Absolute Conic

1. Conic

  • Conic is a curve on a plance which is of form ax2+bxy+cy2+dx+ey+f=0ax^2 + bxy + cy^2 + dx + ey + f = 0.
  • Let (x1,x2,x3)t(x1x3,x2x3,1)t(x_1, x_2, x_3)^t \sim (\frac{x_1}{x_3}, \frac{x_2}{x_3}, 1)^t be the homogeneous coordinates of a point, then ax12+bx1x2+cx22+dx1x3+ex2x3+fx32=0ax^2_1 + bx_1x_2 + cx^2_2 + dx_1x_3 + ex_2x_3 + fx^2_3 = 0.
  • For matrix representation, it is of form xtCx=0x^t C x = 0.
(x1x2x3)(ab2d2b2ce2d2e2f)(x1x2x3)=0    xtCx=0\begin{equation} (x_1 x_2 x_3) \left(\begin{array}{ccc} a & \frac{b}{2} & \frac{d}{2} \\ \frac{b}{2} & c & \frac{e}{2} \\ \frac{d}{2} & \frac{e}{2} & f \end{array}\right) \left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right) = 0 \implies x^t C x = 0 \end{equation}
  • CC is the conic as matrix representation which is symmetric. In addition, all kC(k0)kC(k \not =0) are the same conic.

2. The tangent to the conic

  • Let the point xx be on the conic CC and on the line ll. It means xtCx=0x^t C x = 0 and ltx=0l^t x = 0. If there exists another point yy such as xx, ytCy=0y^t C y = 0 and lty=0l^t y = 0.
  • For any α>0\alpha > 0, (x+αy)tC(x+αy)=xtCx+α2ytCy+αytCx+αxtCy=0+0+0+0=0(x + \alpha y)^t C (x + \alpha y) = x^t C x + \alpha^2 y^t C y + \alpha y^t C x + \alpha x^t C y = 0 + 0 + 0 + 0 = 0. It implies that all the points on the line passing through points xx and yy are also on CC, so it is a contradiction. Therefore, ll is the tangent to the conic and l=Cxl = Cx.

3. Degenerate conic

DegenerateConic

  • When a double conic is divided by a plane passing through its center, its cross section is two lines, not parabola, hyperbola, circle, or ellipse. These two lines ll and mm are a degenerate conic CC_{\infty} such that C=lmt+mltC_{\infty} = lm^t + ml^t.
  • Since xtCx=0x^t C_{\infty} x = 0, ltx=0l^t x = 0, and mtx=0m^t x = 0, C=lmt+mltC_{\infty} = lm^t + ml^t and rank(C)=2rank(C_{\infty}) = 2.

4. Dual conic

DualConic

  • A conic on a plane is a set of points, but it can be viewed as a set of tangents to each point which is called dual conic.
  • The conic CC satisfies xtCx=0x^t C x = 0 for a point xx on CC, and the dual conic CC^{*} satisfies ltCl=0l^t C^{*} l = 0 for the tangent to the xx.
  • If a point xx is on the conic CC, then the tangent ll to xx is l=Cxl = Cx, so x=C1lx = C^{-1}l.
  • xt=ltCt=ltC1x^t = l^t C^{-t} = l^t C^{-1} since CC is symmetric. It implies that xtCx=(ltC1)l=0x^t C x = (l^t C^{-1})l = 0. Therefore, C=C1C^{*} = C^{-1}.

5. Homography of conic

  • Assume that a point xx is on the conic CC and a line ll is on the dual conic CC^{*} of CC. These are transformed to xx', CC', ll' and CC'^{*} by a homography HH.
  • From xtCxx^t C x and x=Hxx' = Hx, xtHtCH1x=0x'^t H^{-t} CH^{-1} x' = 0, so C=HtCH1C' = H^{-t} C H^{-1}.
  • The tangent ll to xx satisfies ltx=0l^t x = 0, so ltx=lt(H1x)=(ltH1)x=0=ltxl^t x = l^t (H^{-1}x') = (l^t H^{-1})x' = 0 = l'^t x'. It means l=Htll' = H^{-t}l.
  • From ltCl=0l^t C^{*} l = 0 and l=Htll' = H^{-t}l, ltHCHtl=0l'^t H C^{*} H^t l' = 0, so C=HCHtC'^{*} = H C^{*} H^t.

6. Circular points: all the circles intersects with ll_{\infty} at two points

  • All the circles have b=0b = 0 and a=ca = c from the conic equation ax12+bx1x2+cx22+dx1x3+ex2x3+fx32=0ax^2_1 + bx_1x_2 + cx^2_2 + dx_1x_3 + ex_2x_3 + fx^2_3 = 0.
  • If we set a=1a = 1, x12+x22+dx1x3+ex2x3+fx32=0x^2_1 + x^2_2 + dx_1x_3 + ex_2x_3 + fx^2_3 = 0.
  • The intersection point with ll_{\infty} is at infinity, so x3=0x_3 = 0 and it reduces to x12+x22=0x^2_1 + x^2_2 = 0.
  • In homogeneous coordinates, this has two solutions, (x1,x2)={(1,i),(1,i)}(x_1, x_2) = \{ (1, i), (1, -i) \}.
  • Therefore, the circular points are (1,i,0)t(1, i, 0)^t and (1,i,0)t(1, -i, 0)^t.

7. Dual degenerate conic

  • It is the dual of degenerate conic C=lmt+mltC_{\infty} = l m^t + m l^t.
  • Degenerate conic is defined by lines, so its dual is defined by points.
  • Let the circular points be UU and VV, the dual degenerate conic CC^{*}_{\infty} satisfies C=UVt+VUtC^{*}_{\infty} = UV^t + VU^t.
  • As the dual of xtCx=0x^t C x = 0 form is ltCl=0l^t C^{*} l = 0, CC^{*}_{\infty} consists of the lines such that ltCl=0l^t C^{*}_{\infty} l = 0.
  • CC^{*}_{\infty} as matrix representation is the following:
C=UVt+VUt=(1i0)(1i0)+(1i0)(1i0)=(1i0i10000)+(1i0i10000)(100010000)=(I0undefined0undefined0)\begin{aligned} C^{*}_{\infty} &= UV^t + VU^t = \left(\begin{array}{c} 1 \\ i \\ 0 \end{array}\right) \left(\begin{array}{ccc} 1 & -i & 0 \end{array}\right) + \left(\begin{array}{c} 1 \\ -i \\ 0 \end{array}\right) \left(\begin{array}{ccc} 1 & i & 0 \end{array}\right) \\ &= \left(\begin{array}{ccc} 1 & -i & 0 \\ i & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) + \left(\begin{array}{ccc} 1 & i & 0 \\ -i & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) \equiv \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) = \left(\begin{array}{cc} I & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) \end{aligned}
  • CC^{*}_{\infty} is invariant to similarity homography. This homography HH is
H=(Avundefined0undefined1)\begin{aligned} H = \left(\begin{array}{cc} A & \overrightarrow{v} \\ \overrightarrow{0} & 1 \end{array}\right) \end{aligned}

where AAt=λ2I2AA^t = \lambda^2 I^2. Since C=HCHtC'^{*}_{\infty} = H C^{*}_{\infty} H^t, C=(Avundefined0undefined1)(I0undefined0undefined0)(At0undefinedvundefinedt1)=(Avundefined0undefined1)(At0undefined0undefined0)=(λ2I0undefined0undefined0)(I0undefined0undefined0)=C\begin{aligned} C'^{*}_{\infty} &= \left(\begin{array}{cc} A & \overrightarrow{v} \\ \overrightarrow{0} & 1 \end{array}\right) \left(\begin{array}{cc} I & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) \left(\begin{array}{cc} A^t & \overrightarrow{0} \\ \overrightarrow{v}^t & 1 \end{array}\right) = \left(\begin{array}{cc} A & \overrightarrow{v} \\ \overrightarrow{0} & 1 \end{array}\right) \left(\begin{array}{cc} A^t & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) = \left(\begin{array}{cc} \lambda^2 I & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) \\ &\equiv \left(\begin{array}{cc} I & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) = C^{*}_{\infty} \end{aligned}

8. Quadric

  • Quadric is similar to conic and is defined in one more higher dimension than that of conic.
  • Quadric QQ is a 4×44 \times 4 symmetric matrix and xtQx=0x^t Q x = 0 for the point xx on QQ.
  • As rank(C)=3rank(C) = 3 for a conic CC in general except for the degenerate conic whose rank(C)=2rank(C_{\infty}) = 2, quadric QQ is, in general, rank(Q)=4rank(Q) = 4 except for the degenerate quadric whose rank(Q)=3rank(Q_{\infty}) = 3.

9. The intersection of quadric and plane is a conic

  • For the non-colinear points AA, BB, and CC are on a plane Π\Pi, a point xx on Π\Pi is represented as x=uA+vB+wCx = uA + vB + wC for some uu, vv, cRc \in \R. In other words, x=(A,B,C)(u,v,w)t=Mpundefinedx = (A, B, C)(u, v, w)^t = M\overrightarrow{p} where MM is a 4×34 \times 3 matrix and pundefined\overrightarrow{p} is a 3×13 \times 1 matrix.
  • If xx is on the intersection area of a quadric QQ and Π\Pi, xtQx=0x^t Q x = 0 and pundefinedtMtQMpundefined=0\overrightarrow{p}^t M^t Q M \overrightarrow{p} = 0.

QuadricIntersection

  • In fact, pundefined\overrightarrow{p} can be represented in homogeneous coordinates which is composed of AA, BB, and CC. It means that x=me1undefined+ne2undefined=m(AC)+n(BC)=mA+nB(m+n)Cx = m \overrightarrow{e_1} + n \overrightarrow{e_2} = m(A - C) + n(B - C) = mA + nB - (m + n)C where m=um = u, n=vn = v, and w=mnw = -m-n.
  • Therefore, pundefinedt(MtQM)pundefined\overrightarrow{p}^t (M^t Q M) \overrightarrow{p} can be considered as the conic CC where C=MtQMC = M^t Q M for the point pundefined\overrightarrow{p} on CC.

10. The tangent to the quadric

  • For the point xx on the tangent plane Π\Pi to the quadric QQ, xtQx=0x^t Q x = 0 and Πtx=0\Pi^t x = 0. It implies that Π=Qtx=Qx\Pi = Q^t x = Qx. This property is corresponding to that of conic.

11. Dual quadric

  • A quadric is a set of points, but it can be viewed as a set of tangent planes to each point which is called dual quadric.
  • The quadric QQ satisfies xtQx=0x^t Q x = 0 for a point xx on QQ, and the dual quadric QQ^{*} satisfies ΠtQΠ=0\Pi^t Q^{*} \Pi = 0 for the tangent plane to the xx.
  • The tangent plane Π\Pi on the quadric QQ satisfies Π=Qx\Pi = Qx, so xtQx=ΠtQtQQ1Π=0x^t Q x = \Pi^t Q^{-t}Q Q^{-1}\Pi = 0. It means Q=Qt=Q1Q^{*} = Q^{-t} = Q^{-1}. This property is corresponding to that of conic.

12. Homography of quadric

  • Assume that a point xx is on the quadric QQ and a plane Π\Pi is on the dual quadric QQ^{*} of QQ. These are transformed to xx', QQ', Π\Pi', and Π\Pi'^{*} by a homography HH.
  • From xtQxx^tQx and x=Hxx' = Hx, xtHtQH1x=0x'^t H^{-t} Q H^{-1} x' = 0, so Q=HtQH1Q' = H^{-t} Q H^{-1}.
  • From ΠtQΠ=0\Pi^t Q^{*} \Pi = 0 and Π=HtΠ\Pi' = H^{-t} \Pi, ΠtHQHtΠ=0\Pi'^t H Q^{*} H^t \Pi' = 0, so Q=HQHtQ'^{*} = H Q^{*} H^t.

13. What QxQx stands for where the point xx is NOT on the quadric QQ

PolarPlane

  • QxQx means a polar plane when the point xx is not on the quadric QQ.
  • In other words, the polar plane consists of the planes which passes through xx and tangent to QQ. This property is corresponding to that of conic which is called polar line.
  • Assume that the point yy is on QQ and its tangent plane QyQy passes through xx. So, (Qy)tx=0=ytQtx=ytQx(Qy)^t x = 0 = y^t Q^t x = y^t Qx. Since ytQxy^t Qx is a scalar, ytQx=(ytQx)t=xtQty=(Qx)tyy^t Qx = (y^t Qx)^t = x^t Q^t y = (Qx)^t y. Therefore, QxQx is the polar plane.
  • Assume that the point yy is on a conic CC and its tangent line CyCy passes through xx. So, (Cy)tx=0=ytCtx=ytCx(Cy)^t x = 0 = y^t C^t x = y^t Cx. Since ytCxy^t Cx is a scalar, ytCx=(ytCx)t=xtCty=(Cx)tyy^t Cx = (y^t Cx)^t = x^t C^t y = (Cx)^t y. Therefore, CxCx is the polar line.

14. Absolute conic: all the spheres intersect with Π\Pi_{\infty}

  • A sphere is a kind of a quadric, so the intersection of a sphere and Π=(0,0,0,1)t\Pi_{\infty} = (0, 0, 0, 1)^t is a conic.
  • All the spheres satisfy x12+x22+x32+dx1x4+ex2x4+fx3x4+gx42=0x^2_1 + x^2_2 + x^2_3 + d x_1 x_4 + e x_2 x_4 + f x_3 x_4 + g x^2_4 = 0 for the point (x1,x2,x3,x4)t(x_1, x_2, x_3, x_4)^t in homogeneous coordinates.
  • The intersection with Π\Pi_{\infty} is at infinity, so x4=0x_4 = 0 and it reduces to x12+x22+x32=0x^2_1 + x^2_2 + x^2_3 = 0.
  • This form can be changed as (x1,x2,x3)I(x1,x2,x3)t=0(x_1, x_2, x_3)I(x_1, x_2, x_3)^t = 0 which is of conic form. This conic is on Π\Pi_{\infty} and consists of the only imaginary part, which is called the absolute conic.
  • The absolute conic is denoted by Ω\Omega_{\infty}.
  • The dual of Ω\Omega_{\infty} is called absolute dual quadric Ω\Omega^{*}_{\infty},
(1000010000100000)\begin{aligned} \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{aligned}
  • For all the plances Π=(x1,x2,x3,x4)t\Pi = (x_1, x_2, x_3, x_4)^t such that (x1,x2,x3)(x1,x2,x3)t=0(x_1, x_2, x_3)(x_1, x_2, x_3)^t = 0, Π\Pi is tangent to the Ω\Omega_{\infty}. Moreover, this Π\Pi satisfies ΠtΩΠ=0\Pi^t \Omega^{*}_{\infty} \Pi = 0, which is the dual quadric form.
  • Geometrically, Ω\Omega^{*}_{\infty} consists of all the tangent planes of Ω\Omega_{\infty}.

AbsoluteConic

15. Projection of Ω\Omega_{\infty}

  • When the point x=(x1,x2,x3,0)tx = (x_1, x_2, x_3, 0)^t on Π\Pi_{\infty} is projected by PP, the image of xx is
PX=KR[IE](x1x2x30)=KR(x1x2x3)=Hx\begin{aligned} PX = KR[I \vert -E] \left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ 0 \end{array}\right) = KR \left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right) = H \overline{x} \end{aligned}

where KK is the intrinsic matrix of the camera, RR is the rotation matrix, EE is the camera center, H=KRH=KR and x=(x1,x2,x3)t\overline{x} = (x_1, x_2, x_3)^t. It means that x\overline{x} is the direction to the intersection with Π\Pi_{\infty}.

  • Ω\Omega_{\infty} is defined by this xx which satisfies xtΩx\overline{x}^t \Omega_{\infty} \overline{x}. Moreover, Ω\Omega_{\infty} can be transformed by this HH as follows:
HtΩH1=HtIH1=(KR)t(KR)1=(RtKt)1(KR)1=KtRtR1K1=Kt(RRt)1K1=KtK1=ω\begin{aligned} H^{-t} \Omega_{\infty} H^{-1} &= H^{-t}IH^{-1} = (KR)^{-t} (KR)^{-1} = (R^t K^t)^{-1} (KR)^{-1} \\ &= K^{-t}R^{-t}R^{-1}K^{-1} = K^{-t}(RR^{-t})^{-1}K^{-1} = K^{-t}K^{-1} = \omega \end{aligned}
  • x\overline{x} and Ω\Omega_{\infty} are transformed to ω\omega and HxH\overline{x} by HH. Since x\overline{x} is on the Ω\Omega_{\infty}, HxH\overline{x} should be on ω\omega which means (Hx)tω(Hx)(H\overline{x})^t \omega (H\overline{x}).
  • (Hx)tω(Hx)=(Hx)t(KtK1)(Hx)>0(H\overline{x})^t \omega (H\overline{x}) = (H\overline{x})^t (K^{-t} K^{-1}) (H\overline{x}) > 0 yields that KtK1K^{-t} K^{-1} is positive definite. Therefore, all this kind of HxH\overline{x} are imaginary.

16. All the planes intersect with Ω\Omega_{\infty} at circular points

AbsolutePoint

  • Since Ω\Omega_{\infty} is on Π\Pi_{\infty}, ll_{\infty} is also on Π\Pi_{\infty}. An arbitrary plane Π\Pi includes any circles, so these circles intersect with ll_{\infty} at circular points. It means that this Π\Pi includes ll_{\infty}.
  • For a circle on Π\Pi, there exists the sphere including this circle. This sphere also intersects with Π\Pi_{\infty} because all the spheres intersect with Π\Pi_{\infty} at absolute conic Ω\Omega_{\infty}. Therefore, this sphere includes Ω\Omega_{\infty}. As a result, the intersection of Ω\Omega_{\infty} and ll_{\infty} is circular points.

Reference

[1] https://engineering.purdue.edu/kak/computervision/ECE661Folder/Index.html

[2] Hartley, R. and Zisserman, A. (2003) Multiple View Geometry in Computer Vision. 2nd Edition, Cambridge University Press, Cambridge.


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