PCA on manifolds: application to spaces of landmarks Dr Sergey - - PowerPoint PPT Presentation

▶
pca on manifolds application to spaces of landmarks
SMART_READER_LITE
LIVE PREVIEW

PCA on manifolds: application to spaces of landmarks Dr Sergey - - PowerPoint PPT Presentation

Mar 22, 2024 295 likes •714 views

PCA on manifolds: application to spaces of landmarks Dr Sergey Kushnarev Singapore University of Technology and Design Infinite-dimensional Riemannian geometry with applications to image matching and shape analysis. ESI, Vienna January 16,

slide-1
SLIDE 1

PCA on manifolds: application to spaces of landmarks

Dr Sergey Kushnarev

Singapore University of Technology and Design

Infinite-dimensional Riemannian geometry with applications to image matching and shape analysis. ESI, Vienna

January 16, 2015

slide-2
SLIDE 2

Principal Component Analysis

Goal of PCA: find the sequence of linear subspaces Vk that best represent the variability of the data.

slide-3
SLIDE 3

Principal Component Analysis

Goal of PCA: find the sequence of linear subspaces Vk that best represent the variability of the data. x1, . . . , xn ∈ Rd. Vk = span{v1, v2, ..., vk}.

◮ v1 = arg max v=1 n

  • i=1

(v · xi)2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

(v1 · xi)2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

(vj · xi)2 + (v · xi)2.

slide-4
SLIDE 4

Principal Component Analysis

Goal of PCA: find the sequence of linear subspaces Vk that best represent the variability of the data. x1, . . . , xn ∈ Rd. Vk = span{v1, v2, ..., vk}.

◮ v1 = arg max v=1 n

  • i=1

(v · xi)2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

(v1 · xi)2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

(vj · xi)2 + (v · xi)2.

slide-5
SLIDE 5

Principal Component Analysis

Goal of PCA: find the sequence of linear subspaces Vk that best represent the variability of the data. x1, . . . , xn ∈ Rd. Vk = span{v1, v2, ..., vk}.

◮ v1 = arg max v=1 n

  • i=1

(v · xi)2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

(v1 · xi)2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

(vj · xi)2 + (v · xi)2.

slide-6
SLIDE 6

Principal Component Analysis

Goal of PCA: find the sequence of linear subspaces Vk that best represent the variability of the data. x1, . . . , xn ∈ Rd. Vk = span{v1, v2, ..., vk}.

◮ v1 = arg max v=1 n

  • i=1

(v · xi)2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

(v1 · xi)2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

(vj · xi)2 + (v · xi)2.

slide-7
SLIDE 7

Principal Component Analysis

Goal of PCA: find the sequence of linear subspaces Vk that best represent the variability of the data. x1, . . . , xn ∈ Rd. Vk = span{v1, v2, ..., vk}.

◮ v1 = arg max v=1 n

  • i=1

(v · xi)2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

(v1 · xi)2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

(vj · xi)2 + (v · xi)2.

slide-8
SLIDE 8

Principal Component Analysis

Maximizing projected variance: vk = arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

(vj · xi)2 + (v · xi)2 Or, equivalently, minimizing residuals: vk = arg min

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

xi − (vj · xi)vj2 + xi − (v · xi)v2

slide-9
SLIDE 9

PCA on a manifold?

M xk

slide-10
SLIDE 10

Tangent PCA

Karcher mean µ = arg min

y∈M N

  • k=1

d(xk, y)2

M xk µ

slide-11
SLIDE 11

Tangent PCA

Karcher mean µ = arg min

y∈M N

  • k=1

d(xk, y)2

M µ TµM vk xk

slide-12
SLIDE 12

Tangent PCA

Shape space linearization via vk ∈ TµM

M µ TµM vk xk

Tangent PCA: finding the sequence of linear subspaces Vk ∈ TµM that best represent the variability of the data.

slide-13
SLIDE 13

Tangent PCA

Shape space linearization via vk ∈ TµM

M µ TµM vk xk

Tangent PCA: finding the sequence of linear subspaces Vk ∈ TµM that best represent the variability of the data.

slide-14
SLIDE 14

Problems with linearization

dist(s1, s2)2 = dist(expµ(v1), expµ(v2))2 = v1 − v22 − 1 3K(v1, v2) + o(v4).

If K(v1, v2) < 0, then dist(expµ(v1), expµ(v2)) > v1 − v2 If K(v1, v2) > 0, then dist(expµ(v1), expµ(v2)) < v1 − v2

v1 v2 s1 s2 µ K > 0 v1 v2 s1 s2 µ K = 0 v1 v2 s1 s2 µ K < 0

slide-15
SLIDE 15

PCA on a manifold, effects of curvature

M = R2, ds2 = (1 + y2)(dx2 + dy2), K = y2 − 1 (y2 + 1)3 . Points uniformly distributed in T0M.

slide-16
SLIDE 16

PCA on a manifold, effects of curvature

ds2 = (1 + y2)(dx2 + dy2), K = y2 − 1 (y2 + 1)3 . Points uniformly distributed in M.

slide-17
SLIDE 17

Geodesic PCA (Fletcher)

Geodesic PCA: finding the sequence of geodesic subspaces Sk ∈ M that best represent the variability of the data.

M xk µ S1 S2

slide-18
SLIDE 18

Geodesic PCA (Fletcher)

Projection π : M → H. πH(x) = arg min

y∈H

d(x, y)2.

Logx y Logµ x Logµ y µ x y H y π(x)

slide-19
SLIDE 19

Geodesic PCA (Fletcher)

Geodesic subspace S = Expµ V , V ⊂ TµM. Projection π : M → H. πH(x) = arg min

y∈H

d(x, y)2.

◮ v1 = arg max v=1 n

  • i=1

d(µ, πSv (xi))2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

d(µ, πSv1(xi))2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

d(µ, πSvj (xi))2 + (v · xi)2.

slide-20
SLIDE 20

Geodesic PCA (Fletcher)

Geodesic subspace S = Expµ V , V ⊂ TµM. Projection π : M → H. πH(x) = arg min

y∈H

d(x, y)2.

◮ v1 = arg max v=1 n

  • i=1

d(µ, πSv (xi))2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

d(µ, πSv1(xi))2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

d(µ, πSvj (xi))2 + (v · xi)2.

slide-21
SLIDE 21

Geodesic PCA (Fletcher)

Geodesic subspace S = Expµ V , V ⊂ TµM. Projection π : M → H. πH(x) = arg min

y∈H

d(x, y)2.

◮ v1 = arg max v=1 n

  • i=1

d(µ, πSv (xi))2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

d(µ, πSv1(xi))2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

d(µ, πSvj (xi))2 + (v · xi)2.

slide-22
SLIDE 22

Geodesic PCA (Fletcher)

Geodesic subspace S = Expµ V , V ⊂ TµM. Projection π : M → H. πH(x) = arg min

y∈H

d(x, y)2.

◮ v1 = arg max v=1 n

  • i=1

d(µ, πSv (xi))2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

d(µ, πSv1(xi))2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

d(µ, πSvj (xi))2 + (v · xi)2.

slide-23
SLIDE 23

Geodesic PCA (Fletcher)

Geodesic subspace S = Expµ V , V ⊂ TµM. Projection π : M → H. πH(x) = arg min

y∈H

d(x, y)2.

◮ v1 = arg max v=1 n

  • i=1

d(µ, πSv (xi))2,

◮ v2 =

arg max

v=1,v⊥V1 n

  • i=1

d(µ, πSv1(xi))2 + (v · xi)2,

◮ . . . ◮ vk =

arg max

v=1,v⊥Vk−1 n

  • i=1

k−1

  • j=1

d(µ, πSvj (xi))2 + (v · xi)2.

slide-24
SLIDE 24

Variance vs residuals

Maximizing variance: vi = arg max

v=1,v∈V ⊥

i−1

N

  • j=1

d(µ, πSv (xj))2, Minimizing residuals vi = arg min

v=1,v∈V ⊥

i−1

N

  • j=1

d(xj, πSv (xj))2,

slide-25
SLIDE 25

What the PCA?!

◮ Tangent PCA (linearization) ◮ Geodesic PCA (computationally intensive) ◮ Let’s try: Curvature PCA (quadratic PCA?)

slide-26
SLIDE 26

What the PCA?!

◮ Tangent PCA (linearization) ◮ Geodesic PCA (computationally intensive) ◮ Let’s try: Curvature PCA (quadratic PCA?)

slide-27
SLIDE 27

What the PCA?!

◮ Tangent PCA (linearization) ◮ Geodesic PCA (computationally intensive) ◮ Let’s try: Curvature PCA (quadratic PCA?)

slide-28
SLIDE 28

Dealing with curvature

Sphere of radius 1/ √ K (sectional curvature is K). cos β = tan √ Kc tan √ Ka , sin β = sin √ Kb sin √ Ka Note: if K < 0, cos β = tanh i √ Kc tanh i √ Ka , which is the same as above, since tanh ix = i tan x.

slide-29
SLIDE 29

Dealing with curvature

Sphere of radius 1/ √ K (sectional curvature is K). cos β = tan √ Kc tan √ Ka , sin β = sin √ Kb sin √ Ka Note: if K < 0, cos β = tanh i √ Kc tanh i √ Ka , which is the same as above, since tanh ix = i tan x.

slide-30
SLIDE 30

Dealing with curvature

Projected variance c = 1 √ K arctan

  • tan(

√ Ka) cos β

  • ,

Residual b = 1 √ K arcsin

  • sin(

√ Ka) sin β

slide-31
SLIDE 31

Dealing with curvature

Looking infinitesimally Projected variance c2 = a2 cos2 β + K 6 a4 sin2(2β) + o(K) Residual b2 = a2 sin2 β − K 3 a4 sin4 β + o(K)

slide-32
SLIDE 32

For vectors {a1, a2, . . . , an} ∈ TpM representing the data {x1, x2, . . . , xn}, we seek direction v, s.t. the projected variance in maximized: v1 = arg max

v,v=1 n

  • k=1
  • ak2 cos2 β + K(ak, v)

6 ak4 sin2(2β)

  • ,

where cos βk = ak, v/akv. Or, not equivalently, the residuals are minimized: v1 = arg min

v,v=1 n

  • k=1
  • ak2 sin2 β − K(ak, v)

3 ak4 sin4 β

slide-33
SLIDE 33

For vectors {a1, a2, . . . , an} ∈ TpM representing the data {x1, x2, . . . , xn}, we seek direction v, s.t. the projected variance in maximized: v1 = arg max

v,v=1 n

  • k=1
  • ak2 cos2 β + K(ak, v)

6 ak4 sin2(2β)

  • ,

where cos βk = ak, v/akv. Or, not equivalently, the residuals are minimized: v1 = arg min

v,v=1 n

  • k=1
  • ak2 sin2 β − K(ak, v)

3 ak4 sin4 β

slide-34
SLIDE 34

Mario’s formula: work in progress

slide-35
SLIDE 35

Two landmarks

slide-36
SLIDE 36

Hippocampus data

JHU data: 60 hippocampi (left, right), 22 landmarks each.

slide-37
SLIDE 37

Danke Sch¨

  • n!
slide-38
SLIDE 38

Where is this?

slide-39
SLIDE 39

Workshop “Mathematics of Shapes and Applications”

July 2016

slide-40
SLIDE 40

Workshop “Mathematics of Shapes and Applications”

July 2016

Where: Singapore. When: July 2016. What:

◮ Tutorial on Applications: Shapes in Multiples Disciplines: 4 -

8 July 2016

◮ Summer School on Mathematics of Shapes: 11 - 22 July 2016 ◮ Workshop on State-of-the-Art Shape Research and its

Applications: 25 - 29 July 2016 Organizers: Sergey Kushnarev (SUTD), Laurent Younes (JHU), Alvina Goh (NUS), Anqi Qiu (NUS), Ji Hui (NUS).