viewing direction estimation in cryo-em using …saharon/bigdata2015/yoel.pdfviewing direction...

Viewing direction estimation in cryo-EM usingsynchronization

Yoel Shkolnisky

Tel-Aviv University

Joint work with Amit Singer (Princeton)

July 08, 2014

Yoel Shkolnisky (Tel-Aviv University) Synchronization in Cryo-EM July 08, 2014 1 / 48

Structural Biology

GoalUnderstand structure and function of molecular structures.

ReasonUnderstand biological processes, understand diseases, find cures.


Orders of Magnitude

Optical microscopy → cannot see too small (wavelength ≈ 550nm).

Resolving power of optical microscope ≈ 100nm. Magnification < ×2000.

Magnification of an electron microscope is hundreds of thousands.


Electron microscope


Electron microscope

SEM 20, 000×Toumey, Nature Nanotechnology


Electron microscope

Light microscope 200×Toumey, Nature Nanotechnology


Electron microscope


The Experiment


The Experiment (cont.)


Cryo-Electron Microscopy

Tomographic image 3D structure

?


Physical setting – CryoEM imaging

Electron source

Copy 1

Projection 1

Electron source

Copy 2

Projection 2

Electron source

Copy 3

Projection 3

g1∈SO(3) g2∈SO(3) g3∈SO(3)

CryoEM: Put sample in liquidmedium, freeze, take images usingan Electron Microscope.

Many images: each imagecorresponds to a different copy ofthe molecule in a different spatialorientation.

Orientations are random andunknown: there might be apreferred orientation!

Goal: Reconstruct the 3D molecule from its projections taken atrandom unknown directions


Workflow

1 Particle selection – manual or automatic image segmentation.

2 Preprocessing – centering, masking, normalization, CTF correction,. . .


Workflow (cont.)

3 Class averaging – find images with similar viewing direction, register andaverage to improve their SNR.

4 Orientations assignment.

5 Reconstructing an initial model – generate a 3D volume by a tomographicinversion algorithm.

6 Iterative refinement.


Iterative refinement


EM imaging – mathematical modeling

Projection Pi

Molecule φ

Electronsource

Ri =

| | |

R(1)i

R(2)i

R(3)i

| | |

φ(r) is the electric potential of the molecule(r = (x , y , z)T ∈ R

3).

The molecule is rotated byRi ∈ SO(3): φi (r) = φ(Ri r).

A projection image is

Pi (x,y) =

∫ ∞

−∞

φ (Ri r) dz

=

∫ ∞

−∞

φ(

xR(1)i + yR

(2)i + zR

(3)i

)

dz .

(known as Radon transform)

Each projection corresponds to an unknownrotation Ri ∈ SO(3).

Goal: Given n images P1, . . . ,Pn, estimate R1, . . . ,Rn.


Projection Images: Toy Example


Projection Images: Real Data

Courtesy of Dr. Joachim Frank


The Fourier projection-slice theorem

The Radon transform and the Fourier transform are related by theprojection-slice theorem.

The 2D FT of the projection image is the double integral

P̂i (ωx , ωy) =

∫

R2

e−ı(xωx+yωy )Pi(x , y) dx dy .

The 3D FT of the molecule is the triple integral

φ̂(ωx , ωy , ωz) =

∫

R3

e−ı(xωx+yωy+zωz )φ(x , y , z) dx dy dz .


Fourier projection-slice theorem (cont.)

P̂i (ωx , ωy) = φ̂(

ωxR(1)i + ωyR

(2)i

)



P̂i (ωx , ωy) = φ̂(

ωxR(1)i + ωyR

(2)i

)

The Fourier transform P̂i of the projection Pi equals φ̂ on the plane

ω = ωxR(1)i + ωyR

(2)i (recall: R

(1)i , R

(2)i , R

(3)i are the columns of Ri).



P̂i (ωx , ωy) = φ̂(

ωxR(1)i + ωyR

(2)i

)



(2)i (recall: R

(1)i , R

(2)i , R


Projection Fourier transform 3D Fourierspace

3D Fourierspace

Fourier transformProjection

P0 P̂0

P1 P̂1



P̂i (ωx , ωy) = φ̂(

ωxR(1)i + ωyR

(2)i

)



(2)i (recall: R

(1)i , R

(2)i , R


Projection Fourier transform 3D Fourierspace

3D Fourierspace

Fourier transformProjection

P0 P̂0

P1 P̂1

Any two planes intersect at a common line ⇒ any two images have acommon line.


Angular Reconstitution (Van Heel 1987, Vainshtein & Goncharov 1986)

Why is reconstruction possible at all?

Problems: requires accurate detection of common lines (noise), sensitivity toerrors (depends on the three initial images).


Detection rates

Each projection is 129x129 pixels. Additive Gaussian white noise.

SNR =Var(Signal)

Var(Noise),

(a) Clean (b) SNR=1 (c) SNR=1/2 (d) SNR=1/4 (e) SNR=1/8

(f) SNR=1/16 (g) SNR=1/32 (h) SNR=1/64 (i) SNR=1/128 (j) SNR=1/256


Detection rates (cont.)

Correctly identified common line – within 10◦ of its true value (known fromsimulation).

(a) n = 100

SNR p

clean 0.9971 0.968

1/2 0.9301/4 0.8281/8 0.6531/16 0.4441/32 0.2471/64 0.1081/128 0.0461/256 0.0231/512 0.017

(b) n = 500

SNR p

clean 0.9971 0.967

1/2 0.9221/4 0.8171/8 0.6391/16 0.4331/32 0.2481/64 0.1131/128 0.0461/256 0.0231/512 0.015


Main idea (but slightly wrong)

Use common lines information to compute all matrices R−1i Rj ,

i , j = 1, . . . ,N .Define the 3N × 3N matrix S , whose (i , j) block of size 3× 3 is R−1

i Rj .S is given by

S =

...· · · R−1

i Rj · · ·...

=

RT1

...

RTN

R1 · · · RN

Extract rotations from SVD of S .Yoel Shkolnisky (Tel-Aviv University) Synchronization in Cryo-EM July 08, 2014 20 / 48

Resolving handedness ambiguity

Given only common lines between three images, we can recover either R−1i Rj

or JR−1i RjJ – No way to distinguish between the two! J = diag(1, 1,−1)

The sum R−1i Rj + JR−1

i RjJ is invariant under this ambiguity: if we haverecovered JR−1

i RjJ then (J = J−1, J2 = I )

JR−1i RjJ + J

(JR−1

i RjJ)J = JR−1

i RjJ + R−1i Rj .

No matter if we recover the product R−1i Rj or JR

−1i RjJ, the sum

R−1i Rj + JR−1

i RjJ remains the same.

Write the rotation matrix R−1i Rj explicitly as

R−1i Rj=

r(ij)11 r

(ij)12 r

(ij)13

r(ij)21 r

(ij)22 r

(ij)23

r(ij)31 r

(ij)32 r

(ij)33

⇒ R−1i Rj+JR

−1i RjJ=

2r(ij)11 2r

(ij)12 0

2r(ij)21 2r

(ij)22 0

0 0 2r(ij)33

.


Synchronization matrix

Take the matrix B(k)ij to be the upper-left 2× 2 block of

(R−1i Rj + JR−1

i RjJ)/2

B(k)ij =

(

r(ij)11 r

(ij)12

r(ij)21 r

(ij)22

)

=

R

(1)i

T

R(1)j R

(1)i

T

R(2)j

R(2)i

TR

(1)j R

(2)i

TR

(2)j

=

(

— R(1)i

T—

— R(2)i

T—

)

| |

R(1)j R

(2)j

| |

.

(R−1i Rj)

T = R−1j Ri ⇒ B

(k)ji =

(

B(k)ij

)T

.

Due to common line misidentifications, each block B(k)ij might contain some

error – improve estimate by averaging over all k

Bij =1

N − 2

∑

k 6=i ,j

B(k)ij .

Define Bii = I2 (I2 is the 2× 2 identity matrix).


Synchronization matrix (cont.)

The synchronization matrix S is defined as the 2N × 2N matrix whose (i , j)block of size 2× 2 is Bij

Sij = Bij , i , j = 1, . . . ,N .

S is symmetric and admits the decomposition

S =

...

· · · Bij · · ·

...

=

HT

︷︸︸︷

— R(1)1

T—

— R(2)1

T—

...

— R(1)N

T—

— R(2)N

T—

H︷︸︸︷

| | | |

R(1)1 R

(2)1 · · · R

(1)N R

(2)N

| | | |


Synchronization matrix (cont.)

Reason: The (i , j) 2× 2 block of HTH is give by

...

— R(1)i

T—

— R(2)i

T—

...

| |

· · · R(1)j R

(2)j · · ·

| |

=

...

· · · Bij · · ·

...

.

S is of rank 3.

Three nontrivial eigenvectors of S are linear combinations the columns of HT

(the column space of S).


Unmixing

The nontrivial eigenvectors of S , denoted V = (v1, v2, v3), vi ∈ R2N , span

the same space as the columns of HT .

Need to unmix the columns of HT from these eigenvectors – find A ∈ R3×3

such thatVAT = HT ,

or equivalently,

A

— vT1 —

— vT2 —

— vT3 —

︸︷︷︸

VT

=

| | | |

R(1)1 R

(2)1 · · · R

(1)N R

(2)N

| | | |

︸︷︷︸

H

.

Gives first two columns of the rotation matrices.

The third column of each rotation matrix is then given by the cross productof the first two columns.


Unmixing (cont.)

Split V T into two sets

A

v(1)1 v

(3)1 · · · v

(2N−1)1

v(1)2 v

(3)2 · · · v

(2N−1)2

v(1)3 v

(3)3 · · · v

(2N−1)3

︸︷︷︸

V1

=

| | | |

R(1)1 R

(1)2 · · · R

(1)N

| | | |

.

A

v(2)1 v

(4)1 · · · v

(2N)1

v(2)2 v

(4)2 · · · v

(2N)2

v(2)3 v

(4)3 · · · v

(2N)3

︸︷︷︸

V2

=

| | | |

R(2)1 R

(2)2 · · · R

(2)N

| | | |

.

From the orthogonality conditions, for i = 1, . . . ,N,(

R(1)i

)T

R(1)i = 1,

(

R(2)i

)T

R(2)i = 1,

(

R(1)i

)T

R(2)i = 0,

we get the 3N linear equations

(V T1 A

TAV1)ii = 1, (V T

2 ATAV2)ii = 1, (V T

1 ATAV2)ii = 0,

for the 6 unknowns of ATA.

Get A from ATA using SVD.


Algorithm

Require: Common line matrix C = (cij) of size N × N .

1: for i , j = 1, . . . ,N do

2: for k = 1, . . . ,N do

3: if i 6= j 6= k then

4: Gijk =

(1 〈cij ,cik〉〈cji ,cjk〉

〈cik ,cij 〉 1 〈cki ,ckj〉〈cjk ,cji〉〈ckj ,cki〉 1

)

.

5: Gijk = QTijkQijk , Qijk = (qij qik qjk ) ∈ R

3×3.6: Ci = (cij cik cij × cik) , Qi = (qij qik qij × qik).7: Cj = (cji cjk cji × cjk ) , Qj = (qji qjk qji × qjk ) .8: Rij = CiQ

−1i QjC

−1j . //Now Rij = R−1

i Rj or Rij = JR−1i RjJ.

9: B(k)ij = upper-left 2× 2 block of Rij .

10: end if

11: end for

12: Bij =1

N−2

∑

k 6=i ,j B(k)ij , Bii = I2, Sij = Bij .

13: end for

14: Compute the to three eigenvectors of S – v1, v2, v3.

15: Unmix rotation matrices from v1, v2, v3.


The limit of S

Take a vector f = (f1, . . . , f2N)T, viewed as the samples of some function

f : SO(3) → R2, that is f (Ri) = (f2i−1, f2i )

T , i = 1, . . . ,N .The operation of S on f is given by

Sf =

...· · · Bij · · ·

...

f1...

f2N

,

or equivalently, the ith 2× 1 block of the product is

Sf (Ri ) =N∑

j=1

Bij

(f2j−1

f2j

)

=N∑

j=1

Bij f (Rj).

Bij is the upper-left 2× 2 block of R−1i Rj , thus

S(Ri ,Rj) = Bij =

(1 0 00 1 0

)

R−1i Rj

(1 0 00 1 0

)T

, for i 6= j ,

S(Ri ,Ri) = Bii =

(1 00 1

)

, for i = j .


The limit of S (cont.)

Using the kernel S(Ri ,Rj), S is given by

Sf (Ri ) =

N∑

j=1

S(Ri ,Rj)f (Rj).

Assuming Ri are sampled from SO(3) uniformly at random

limN→∞

1

N(Sf ) (R) = E [S(R ,U)f (U)] =

∫

SO(3)

S(R ,U)f (U)dU ,

where dU is the Haar measure.

We will compute the eigenvalues of the operator

(Sf ) (R) =

∫

SO(3)

S(R ,U)f (U)dU .

Those are approximations to the eigenvalues of the matrix S .



In fact, we will show that the functions g : SO(3) → R2×3 given by

g(R) =

(1 0 00 1 0

)

R−1.

are eigenfunctions corresponding to the nontrivial eigenvalue.

g(R) extracts the first two rows of the rotation matrix R−1, that is, the firsttwo columns of R .

In other words, the first two columns of all rotation matrices areeigenfunctions of S.

We have constructed an operator whose eigenfunctions are the unknownrotations.



By direct evaluation

S(R ,U)g(U) =

(1 0 00 1 0

)

R−1U

(1 0 00 1 0

)T (1 0 00 1 0

)

U−1

=

(1 0 00 1 0

)

R−1U

1 0 00 1 00 0 0

U−1

=

(1 0 00 1 0

)

R−1(

I − U(3)U(3)T)

,

where U(3) is the third column of the matrix U .

Write U(3) in coordinates as

U(3) = (x , y , z)T, x2 + y2 + z2 = 1,

and so

U(3)U(3)T =

x2 xy xz

xy y2 yz

xz yz z2

.



Substitute in S

(Sg) (R) =

∫

SO(3)

(1 0 00 1 0

)

R−1(

I − U(3)U(3)T)

dU

=

(1 0 00 1 0

)

R−1

∫

SO(3)

(

I − U(3)U(3)T)

dU

=

(1 0 00 1 0

)

R−1

∫

SO(3)

(1−x2 −xy −xz

−xy 1−y2 −yz

−xz −yz 1−z2

)

dU .

Parameterize SO(3) as an axis of rotation and a rotation around that axis –SO(3) as S2 × [0, 2π) ⇒ the last integral depends only on the axis ofrotation; the integral collapses to an integral over S2.

From symmetry of S2

∫

S2

x2dµ =

∫

S2

y2dµ =

∫

S2

z2dµ.



Since∫

S2

x2dµ+

∫

S2

y2dµ+

∫

S2

z2dµ =

∫

S2

(x2 + y2 + z2

)dµ =

∫

S2

dµ = 1,

we get that ∫

S2

x2dµ =

∫

S2

y2dµ =

∫

S2

z2dµ = 1/3.

Again from symmetry of S2

∫

S2

xy dµ =

∫

S2

xz dµ =

∫

S2

yz dµ = 0,

We conclude that∫

SO(3)

(1−x2 −xy −xz

−xy 1−y2 −yz

−xz −yz 1−z2

)

dU =

(2/3

2/32/3

)

,

and so,

(Sg) (R) =2

3

(1 0 00 1 0

)

R−1 =2

3g(R),

The eigenvalues of the matrix S converge to 23N .


Examples: noiseless projections – data set

N = 10, 100, 1000 projections from the density map of the 50S ribosomalsubunit. Each of size 129× 129.

Rotations drawn uniformly at random from the uniform distribution on SO(3).

Compute L = 360 Fourier rays in each projection.

Find common lines using correlation.

Construct the matrix S .


Examples: noiseless projections – spectrum of S

1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

i

λi

N = 10

1 2 3 4 5 6 7 8 9 100

10

20

30

40

50

60

70

i

λi

N = 100

1 2 3 4 5 6 7 8 9 100

100

200

300

400

500

600

700

i

λi

N = 1000

Three dominant eigenvalues.

Fourth and higher eigenvalues are not exactly zero due to the discretizationof Fourier space using L = 360 (but much smaller than the first three ones).

Top eigenvalues approximately equal 2N/3 (688, 656, and 655 for N = 1000).


Examples: noiseless projections – estimation error

Estimated rotations – R̃1, . . . , R̃N .

True rotations – R1, . . . ,RN .

Split each projection Pi into L = 360 rays c(l)i = (cos 2πl/L, sin 2πl/L, 0),

l = 0, . . . , L− 1.

Projection Pi

Projection Pj

P̂i

P̂j

3D Fourier space

3D Fourier space

(cosαij , sinαij )

(cosαji , sinαji )

Ricij = qij

Ricij = Rjcji = qij

Estimation error for ray l of projection i : 〈Ric(l)i , R̃ic

(l)i 〉.


Examples: noiseless projections – estimation error (cont.)

Histogram of estimation errors (after registration) in degrees:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

50

100

150

200

250

N = 100 0.05 0.1 0.15 0.2 0.25 0.3 0.35

0

500

1000

1500

2000

2500

N = 1000 0.02 0.04 0.06 0.08 0.1 0.12

0

0.5

1

1.5

2

2.5x 10

4

N = 1000

Error is very small even for N = 10, and decreases as N increases.

Error is not zero even in the noiseless case due to discretization of Fourierspace into L = 360.


Examples: noisy projections – data set

500 projections from the density map of the 50S ribosomal subunit.

corrupted them with additive Gaussian white noise with SNR = 1/16 andSNR = 1/32.

Each image of size 129× 129.

Clean

SNR = 1/16

SNR = 1/32


Examples: noisy projections – spectrum of S

SNR = 1/16

1 2 3 4 5 6 7 8 9 100

50

100

150

200

250

300

i

λi

−50 0 50 100 150 200 250 300 3500

50

100

150

200

250

300

350

SNR = 1/32

1 2 3 4 5 6 7 8 9 100

50

100

150

200

250

i

λi

−50 0 50 100 150 200 2500

20

40

60

80

100

120

140

160

180

200

S still has three dominant eigenvalues.

Remaining eigenvalues are no longer small due to misidentifications ofcommon lines.

Still, there is a significant gap between the third and the fourth eigenvalues.

Gap gets smaller as the SNR decreases.


Examples: noisy projections – estimation error

Angle estimation errors (in degrees):

0 5 10 15 20 25 300

2000

4000

6000

8000

10000

12000

14000

SNR = 1/16Mean error 5.73◦

90% less than 10◦

0 10 20 30 40 50 60 70 80 900

0.5

1

1.5

2

2.5x 10

4

SNR = 1/32Mean error 10.64◦

60% less than 10◦

For SNR = 1/16 only 28% of the common lines were identified correctly(deviates from true common line by up to 5 degrees).

At least 72% of the blocks Bij are completely wrong (the ones for which thecommon line between Pi and Pj was misidentified).

For SNR = 1/32 only 14% of the common lines were identified correctly.


Examples: noisy projections – reconstructions

Reference SNR = 1/16 SNR = 1/32

Reconstructed from noisy images and estimated orientations


Examples: real data

40, 338 images of 70S split randomly into two group (20, 169 images each).

Each image 247× 247 pixels.

Class averaging - average each image with its 5 nearest neighbors.

Choose 1450 best class averages.

Pipeline implemented with ASPIRE (http://spr.math.princeton.edu/).

Raw images (provided by Dr. Joachim Frank)


http://spr.math.princeton.edu/

Examples: real data

40, 338 images of 70S split randomly into two group (20, 169 images each).

Each image 247× 247 pixels.

Class averaging - average each image with its 5 nearest neighbors.

Choose 1450 best class averages.

Pipeline implemented with ASPIRE (http://spr.math.princeton.edu/).

Class averages



Examples: real data (cont.)

Group 1 Group 2


Examples: real data (cont.)

18.3 9.2 6.1 4.6 3.7 3.1

Resolution=24.90A

0.055 0.109 0.164 0.219 0.273 0.3280

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1/A

Fourier shell correlation


What’s next?


What’s next?

Details inY. Shkolnisky, A. Singer, Viewing Direction Estimation in Cryo-EM UsingSynchronization, SIAM Journal on Imaging Sciences, 5 (3), pp. 1088-1110(2012).


What’s next?


3N × 3N


What’s next?


3N × 3N

Improved common lines detection


What’s next?


3N × 3N


Multiple common line candidates


What’s next?


3N × 3N


Multiple common line candidates

common line algorithms for molecules with symmetries


What else (1)?

Common-Lines Based Approaches for 3D Ab-Initio Modeling

◮ R. R. Coifman, Y. Shkolnisky, F. J. Sigworth, A. Singer, Reference FreeStructure Determination through Eigenvectors of Center of Mass Operators,Applied and Computational Harmonic Analysis, 28 (3), pp. 296-312 (2010).

◮ A. Singer, R. R. Coifman, F. J. Sigworth, D. W. Chester, Y. Shkolnisky,Detecting Consistent Common Lines in Cryo-EM by Voting, Journal ofStructural Biology, 169 (3), pp. 312-322 (2010).

◮ A. Singer, Y. Shkolnisky, Three-Dimensional Structure Determination fromCommon Lines in Cryo-EM by Eigenvectors and Semidefinite Programming,SIAM Journal on Imaging Sciences, 4 (2), pp. 543-572 (2011).

◮ Y. Shkolnisky, A. Singer, Viewing Direction Estimation in Cryo-EM UsingSynchronization, SIAM Journal on Imaging Sciences, 5 (3), pp. 1088-1110(2012).

◮ L. Wang, A. Singer, Z. Wen, Orientation Determination from Cryo-EM imagesusing Least Unsquared Deviations, SIAM Journal on Imaging Sciences, 6 (4),pp. 2450-2483 (2013).


What else (2)?

2D Class Averaging

◮ A. Singer, Z. Zhao, Y. Shkolnisky, R. Hadani, Viewing Angle Classification ofCryo-Electron Microscopy Images using Eigenvectors, SIAM Journal onImaging Sciences, 4 (2), pp. 543-572 (2011).

◮ Z. Zhao, A. Singer, Rotationally Invariant Image Representation for ViewingDirection Classification in Cryo-EM, Journal of Structural Biology, 186 (1), pp.153-166 (2014).

◮ A. Singer, H.-T. Wu, Vector Diffusion Maps and the Connection Laplacian,Communications on Pure and Applied Mathematics (CPAM), 65 (8), pp.1067-1144 (2012).

Code available in ASPIRE (http://spr.math.princeton.edu/).



Thank You!


Computing the ratios R−1i Rj

qij – unit vector in the direction of intersection of the planes.Since Ricij = Rjcji = qij and Ri are orthogonal

Gijk =

— qTij —

— qTik —

— qTjk —

| | |qij qik qjk| | |




Gijk =

— qTij —

— qTik —

— qTjk —


=

1 〈qij , qik〉〈qij , qjk〉〈qik , qij〉 1 〈qik , qjk〉〈qjk , qij〉〈qjk , qik〉 1




Gijk =

— qTij —

— qTik —

— qTjk —


=


=

1 〈Ricij ,Ri cik〉〈Rj cji ,Rjcjk 〉〈Ricik ,Ri cij 〉 1 〈Rkcki , Rkckj〉〈Rjcjk ,Rjcji 〉〈Rkckj ,Rkcki 〉 1




Gijk =

— qTij —

— qTik —

— qTjk —


=


=

1 〈Ricij ,Ri cik〉〈Rj cji ,Rjcjk 〉〈Ricik ,Ri cij 〉 1 〈Rkcki , Rkckj〉〈Rjcjk ,Rjcji 〉〈Rkckj ,Rkcki 〉 1

=

1 〈cij , cik 〉〈cji , cjk 〉〈cik , cij〉 1 〈cki , ckj 〉〈cjk , cji 〉〈ckj , cki〉 1

.


Computing the ratios R−1i Rj (cont.)

Gijk =

— qTij —

— qTik —

— qTjk —

︸︷︷︸

QTijk


︸︷︷︸

Qijk

=

1 〈cij , cik〉〈cji , cjk 〉〈cik , cij〉 1 〈cki , ckj 〉〈cjk , cji 〉〈ckj , cki〉 1

.

Extract Qijk = (qij qik qjk) from Gijk by SVD.



Gijk =

— qTij —

— qTik —

— qTjk —

︸︷︷︸

QTijk


︸︷︷︸

Qijk

=


.


Actually extract Q̃ijk = OijkQijk , for some arbitrary orthogonal matrix Oijk



Gijk =

— qTij —

— qTik —

— qTjk —

︸︷︷︸

QTijk


︸︷︷︸

Qijk

=


.


Actually extract Q̃ijk = OijkQijk , for some arbitrary orthogonal matrix Oijk

(since Q̃Tijk Q̃ijk = (OijkQijk)

TOijkQijk = QT

ijkOTijkOijkQijk = QT

ijkQijk = Gijk).



From the triplet of images Pi , Pj , Pk we have computed the matrixQijk = (qij qik qjk ).



From the triplet of images Pi , Pj , Pk we have computed the matrixQijk = (qij qik qjk ).Construct the 3× 3 matrices (using the computed Qijk)

Ci = (cij cik cij × cik) , Qi = (qij qik qij × qik) ,



From the triplet of images Pi , Pj , Pk we have computed the matrixQijk = (qij qik qjk).Construct the 3× 3 matrices (using the computed Qijk)




From the triplet of images Pi , Pj , Pk we have computed the matrixQijk = (qij qik qjk).Construct the 3× 3 matrices (using the computed Qijk)


Cj = (cji cjk cji × cjk) , Qj = (qji qjk qji × qjk) .







RiCi = Ri


)

=

(| | |


)

=

(| | |qij qik qij×qik| | |

)







RiCi = Ri


)

=

(| | |


)

=


)

= Qi .







RiCi = Ri


)

=

(| | |


)

=


)

= Qi .

Similarly, RjCj = Qj .







RiCi = Ri


)

=

(| | |


)

=


)

= Qi .


w�

Ri = QiC−1i , Rj = QjC

−1j







RiCi = Ri


)

=

(| | |


)

=


)

= Qi .


w�

Ri = QiC−1i , Rj = QjC

−1j

w�

R−1i Rj =

(QiC

−1i

)−1QjC

−1j = CiQ

−1i QjC

−1j .



We actually have Q̃ijk = OijkQijk for some arbitrary orthogonal Oijk .




If detOijk = 1, we get the matrices

Qi = ( qij , qik , qij × qik)

Qj = ( qji , qjk , qji × qjk)





Q̃i = (Oijkqij ,Oijkqik ,Oijkqij × Oijkqik)

Q̃j = (Oijkqji ,Oijkqjk ,Oijkqji × Oijkqjk)





Q̃i = (Oijkqij ,Oijkqik ,Oijkqij × Oijkqik) = Oijk (qij , qik , qij × qik)

Q̃j = (Oijkqji ,Oijkqjk ,Oijkqji × Oijkqjk) = Oijk (qji , qjk , qji × qjk )





Q̃i = (Oijkqij ,Oijkqik ,Oijkqij × Oijkqik) = Oijk (qij , qik , qij × qik) = OijkQi ,

Q̃j = (Oijkqji ,Oijkqjk ,Oijkqji × Oijkqjk) = Oijk (qji , qjk , qji × qjk ) = OijkQj .







Then, instead of Ri = QiC−1i and Rj = QjC

−1j we recover

R̃i = Q̃iC−1i = OijkQiC

−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,








−1j we recover


−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,

and

R̃−1i R̃j =








−1j we recover


−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,

and

R̃−1i R̃j =

(OijkQiC

−1i

)−1OijkQjC

−1j








−1j we recover


−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,

and

R̃−1i R̃j =

(OijkQiC

−1i

)−1OijkQjC

−1j = CiQ

−1i O−1

ijk OijkQjC−1j








−1j we recover


−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,

and

R̃−1i R̃j =

(OijkQiC

−1i

)−1OijkQjC

−1j = CiQ

−1i O−1

ijk OijkQjC−1j

= CiQ−1i QjC

−1j








−1j we recover


−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,

and

R̃−1i R̃j =

(OijkQiC

−1i

)−1OijkQjC

−1j = CiQ

−1i O−1

ijk OijkQjC−1j

= CiQ−1i QjC

−1j = R−1

i Rj .








−1j we recover


−1i , R̃j = Q̃jC

−1j = OijkQjC

−1j ,

and

R̃−1i R̃j =

(OijkQiC

−1i

)−1OijkQjC

−1j = CiQ

−1i O−1

ijk OijkQjC−1j

= CiQ−1i QjC

−1j = R−1

i Rj .

If detOijk = 1 then R−1i Rj is independent of Oijk .



If detOijk = −1, we get






= (Oijkqij ,Oijkqik ,−Oijk (qij × qik))






= Oijk (qij , qik ,− (qij × qik))







= OijkQiJ, J = diag(1, 1,−1).








Similarly Q̃j = OijkQjJ.









The matrices R̃i and R̃j are given by

R̃i = OijkQiJC−1i , R̃j = OijkQjJC

−1j .











−1j .

Since the third coordinate in all vectors cij is zero, by a direct calculation weget that CiJ = JCi , and JC−1

i = C−1i J.











−1j .


i = C−1i J. Thus,

R̃−1i R̃j =











−1j .


i = C−1i J. Thus,

R̃−1i R̃j =

(OijkQiJC

−1i

)−1OijkQjJC

−1j











−1j .


i = C−1i J. Thus,

R̃−1i R̃j =

(OijkQiJC

−1i

)−1OijkQjJC

−1j

= CiJQ−1i QjJCj











−1j .


i = C−1i J. Thus,

R̃−1i R̃j =

(OijkQiJC

−1i

)−1OijkQjJC

−1j

= CiJQ−1i QjJCj = JCiQ

−1i QjC

−1j J (since CiJ = JCi )











−1j .


i = C−1i J. Thus,

R̃−1i R̃j =

(OijkQiJC

−1i

)−1OijkQjJC

−1j

= CiJQ−1i QjJCj = JCiQ

−1i QjC

−1j J (since CiJ = JCi )

= JR−1i RjJ.


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