Supplementary information:
Exact Solution for Elastic Networks on Curved Surfaces

We will consider a dimensionless free energy normalized per particle, that is

$$f\equiv\frac{F}{Y\bar{A}}=\frac{2F}{\sqrt{3}NYa^{2}_{L}}\ ,$$

(S1)

hence the area in reference space, see Eq. 3. This area is given by

$$\hat{A}=\pi\left(1-\frac{q_{i}}{6}\right)\rho_{0}^{2}\ .$$

(S2)

Given two systems with the same number of particles, the one with the smallest free energy per particle, Eq. S1 is the stable minimum.

The free energy, see Eq. 1 is

$$F=\int d^{2}{\bm{x}}\sqrt{g}\left[{\cal F}^{elastic}+{\cal F}^{bending}\right]+F^{abs}+F^{line}\ .$$

(S3)

Note that the stress tensor, Eq. A6 is given by

$$\sigma^{\alpha\beta}=\frac{1}{\sqrt{g}}\frac{\delta F}{\delta u_{\alpha\beta}}=A^{\alpha\beta\gamma\delta}u_{\gamma\delta}=Y\times\left(\mbox{Terms that Depend on }\nu_{p}\right)$$

(S4)

Note also, that the ratio of the Young modulus and the line tension defines a coefficient $l_{A}$ with units of length

$$\frac{\tau}{Y}\equiv l_{A}$$

(S5)

Therefore, through the boundary conditions Eq. 8, the quantity

$$\frac{\sigma^{\alpha\beta}}{Y}=h(\nu_{p},l_{A})\ ,$$

(S6)

does not directly depend on the Young modulus.

Also,

$$F^{abs}=-N\Delta F=-\frac{2\Delta F}{\sqrt{3}a^{2}_{L}}\frac{\sqrt{3}}{2}Na^{2}_{L}=-\Pi\hat{A}\mbox{ with }\hat{A}=\int d^{2}{\bm{r}}\sqrt{\bar{g}({\bm{x}})}\ .$$

(S7)

Therefore $\Pi=\frac{2\Delta F}{\sqrt{3}a^{2}_{L}}$, and $\hat{A}$ is the area in reference space.

The line tension term is a function of the perimeter ($P$), given by

$$\frac{F^{line}}{YL^{2}}=\frac{\tau}{YL^{2}}\oint_{\partial D}ds\equiv\frac{\tau P}{YL^{2}}\ .$$

(S8)

Finally, the free energy Eq. S3 is

$$\frac{F}{L^{2}Y}=f(\nu_{p},\frac{l_{A}}{L})+\frac{\kappa}{YL^{2}}\int d^{2}{\bm{x}}\sqrt{g}\left[\left(\frac{1}{R_{1}}-H_{0}\right)^{2}+\left(\frac{1}{R_{2}}-H_{0}\right)^{2}\right]+\frac{\Pi}{Y}\frac{\hat{A}}{L^{2}}+\frac{\tau P}{YL^{2}}\ ,$$

(S9)

where $L$ is a characteristic dimension of the system. In general we will choose $L^{2}=\hat{A}$, so

$$\frac{F}{Y\hat{A}}=f(\nu_{p},\frac{l_{A}}{\sqrt{\hat{A}}})+\frac{\kappa}{Y\hat{A}}\int d^{2}{\bm{x}}\sqrt{g}\left[\left(\frac{1}{R_{1}}-H_{0}\right)^{2}+\left(\frac{1}{R_{2}}-H_{0}\right)^{2}\right]+\frac{\Pi}{Y}+\frac{\tau P}{Y\hat{A}}\ ,$$

(S10)

which defines the effective linear tension $\hat{\tau}=\frac{\tau a_{L}}{Y\hat{A}}$ and dimensionless area $\hat{A}/a_{L}^{2}$, so that all lengths are expressed in terms of the lattice constant $a_{L}$.

Here we show that the covariant formalism defined by Eq. 1, Eq. 2 reduce to the known formulas from elasticity theory when the displacements are small. Within elasticity theory, the reference metric (without disclinations)

$${\bar{g}}_{\alpha\beta}=\delta_{\alpha\beta}$$

(S11)

The surface is described in the Monge gauge,

$$z=h(x,y)\ .$$

(S12)

The mapping ${\bm{x}}={\cal U}({\bar{x}})$ is given by

$${\bm{x}}=\bar{{\bm{x}}}+{\bm{u}}(\bar{{\bm{x}}})\ ,$$

(S13)

where ${\bm{u}}$ is the displacement. Then, the actual metric becomes

	$\displaystyle g_{\alpha\beta}(\bar{\bm{x}})$	$\displaystyle=$	$\displaystyle\bar{\partial}_{\alpha}{\vec{r}}({\bar{\bm{x}}})\bar{\partial}_{\beta}{\vec{r}}({\bar{\bm{x}}})=\delta_{\alpha\beta}+\bar{\partial}_{\alpha}u_{\beta}+\bar{\partial}_{\beta}u_{\alpha}+\bar{\partial}_{\alpha}u_{\gamma}\bar{\partial}_{\beta}u_{\gamma}+{\bar{\partial}}_{\rho}h{\bar{\partial}}_{\gamma}h\left(\delta_{\alpha\rho}\delta_{\gamma\beta}+\delta_{\alpha\rho}\bar{\partial}_{\beta}u_{\gamma}+\bar{\partial}_{\alpha}u_{\rho}\delta_{\beta\lambda}+\bar{\partial}_{\alpha}u_{\rho}\bar{\partial}_{\beta}u_{\gamma}\right)$		(S14)
		$\displaystyle\approx$	$\displaystyle\delta_{\alpha\beta}+\bar{\partial}_{\alpha}u_{\beta}+\bar{\partial}_{\beta}u_{\alpha}+{\bar{\partial}}_{\alpha}h{\bar{\partial}}_{\beta}h$

If only linear terms in ${\bm{u}}$ and the leading term in $h$ are kept the strain tensor Eq. 2 becomes

$$u_{\alpha\beta}=\frac{1}{2}\left(\bar{\partial}_{\alpha}u_{\beta}+\bar{\partial}_{\beta}u_{\alpha}+{\bar{\partial}}_{\alpha}h{\bar{\partial}}_{\beta}h\right)\ .$$

(S15)

The actual metric is

$$g_{\alpha\beta}({\bar{\bm{x}}})=\bar{g}_{\alpha\beta}({\bar{\bm{x}}})+2u_{\alpha\beta}({\bar{\bm{x}}})$$

(S16)

The leading elastic part of the free energy, consistent with the expansion Eq. S15 becomes

	$\displaystyle{\cal F}^{elastic}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{Y}{1-\nu_{p}^{2}}\left(\nu_{p}(u_{\alpha\alpha}^{2}+(1-\nu_{p})u_{\alpha\beta}u_{\alpha\beta}\right)$		(S17)
		$\displaystyle=$	$\displaystyle\frac{1}{2}\left(2\mu(u_{\alpha\beta})^{2}+\lambda(u_{\alpha\alpha})^{2}\right)$

expressed in terms of the Lame coefficients $\lambda,\mu$ instead of the Young modulus $Y=\frac{4\mu(\mu+\lambda)}{2\mu+\lambda}$ and Poisson ratio $\nu_{p}=\frac{\lambda}{2\mu+\lambda}$. For fixed geometry, that is for a given $f$, this is exactly the same free energy and strains as used in linear elasticity theory, see for example Ref. [1]. The Airy function is the solution to the equation,

$$\frac{1}{Y_{0}}{\bar{\Delta}}^{2}\chi(\bar{{\bm{x}}})=s({\bar{\bm{x}}})-K({\bar{\bm{x}}})\ ,$$

(S18)

see Ref. [1] for a full derivation. The laplacian ${\bar{\Delta}}$ refers to a flat metric. Adding an arbitrary disclination density is done by introducing singularities in the reference metric, as discussed for a central disclination in the main text. It is

$$s({\bar{\bm{x}}})=\frac{\pi}{3}\sum_{i=1}^{N}q_{i}\delta({\bar{\bm{x}}}-{\bar{\bm{x}}}_{i})$$

(S19)

where ${\bar{x}}_{i}$ are the positions of the $N$ disclinations. The Gaussian curvature is obtained by expanding Eq. A3 to leading order, consistent with the expansion in the actual metric Eq. S14. Therefore

$$K({\bar{\bm{x}}})=-\frac{1}{2}\varepsilon_{\alpha\beta}\varepsilon_{\gamma\rho}\bar{\partial}_{\beta}\bar{\partial}_{\rho}\left(\bar{\partial}_{\alpha}h\bar{\partial}_{\gamma}h\right)\ .$$

(S20)

We remark that Eq. S18 is written in terms of a flat metric, and the only contribution from the curved surface is through the approximated Gaussian curvature Eq. S20. Eq. S18 has the physical interpretation of the Gaussian curvature screening the disclination density.

The explicit form of the stress tensor is obtained from the Airy function as

	$\displaystyle\sigma^{\rho\rho}(\rho)$	$\displaystyle=$	$\displaystyle\frac{1}{\rho}\frac{d\chi(\rho)}{d\rho}$
		$\displaystyle=$	$\displaystyle\frac{Y}{2\rho^{2}}\left(\frac{q_{i}}{6}\rho^{2}\log\left(\frac{\rho}{\rho_{0}}\right)+\int^{\rho}_{0}dvv\int_{v}^{\rho_{0}}\frac{du}{u}\left(\frac{df}{du}\right)^{2}-\frac{\rho^{2}}{\rho_{0}^{2}}\int^{\rho_{0}}_{0}dvv\int_{v}^{\rho_{0}}\frac{du}{u}\left(\frac{df}{du}\right)^{2}\right)$
	$\displaystyle\sigma^{\theta\theta}(\rho)$	$\displaystyle=$	$\displaystyle\frac{1}{\rho^{2}}\frac{d^{2}\chi(\rho)}{d\rho^{2}}=\frac{1}{\rho^{2}}\frac{d(\rho\sigma^{\rho\rho}(\rho))}{d\rho}$
		$\displaystyle=$	$\displaystyle\frac{Y}{2\rho^{2}}\left(\frac{q_{i}}{6}\log\left(e\frac{\rho}{\rho_{0}}\right)+\int_{\rho}^{\rho_{0}}\frac{du}{u}\left(\frac{df}{du}\right)^{2}-\frac{1}{\rho^{2}}\int^{\rho}_{0}dvv\int_{v}^{\rho_{0}}\frac{du}{u}\left(\frac{df}{du}\right)^{2}-\frac{1}{\rho_{0}^{2}}\int^{\rho_{0}}_{0}dvv\int_{v}^{\rho_{0}}\frac{du}{u}\left(\frac{df}{du}\right)^{2}\right)\ .$

The strain tensor is

	$\displaystyle u_{\rho\rho}$	$\displaystyle=$	$\displaystyle\frac{1}{Y}\left(\sigma^{\rho\rho}-\nu_{p}\rho^{2}\sigma^{\theta\theta}\right)=\frac{1}{Y}\left(\frac{1}{\rho}\frac{d}{d\rho}-\nu_{p}\frac{d^{2}}{d^{2}\rho}\right)\chi$
	$\displaystyle u_{\theta\theta}$	$\displaystyle=$	$\displaystyle\frac{\rho^{2}}{Y}\left(\rho^{2}\sigma^{\theta\theta}-\nu_{p}\sigma^{\rho\rho}\right)=\frac{\rho^{2}}{Y}\left(\frac{d^{2}}{d^{2}\rho}-\frac{\nu_{p}}{\rho}\frac{d}{d\rho}\right)\chi\ .$		(S22)

Note that these equations are also equivalent to Eq. S15

	$\displaystyle u_{\rho\rho}$	$\displaystyle=$	$\displaystyle\frac{du_{\rho}}{d\rho}+\frac{1}{2}\partial_{\rho}h\partial_{\rho}h$		(S23)
	$\displaystyle\frac{u_{\theta\theta}}{\rho^{2}}$	$\displaystyle=$	$\displaystyle-{\Gamma}^{\rho}_{\theta\theta}u_{\rho}=\frac{u_{\rho}}{\rho}$

with $r(\rho)=\rho+u_{\rho}(\rho)$. The free energy is

$$F=\pi\int_{0}^{\rho_{0}}d\rho\rho\left(\sigma^{\rho\rho}u_{\rho\rho}+\sigma^{\theta\theta}u_{\theta\theta}\right)\ .$$

(S24)

Finally, the mapping $r(\rho)$ (or $\rho(r))$ can be obtained from solving either equation

	$\displaystyle 2u_{\rho\rho}(\rho)$	$\displaystyle=$	$\displaystyle(1+f^{\prime}(r)^{2})\left(\frac{dr}{d\rho}\right)^{2}-1$		(S25)
	$\displaystyle 2u_{\theta\theta}(\rho)$	$\displaystyle=$	$\displaystyle r^{2}(\rho)-\alpha^{2}\rho^{2}\ .$

The explicit solution of the previous equation, consistent at linear order is

$$r(\rho)=\rho-(1-\alpha)\rho+\frac{u_{\theta\theta}(\rho)}{\rho}=\rho\left(1-\frac{q_{i}}{6}+\frac{1}{Y}\left[\frac{d^{2}}{d\rho^{2}}-\frac{\nu_{p}}{\rho}\frac{d}{d\rho}\right]\chi\right)\ ,$$

(S26)

Note that had we used the first equation, the solution

	$\displaystyle\int_{0}^{r}dr\sqrt{1+f^{\prime}(r)^{2}}dr$	$\displaystyle=$	$\displaystyle\int_{0}^{\rho}d\rho\sqrt{1+2u_{\rho\rho}}$
	$\displaystyle r+\frac{1}{2}\int_{0}^{r}f^{\prime}(r)^{2}$	$\displaystyle=$	$\displaystyle\rho+\int_{0}^{\rho}d\rho u_{\rho\rho}$
	$\displaystyle r$	$\displaystyle=$	$\displaystyle\rho+\int_{0}^{\rho}d\rho\frac{du_{\rho}}{d\rho}=\rho+u_{\rho}\ ,$		(S27)

where Eq. S23 has been used.

Addition of a line tension just adds the boundary condition

$$\sigma^{\rho\rho}=-\frac{\tau}{\rho_{0}}$$

(S28)

to the stress tensor, and the additional free energy contribution

$$F^{line}=2\pi\tau r(\rho_{0})\equiv 2\pi\tau r_{0}\ .$$

(S29)

Analytical formulas for the Airy function, stress tensor, strain tensor, and free energy for two different surfaces, the spheroid and sombrero are given below.

We finally note that Eq. S18 can be promoted to an equation

$$\Delta^{2}\chi({\bm{x}})=s({\bm{x}})-K({\bm{x}})$$

(S47)

in terms of the actual metric. In this case, the values used for the Gaussian curvature and the disclination density are covariant and exact. This is the starting point of the effective theory of defects. For rotational symmetric cases, it is possible to solve the equation exactly, at least by numerical integration.