Lie Group Machine Learning’s Axiom
Hypothesizes
Huan Xu, Fan-zhang Li
Abstract—Lie group is the combination of algebra and
geometry structure by nature and has plentiful structures and
profound inherent theory. [5] It just can meet the needs of machine
learning and describe the procedure of machine learning clearly.
So Lie group machine learning is formed. This paper is based on
the basic conceptions of machine learning and gives the
generalization hypothesis axiom; the partition independence
hypothesis axiom; the duality hypothesis axiom and the learning
compatibility hypothesis axiom of Lie group machine learning.
Index terms—Lie group; machine learning; Lie group machine
leaning; hypothesis axiom
I. INTRODUCTION
A
ssume that W is a set of the given world’s all limited or unlimited
observation objects, because of the limitation of our ability of
observation, we can only acquire a limited subset of the world G
⊂ W and it is the sample set. Just according to this sample set,
machine learning speculates the model W to make it true to the world.
So it should satisfy: (1) Compatibility hypothesis: assuming that the
world W and the sample set G have the same properties; (2) Partition
independence hypothesis: put the sample set G into n-dimensional
space and find the super plane (equivalence relation) to make the
problem-decided different objects be divided in the non-intersected
areas or map n-dimensional non-linear problems to linear ones to form
a solvable linear structure; (3) Universal ability hypothesis: set up the
world observation model from the limited sample set and the universal
ability is the index of the degree of whether the world is true; (4)Dual
space hypothesis. According to these four hypothesizes of machine
learning, early researches are concentrate on partition hypothesis.
Recent researches are mainly concentrated on the basis of efficient
partition hypothesis, which must satisfy the conditions of the universal
ability hypothesis too. The compatibility hypothesis is less considerate
and it’s the problem that future machine learning normal forms must
consider. Lie group, a mathematic tool with excellent structure, just
can satisfy these hypothesizes; describes the procedure of machine
learning clearly; transact the large quantity of data waiting to be
transacted now, which are higher dimensional, non-linear and
higher-order. The document [2-3] give the basic conceptions of
differential machine learning from the view of differentiate to solve
the data’s higher dimension. The documents [1], [4] give basic
conceptions of Lie group machine learning and advance the above
hypothesizes. This paper will give further research to these
Manuscript received December 20, 2005. This work was supported by
Development 863 Program of China (Grant No.2002AA881030); the Nature
Science Foundation of Jiangsu Province (BK2005027) and the 211 Foundation
of Soochow University.
Huan Xu is with the Computer Science & Technology School of Soochow
University, Suzhou 215006 P.R. China (e-mail: 210413002@ suda.edu.cn).
Fan-zhang Li is with the Computer Science & Technology School of
Soochow University, Suzhou 215006 P.R. China (e-mail: lfzh@ suda.edu.cn).
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2006IEEE
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401
hypothesizes and give its concrete content. In the following, we will
give the generalization hypothesis axiom; the partition independence
hypothesis axiom; the duality hypothesis axiom and the learning
compatibility hypothesis axiom of Lie group machine learning
separately.
II. LIE GROUP MACHINE LEARNING’S AXIOM HYPOTHESIZES
In this section, Lie group machine learning’s axiom hypothesizes
are mainly discussed.
Axiom hypothesis 1 Lie group machine learning’s ability of
generalization hypothesis axiom. It contains 1), 2), 3):
1) Assume that g is a Lie group; to G’s Lie algebra g’s any Lie
sub-algebra h, there is only one connected Lie subgroup H of G, which
makes the following diagram exchangeable.
h ⊂ g
exp↓ ↓exp
H ⊂G
2) Assume that V is n-dimensional linear space, G is GL (V)’s Lie
subgroup, if there is (g(x),g(y))=(x,y) ∀ x,y∈V, g∈G, then the
inner product (x, y) in V is called G lower invariance.
3) Assume that left-and-right invariant Riemannian structure whose
volume is 1 is defined in compact connected Lie group G, f is any
central function on it, and then there exists the following Weyl integral
formula:
∫G f (g )dg =
1
w
∫T f ( t ) Q ( t )
2
dt
（*）
|W| in it represents the number of elements in W. When t=expH(H∈h),
Q(t) can be represented by the following equation:
（*’）
Q (expH)=
si gn(σ )e 2 xi (σ ( δ ),H)
∑
σ ∈W
δ in it is half of the sum of all positive roots and namely
δ=
1
∑ α
2 α ∈∆+ (G)
Proof. We prove Axiom 1) first: Assume that X1, X2,……Xm is a
random group of bases of the vector subspace h, we now use them as
the left invariant vector field in G, then they extends a subspace in
every point g’s tangent space, namely D (g)=span{X1(g),…,Xm(g)} , in
it Xi(g) represents the value of vector field Xi at g. This defined
m-dimensional subspace in every point’s tangent space is called
m-dimensional distribution. It can be regarded as the vector field’s
higher dimensional extension. Because Xi(1≤i≤m) are all left
invariant. Namely dla(X1(g))= Xi(ag), the distribution D is extended by
them is left invariant too. Namely dla(D(g))= D (ag). What’s more,
because h =span {X1,…,Xm} is a Lie subgroup, [Xi，Xj](1≤i,j≤m) can
always be represented by suitable linear combinations of X1,…,Xm.
This shows that the distribution D above satisfies the condition of
intergrability in Frobenius Axiom. So there exists an exclusive
maximal integrated submanifold corresponding with D, it is
represented by H through the unit element e.
Now we prove H is a connected subgroup. Assume that h is a
random element in H. Because is D left invariant, lh-1·H=h-1·H of cause
is a D ’s maximal integrated submanifold. From the definition,
e∈h-1·H, namely h-1·H also goes through point e. From the exclusivity
of the maximal integrated submanifold, we get h-1·H=H. But h is
random, so H-1·H=
h -1 ·H =h
U
h∈H
This just proves that H is a connected subgroup.
3) Proof. It can be proved as the following steps: ①From the analysis
of the orbital geometry of A% d (adjoint transformation) on G, we know
that T is a totally geodesic submanifold and is orthogonal with every
% d is an excellent basic
conjugate class. Further more, C0 ≅ T/W≅ G/A
field, the distance of two random points t1 and t2 on it in C0 is the
shortest distance of conjugate classes G (t1) and G (t2) in G, it can be
called orbital distance for short.
②To any randomly given inner point t∈C0, its type of orbit is
always G/T, but to any randomly given boundary point
t ∈ ∂, C0 = C0 \ C0 . Then the dimension of its orbit is always less than
dimG/T, so it’s not hard to see that all union sets of the non-G/T-type
orbits, namely the measure of G(∂C0) =
G(x)
U
x∈∂C 0
③Assume that t is a free point on C0, m=dimG/T=dimG-dimT, then
G(t) is G\T-type orbit and its m-dimensional volume is a function
which domain of definition is C0, it is called (principle type) orbit
volume function and is remarked with symbol v(t).
Because every orbit G (t) is a G\T-type homogeneous Riemannian
space, the proportion of their volumes is equal to that of their volume
elements. We can select a reference G\T-type homogeneous
Riemannian space. To every G\T-type orbit G (t), the coset g·T
mapping to g (t) is a G-equivariant mapping E between the same type
of homogeneous Riemannian spaces: E: G/T→G(t), g·t|→g(t)=gtg-1,
Thus the proportion of the “volume elements” of the two above is the
above mapping Et’s Jacobi determinant. That is to say, the volume
function to be determined v (t) can be calculated by the following
equation:
v(t)=c·det(dEt|0),
(1)
c in is an underdetermined proportional constant and dEt| 0 is the linear
mapping of the base point’s tangent space.
(2)
dEt|0 : ⊕
R 2 → ℑ (G(t))
a∈∆
+
( ±a)
0
④Because Et is G-equivariant, dEt|0 is T- equivariant (because the
base point is a fixed point under the effect of T). So dEt|0 can be
decomposed as the linear direct sum of the following two-dimensional
(3)
invariant subspaces: dE t | 0, a : R 2( ± a ) → ℑ0 (G 0 (t))
2
can use the following diagram
Ei ,a
%
(5)
Ga / T  →
Ga (t) ⊂ Ga / Ta = G
∩
∩
t
G / T E
→ G(t) ⊂ G
to aggregate the calculation of det(dEt|0,a) to the S2-type (principle)
2
%
In it, Q(expH)
=
∏ (e
πi(a,H)
(7)
− e− πi(a,H) ) .
a∈∆+
%
in the equation above is equal
⑤At last, we will prove Q(expH)
to Q(expH) in (*’). The proof is as follows:
Proof. Assume that {a1,…,ak} are prime roots towards Weyl house
c0, then through the definition of reflecting symmetry we can know
that
rα i (α i ) =-ai, it is equal to the selected change of the two half
spaces’ positive and negative sides divided by < ai >┴, it naturally
has no influence to selection of the two half spaces’ positive and
negative sides divided by < ai >┴ of other positive root a. So under
ra i , △+\{ai} still change to be △+\{ai}, so we have
rα i ( ∆ + ) = (∆ + \{a i }) ∪ {-ai} (1≤i≤k)
rαi (δ) = δ-α i and have
(8)
ra i ( δ )= δ − 2(δ,a ) a , 1≤i≤k.
i
(a i ,a i )
i
2(δ,a i ) =1
(a i ,a i )
So
(9)
Moreover, it can be speculated immediately by the expression of
%
and (8)
Q(expH)
%
%
Q(expr
a i (H)) = ( −1) · Q(expH)
(10)
%
is an odd function. Namely
Therefore, to the effect of W, Q(expH)
%
%
Q(exp
σ H) = sign(σ ) · Q(expH)
σ ∈W
(11)
sign(σ ) in it is the symbol of the orthogonal transformation’s
determinant on σ as h.
%
There is a principle item in the expansion in Q(expH)
and it
%
should
is e πi( δ,H ) . From (11) we know that the expansion of Q(expH)
2πi (σ (δ),H )
contain all the items sign(σ ) · e
, (σ ∈ W) . Moreover,
from (9) it is not hard to prove that there can’t be other items in its
%
expansion and only be Q(expH)
= Q(expH) .
is being determined, we first quadrature along G-orbit.
Because f gets the same value on every G-orbit G(t), we have
1
f (g )dg =
G
c0 f ( t ) v( t )dt = w T f ( t )v ( t )dt
∫
∫
=
∫
1
w
∫T f ( t ) Q( t )
2
dt
(12)
In it c is an underdetermined constant, |W| is the number of W’s orders
(T is divided into |W| Weyl houses); then the condition f (g)≡1 is
substituted into (12) and it’s easy to determine that the
underdetermined constant discussed above must be 1.
% again(because G
% is isomorphic with S3,
orbit volume function in G
a
a
% .Cartan-H.Weyl) The highest weight ∧ φ’s
Theorem 1 ( E
multiplicity of a complex uncountable representation φ of G must be 1
and its characteristic function xφ can be briefly expressed as the
following formula using its highest weight
% has the same volume element with S . So G
% in
or with SO(3), G
a
a
3
the above problem can be regarded as S ), so we have
3
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%
|
π (a, H) = c · | Q(expH)
∫G f (g)dg
On the other side, det(dEt|0) is of direct proportion with the S -type
% ≅ S3 .In other words, we
(principle) orbit volume function in Ga/Ta= G
a
det(dE t | 0, a) = c ' · sin 2 π (α , H)
a∈∆ +
2
⑥ Assume that f is any central function defined on G. When
So det(dEt|0) can be calculated by the following equation, namely:
(4)
det(dEt|0)=
∏ det(dE t | 0, a)
a∈∆ +
∏ sin
Assume that δ= 1
∑ a , from (8) we can know that
2 a∈∆+
is 0. In other words, when the integration is determined on G, the
subset G(∂C0 ) above can be omitted and only calculate it on G (C0).
∑
H in it satisfies t=expH (Attention, the positive root of S3 is 2Ө). Unite
(1), (4), (5), we can conclude v(expH) = c '' ·
(6)
402
∑
xφ (expH) = σ ∈ W
∑
sign ( σ ) e 2 π i ( σ ( ∧ ϕ + δ ),H )
σ ∈W
sign ( σ )e 2 π i ( σ ( δ ),H )
, H∈h
Therefore, the sufficient and necessary condition to be equivalent of
G’s two complex irreducible expressions φ1 and φ2 is their highest
weight ∧φ1 and ∧φ2 are the same.
Deduction The dimension dimφ of the complex irreducible expression
discussed above can be directly calculated with the following formula
and the highest weight.
∏+ (∧ϕ + δ ,α ) .
dim ϕ = α ∈∆
∏ (δ ,α )
α∈∆ +
Axiom hypothesis 2 The duality hypothesis axiom of Lie group
machine learning:
In a learning system, the learning results are learned according to
the knowledge in the knowledge database and should be dual with
original knowledge. So we introduce the following conception to built
dual space of the learning system.
Definition 1 Assume that P is a point on Riemannian manifold Mn. If
there is an involutory length-preserved transformation σP on Mn: Mn
1) M is a complete Riemannian manifold；
2) M is a homogeneous space, and namely the functions of G(M) on M
is transferable；
3) Assume that K is a structure maintenance transformation subgroup,
which makes some certain given base point P0∈M fixed. Thus K is a
compact uniform subgroup of G (M) and a Lie group.
4) To any geodesic r in M: R → M, there all exist an exclusive
one-parameter subgroup Φr in G (M) ：R→G (M) conforms to the
following conditions: Φr (t1)·r (t2) = r (t1+t2)
dΦr (t1)| TMr(t2)= // r(t2,t1+t2)
Proof. 1) To proof M’s completeness, we only need to illustrate any
geodesic can be unlimitedly extended. Assume that γ[a, b]→M is a:
randomly given geodesic in symmetry space. Let ε< 1 (b-a).
2
Assume that s is centrally symmetry of M to point γ (b-ε), and then
γ and s·γ have a connection which length is 2ε. It explains that γ can
extend a length b-a-2ε (as Fig.1). And it explains that any geodesic
can be unlimitedly extended. So from Theorem Hopf-Rinow we
know that M is perfect.
sγ(b)
γ(b-c)
γ(b)
2
→Mn， σ P ＝Id, which makes P a insular fixed point on it, we will
call Mn is centrally symmetrical with P and σ P is centrally
symmetrical with P; X and σP(X)(X∈Mn)are called symmetrical
points to P. If Mn is centrally symmetrical to all points in it, we will
call Mn a symmetrical space. In other words, a symmetrical space is a
Riemannian manifold centrally symmetrical with any point P on it.
1) In a Riemannian manifold, what the simplest and the most basic is a
conjecture horizontal line and the corresponding sets to the most usual
knowledge in the knowledge database.
2) Assume P1, P2, P3 are three random points in a connected
Riemannian manifold Mn and conform to “triangle inequity”
d(p1,p2)+d(p2,p3)≥d(p1,p3)
We measure between knowledge items using this relation.
3) In the knowledge metric space X, the necessary condition that
makes points series {Pn; n∈N} be convergent to a certain fixed point a
(namely regard a as the limitation) is it accords to the usual Cauchy
conditions, namely ∀ ε>0，constantly existing a big enough N，
which makes when m,n≥N , d (pm,pn)<ε. If these Cauchy conditions
are also the sufficient ones of the existence of the limit point of points
series in X ， then we call X is a complete metric space. If a
Riemannian manifold Mn is complete to the metric structure on it, then
we call it complete Riemannian manifold.
4) In Parallelism and covariant differentiation.
Theorem 1 In any randomly given Riemannian manifold Mn
exclusively exists a covariant differentiation that conforms to the
following operation laws:
∇ f1x1+f2x2Y = f1 ∇ x1Y+f2 ∇ x2Y
（1） ∇ x(f1Y1+f2Y2) = (Xf1)Y1+f1 ∇ xY1+(Xf2)Y2+f2 ∇ xY2
∇ xY- ∇ YX = [X, Y]
∇ Z (<X.Y>) = < ∇ ZX, Y>+<X, ∇ ZY>
n
<X, Y> = g (X, Y) =
∑g
ij ξ i η j
i, j =1
From these we can see, the left side of the equation above decides
the value of ∇ Z Y in every point’s tangent space. So the covariant
differentiation’s existence and exclusiveness are proved.
From it we can get, that the duality hypothesis axiom of Lie group
machine learning contain the following 1°, 2°:
1° Assume that M is a couple of dual space, G (M) is M’s structure
maintenance transformation group, so
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403
Fig.1
2) Assume that p and q are two randomly given points on M. From
M’s completeness we can know that in M there is a shortest geodesic γ
that connects point p, q: γ: [0,d(p, q)]→M, which makes γ(0)=p,
γ(d(p,q))=q. Assume that s is central symmetry of M to the point
γ( 1 d(p,q)) on γ. Obviously, we can have s(p)=q, s(q)=p.
2
So the effect of G (M) on M is transmittable! Namely M is
homogenous.
3) Assume that K is a length-preserving transformation subgroup
which makes the given base point p0∈M to be fixed; O (n) is the
length-preserving transformation group of TM p . It is not hard to
0
find that every element in K is exclusively determined by the
length-preserving transformation induced by it on TM p . From it, we
0
know that
K
O(n)
k
dk| p 0
is an isomorphism and K “likes” a closed subset of O(n). Therefore, K
itself is a compact subset and a Lie group.
2°Assume that M is a symmetrical space, P0 is a random base point
on it, S0 is central symmetry to base point P0, G (M) is the
structure-preserving transformation group of M, K is
structure-preserving subgroup that makes P0 fixed. Then we have:
1) M≌G (M)/K, K is a contact sub-Lie group;
2) σ: G (M) → G (M) is defined as σ (g) = S0gS0 ∀ g∈G(M)
Then σis one of G (M)’s dual automorphism.
3) Assume that F (σ) is a fixed point subset of σ, namely:
F (σ) = {g∈G(M)|σ(g)=g}
Using F°(σ) to represent the unit connected region of F (σ), then F
(σ) and F°(σ) are all closed subgroup of G (M) and:
F°(σ) ⊆ K ⊆ F (σ)
4) Assume that g and k are G (M) and K’s Lie algebra separately, thus
g = k⊕p, k = ker (dσ-id), P = ker(dσ+id)
5) To any X∈P,r(t)=Exp , X(P0) is a geodesic line through P0 in M
and ExptX =Φr (t) (Φr (t) is one-parameter subgroup of G(M))
Proof. 1) G (M) is a transmittable Lie communicative group effecting
on M, K is a fixed point subgroup which makes point p0 fixed. The
induced effect of K on
TM p0 is a closed subset of O (n), so K is a
compact Lie group and M  G(M) / K . Moreover, ∀g ∈ G(M) ,
∀g ∈ G(M), σ (g) = s 0gs 0 obviously G (M) is a dual automorphism.
From it we can see that, its fixed point subset F(σ ) is a closed subset,
so its unit connected field F°(σ) is also a closed subset.
in which Tn-k=R/Z ⊗ R/Z ⊗ … ⊗ R/Z
n-k
is an n-k-dimensional torus.
Definition 2 If there is an n2-variate polynomial set S which makes the
sufficient and necessary condition of g=(gij)∈G is
P (g)=P (gij)=0, ∀ p∈S
2) Assume that I: K→O (n) is i(k)=dk| p , k∈K. This is a naturally
Then general linear group (GL (n.R), GL (n. C))’s subgroup G is
isomorphic mapping and i(s0)=-id∈O(n).
have
Therefore,
to
any
randomly
given
k∈K,
i(s0ks0)=(-id)t(k)(-id)=t(k). Namely s0ks0=k, So K ⊆ F(σ ) . What’s
more, assume that exptX is a one-parameter subset in G(M) satisfying:
s0(exptX)s0=exptX
Namely (exptX)·s0·exp (-tX)=s0 ∀ t∈R,
And have s0 (exptX·p0)=(exptX) ·s0 exp (-tX)(exptX·p0)=exptX·p0
In other words, {exptX·p0；t∈R} is s0’s fixed point subset containing
p0, but point p0 is s0’s isolated fixed point. Therefore, exptX·p0=p0
Namely exptX∈K. It is proved that F0 (σ ) ⊆ K .
3) From the Proposition and 1） above, we know that dσ ：g→g is a
g’s dual automorphism so g=k ⊕ p, k=ker( dσ -id), p=ker( dσ +id)
called algebra group.
If there is a 2n2-variate polynomial set S which makes the sufficient
and necessary condition of g=(aij+bij −1 )∈G (aij,bij ∈ R) is
0
In it the proof of 2）has proved that k is K (namely
F0 (σ ) )’s Lie
algebra. Moreover, to X∈p, exptX’s character is s0(exptX)s0=exp(-tX)
Assume that X0 is a tangent vector at point p0 of the curve {exptX(p0)
；t∈R }, γ is a geodesic with initial speed vector X0, then ϕγ (t) =s1/2
∀ p∈S
, then GL(n, C)’s subgroup G is called
pseudo-algebra group.
Theorem 2 The algebra group and pseudo-algebra group of a general
linear group are both closed subset and also a Lie sub group.
Assume that F= R or C, we regard GL (U. F)’s Lie algebra as
gl(n, F), then X∈gl(n, F) has exptx=etx
If H is G’s closed subgroup and h is H’s Lie algebra, then
h ={x∈gl(n, F)|eπx∈H, ∀ x∈F}
III. CONCLUSION
·s0 obviously is G(M)’s one-parameter subgroup.
s 0ϕγ (t)s 0 = s0 (s1/2·s0)= s0s1/2 = ϕγ (t) −1 = ϕγ (− t)
Therefore, exptX = ϕγ (t) , γ(t) = exp tX(p0 ) .
This theorem tell us that the tight relation of dual space, Lie group
and Lie algebra and can regard dual space as a special Lie group and
Lie algebra to discuss.
Axiom hypothesis 3 The partition independence hypothesis axiom of
Lie group machine learning:
1° Assume that g is a real Lie algebra, σ is one of its dual
automorphism, g decomposes σ
g = k⊕P
k = ker(σ-id)
P = ker(σ+id)
If adp：k → gl(p)，ad，(X)·Y = [X,Y]，X∈K，Y∈P are
correspond one to one， and there is an ad on P. K’s fixed inner
product Q，namely to any random X∈K，Y,Z∈P，has
Q ([X, Y], Z)+Q (Y, [X, Z]) ≡0
constantly true, we will call（g,σ,Q）is a orthogonal dual Lie algebra.
2°Any orthogonal dual Lie algebra（g,σ,Q）can be exclusively
decomposed to the following equation：
（g, σ, Q）= （g0,σ0,Q0）⊕（g+,σ+,Q+）+（g-,σ-,Q-）
in which
g0 = k0⊕P0, g+ = k+⊕P+, g- = k-⊕P-,
σ0 =σ| g0, σ+ =σ| g+, σ- =σ| g-,
Q0 =Q| P0, Q0+ =Q| P+, Q- =Q| P-,
and B|P+, B|P- separately represent positive and negative definite.
Axiom hypothesis 4 The learning compatibility hypothesis axiom of
Lie group machine learning:
1° Assume that Q is the homomorphism of Lie group G1 to Lie group
G2, then Q-1(l2) is G1’s closed normal subgroup and assume that π is
natural homomorphism of G1 to G1(Q-1(l2)), then we have the Lie
group’s homomorphism φ of G1/Q-1(l2) to G2 which makes Q=φπ.
2° Assume that G is an n-dimensional connected communicative
group(complex group) , then G ≅ Rk ⊗ Tn-k
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P(g)=P(aij,bij)=0,
404
We use Lie group to research machine learning; assume machine
learning on the basis of the sample set G (G ⊂ W); judge the truth of
W through G and understand nonlinear, higher dimensional,
higher-order data and linearize, reduce dimension and order of them.
The documents [1-4] have given the basic conceptions of Lie group
machine learning. So we go further in this paper and research the
axiom hypothesizes of Lie group machine learning, including the
generalization hypothesis axiom; the partition independence
hypothesis axiom; the duality hypothesis axiom and the learning
compatibility hypothesis axiom. Based on these hypothesizes, in the
future, we will do further research and give the learning algorithms
and complete an example system.
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