Approach to Cryptography from Differential Geometry with Example

Abstract

We propose a public-key encryption scheme that arise from a kind of differential geometry called Finsler geometry. Our approach is first to observe a map of a tangent space to another tangent space, and find asymmetricity of linear parallel displacement, which is easy to compute but hard to invert. Then we construct an example of the map over the real numbers. By quantization, we propose a public-key encryption scheme. The scheme is proved to be IND-CCA2 secure under the new assumption of the decisional linear parallel displacement problem.

6 Appendix

(1) Calculation of (2) (in p.3)

$$\begin{aligned} N^i_j(x,y)=\sum _{r}\gamma ^i_{rj}(x,y)y^r-\sum _{p,q,r}C^i_{jr}(x,y)\gamma ^r_{pq}(x,y)y^py^q, \end{aligned}$$

where

$$\begin{aligned}&\gamma ^i_{pq}(x,y)=\sum _{h}\frac{1}{2}g^{hi}\left( \frac{\partial g_{ph}}{\partial x^q}+\frac{\partial g_{hq}}{\partial x^p}-\frac{\partial g_{pq}}{\partial x^h}\right) ,\\&\qquad \qquad \;\; C^i_{jr}(x,y)=\sum _{h}\frac{1}{2}g^{hi}\frac{\partial g_{jh}}{\partial y^r}. \end{aligned}$$

Calculation of (3) (in p.3)

$$\begin{aligned} F^i_{jr}(x,y)=\sum _{h}\frac{1}{2}g^{hi}\left( \frac{\delta g_{jh}}{\delta x^r}+\frac{\delta g_{hr}}{\delta x^j}-\frac{\delta g_{jr}}{\delta x^h}\right) , \end{aligned}$$

where

$$\begin{aligned} \frac{\delta }{\delta x^i}=\frac{\partial }{\partial x^i}-\sum _{r}N^r_i\frac{\partial }{\partial y^r}. \end{aligned}$$

Here the indices $h,i,j,\cdots ,p,q,r,\cdots$ of $\sum$ run from 1 to $n(=dimM)$.

(2) Metric tensor field: $g_{ij}(c,\dot{c})$ ( in p.8.)

(3) Nonlinear Connection: $N^i_j(c,\dot{c})=\sum _rF^i_{jr}\dot{c}^r$ ( in p.8)

(4) The components $B_1^1, B_1^2, B_2^1, B_2^2$ of $\varPi _{c_m}(t)$ ( in p.8)

(5) The components $E_0, E_1, E_2$ of the energy $E(v_1)$ ( in p.9)

(6) The components $V_3^1, V_3^2$ of $V_3=\varPi _c(\tau )V=(V_3^1,V_3^2)$ ( in p.9)

(7) We explain the concrete example as below

Let $v=(6806, 2346)$ be a plaintext.

Encryption

$v_0=d+dv=(8040,7778)$,

$\tau =3, \beta _0=1$ then $ct_0=\{5.58763\times 10^{19}, 2.64606\times 10^{18}, 1.47664\times 10^{18}\}$,

$\tau =3, \beta _1=2$ then $ct_1=\{8.73996\times 10^{17}, 4.13985\times 10^{16}, 2.31025\times 10^{16}\}$,

$\tau =3, \beta _2=3$ then $ct_2=\{7.68363\times 10^{16}, 3.64162\times 10^{15}, 2.03221\times 10^{15}\}$,

then we obtain the $\text {ciphertext}=\{ct_0,ct_1,ct_2\}$.

Decryption

First, from $ct_0$,

$$\begin{aligned} \begin{aligned}V=\varPi _c^{-1}(\tau )V_3&=(-3.10237\times 10^{15} - 1.14863\times 10^{13} \tau + 3.06421\times 10^{14} \tau ^2,\\&-1.72354\times 10^{15} + 6.4659\times 10^{13} \tau + 1.70234\times 10^{14} \tau ^2)\end{aligned} \end{aligned}$$

Further, from $ct_1$ and $ct_2$, we have others V. Next, from $SKX=\{\frac{4}{3}, \frac{24570}{499}X^1, 3822X^2\}$, we can construct $ct_i\cdot SKX$ as follows:

then we have the following linear system from (7)

and this system is solved unique and the solution $(X^1,X^2)$ is

(17)

Next, input the above $(X^1,X^2)$ to the Equations (8) and (16), then we have the equation of $E(v_1)$

$$\begin{aligned}&-1.73632\times 10^{44} + 3.51579\times 10^{44} \tau - 5.99366\times 10^{44} \tau ^2 - 2.21611\times 10^{43} \tau ^3\\&+ 1.20382\times 10^{44} \tau ^4 - 1.20558\times 10^{42} \tau ^5 - 5.9622\times 10^{42} \tau ^6=0. \end{aligned}$$

If we solve this equation, finally we have the integer $\alpha$ as the value of $\tau$ as follows

$$\begin{aligned} \alpha =3, \end{aligned}$$

where $\tau = 3.00000,\ 2.98760$. By using the solution $\alpha$ to (17), we can obtain the value of $v_0=(v_0^1,v_0^2)$ and the plaintext $v=(v^1,v^2)$ as follows:

$$\begin{aligned} v_0^1=X^1=8039.99999 \sim 8040,\ v_0^2=X^2=7778.00000\sim 7778, \end{aligned}$$

$$\begin{aligned} v^1=v_0^1-dv^1= 8040-1234=6806,\ v^2=v_0^2-dv^2= 7778-5432=2346.\ \ \ \ \Box \end{aligned}$$