The Hessians ${\bf H}_P$, ${\bf H}_z$

We now calculate the Hessian of the functional $-E_P$. For fixed $\Lambda_X$ one finds the Hessian by differentiating again the gradient (166) of $-E_P$

$${\bf H}_P(P) (x,y; x^\prime y^\prime ) = -{{\bf K}}(x^\prim... ...rime ) \sum_i \frac{\delta (x-x_i )\delta (y-y_i)}{P^2(x,y)},$$ (181)

i.e.,

$${\bf H}_P = -{{\bf K}} - {\bf P}^{-2} {\bf N}.$$ (182)

Here the diagonal matrix ${\bf P}^{-2} {\bf N}$ is non-zero only at data points.

Including the dependence of $\Lambda_X$ on $P$ one obtains for the Hessian of $-E_z$ in (175) by calculating the derivative of the gradient in (180)

$${\bf H}_z (x,y;x^\prime,y\prime) = -\frac{1}{Z_X(x)} \Big[ {{\bf K}} (x,y;x^\prime,y\prime)$$

$$- \int\! dy^{\prime\prime} \Big( p(x,y^{\prime\prime}){{\... ...x^\prime,y^{\prime\prime}) p(x^\prime,y^{\prime\prime}) \Big)$$

$$+ \int\! dy^{\prime\prime} dy^{\prime\prime\prime} p(x,y^{\... ...,y^{\prime\prime\prime}) p(x^\prime,y^{\prime\prime\prime})$$

$$+ \delta (x-x^\prime) \delta (y-y^\prime) \sum_i \frac{ \de... ...y_i) }{ p^2(x,y) } - \delta (x-x^\prime) \sum_i \delta (x-x_i)$$

$$-\delta (x-x^\prime) \int\! dx^{\prime\prime} dy^{\prime\p... ...}}(x^{\prime\prime},y^{\prime\prime};x^\prime,y^\prime) \Big)$$

$$+ 2\, \delta (x-x^\prime) \int\! dy^{\prime\prime} dx^{\pr... ...e},y^{\prime\prime\prime}) \Big] \frac{1}{ Z_X (x^\prime) },$$ (183)

i.e.,

$${\bf H}_z = {\bf Z}_X^{-1} \left( {\bf I}-{\bf I}_X {\bf P} \right) \left(-{{... ...}^{-2} {\bf N} \right) \left( {\bf I}- {\bf P} {\bf I}_X \right) {\bf Z}_X^{-1}$$

$$- {\bf Z}_X^{-1} \left( {\bf I}_X \left( {\bf G}_P - {\bf\Lambda}... ...) + \left( {\bf G}_P - {\bf\Lambda}_X \right) {\bf I}_X \right) {\bf Z}_X^{-1},$$ (184)

$$= - {\bf Z}_X^{-1} \Big[ \left( {\bf I}-{\bf I}_X {\bf P} \right) {{\bf K}} \left( {\bf I} - {\bf P} {\bf I}_X \right) +{\bf P}^{-2} {\bf N}$$

$$- {\bf I}_X {\bf P}^{-1} {\bf N} - {\bf N} {\bf P}^{-1} {\bf I}_X +{\bf I}_X {\bf N} {\bf I}_X$$

$$+ {\bf I}_X {\bf G}_P + {\bf G}_P {\bf I}_X - 2\, {\bf I}_X {\bf\Lambda}_X \Big] {\bf Z}_X^{-1}.$$ (185)

Here we used $[{\bf\Lambda}_X,{\bf I}_X]$ = 0. It follows from the normalization $\int \! dy \, P(x,y) = 1$ that any $y$-independent function is right eigenvector of $\left( {\bf I}-{\bf I}_X {\bf P} \right)$ with zero eigenvalue. Because $\Lambda_X$ = ${\bf I}_X {\bf P} G_P$ this factor or its transpose is also contained in the second line of Eq. (184), which means that ${\bf H}_z$ has a zero mode. Indeed, functional $E_z$ is invariant under multiplication of $z$ with a $y$-independent factor. The zero modes can be projected out or removed by including additional conditions, e.g. by fixing one value of $z$ for every $x$.