)

or

$${\bf H}_L = -{{\bf K}} - {\bf\Lambda}_X {\bf e^L}.$$ (156)

The addition of the diagonal matrix ${\bf\Lambda}_X {\bf e^L}$ = ${\bf e^L} {\bf\Lambda}_X$ can result in a negative definite ${\bf H}$ even if ${{\bf K}}$ has zero modes, like for a differential operator ${{\bf K}}$ with periodic boundary conditions. Note, however, that ${\bf\Lambda}_X {\bf e^L}$ is diagonal and therefore symmetric, but not necessarily positive definite, because $\Lambda_X (x)$ can be negative for some $x$. Depending on the sign of $\Lambda_X (x)$ the normalization condition $Z_X(x)=1$ for that $x$ can be replaced by the inequality $Z_X(x)\le 1$ or $Z_X(x)\ge 1$. Including the $L$-dependence of $\Lambda_X$ and with

$$\frac{\delta e^{L(x^\prime,y^\prime)} }{\delta g(x,y)} = \de... ...} - \delta (x-x^\prime) e^{L(x,y)} e^{L(x^\prime,y^\prime )} ,$$ (157)

i.e.,

$$\frac{\delta e^{L} }{\delta g} = \left( {\bf I} - {\bf e^L}\... ...ight) {\bf e^L} = {\bf e^L} - {\bf e^L}\, {\bf I}_X {\bf e^L},$$ (158)

or

$$\frac{\delta P }{\delta g} = {\bf P} - {\bf P}\, {\bf I}_X {\bf P},$$ (159)

we find, written in terms of $L$,

$${\bf H}_g (L)(x,y;x^\prime,y^\prime ) = \frac{\delta^2 (-E_g)} {\delta g (x,y)\delta g(x^\prime ,y^\prime )}\Bigg\vert _{L}$$

$$= \!\! \int \!\! dx^{\prime\prime} dy^{\prime\prime} \! \lef... ...g (x,y)\delta g (x^\prime,y^\prime) } \right)\!\Bigg\vert _{L}$$

$$= -{{\bf K}}(x,y;x^\prime,y^\prime ) - e^{L(x^\prime, y^\prime)}e^{... ...\prime\prime} {{\bf K}}( x^\prime,y^{\prime\prime} ; x,y^{\prime\prime\prime} )$$

$$+ e^{L(x^\prime, y^\prime)} \int \!dy^{\prime\prime} {{\bf K}}(x^... ...(x,y)} \int \!dy^{\prime\prime} {{\bf K}}(x^\prime,y^\prime;x,y^{\prime\prime})$$

$$-\delta (x-x^\prime) \delta (y-y^\prime) e^{L(x,y)} \left( N_X(x ) -\int\!dy^{\prime \prime} ({{\bf K}} L)(x,y^{\prime\prime}) \right)$$

$$+ \delta(x- x^\prime) e^{L(x,y)} e^{L(x^\prime,y^\prime)} \left( N_X(x)-\int \!dy^{\prime \prime} ({{\bf K}} L)(x,y^{\prime\prime})\right).$$ (160)

The last term, diagonal in $X$, has dyadic structure in $Y$, and therefore for fixed $x$ at most one non-zero eigenvalue. In matrix notation the Hessian becomes

$${\bf H}_g = - \left( {\bf I} - {\bf e^L} {\bf I}_X \right) {{\bf K}} \left( {... ...\right) - \left( {\bf I} - {\bf e^L} {\bf I}_X \right) {\bf\Lambda}_X {\bf e^L}$$

$$= - \left( {\bf I} - {\bf P} {\bf I}_X \right) \left[ {{\bf K}} \left( {\bf I} - {\bf I}_X {\bf P} \right) + {\bf\Lambda}_X {\bf P} \right] ,$$ (161)

the second line written in terms of the probability matrix. The expression is symmetric under $x\leftrightarrow x^\prime$, $y\leftrightarrow y^\prime$, as it must be for a Hessian and as can be verified using the symmetry of ${{\bf K}} = {{\bf K}}^T$ and the fact that ${\bf\Lambda}_X$ and ${\bf I}_X$ commute, i.e., $[{\bf\Lambda}_X , {\bf I}_X] = 0$. Because functional $E_g$ is invariant under a shift transformation, $g(x,y) \rightarrow g^\prime(x,y) + c(x)$, the Hessian has a space of zero modes with the dimension of $X$. Indeed, any $y$-independent function (which can have finite $L^1$-norm only in finite $Y$-spaces) is a left eigenvector of $\left( {\bf I} - {\bf e^L} {\bf I}_X \right)$ with eigenvalue zero. Thus, where necessary, the pseudo inverse of ${\bf H}$ have to be used instead of the inverse. Alternatively, additional constraints on $g$ can be added which remove zero modes, like for example boundary conditions.