El e c t r o n i c
J o
u o
f P
r o b
a b i l i t y
Vol. 3 (1998) Paper no. 4, pages 1–36. Journal URL
http://www.math.washington.edu/˜ejpecp/ Paper URL
http://www.math.washington.edu/˜ejpecp/EjpVol3/paper4.abs.html GEOMETRIC EVOLUTION UNDER
ISOTROPIC STOCHASTIC FLOW
M. Cranston and Y. LeJan
University of Rochester Universit´e de Paris, Sud Rochester, New York, USA Orsay 91405, France cran@math.rochester.edu Yves.LeJan@math.u-psud.fr
Abstract. Consider an embedded hypersurfaceM inR3. For Φt a stochastic flow
of differomorphisms on R3 and x ∈ M, set xt = Φt(x) and Mt = Φt(M). In this
paper we will assume Φt is an isotropic (to be defined below) measure preserving
flow and give an explicit descripton by SDE’s of the evolution of the Gauss and mean curvatures, ofMt at xt. If λ1(t) and λ2(t) are the principal curvatures of Mt at xt
then the vector of mean curvature and Gauss curvature, (λ1(t) +λ2(t), λ1(t)λ2(t)),
is a recurrent diffusion. Neither curvature by itself is a diffusion. In a separate addendum we treat the case of M an embedded codimension one submanifold of Rn. In this case, there aren−1 principal curvaturesλ1(t), . . . , λn−1(t). IfPk, k = 1, n−1 are the elementary symmetric polynomials in λ1, . . . , λn−1, then the vector
(P1(λ1(t), . . . , λn−1(t)), . . . , Pn−1(λ1(t), . . . , λn−1(t)) is a diffusion and we compute
the generator explicitly. Again no projection of this diffusion onto lower dimensions is a diffusion. Our geometric study of isotropic stochastic flows is a natural offshoot of earlier works by Baxendale and Harris (1986), LeJan (1985, 1991) and Harris (1981).
1991Mathematics Subject Classification. 60H10, 60J60.
Key words and phrases. Stochastic flows, Lyapunov exponents, principal curvatures.
Research supported by NSA and Army Office of Research through MSI at Cornell. Submitted to EJP on February 19, 1996. Final version accepted on February 12, 1998.
§1 Isotropic Flows and Geometric Setting.
We begin with a nonnegative measureF(dρ) on [0,+∞) with finite moments of all orders. The pth moment ofF will be denoted by µp. ToF we can associate a
covariance
(1.1) Cij(z) =
Z ∞
0
Z
Sn−1
eiρhz,ti(δji −titj)σn−1(dt)F(dρ),
where σn−1 is normalized Lebesgue measure on Sn−1. By a result of Kolmogorov,
toCij is associated a smooth (inx) vector-field valued Brownian motionUt(x) such
that
(1.2) EUti(x)Usj(y) = (t∧s)Cij(x−y), 1≤i, j ≤n.
A Brownian flow on Rn is constructed by solving the equation
(1.3) Φt(x) =x+
Z t
0
∂Us(Φs(x))
(∂ denotes Stratonovich differential.) Existence and uniqueness of such flows was proved in Baxendale (1984), LeJan and Watanabe (1984). Then ifTa is translation
by a∈Rn, TaΦtT−a and Φt are identical in law so Φ is stationary. IfR is unitary, RΦtR−1 and Φt are identical in law. When these two properties hold we say Φ is
isotropic. These properties both follow from the nature ofCij so our Φ is isotropic.
The field Ut(x) is smooth so we differentiate it writing
(1.4) Wji = ∂
∂xjU i
(1.5) Bjki = ∂
2
∂xj∂xkU i.
Then we have as direct consequence of (1.1) and (1.2) (using h,i to denote both Euclidean inner product as well as quadratic variation for martingales,dwill denote the Itˆo differential.)
(1.6) hdWi
j(t, y), dWℓk(t, y)i = µ2
n(n+ 2)[(n+ 1)δ
i kδ
j
ℓ −δijδℓk−δiℓδjk]dt
(1.7) hdBjki (t, y), dWqp(t, y)i= 0 and for vectors u, v ∈Rn,
(1.8)
hhdB(u, u), vi,hdB(u, u), vii = 3µ4
n(n+ 2)(n+ 4)[(n+ 3)||u||
By isotropy, (1.6), (1.7) and (1.8) remain valid under any unitary change of coor-dinates. We define,
Cjℓikdt=hdWji, dWℓki= µ2
n(n+ 2)[(n+ 1)δ
i kδ
j ℓ −δ
i
jδℓk−δiℓδjk]dt
where both dWi
j, dWℓk are evaluated at (t, y), and
Ckℓpqij dt=hdBi(ek, eℓ), dBj(ep, eq)i
again both dB terms are evaluated at (t, y). A quick calculation shows
E Pn
i=1dWii
2
= 0 so div U = 0 and therefore Φ is a measure-preserving flow. Also, it’s rather easy to check that Cnn
1111= 3C1122nn and C1122nn = n(n+2)(n+4)(n+3)µ4 .
The most general form of hdWi
j,(t, y), dWℓk(t, y)i = Cjℓikdt for isotropic flows is Cik
jℓ = aδi,kδj,ℓ+bδjiδkℓ +cδiℓδjk where a+c, a−c, a+c+db are nonnegative (see
LeJan (1985) and the references therein). We retrict our attention to the present covariance structure;a= (n+1) µ2
n(n+2),b=− µ2
n(n+2),c=− µ2
n(n+2), as computations
are massively simplified by the fact that certain Ito correction terms vanish. The works of Baxendale and Harris (1986) and LeJan (1985) focused on first order properties of the general isotropic flows in Euclidean space. That is, on properties of the derivative flow DΦt(x). Perhaps the most fundamental information about
the derivative flow is its Lyapunov spectrum. To describe the Lyapunov spectrum, first extend DΦt to p-forms α= v1∧ · · · ∧vp (we’re in Euclidean space so we can
identify forms and vectors) by defining DΦt(x)α = DΦt(x)v1 ∧ · · · ∧DΦt(x)vp.
Then if v1, . . . , vp are linearly independent, a.s.
lim
t→∞
1
t logkDΦt(x)αk=γ1+· · ·+γp
is the sum of the firstp Lyapunov exponents. In the case of the isotropic, measure-preserving flow on Rn, γ
1+· · ·+γp = np(n2(n+2)−p)µ2 (see LeJan 1985). The problem
dealt with in this work involves second order behavior of the flow. The evolving curvature of a submanifold moving under the flow would depend on the properties of the second order flow,
(x, u, v)→(Φt(x), DΦt(x)u, D2Φt(x)(DΦt(x)u, DΦt(x)u) +DΦt(x)v).
In particular, our results rely on the properties given by (1.6), (1.7) and (1.8). Initially, we will considerM an embedded hypersurface inRn, later specializing to the case n = 3. Take Πt : TxtRn → TxtMt to be the orthogonal projection
and ∇ to be the canonical connection on Euclidean space. Then St, the second
fundamental form ofMt atxt, is defined foru, v ∈TxtMt bySt(u, v) = (I−Πt)∇uv
where v is extended in an arbitrary but smooth manner in a neighborhood of xt.
for k = 1,2, . . . , n −1. This arises as follows. First observe that since we are in Euclidean space we can identify vectors and forms. Next take νt ∈ Txt⊥Mt to
be a unit vector (there are two choices and we may select the process νt to be
continuous int.) Now in the casek = 1,u→ hSt(u,·), νtiis a map from Λ1(TxtMt)
to Λ1(TxtMt). When k >1 we need to introduce the index set
Ik ={ ⇀
m∈ {1, . . . , n−1}k :m1 < m2 <· · ·< mk}
and for σ ∈Sk, the set of permutations on k letters, define (−1)σ to be the sign of σ. Then for u1, . . . , uk, v1, . . . , vk ∈TxtMt, set
S(k)(u1∧ · · · ∧uk, v1∧ · · · ∧vk) =
X
σ∈Sk
(−1)σ k
Y
j=1
hSt(uσ(j), vj), νti
This gives rise to the linear map
u1∧ · · · ∧uk →S(k)(u1∧ · · · ∧uk, ·)
from Λk(T
xtMt)→Λk(TxtMt).
The object of our study will be the trace ofS(k), T rS(k), for k= 1,2, . . . , n−1. This is obtained via contraction on indices (see Bishop and Goldberg (1980)). Take {u1, . . . , un−1} to be any basis for TxtMt. For k ∈ {1, . . . , n−1}, ~ℓ ∈ Ik, set
|⇀ℓ|=ℓ1+· · ·+ℓk, α⇀
ℓ =uℓ1∧uℓ2∧ · · · ∧uℓk α
⇀ ℓ = (
−1)|
⇀ ℓ|+ku
1∧ · · · ∧uˆℓ1 ∧ · · · ∧uˆℓk ∧ · · · ∧un−1
( ˆ indicates the indicated vector is omitted from the wedge product)
α=u1∧ · · · ∧un−1.
Define an inner product
hα⇀ ℓ, α
⇀
mi =dethuℓi, umji.
Then we may express T rS(k) as
(1.9) T rS(k) =S(k)(α⇀ ℓ, α
⇀ m)hα
⇀ ℓ, αm⇀
i.||α||−2,
where the sum (Einstein’s convention) is over⇀ℓ ,m⇀ ∈Ik. Now this is invariant under
change of basis so we may select {u1, . . . , un−1} to be the unit principal directions
Λ1(T
xtMt). Let λ1, . . . , λn−1 be the corresponding eigenvalues, otherwise known
as the principal curvatures. Now the ui are orthonormal so
hα ⇀
ℓ, αm⇀
i||α||−2 =δ ⇀
ℓ ,⇀m
and as hSt(uj, v), νti=λjhuj, vi
X
σ∈Sk
(−1)σ
k
Y
j=1
hSt(uℓσ(j), uℓj), νti=λℓ1. . . λℓk ≡λ⇀
ℓ
Thus
(1.10) T rS(k) = X
⇀ ℓ∈Ik
λ⇀
ℓ ≡Pk(λ1, . . . , λn−1). Pk is the elementary symmetric polynomial of degree k.
The process ξt = (xt,Πt, νt, P1(λ1, . . . , λn−1),· · ·Pn−1(λ1, . . . , λn−1)) will, in
our case be a diffusion. By the translation invariance part of isotropy, it is clear that xt is not needed, i.e., (Πt, νt, P1(λ1, . . . , λn−1),· · · , Pn−1(λ1, . . . , λn−1)) has
the Markov property. By the invariance of the law of the flow under unitary actions, it follows that Πt and νt are also superfluous. Thus, for isotropic flows the vector
(P1(λ1, . . . , λn−1), . . . , Pn−1(λ1, . . . , λn−1)) is a diffusion. The point of the present
article is to describe this diffusion explicitly in the measure-preserving isotropic case. We can derive consequences about the behavior of this diffusion from the explicit description. These properties suggest what sort of behavior one might have with more general flows. What we expect for more general flows is that the process of curvatures for a k-dimensional submanifold evolving under a general flow will be recurrent (though not necessarily a diffusion) if γk >0. That is, generically the k
dimensional tangent space TxtMt will ‘see’ only stretching directions and this will
tend to smooth out the curvatures. This condition may even be too strong.
§2 Preliminary Results.
Recall the correlations ofdW given by (1.6). The flow Φt :Rn→Rn is aC∞
dif-feomorphism whose first derivative satisfies the Stratonovich stochastic differential equation, where x∈M,
(2.1) dDΦt(x) =∂W(xt)DΦt(x)
Given u∈TxM, DΦt(x)u =u(t)∈TxtMt satisfies
(2.2) du=∂W u
where we have suppressed the variable (xt) in∂W. In fact, due to the choice of C
in (1.1)
that is, the Itˆo corrections vanish. Furthermore, if we select a basis{u1, . . . , un−1}
for TxM, then {u1(t), . . . , un−1(t)}, where uj(t) = DΦt(x)uj, forms a basis for TxtMt. We will write uj for uj(t) for simplicity of notation.
We now proceed to derive a Stratonovich equation for the second fundamental form St of Mt at xt. Given vectors u, v ∈ TxM move them via the flow, u(t) = DΦt(x)u, v(t) =DΦt(x)v. Extend v to a smooth vector field V in a neighborhood
of x. Set Vt(y) =DΦt(Φ−t 1(y))V(Φ−t 1(y)) so that Vt is a vector field near xt such
that Vt(xt) =v(t). Then if Zt ≡ ∇u(t)Vt(xt) we claim
(2.4) dZ(u, v) =∂W Z(u, v) +∂B(u, v).
For proof of (2.4), take γ to be a curve on M with γ(0) = x, γ′(0) = u and define
γt = Φt(γ) so that γt′(0) =u(t) =DΦtu. Then,
∇utVt = lim s→0s
−1(V
t(γt(s))−Vt(γt(0))
= lim
s→0s
−1(DΦ
t(γ(s))V(γ(s))−DΦt(x)v(x))
= lim
s→0s
−1[(DΦ
t(γ(s))−DΦt(x))V(γ(s))
+DΦt(x)(V(γ(s))−v(x))]
=D2Φt(x)(u, v) +DΦt(x)∇uv
Since
Φt(x) =x+
Z t
0
∂U(Φs(x))
DΦt(x) =I +
Z t
0
∂W(Φs(x))DΦs(x)
D2Φt(x)(·,·) =
Z t
0
∂B(Φs(x))(DΦs(x)·, DΦs(x)·)
+
Z t
0
∂W(Φs(x))D2Φs(x)(·,·)
(2.4) follows. A word about notation: ∂WtZ is an abbreviation for∂Wji(Φt(x))(Ztj)
and ∂B(u, v) is short for (∂Bi
jk(Φt(x))uj(t)vk(t).
Next we need an equation for Πt : TxtRn → TxtMt. For any ξ ∈ TxtMt, let
{u1, . . . , un−1} be the moving frame mentioned above, setα=u1∧ · · · ∧un−1 and
defining
(2.5) αi = (−1)i+1u1∧ · · · ∧uˆi∧ · · · ∧un−1
we have
(2.6) Πtξ =
n−1
X
i=1
hξ∧αi, αi||α||−2ui.
Lemma 2.1. If ξt is a TxtMt-valued continuous semimartingale, then
(I −Πt)·∂W ξt =∂Πtξt.
Proof. Suppose Πtξt =ξt is as in the statement of the lemma withξt =xitui. Then
(I −Πt)∂W ξt −∂Πtξt =∂W ξt −Πt∂W ξt
+ 2h∂αt, αti ||αt||2
ξt −∂W ξt
−hξt ∧α
i t, ∂αti
||αt||2
ui− h
ξt∧∂αit, αti
||αt||2 ui
=−Πt∂W ξt+ 2h
∂αt, αti
||αt||2 ξt
−xithαt, ∂αti
||αt||2
ui−xith
ui∧∂αi, αti
||αt||2 ui
Since
dαt =−ui∧∂αi+∂W ui∧αi
the last term becomes
−xithui∧∂α i, αi
||αt||2 ui
=−h∂αt, αti ||αt||2
ξt + h
∂W ui∧αit, αti
||αt||2
xitui
=−h∂αt, αti ||αt||2
ξt + Πt∂W ξt
From this we get the conclusion of the lemma.
Lemma 2.2. If ηt ∈TxtMt⊥ is a continuous semimartingale, then ∂Πtηt = Πt∂W η˜ t
where W˜ is the transpose of W.
Proof. Recalling that
∂Πt(v) =−2h
∂αt, αti
||αt||2
Πtv+∂WΠtv
+ hv∧α
i t, ∂αti
||αt||2
ui+ h
v∧∂αi t, αti
we see that since Πtηt = 0,
∂Π(ηt) = hηt∧α i t, ∂αti
||αt||2
ui.
But
∂αt =∂W uk∧αkt
so
∂Πt(ηt) = h
ηt∧αit, ∂W uk∧αkti
||αt||2
ui
= h∂W η˜ t, uk(t)ihα
k
t(t), αkt(t)iui(t)
||α(t)||2
= h∂W η˜ t∧α
i t, αti
||αt||2
uit
= Πt∂W η˜ t
and the lemma is proved.
UsingS = (I−Π)Z together with Lemmas 2.1 and 2.2 we obtain the following. Proposition 2.3. For u, v ∈TxM, u(t) =DΦt(x)u, v(t) =DΦt(x)v,
(2.7) dS(u, v) = (I −Π)dB(u, v) +∂W S(u, v)−Π(∂W +∂W˜)S(u, v)
where W˜ denotes the transpose of W.
Proof. We shall abbreviate by suppressing the dependence onu andv. So,
dS = (I−Π)∂Z−∂ΠZ
= (I−Π)dB+ (I −Π)∂W Z −∂ΠZ, by (2.4),
= (I−Π)dB+ (I −Π)∂W S
+ (I−Π)∂W R−∂ΠR−∂ΠS ,ΠZ ≡R
= (I−Π)dB+ (I −Π)∂W S −∂ΠS, by Lemma 2.1 = (I−Π)dB+∂W S−Π(∂W +∂W˜)S,by Lemma 2.2.
The equation (2.7) can be simplified. Set
dP = (I−Π)dW dQ= ΠdW˜
(2.8)
and define λ and µ so that
(I −Π)∂W =dP +dλ
Π∂W˜ =dQ+dµ.
Lemma 2.4.
(2.10) dΠ =dPΠ +dλΠ +dQ(I −Π) +dµ(I −Π).
dS(u, v) = (I −Π)dB(u, v) + (dP −dQ)S(u, v) + 1
2(dP −dQ)
2S(u, v)
+d(λ−µ)S(u, v) (2.11)
Proof. Let ζt be an Rn-valued semimartingale. Then
(2.12) dΠ = (I−Π)∂WΠ + Π∂W˜(I −Π) holds since
∂Πtζt =∂Πt(Πtζt) +∂Πt((I −Πt)ζt)
= (I−Πt)∂WtΠtζt+ Πt∂W˜t(I−Πt)ζt,
by Lemmas 2.1 and 2.2. Thus, we have the following quadratic variations,
dλ= 1
2(−(I−Π)dWΠdW −ΠdW˜(I −Π)dW) = 1
2(−dPΠdW −dQ(I−Π)dW),
dµ= 1
2((I−Π)dWΠdW˜ + ΠdW˜ (I−Π)dW˜) = 1
2(dPΠdW˜ +dQ(I −Π)dW˜). Consequently, performing Itˆo corrections on (2.12), we get
dΠ =dPΠ +dλΠ +dQ(I−Π) +dµ(I−Π) which proves (2.10). In order to derive (2.11), rewrite (2.7) as
dS(u, v) = (I−Π)dB(u, v) + (I −Π)∂W S(u, v)−Π∂W S˜ (u, v).
Then (2.11) follows from the definitions of dP, dQ, dλ and dµ in (2.9).
Lemma 2.5. If ζt is a TxtMt⊥-valued semi-martingale, then
(2.12) dλζ = (n−1)µ2
(2.13) dµζ = (n−1)(n+ 1)µ2 2n(n+ 2) ζdt
(2.14) 1
2(dP −dQ)
2ζ =
−(2nn−(n1)+ 2)nµ2ζdt
Proof. Select an orthonormal basis{e1, . . . , en} for TxtMt such that en ∈TxtMt⊥.
Then,
2dλζ =−dΠdW ζ−dPΠdW ζ−dQ(I−Π)dW ζ
=−(I −Π)dWΠdW ζ −ΠdW˜(I−Π)dW ζ
=−(
n−1
X
i=1 Cin
ni)ζdt, since hdWin, dWnni= 0,1≤i≤n−1,
= (n−1)µ2
n(n+ 2) ζdt, by (1.6) 2dµζ =dPΠdW ζ˜ +dQ(I −Π)dW ζ˜
= (
n−1
X
i=1
Cnnii )ζdt
= (n−1)(n+ 1)µ2
n(n+ 2) ζdt, by (1.6). Thus,
1
2(dP −dQ)
2ζ = 1
2(dP −dQ)((I −Π)dW ζ−ΠdW ζ˜ ) = 1
2(dP −dQ)(dW
n nζ −
n−1
X
i=1
dW˜ineihζ, eni)
= 1 2(dW
n
ndWnnζ− n−1
X
i=1
dWnidW˜inζ)
= 1 2(C
nn nnζ−(
n−1
X
i=1
Cnnii )ζ)dt
= µ2
2n(n+ 2)((n−1)−(n−1)(n+ 1))ζdt , by (1.6), =−(n−1)µ2
2(n+ 2) ζdt.
Theorem 2.6. If ν(t) denotes the unit vector ν = ||(I(I−−Π)unΠ)un|| perpendicular to Mt
at xt, then
dν =−dQν− (n
2−1)µ 2
2n(n+ 2)νdt.
Proof. Setvn = (I −Π)un. Then by Lemmas 2.2 and 2.3, dvn =−Π∂W v˜ n+ (I −Π)∂W vn
= (dP −dQ)vn+
1
2(dP −dQ)
2v
n+ (dλ−dµ)vn
= (dP −dQ)vn−
(n−1)µ2
(n+ 2) vndt, by Lemma 2.5 Then, applying
d||vn||2 = 2hvn, dvni+h(dP −dQ)vni
= 2hvn,(dP −dQ)vni − 2(n−1)µ2
(n+ 2) ||vn||
2dt+h(dP −dQ)v ni
= 2||vn||2dWnn−
2(n−1)µ2
(n+ 2) ||vn||
2dt+ (n−1)µ2 n ||vn||
2dt
= 2||vn||2dWnn−
(n−1)(n−2)µ2 n(n+ 2) ||vn||
2dt
and so
d||vn||−1 =−||vn||−1dWnn+
(n−1)(n+ 1)µ2
2n(n+ 2) ||vn||
−1dt.
Consequently, as ν = ||vnvn||,
dν = (dP −dQ)ν− (n−1)µ2
(n+ 2) νdt−(I −Π)dW ν+
(n−1)(n+ 1)µ2
2n(n+ 2) νdt − h(dP −dQ)ν, dWnni
=−dQν − (n−1) 2µ
2
2n(n+ 2)νdt−
(n−1)µ2 n(n+ 2) νdt =dQν − (n+ 1)(n−1)µ2
2n(n+ 2) νdt , as desired.
Theorem 2.7. Suppose u(t) = DΦtu, v(t) = DΦtv and ν is as in Theorem 2.6.
Then
dhS(u, v), νi= hdB(u, v), νi+hS(u, v), νihdW ν, νi − (n−1) 2µ
2
2n(n+ 2) hS(u, v), νidt.
Moreover, if ui(t) =DΦtui, then
hdhS(ui, uj), νi, dhS(uk, uℓ), νii=C112233 [hui, ujihuk, uℓi
+hui, ukihuj, uℓi+hui, uℓihuj, uki]dt
+ µ2(n−1)
n(n+ 2) hS(ui, uj), νihS(uk, uℓi, νidt.
Proof. From Lemmas 2.4 and 2.5 we have
dS(u, v) = (I −Π)dB(u, v) + (dP −dQ)S(u, v)− (n−1)µ2
(n+ 2) S(u, v)dt. Also
−h(dP −dQ)S(u, v), dQνi=hdQS(u, v), dQνi
=hS(u, v), νi n−1
X
i=1
Cnnii dt
= (n
2−1)µ 2
n(n+ 2) hS(u, v), νidt. So,
dhS(u, v), νi=hdS(u, v), νi+hS(u, v), dνi+hdS(u, v), dνi
=hdB(u, v), νi+hS(u, v), νidWnn−
(n−1)µ2
(n+ 2) hS(u, v), νidt − (n
2 −1)µ 2
2n(n+ 2)hS(u, v), νidt− h(dP −dQ)S(u, v), dQνi =hdB(u, v), νi+hS(u, v), νidWnn− (n−1)(n−1)
2µ 2
2n(n+ 2) hS(u, v), νidt For the quadratic variation term, since B and W are independent and Cnn
111 =
3Cnn 1122,
hdS(ui, uj), νihdS(uk, uℓ), νi=hhdB(ui, uj), νihdB(uk, uℓ), νii
+hS(ui, uj), νihS(uk, uℓ), νiCnnnndt
=C1122nn [hui, ujihuk, uℓi+hui, ukihuj, uℓi
+hui, uℓihuj, uki]dt
+ (n−1)µ2
We now split the exposition. The case of a hypersurface inR3 will be developed below. The case of a hypersurface in Rn, n > 3, will be treated in an addendum. The latter involves lengthy computations together with combinatorial arguments, whereas the former is more readable.
Theorem 2.8. Let u, v ∈ TxM with u and v linearly independent and define u(t) =DΦtu, v(t) = DΦtv, (however in our computations we shall suppress the t
dependence). The process (T r S(1), T r S(2)) satisfies the equations
d T r S(1)= [hdB(u, u), νikvk2+hdB(v, v), νikuk2−2hdB(u, v), νihu, vi]kαk−2
+T r S(1)(dW33−2(dW11+dW22))
+ 2[S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)]kαk−2
+ 4µ2 15 T r S
(1)dt
d T r S(2)= [S(v, v)hdB(u, u), νi+S(u, u)hdB(v, v), νi −2S(u, v)hdB(u, v), νi]kαk−2
+ 2T r S(2)(dW33−(dW11+dW22)) + 4µ2
15 T r S
(2)dt .
hd T r S(1)i=
16µ4
35 +
µ2
15[14(T r S
(1))2
−24T r S(2)]
dt
hd T r S(2)i=
2µ4
35 (3(T r S
(1))2−4T r S(2)) + 32µ2
15 (T r S
(2))2
dt
hd T r S(1), dT r S(2)i=
8µ4
35 T r S
(1)+ 16µ2
15 T r S
(1)T r S(2)
dt .
Proof. Recall we have selected an orthomal basis {e1, e2, e3} for TxtMt such that ν =e3. Then with α=u∧v, from LeJan (1986) we have
(2.15) dkαk
−2
kαk−2 =−2(dW 1
1 +dW22)−
2µ2
15 dt . Using (1.6), it follows that
(2.16) dhu, vi=hdW u, vi+hu, dW vi+ 10µ2
15 hu, vidt . Now by (1.9), the mean curvature is
so by Itˆo’s formula,
d T r S(1)= [kvk2dS(u, u) +kuk2dS(v, v)−2hu, vidS(u, v) +S(u, u)dkvk2+S(v, v)dkuk2−2S(u, v)dhu, vi
+hdS(u, u), dkvk2i+hdS(v, v), dkuk2i −2hdS(u, v), dh(u, v)i]kαk−2
+T r S(1)dkαk
−2
kαk−2 +hd(T r S
(1)kαk2), dkαk−2i
= [hdB(u, u), νikvk2+hdB(v, v), νikuk2−2hdB(u, v), νihu, vi]kαk−2
+T r S(1)dW33− 2µ2
15 T r S
(1)dt
+ [S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)]kαk−2
+ 10µ2 15 T r S
(1)dt− 2µ2
15 T r S
(1)dt
−2T r S(1)(dW11+dW22)− 2µ2
15 T r S
(1)dt ,
using Theorem 2.7, (2.15), (2.16) and the fact that
hd(T r S(1)kαk2), dkαk−2i=−2kαk−2[hkαk2T r S(1)dW33, dW11+dW22i
+ 2hS(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi), dW11+dW22i] =−2
h−2µ
2
15 + 2µ2
15
i
T r S(1)dt
= 0. Thus,
d T r S(1)= [hdB(u, u), νikvk2 +hdB(v, v), νikuk2−2hdB(u, v), νi hu, vi]kαk−2
(2.18)
+T r S(1)(dW33−2(dW11+dW22))
+ 2[S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi
+hu, dW vi)]kαk−2
+ 4µ2 15 T r S
(1)dt .
Consequently, the quadratic variation ofT r S(1) is given by
hd T r S(1)i= [hdB(u, u), νikvk2+hdB(v, v), νikuk2−2hdB(u, v), νi hu, vi]kαk−4
(2.19)
+ (T r S(1))2hdW33−2(dW11+dW22)i
+ 4[S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)]kαk−4
+ 4T r S(1)hdW33−2(dW11+dW22), S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)ikαk−2
Setting
qn =
(n+ 3)µ4
n(n+ 2) (n+ 4) =C
and if ui, uj, uk, uℓ are all orthogonal to ν, then
hdB(ui, uj), νi hdB(uk, uℓ), νi=qn[hui, uji huk, uℓi+hui, uki huj, uℓi
(2.20)
+hui, uℓi huj, uki]dt .
Then
h hdB(u, u), νikvk2+hdB(v, v), νikuk2−2hdB(u, v), νi hu, viikαk−4
(2.21)
= q3[3kuk4kvk4+ 3kvk4kuk4+ 4(2hu, vi2+kuk2kvk2)hu, vi2
+ 2(kuk2kvk2+ 2hu, vi2)kuk2kvk2−4(3kuk2hu, vi)kvk2hu, vi
−4(3kvk2hu, vi)kuk2hu, vk]kαk−4dt
= q3[8kuk4−16kuk2kvk2hu, vi2 + 8hu, vi4]kαk−4dt
= 8q3dt .
Next, observe
hdW33−2(dW11+dW22)i= hdW33i −4hW33, dW11+dW22i+ 4hdW11 +dW22i
= 18µ2 15 dt
and
hhdW v, vii = 2µ2 15 kvk
4dt
hhdW v, vi hdW u, uii = µ2
15(3hu, vi
2− kuk2kvk2)dt
hhdW u, vi,hu, dW vii = µ2
15(3hu, vi
2
− kvk2)dt
hhdW u, vi,hdW u, vii = µ2 15(4kuk
2kvk2−2hu, vi2)dt
hhdW u, vi,hdW v, vii = 2µ2
15 hu, vikvk
2dt
hhdW v, ui,hdW u, uii = 2µ2
15 hu, vikuk
2dt
hdW v, ui hv, vi = 2µ2
15 hu, vikvk
Thus,
[S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)] = µ2
15[2S(u, u)
2kvk4+ 2S(v, v)2kuk4
+ 2S(u, v)2(3kuk2kvk2+hu, vi2) + 2S(u, u)S(v, v)(3hu, vi2− kuk2kvk2)
−2S(u, u)S(u, v)(1hu, vikuk2+ 2hu, vikuk2)kvk2 −2S(v, v)S(u, v)(2hu, vikuk2+ 2hu, vikuk2)kuk2]dt
= µ2
15[2{S(u, u)
2kvk4+S(v, v)2kuk4+ 2S(u, u)S(v, v)kuk2kvk2 −4S(u, u)S(u, v)kuk2kvk2hu, vi
−4S(v, v)S(u, v)kuk2kvk2hu, vi
4S(u, v)2hu, vi2}
−6(S(u, u)S(v, v)−S(u, v)2)kαk2]dt
= µ2
15[2(T r S
(1))2−6T r S(2)]kαk4dt
It’s easy to verify that
hdW33,hdW v, vii=−µ2 15kvk
2dt
hdW33,hdW u, vii=hdW33,hu, dW vii=−µ2
15hu, vidt hdW11+dW22,hdW u, vii= µ2
15hu, vidt so
hdW33, S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)i =−µ2
15[S(u, u)kvk
2+S(v, v)kuk2−2S(u, v)hu, vi]dt
=−µ2 15T r S
(1)
kαk2dt ,
hdW11 +dW22, S(u, u)hdW v, vi+S(v, v)hdW u, ui −S(u, v)(hdW u, vi+hu, dW vi)i = µ2
15[S(u, u)kvk
2+S(v, v)
kuk2−2S(u, v)hu, vi]dt
= µ2 15T r S
(1)kαk2dt .
Therefore
4T r S(1)kαk−2hdW33−2(dW11+dW22), S(u, u)hdW v, vi+S(v, v)hdW u, ui
−S(u, v)(hdW u, vi+hu, dW vi)i (2.23)
= 4(T r S(1))2
−µ152 − 215µ2
dt
=−12µ2 15 (T r S
Combining (2.20), (2.21) and (2.22) we arrive at
(2.24) hd T rS(1)i= 16µ4 35 dt+
µ2
15[14(T r S
(1))2
−24T r S(2)]dt .
Turning now to the Gauss curvature, by (1.9) we have,
(2.25) T r S(2)= [S(u, u)S(v, v)−S(u, v)2]kαk−2 .
So, by Itˆo’s formula,
d T r S(2)= (S(v, v)dS(u, u) +S(u, u)dS(v, v)−2S(u, v)dS(u, v) +hdS(u, u), dS(v, v)i − hdS(u, v), dS(u, v)i]kαk−2
+ [S(u, u)S(v, v)−S(u, v)2]dkαk−2
+hS(v, v)dS(u, u) +S(u, u)dS(v, v)−2S(u, v)dS(u, v), dkαk−2i
= [S(v, v)hdB(u, u), νi+S(u, u)hdB(v, v), νi
−2S(u, v)hdB(u, v), νi]kαk−2
+ 2T r S(2)dW33− 4µ2
15 T r S
(2)dt
+ [hhdB(u, u), νi,hdB(v, v), νii − hdB(u, v), νi,hdB(u, v), νii]kαk−2dt
+ 2µ2 15 T r S
(2)dt
−2T r S(2)(dW11+dW22)− 2µ2 15 T r S
(2)dt
+ 8µ2 15 T r S
(2)dt .
Thus,
dT r S(2)= [S(v, v)hdB(u, u), νi+S(u, u)hdB(v, v), νi
−2S(u, v)hdB(u, v), νi]kαk−2
(2.26)
+ 2T r S(2)(dW33−(dW11+dW22)) + 4µ2 15 T r S
(2)dt ,
Using (2.20), (2.26) and our remarks above on W correlations, we get
hd T r S(2)i=hS(v, v)hdB(u, u), νi+S(u, u)hdB(v, v), νi
(2.27)
−2S(u, v)hdB(u, v), νiikαk−4+ 4(T r S(2))2hdW3 −(dW11+dW22)i =q3[3S(v, v)2kuk4+ 3S(u, u)2kvk4+ 4S(u, v)2(kuk2kvk2+ 2hu, vi2)
+ 2S(u, u)S(v, v)(kuk2kvk2+ 2hu, vi2)
−12S(v, v)S(u, v)kuk2hu, vi −12S(u, u)S(u, v)kvk2hu, vi]kαk−2dt
+ 32µ2 15 (T r S
(2))2dt
= 3q3(T r S(1))2dt−4q3T r S(2)dt+
32µ2
15 (T r S
(2))2dt
= 2µ4
35 (3(T r S
(1))2−4T r S(2))dt+ 32µ2
15 (T r S
(2))2dt .
Finally we need
hdT r S(1), dT r S(2)i= 2T r S(1)T r S(2)hdW33−2(dW11+dW22), dW33
(2.28)
−(dW11+dW22)i+ 4T r S(2)hS(u, u)hdW v, v) +S(v, v)hW u, ui −S(u, v)(hdW u, vi
+hu, dW vi), dW33−(dW11+dW22)ikαk−2
+kαk−4hhdB(u, u), νikvk2+hdB(v, v), νikuk2 −2hdB(u, v), νihu, vi, S(v, v)hdB(u, u), νi
+S(u, u)hdB(v, v), νi −2S(u, v)hdB(u, v), νii
= 24µ2 15 T r S
(1)T r S(2)dt−8µ2
15 T r S
(1)T r S(2)dt+ 4q
3T r S(1)dt
= 16µ2 15 T r S
(1)T r S(2)dt+8µ4
35 T r S
(1)dt .
This completes the proof.
Theorem 2.9. The process (T r St(1), T r St(2)) is a recurrent diffusion with genera-tor
Lf(κ, m) = 1 2
µ2
15(14κ
2
−24m) + 16µ4 35
∂2f
∂κ2(κ, m)
+
16µ2
15 κm+ 8µ4
35 κ
∂2f
∂κ ∂m(κ, m)
+ 1 2
32µ2
15 m
2+ 2µ4
35 (3κ
2−4m)
∂2f
∂m2(κ, m)
+ 4µ2 15 κ
∂f
∂κ(κ, m) +
4µ2
15 m
∂f
∂m(κ, m).
Remarks.
(1) Inn= 3, the three Lyapunov exponents are (see LeJan 1986)γ1 = 3,γ2 = 0, γ3 =−3. Thus there is always a stretching direction, a neutral direction and
a compressing direction. A curve will always “see” the stretching direction. This roughly explains the positive recurrence of the curvature of a curve proved in LeJan 1986. In the case of a surface, the tangent plane will “see” the stretching and the neutral direction. This also explains recurrence. The authors plan a complete account of the situation in higher dimensions in a forthcoming paper.
(2) SinceX =T r S(1)=λ1+λ2andY =T r S(2)=λ1λ2, it follows that (X, Y)
stays in the region κ2 ≥ 4m. The boundary of this region corresponds to λ1 =λ2. That is κ2 = 4mwhen the surface in question has an umbilic (i.e.
a point where λ1 = λ2). Thus, x may be an umbilic for M but xt, t > 0,
will never be an umbilic for Mt.
One easily checks that the matrix a appearing in the generator for this diffusion (2nd order derivative coefficients) degenerates on the curve κ2 −
4m = 0. Moreover, the two eigenvectors of this matrix are, respectively, perpendicular to and tangent to the curve κ2−4m = 0. The eigenvector perpendicular to this curve has eigenvalue zero. Notice the drift vector
4µ2
15 κ , 4µ2
15 m
points away from the region κ2 −4m < 0. Thus, when the pointx on the initial manifold M is an umbilic point, the diffusion (Xt, Yt)
has diffusion component only tangential the curveκ2 = 4m and drifts away from this curve never to return.
(3) Neither T r S(1) nor T r S(2) is a diffusion by itself.
(4) In the casen= 3 one can assert that (Xt, Yt) does not spend a lot of time in
the region λ = {(κ, m) :κ > M1, m > M2} with M1 and M2 large positive
constants. Recall the following results of LeJan (1986): draw a curve γ on the initial manifold M with γ(0) =x. Set γt = Φt(γ). Then the curvature τt ofγt atγt(0) is a positive recurrent diffusion. Thus, the average amount
of time that τt spends above a large positive constant is small. Since it
is impossible to draw a curve on Mt with zero curvature (or even small
curvature) when (Xt, Yt) ∈ Λ, (Xt, Yt) can not stay for long in Λ. More
precise statements can be made by referring to the article of LeJan where the exact form of the invariant measure for τt is given.
Proof (of Theorem 2.9). The claim that (Xt, Yt) is a diffusion follows immediately
from Theorem 2.8. Define h = √κ2−4m. Then with a= µ2
15, b = µ4
35, in (m, h
)-coordinates
Lf(κ, h) = (4aκ2+ 3ah2+ 8b)∂
2f
∂κ2 + 14aκh ∂2f ∂κ∂h
+ (3aκ2 + 4ah2+ 4b)∂
2f ∂h2
+ 4aκ∂f ∂κ +h
We prove the claim regarding lack of umbilics first as it eases a technical point which might arise in the proof of the recurrence. For this, it suffices to show that
Ht does not hit zero a.s. where (Xt, Ht) is diffusion given in (κ, h) coordinates.
First defineηt = 3aXt2+ 4aHt2+ 4b and then τt =
Rt
0 dx
ηs. Then
Hτt =h+bt+
Z τt
0
Hs−1(3aXs2+aHs2+ 4b)ds
=h+bt+
Z t
0
Hτs−1(3aXτs2 +aHτs2 + 4b)ητs−1ds
with bt a one-dimensional Brownian motion. Fix now anǫ ∈ (0,1/6) and suppose
0< h <1/6 as well. Set
ρǫt =h+bt+
1 2 +ǫ
Z t
0 ds ρǫ
s .
If σǫ = inf{t > 0 : Hτt > 2
q
(1 2−ǫ)b
(1+4ǫ)a}, then since for h ≤
q
(1 2−ǫ)b
(1+4ǫ)a, one has
(21+ǫ)< 3aκ3aκ22+4ah+ah22+4b+4b for allκ. By an elementary comparison theorem (see Ikeda-Watanabe, 1989), it follows that ρǫ
t ≤ Hτt for t < σǫ. Since ρǫt can not hit zero (it
is a Bessel process above the critical dimension),Ht can’t hit zero either.
For recurrence, we shall show (Xt, Ht) is recurrent which will suffice. Set f(κ, h) =ℓn(κ6+h6). Then
fκ =
6κ5
κ6+h6 , fh =
6h5 κ6+h6 fκκ =
30κ4h6−6κ10
(κ6 +h6)2 , fκh =−
36κ5h5
(κ6+h6)2 ; fhh =
30κ6h4−6h10
(κ6+h6)2
and
Lf = 6(κ6+h6)−2[(4aκ2+ 3ah2 + 8b)(5κ4h6−κ10)−84aκ6h6
+ (3aκ2+ 4ah2 + 4b)(5κ6h4−h10) + 4aκ12+ 4aκ6h6
+h4(3aκ2+ah2+ 4b)(κ6+h6)]
= 6(κ6+h6)−2[−a(3κ10h2−18κ8h4+ 39κ6h6−15κ4h8+ 3h12) −b(8κ10−24κ6h4−40κ4h6)]
κ=uh
=−6(κ6+h6)−2h10[3ah2(u10−6u8+ 13u6−5u4+ 1) + 8bu4(u6−3u2−5)]
Also, there is a u0 > 0 such that Q(u) = u6−3u2−5 ≥ 0 whenever |u| ≥ u0.
Thus,
Lf =−6(κ6+h6)−2h10[3ah2P(u) + 8bu4Q(u)] has
{Lf ≤0} ⊇ {(κ, h) :|κ| ≥u0h}.
Also, denote M = supu Q(u)P(u), one sees
{Lf ≤0} ⊇
(
(κ, h) :h≥
r
86M
3a
)
Consequently, off of the compact triangle
T ={(κ, h) :|κ| ≤u0h} ∩
(
(κ, h) :h ≤
r
86M
3a
)
we have Lf ≤0. Now, define
C(r) ={(κ, h) :h >0, κ6+h6 < r}
and select r0 large enough so that C(r0)⊃ T. Put for R > r0
uR(κ, h) =
ℓn(κ6+h6)−ℓnR ℓnr0−ℓnR
.
Then on C(R)\C(r0), LuR ≥ 0 and uR
∂C(r0)∩{h>0} = 1, uR
∂C(R)∩{h>0} = 0.
Thus, if τr = inf{t >0 : (Xt, Ht)∈ ∂C(r)} (recall, Ht never hits 0), then as uR is L-subharmonic and has the same boundary values as P(κ,w)(τr0 < TR)
uR(κ, h)≤P(κ,h)(τr0 < τR).
But limR↑∞uR(κ, h) = 1 which implies the recurrence of (Xt, Ht) and therefore the
References
P. Baxendale, (1984), Brownian motions in the diffeomorphism Group. I,
Compositio Math. 53, 19–50.
P. Baxendale and T.Harris, (1986), Isotropic stochastic flows, Annals of Prob.,Vol 14, 1155–1179.
R. Bishop and S. Goldberg, (1980), Tensor Analysis on Manifolds, Dover, Toronto.
Hardy, Littlewood, Polya, (1973),Inequalities, Cambridge University Press, Cambridge.
T. Harris, (1981), Brownian motions on the homeomorphisms of the plane,
Annals of Prob. Vol. 9, 232–254.
N. Ikeda, S. Watanabe, (1989), Stochastic Differential Equations and Dif-fusion Processes, North-Holland, New York.
Y. LeJan, (1985), On isotropic Brownian motions, Zeit. f¨ur Wahr., 70, 609–620.
Y. LeJan, (1991), Asymototic properties of isotropic Brownian flows, in Spatial Stochastic Processes, ed. K. Alexander, J. Watkins, Birkh¨auser, Boston, 219–232.
Y. LeJan, S. Watanabe (1984), Stochastic flows of diffeomorphisms, Proc. of the Taniguchi Symposium, 307–332, North Holland, New York.
Addendum. Higher dimensional hypersurfaces.
Recall α~ℓ =uℓ1 ∧ · · · ∧uℓk, αm~ =um1 ∧ · · · ∧umk, for ~ℓ, ~m∈Ik. Define α~ℓ
p = (−1)
p+1u
ℓ1∧ · · · ∧uˆℓp∧ · · · ∧uℓk, αm~i = (−1)
i+1u
m1∧ · · · ∧uˆmi∧ · · · ∧umk.
Thenα~ℓp = (−1)p+1u(p)ℓ1 ∧· · · ∧u
p
ℓk−1 α
(i) ~
mi = (−1)p+1u (i)
m1∧· · · ∧u
(i)
mk−1 definesu (p) ℓ
and u(i)m, and set S(0)(·,·)≡1.
Theorem A.1.
dS(k)(α~ℓ, αm~) = k
X
i,p=1
S(k−1)(α~ℓp, αmi~ )hdB(uℓp, umi), νi
+kS(k)(α~ℓ, αm~)dWnn−
k(n−k)(n−1)µ2
2n(n+ 2) S
(k)(α ~
ℓ, αm~)dt ,
or, in a more synthetic form
dS(α)(k) =S(k−1)∧dBν(α) +kS(k)(α)hdhW ν, νi −−k(n−k)(n−1)
2n(n+ 2) S
(k)(α),
with hBνu, vi=hB(u, v), νi.
Proof. By Theorem 2.7 and Itˆo’s formula,
dS(k)(α~ℓ, αm~) = X
σ∈Sk (−1)σ
k X i=1 [ k Y j=1 j6=i
hS(uℓσ(j), umj), νi]hdB(uℓσ(i), umi), νi
+kS(k)(α~ℓ, αm~)dWnn− k(n−1)(n−1)µ2
2n(n+ 2) S
(k)(α ~ ℓ, αm~)dt
+1 2
X
σ∈Sk
(−1)σ X
1≤i6=p≤k
[
k
Y
j=1 j6={i,p}
hS(uℓ
σ(j), umj), νi][dBn(uℓσ(i), umi), dBn(uℓσ(p), ump)i
+hdBn(uℓσ(i), ump),dB
n(u
ℓσ(p), umi)i+hdB
n(u
ℓσ(i), uℓσ(p)), dB n(u
mi, ump)i] +k(k−1)(n−1)µ2
2n(n+ 2) S
(k)(α ~
ℓ, αm~)dt. The first term may be written
k
X
i,p=1
X
σ∈Sk
σ(i)=p
(−1)σ[
k
Y
j=1 j6=i
hS(uℓ
σ(j), umj), νi]hdB(uℓp, umi), νi
=
k
X
i,p=1
[ X
σ=Sk−1
(−1)p+i(−1)σ[
k
Y
j=1 j6=i
hS(u(p)ℓ
σ(j), u (i)
mj), νi]]hdB(uℓp, umi), νi
=
k
X
i,p=1
S(k−1)(α~ℓ
The third term combines with the fifth term to give
−k(n−2nk(n)(n+ 2)−1)µ2S(k)(α~ℓ, αm~)dt.
Finally, in the fourth term, notice that givenσ ∈Skandi, p ∈ {1, . . . , k}there is
exactly one otherη∈ Sksuch that both {σ(i), σ(p)}={η(i), η(p)}and σ(j) =η(j)
for j /∈ {i, p}. Furthermore, (−1)σ =−(−1)η, and [
k
Y
j=1 j /∈{i,p}
hS(uℓ
σ(j), umj), νi][hBn(uℓσ(i), umi), dBn(uℓσ(p), ump)i
+hdBn(uℓσ(i), ump), dBn(uℓσ(p), umi)i+hdB
n(u
ℓσ(i), uℓσ(p)), dB n(u
mi, ump)i]
=[
k
Y
j=1 j /∈{i,p}
hS(uℓ
η(j), umj), νi][hdBn(uℓη(i), umi), dBn(uℓη(p), ump)i
+hdBn(uℓη(i), ump), dBn(uℓη(p), umi)i +hdBn(uℓ
η(i), uℓη(p)), dB n(u
mi, ump)i]
Thus exchanging the sums in the fourth term reveals the cancellation of terms from these σ−η pairs and so the whole term vanishes. This finishes the proof.
From LeJan (1985) we recall the notation
τℓjβ =ej ∧i(eℓ)β
where i(eℓ)β = Σ(−1)k+1hβ
k, eℓiβ1∧ · · · ∧βˆk∧ · · · ∧βm. The situation for general n is given by the following Theorem which we prove in the Addendum.
Theorem A.2. For 1≤k≤n−1,
dT rS(k) = [
k
X
i,p=1
S(k−1)(α~ℓ
p, αm~i)hdB(uℓp, umi), νi]hα
~
ℓ, αm~i||α||−2
+T rS(k)[kdWnn−2
n−1
X
i=1 dWii]
+S(k)(α~ℓ, αm~)(hτijα~ℓ, αm~i+hα~ℓ, τijαm~i)dWji||α||−2 −(n+ 1)k(n−k)µ2
2n(n+ 2) T rS
(k)dt , or
d T rS(k) =dhS(k)α, αi
kαk2
=hS(k−1)∧dBν(α) +
hdW ν, νi −2hdW α, αi kαk2
S(k)(α)
+S(k)∧dW(α), αi
kαk2− (n+ 1)k(n−k)µ2
2n(n+ 2) hS
(k)(α), αi
Proof. For~ℓ∈Ik,
dhα~ℓ, αm~i= (hτijα~ℓ, αm~i+hα~ℓ, τijαm~i)dWji
(A.1)
+ (k+ 1)(n−1−k)µ2
n hα
~ ℓ, αm~
idt
(A.2) d||α||
−2
||α||−2 =−2 n−1
X
i=1
dWii− (n−2)(n−1)µ2 n(n+ 2) dt
(A.3) d||α||−2 =−2||α||−2dWii+At, At has bounded variation
Since
TrS(k)=S(k)(α~ℓ, αm~)hα ~ ℓ, αm~
i||α||−2
by Itˆo’s formula,
dT rS(k)= (dS(k)(α~ℓ, αm~))hα ~ ℓ, αm~
i||α||−2
+S(k)(α~ℓ, αm~)(dhα ~ ℓ, αm~
i)||α||−2
+T rS(k)d||α||
−2
||α||−2
+hdS(k)(α~ℓ, αm~), dhα ~ ℓ, αm~
ii||α||−2
+hdS(k)(α~ℓ, αm~), d||α||−2ihα ~ ℓ, αm~
i +S(k)(α~ℓ, αm~)hdhα
~ ℓ, αm~
i, d||α||−2i
Proceeding in order through the terms, we have using Theorem A.1
dS(k)(α~ℓ, αm~)hα ~ ℓ, αm~
i||α||−2
(A.4)
= [
k
X
i,p=1
S(k−1)(α~ℓp, αmi~ )hdB(uℓp, umi), νi]hα ~ ℓ, αm~
i||α||−2
+kT rS(k)dWnn− k(2n−k)(n−1)µ2
2n(n+ 2) T rS
(k)dt.
From (A.1) follows,
S(k)(α~ℓ, αm~)dhα ~ ℓ
, αm~i||α||−2
(A.5)
= S(k)(α~ℓ, αm~)(hτijα ~
ℓ, αm~i+hα~ℓ, τj
iαm~i)dWji||α||−2
+ (k+ 1)(n−k−1)µ2
n T rS
By (A.2)
(A.6) T rS(k)d||α||
−2
||α||−2 =−2T rS (k)
n−1
X
i=1
dWii− (n−2)(n−1)µ2 n(n+ 2) T rS
(k)dt.
Combining Theorem A.1 and (A.1),
hdS(k)(α~ℓ, αm~), dhα~ℓ, αm~ii||α||−2
=kS(k)(α~ℓ, αm~)(hτijα~ℓ, αm~i+hα~ℓ, τijαm~i)hdWji, dWnni
= kµ2
n(n+ 2)S
(k)(α ~
ℓ, αm~)(hτ j iα
~
ℓ, αm~i+hα~ℓ, τj
iαm~i)[(n+ 1)δinδjn−δijδnn−δinδnj]dt
=− kµ2 n(n+ 2)S
(k)(α ~
ℓ, αm~)(hτiiα ~
ℓ, αm~i+hα~ℓ, τi iαm~i)dt
=−2k(n−k−1)µ2 n(n+ 2) T rS
(k)dt.
By Theorem A.1 and (A.3)
hdS(k)(α~ℓ, αm~), d||α||−2ihα~ℓ, αm~i
(A.7)
=−2kS(k)(α~ℓ, αm~)hdWnn, dWiiihα~ℓ, αm~i||α||−2
= 2k(n−1)µ2
n(n+ 2) T rS
(k)dt.
From (A.1) and (A.3) we get
S(k)(α~ℓ, αm~)hdhα~ℓ, αm~i, d||α||−2i
(A.8)
=−2S(k)(α~ℓ, αm~)(hτijα~ℓ, αm~i+hα~ℓ, τijαm~i)hdWpp, dWjii||α||−2
= 2µ2
n(n+ 2)S
(k)(α ~
ℓ, αm~)(hτ i iα
~
ℓ, αm~i+hα~ℓ, τi
iαm~i)||α||−2dt
=−4(n−k−1)µ2 n(n+ 2) T rS
(k)dt
Summing (A.4)–(A.8) shows
dTrS(k) = [ k
X
i,p=1
(−1)i+p+1S(k−1)(αℓ~p ℓ , α
mi
~
m )hdB(uℓp, umi), νi]hα
~
ℓ, αm~i||α||−2
+T rS(k)[kdWnn−2dWii]
+S(k)(α~ℓ, αm~)(hτijα~ℓ, αm~i+hα~ℓ, τijαm~i)dWji· kαk−2
+k(n−k)(n+ 1)µ2 2n(n+ 2) T rS
(k)dt
Theorem A.3. For n−1≥k ≥ℓ≥ 1,
hdT rS(k), dT rS(ℓ)i= µ2
n(n+ 2)[(kℓ+ 2(k+ℓ) + 4)(n−1)
−2(k+ 2)(n−ℓ−1)−2(ℓ+ 2)(n−k−1)−4(n−ℓ−1)(n−k−1)]T rS(k)T rS(ℓ)dt
+ (1−δnk−1) 4µ2 (n+ 2)
(n−k−1)∧ℓ
X
j=0
d(n,(k, ℓ);j)T rS(k+j)T rS(ℓ−j)dt
+C1122nn
(n−k)∧(ℓ−1)
X
j=0
a(n,(k, ℓ);j)T rS(k+j−1)T rS(ℓ−j−1)dt
where
d(n,(k, ℓ); 0) = (n−1−k)
d(n,(k, ℓ);j) = (n−1−k−j) k−ℓ+ 2j
j
− j−1
X
r=0
d(n,(k, ℓ);r) k−ℓ+ 2j
j−r
, j≥1
b(n,(k, ℓ); 0) = (n−k)(n−ℓ+ 2)
b(n,(k, ℓ);j) = 3(n−k−j) k−ℓ+ 2j
j
+ (k−ℓ+ 2j)(k−ℓ+ 2j−1) k−ℓ+ 2j−2
j−1
+ (k−ℓ+ 2j)(n−k−j)[ k−ℓ+ 2j−1
j
+ k−ℓ+ 2j−1
j−1
]
+ (n−k−j)(n−k−j−1) k−ℓ+ 2j
j
, j≥1
a(n,(k, ℓ); 0) =b(n,(k, ℓ); 0)
a(n,(k, ℓ);j) =b(n,(k, ℓ);j)− j−1
X
m=0
a(n,(k, ℓ);m) k−ℓ+ 2j
j−m
C1122nn (n2−1) = (n+ 3)µ4
n(n+ 2)(n+ 4).
Proof. We compute the quadratic variation terms in hdT rS(k), dT rS(ℓ)i, k ≥ ℓ, which do not arise from dB. These are
hT rS(k)[kdWnn−2
n−1
X
i=1
dWii] +S(k)(α~ℓ, αm~)(hτhjα~ℓ, αm~i+hα~ℓ, τhjαm~i)||α||−2dWjh,
T rS(ℓ)[ℓdWnn−2
n−1
X
m=1
dWmm] +S(ℓ)(α~r, α~s)(hτqpα~r, α~si+hα~r, τqpα~si)||α||−2dWpqi
unless k = n−1 or both ℓ = k = n−1 in which case, the term S(k)(α ~ ℓ, αm~)
(hτijα~ℓ, αm~i+hα~ℓ, τj
iαm~i)||α||−2dWji = 0 or both this and the corresponding S(ℓ)
term vanishes. Thus we get
T rS(k)T rS(ℓ)hkdWnn−2
n−1
X
i=1
dWii, ℓdWnn−2
n−1
X
+T rS(k)S(ℓ)(α~r, αs~)(hτqpα~r, α~si+hα~r, τqpα~si)||α||−2(1−δℓn−1)
hkdWnn−2
n−1
X
i=1
dWii, dWpqi
+T rS(ℓ)S(k)(α~ℓ, αm~)(hτhjα~ℓ, αm~i+hα~ℓ, τhjαm~i)||α||−2(1−δkn−1)
hℓdWnn−2
n−1
X
i=1
dWii, dWjhi
+S(k)(α~ℓ, αm~)S(ℓ)(α~r, α~s)(hτhjαℓ~, αm~i+hα~ℓ, τhj, αm~i)(hτqpα~r, α~si
+hα~r, τqpα~si)(1−δℓn−1)
·(1−δkn−1)||α||−4hdWjh, dWpqi
= µ2
n(n+ 2)T rS
(k)T rS(ℓ)[k·ℓ(n−1) + 2(k+ℓ)(n−1) + 4(n−1)]dt
+ µ2
n(n+ 2)T rS
(k)S(ℓ)(α ~
r, α~s)(hτppαr~, α~si+hα~r, τppα~si)||α||−2(1−δℓn−1)[kCnpnp−2 n−1
X
i=1 Cipip]dt
+ µ2
n(n+ 2)T rS
(ℓ)S(k)(α ~
ℓ, αm~)(hτ j jα
~
ℓ, αm~i+hα~ℓ, τj
jαm~i)||α||−2(1−δkn−1)[ℓCnjnj−2 n−1
X
i=1 Cijij]dt
+ µ2
n(n+ 2)S
(k)(α ~
ℓ, αm~)S(k)(α~r, α~s)(hτ j hα
~
ℓ, αm~i+hα~ℓ, τj
hαm~i)(hτ p qα~r, α~si
+hα~r, τqpα~si)(1−δkn−1)||α||−4[(n+ 1)δqhδjp−δhjδqp−δphδjq]dt
We now proceed singly through the last three terms. Using the invariance of our expressions under a change of basis we may take the ui to be the unit principal
curvature directions. Then
S(k)(α~ℓ, αm~) =λ~ℓδm~ℓ~
(A.9)
S(ℓ)(α~r, α~s) =λ~rδ~s~r hτqpα~r, α~siδ~r~s =δ
p
qδ~s~r if and only ifp∈~r ,= 0 otherwise, hτhjα~ℓ, αmiδm~ℓ~ =δhjδ~ℓm~ if and only ifj ∈~ℓ ,= 0 otherwise.
Thus,
µ2
n(n+ 2)T rS
(k)S(ℓ)(α ~
r, α~s)(hτppαr~, α~si+hα~r, τppα~si)(1−δnℓ−1)||α||−2[kCnpnp−2 n−1
X
i=1 Cipip]dt
(A.10)
= µ2
n(n+ 2)T rS
(k)λ
~rhτppα~r, α~ri(1−δℓn−1)[k(nδnp −δnnδpp)−2]dt
=−2(k+ 2)(n−ℓ−1)µ2 n(n+ 2) T rS
(k)T rS(ℓ)dt.
Similarly,
µ2
n(n+ 2)T rS
(ℓ)S(k)(α ~
ℓ, αm~)(hτ j jα
~
ℓ, αm~i+hα~ℓ, τj
jαm~i)||α||−2(1−δkn−1)(ℓCnjnj−2 n−1
X
i=1 Cijij)dt
(A.11)
=−2(ℓ+ 2)(n−k−1)µ2
n(n+ 2) T rS
Finally, the last term is
(1−δnk−1) 4µ2
n(n+ 2)λ~ℓλ~rhτ
j hα
~ ℓ, α~ℓihτp
qα~r, α~ri(1−δℓn−1)[(n+ 1)δqhδpj−δjhδpq−δhpδqj]dt
(A.12)
=(1−δnk−1) 4µ2
n(n+ 2)λ~ℓλ~r[(n+ 1)hτ j hα
~ ℓ, α~ℓihτj
hα~r, α~ri − hτ j jα
~ ℓ, α~ℓihτp
pα~r, α~ri
−hτhjα~ℓ, α~ℓihτjhα~r, α~ri]dt
=(1−δnk−1) 4µ2
n(n+ 2)λ~ℓλ~r[nhτ
j jα
~ ℓ, α~ℓihτj
jα~r, α~ri − hτ j jα
~ ℓ, α~ℓihτp
pα~r, α~ri]dt
=(1−δnk−1) 4µ2
n(n+ 2)λ~ℓλ~r[nδj6∈~ℓδj6∈r~−δj6∈~ℓδp6∈~r]dt (δj6∈~ℓ=
1, whenj6∈~ℓ,
0, otherwise ) =(1−δnk−1)( 4nµ2
n(n+ 2)λ~ℓλ~r·δj6∈~ℓδj6∈~rdt−
4(n−k−1)(n−ℓ−1µ2) n(n+ 2) T rS
(k)T rS(ℓ)dt). However, by summing on j we see that
(A.13) λ~ℓλ~rδj6∈~ℓδj6∈~r =
n−ℓ−1
X
m=0 |~ℓc∩~rc|=m
mλ~ℓλ~r
which is a symmetric homogeneous polynomial in λ1, . . . , λn−1 of degree k+ℓ so
we seek an expansion for (A.13) of the form
(A.14)
(n−k−1)∧ℓ
X
j=0
d(n,(k, ℓ);j)T rS(k+j)T rS(ℓ−j).
The principal term inT rS(k+j)T rS(ℓ−j) (which doesn’t appear in T rS(k+m)T rS(ℓ−m) for m > j) is
λ21. . . λ2ℓ−jλℓ−j+1. . . λk+j.
In Pn−ℓ−1
m=0 |~ℓ∩~r|=m
mλ~ℓλ~r this term appears (n−1−k−j) k−ℓ+2jj
times.
In T rS(k+r)T rS(ℓ−r) for r < j this term appears k−ℓ+2j j−r
times. Thus in the expansion for Pn−ℓ−1
m=0 |~ℓc∩~rc|=m
mλ~ℓλ~r
d(n,(k, ℓ); 0) = (n−1−k)
d(n,(k, ℓ);j) = (n−1−k−j) k−ℓ+ 2j
j
− j−1
X
r=0
d(n,(k, ℓ);r) k−ℓ+ 2j
j−r
, j≥1.
Putting (A.10), (A.11) and (A.12) together results in
[ µ2
n(n+ 2)T rS
(k)T rS(ℓ)[(kℓ+ 2(k+ℓ) + 4)(n−1)−2(k+ 2)(n−ℓ−1)
(A.15)
−2(ℓ+ 2)(n−k−1)−4(n−ℓ−1)(n−k−1)]
+ (1−δnk−1) 4µ2
n(n+ 2)
(n−k−1)∧ℓ
X
j=0
The quadratic variation term in hdT rS(k), dT rS(ℓ)i, k ≥ℓ, arising from dB is,
Q(n,(k, ℓ))dt
=h k
X
i,p=1
S(k−1)(α~ℓ
p, αm~i)hα
~
ℓ, αm~ihdB(u
ℓp, umi), νi,
ℓ
X
j,q=1
S(ℓ−1)(α~rq, α~sj)hαr~, α~sihdB(urq, usj), νii||α||−4
=C1122nn k X i,p=1 ℓ X j,q=1
S(k−1)(α~ℓp, αm~i)S
(ℓ−1)(α
~rq, α~sj)[huℓp, umiihurq, usji
+huℓp, urqihumi, usji+huℓp, usjihurq, umii]hα
~
ℓ, αm~ihα~r, α~si||α||−4dt
and taking advantage of the invariance of this expression under a change of basis, we take ui to be the ith unit principal direction of curvature. Then
S(k−1)(α~ℓ
p, αm~i)hα
~
ℓ, αm~i=λ ~ ℓpδ
~ ℓ ~ mδpi Sℓ−1(α~rq, α~sj)hα~r, α~si=λ~rqδ~r~sδjq
[huℓp, umiihurq, usji+huℓp, urqihumi, usji+huℓp, usjihurq, umii]δ
~ ℓ ~ mδ~s~r
= (δℓp
miδ
rq
sj +δ
ℓp
rqδ
mi
sj +δ
ℓp
sjδ
rq
mi)δ
~ ℓ ~ mδ~s~r.
Thus,
Q(n,(k, ℓ)) =C1122nn k X p=1 ℓ X q=1 λ~ℓ
pλ~rq(1 + 2δ
ℓp
rq)dt
=C1122nn (3
k X p=1 ℓ X q=1 ℓp=rq
λ~ℓ
pλ~rq +
k X p=1 ℓ X q=1 ℓp6=rq
λ~ℓ
pλ~rq).
SinceQ(n,(k, ℓ)) is a symmetric homogeneous polynomial of degreek+ℓ−2 in
λ1, . . . , λn−1, it may be written
Q(n,(k, ℓ)) =C1122nn
(n−k)∧(ℓ−1)
X
j=1
a(n,(k, ℓ);j)T rS(k+j−1)T rS(ℓ−j−1).
We now employ some elementary counting arguments to compute the coefficients
a(n,(k, ℓ);j).
In T rS(k+j−1)T rS(ℓ−j−1) the term
but no such term (with ℓ−j −1 squares) appears in T rS(k+m−1)T rS(ℓ−m−1) for m > j. However, if r < j,
(A.17) λ21. . . λ2ℓ−j−1λℓ−j. . . λk+j−1 appears k−ℓ+2j
j−r
times in T rS(k+r−1)T rS(ℓ−r−1).
We now count the number of appearances of (A.16) in Q(n,(k, ℓ)). From 3
k
X
p=1 ℓ
X
q=1 ℓp=rq
λ~ℓpλ~rq we must fill in the empty spaces in λ~r and λ~ℓ
λ~r=λ1. . . λℓ−j−1− · · · −
(A.18)
λ~ℓ=λ1. . . λℓ−j−1− · · · −ℓ· · · −k
with the terms λℓ−j, . . . , λk+j−1, λx, λx, where x = ℓp = rq. Notice λt appears
before λu only if t < u. One λx must go in λ~r the other in λ~ℓ. If the order is not
to be violated, this forces x > k−1 +j andλx must end each of the terms λ~r and λ~ℓ. There are (n−1)−(k−1 +j) =n−k−j such choices for x. This leaves
k−1 +j −(ℓ−j−1)
j
=
k−ℓ+ 2j j
remaining possibilities for filling in the empty spaces in (A.18) in λ~r and λ~ℓ. Note
that if we interpret 00
= 1 this is still correct for k =ℓand j = 0. Thus
(A.19) 3
k
X
p=1 ℓ
X
q=1 ℓp=rq
λ~ℓpλ~rq has 3(n−k−j)
k−ℓ+ 2j j
terms of the form λ2
1. . . λ2ℓ−j−1λℓ−j. . . λk+j−1.
In
k
X
p=1 ℓ
X
q=1 ℓp6=rq
λ~ℓpλ~rq we must fill in the blanks of
λ~r=λ1. . . λℓ−j−1− · · · − λ~ℓ=λ1. . . λℓ−j−1− · · · −ℓ· · · −k
with λℓ−j, . . . , λk+j−1, λx, λy with x=6 y, λx going into λ~r, λy going intoλ~ℓ.
If ℓ−j ≤x, y ≤k+j −1, then λx and λy appear on the list λℓ−j, . . . , λk+j−1.
Since identical terms can’t appear twice in λ~r or λ~ℓ the twin of λx must go in λ~ℓ,
the twin of λy must go inλ~r. There are thus k−ℓ+2jj−1−2
j −1))(k+j −1−(ℓ−j −1)−1) choices for the ordered pair (x, y). Notice that when j = 0 there are no such terms. Thus
ℓ−j≤x, y≤k+j−1 accounts for (A.20)
(k−ℓ+ 2j)(k−ℓ+ 2j−1) k−ℓ+ 2j−2
j−1
terms of the form λ2
1. . . λ2ℓ−j−1λℓ−j. . . λk+j−1 in Q(n,(k, ℓ)) (interpreting z −1
= 0.)
If ℓ−j ≤ x ≤ k+j−1 < y ≤ n−1, then λy must terminate the sequence λ~ℓ
and the twin of λx must go into λ~ℓ. There are then
k−ℓ+ 2j −1
j
ways to fill in the spaces in (A.18) once such a pair (x, y) is given. We interpret this expression as 0 when k =ℓandj = 0. There are (k−ℓ+ 2j)(n−k−j) such pairs so we have
(A.21)
corresponding toℓ−j ≤x≤k+j−1< x≤n−1 there are (k−ℓ+ 2j)(n−k−j)
k−ℓ+ 2j−1
j
terms of the form λ21· · ·λ2ℓ−j−1λℓ−j· · ·λk+j−1 in Q(n,(k, ℓ)) .
If ℓ−j ≤ y ≤ k +j −1 < x ≤ n−1, then the twin of λy must go in λ~r. When j = 0, this is impossible. There are thus
k−ℓ+ 2j −1
j −1
ways to fill the spaces in (A.18) once (x, y) is given and (k−ℓ+ 2j)(n−k−j) such pairs so
(A.22)
corresponding toℓ−j ≤y≤k+j−1< x≤n−1, there are (k−ℓ+ 2j)(n−k−j)
k−ℓ+ 2j−1
j−1
terms of the form λ2
1· · ·λ2ℓ−j−1λℓ−j· · ·λk+j−1 in Q(n,(k, ℓ)), for j ≥0, (interpreting −z1
= 0.)
Finally, when k+j −1< x, y ≤n−1, there are
k−ℓ+ 2j j
ways to fill in the spaces in (A.18) once (x, y) is given and (n−k−j)(n−k−j−1) such pairs (x, y). Thus
(A.23)
corresponding tok+j−1< x, y ≤n−1, there are (n−k−j)(n−k−j−1)
k−ℓ+ 2j j
terms of the form λ2
1· · ·λ2ℓ−j−1λℓ−j· · ·λk+j−1 in Q(n,(k, ℓ)),
interpreting 00
Totalling (A.18)–(A.23), we see
Q(n, k) has (A.24)
b(n,(k, ℓ);j) = 3(n−k−j)
k−ℓ+ 2j j
+ (k−ℓ+ 2j)(k−ℓ+ 2j−1)
k−ℓ+ 2j−2
j−1
+ (k−ℓ+ 2j)(n−k−j)
k−ℓ+ 2j−1
j
+ (k−ℓ+ 2j)(n−k−j)
k−ℓ+ 2j−1
j−1
+ (n−k−j)(n−k−j−1)
k−ℓ+ 2j j
terms with exactly ℓ−j−1 squares. Notice that we interpret −z1
= 0, −01
= 0,
0 0
= 1. Thus, b(n,(k, ℓ); 0) = (n−k)(n−k −2). Upon decomposingQ(n,(k, ℓ)) into sums of expressions with exactly ℓ−j−1 squares summed overj, we are lead to, using (A.16) and (A.23),
Q(n, k) =Cnn 1122
(n−k)∧(ℓ−1)
X
j=0
a(n,(k, ℓ);j) TrS(k+j−1) Tr S(ℓ−j−1)
where
a(n,(k, ℓ);j) =b(n,(k, ℓ);j)−
j−1
