Single Basepoint Subdivision Schemes for Manifold-valued Data: Time-Symmetry Without Space-Symmetry

Abstract

This paper establishes smoothness results for a class of nonlinear subdivision schemes, known as the single basepoint manifold-valued subdivision schemes, which shows up in the construction of wavelet-like transform for manifold-valued data. This class includes the (single basepoint) Log–Exp subdivision scheme as a special case. In these schemes, the exponential map is replaced by a so-called retraction map f from the tangent bundle of a manifold to the manifold. It is known that any choice of retraction map yields a C ² scheme, provided the underlying linear scheme is C ² (this is called “C ² equivalence”). But when the underlying linear scheme is C ³, Navayazdani and Yu have shown that to guarantee C ³ equivalence, a certain tensor P _f associated to f must vanish. They also show that P _f vanishes when the underlying manifold is a symmetric space and f is the exponential map. Their analysis is based on certain “C ^k proximity conditions” which are known to be sufficient for C ^k equivalence.

In the present paper, a geometric interpretation of the tensor P _f is given. Associated to the retraction map f is a torsion-free affine connection, which in turn defines an exponential map. The condition P _f =0 is shown to be equivalent to the condition that f agrees with the exponential map of the connection up to the third order. In particular, when f is the exponential map of a connection, one recovers the original connection and P _f vanishes. It then follows that the condition P _f =0 is satisfied by a wider class of manifolds than was previously known. Under the additional assumption that the subdivision rule satisfies a time-symmetry, it is shown that the vanishing of P _f implies that the C ⁴ proximity conditions hold, thus guaranteeing C ⁴ equivalence. Finally, the analysis in the paper shows that for k≥5, the C ^k proximity conditions imply vanishing curvature. This suggests that vanishing curvature of the connection associated to f is likely to be a necessary condition for C ^k equivalence for k≥5.

Appendices

Appendix A: Proof of Lemma 19

By the definition of S _lin, the sequence is the second square bracket of (4.6) is simply $\prod_{i} S_{\mathrm{lin}} A_{J_{2}^{i}} - S_{\mathrm{lin}}\prod_{i} A_{J_{2}^{i}}$, where A _J is thought of as a sequence on ℤ whose hth entry is $A^{h}_{J}$. Note that A _J is a polynomial sequence of degree |J|. Since $\sum_{i=1}^{\alpha}|J_{2}^{i}| \leq k$, and S _lin leaves the polynomial spaces Π _ℓ , ℓ≤k, invariant, both $\prod_{i} S_{\mathrm{lin}} A_{J_{2}^{i}}$ and $S_{\mathrm{lin}} \prod_{i} A_{J_{2}^{i}}$, and hence also their difference, must have degrees no bigger than ${\sum_{i=1}^{\alpha}|J_{2}^{i}|}$.

We now prove that, in fact, the difference is two degree less than what we expect. The proof of this part merely requires the simple fact ∑ _ℓ a _2ℓ+σ =1. Note that

$$A_j^{h-\ell} = \frac{1}{j!} \biggl(h^j + B_j(\ell) h^{j-1} + \sum_{d<j-1} B_j^d(\ell) h^d \biggr) . $$

If J=(j ₁,…,j _γ ),

$$A_{J}^{h-\ell} = \prod A_{j_i}^{h-\ell} = \frac{1}{j_1!\cdots j_\gamma!} \bigl( h^{|J|} + B_J(\ell) h^{|J|-1} + \cdots \bigr), $$

where $B_{J}(\ell) = B_{j_{1}}(\ell) + \cdots+ B_{j_{\gamma}}(\ell)$. Write J!:=j ₁!⋯j _γ ! and $D:=\sum_{i} |J_{2}^{i}|$. By aggregating terms one more time we have

$$\prod_{i=1}^\alpha A_{J_2^i}^{h-\ell} =\frac{1}{J_2^1!\cdots J_2^\alpha!} \Biggl( h^{D} + \sum_{i=1}^\alpha B_{J_2^i}(\ell) \cdot h^{D-1} + \mbox{lower deg. terms} \Biggr) . $$

It should now be evident that both the h ^D and the h ^D−1 terms in $\prod_{i=1}^{\alpha}\sum_{\ell}a_{2\ell+\sigma} A_{J_{2}^{i}}^{h-\ell}$ and $\sum_{\ell}a_{2\ell+\sigma} \prod_{i=1}^{\alpha}A_{J_{2}^{i}}^{h-\ell}$ are the same, and hence drop off under the difference. The lemma follows.

Appendix B: Proof of Lemma 20

Using only the property that S _lin reproduces Π ₃, by Lemma 19, we already know that for each fixed σ=0 or 1, $\varXi_{i}^{h,\sigma}$ or $\varUpsilon_{i}^{h,\sigma}$ is a quadratic polynomial in h, therefore, so are

$$\varXi_1^{h,\sigma}-\varXi_2^{h,\sigma}, \;\; \varXi_3^{h,\sigma}- \frac{3}{2} \varXi_1^{h,\sigma}, \;\; \varUpsilon_1^{h,\sigma}-\varUpsilon_2^{h,\sigma} . $$

The lemma is proved if we can prove that in each of these three cases, the sequence is actually a single quadratic polynomial sampled at 2h+σ. (Recall Remark 18.) But this is equivalent to showing that

(B.1)

(B.2)

(B.3)

Our goal is to prove that (B.1)–(B.3) hold under the additional assumption that S _lin has a dual time-symmetry.

Preparation

Since S _lim reproduces Π ₃, we have the sum rules

(B.4)

(B.5)

Combining (B.4) with the dual time-symmetry of S _lim, expressed as a _2ℓ+1=a _−2ℓ , we have

$$\sum_{\ell}a_{2\ell+1}\ell=\sum _{\ell}a_{-2\ell}\ell=-\sum_{\ell}a_{2\ell} \ell = -\sum_{\ell}a_{2\ell+1} \biggl( \ell+ \frac{1}{2} \biggr) = -\frac {1}{2}-\sum _{\ell}a_{2\ell+1}\ell , $$

from which we obtain the identity

$$ \sum_{\ell}a_{2\ell+1} \ell=-\frac{1}{4}. $$

(B.6)

We compute as follows using the sum rules:

This gives the identity

$$ \sum_{\ell}a_{2\ell+1} \ell^3=-\frac{3}{4}\sum_{\ell}a_{2\ell +1} \ell^2+\frac{1}{32}. $$

(B.7)

Finally use Eqs. (B.4)–(B.7) to obtain the next three identities:

$$ \sum_\ell a_{2\ell+1}(h-\ell)=h+\frac{1}{4}, $$

(B.8)

$$ \sum_\ell a_{2\ell+1}(h-\ell)^2=h^2-2h\sum _\ell a_{2\ell+1}\ell +\sum _\ell a_{2\ell+1}\ell^2 =h^2+ \frac{1}{2}h+\sum_\ell a_{2\ell+1} \ell^2, $$

(B.9)

and

(B.10)

We are now ready to prove (B.1)–(B.3).

Proof of (B.1)

By definition,

Hence,

Therefore,

$$ \varXi_1^{h+\frac{1}{2},0}- \varXi_1^{h,1}=\frac{1}{2} \biggl[\sum _\ell a_{2\ell+1}(h-\ell)\sum _\ell a_{2\ell+1}(h-\ell)^2-\sum _\ell a_{2\ell+1}(h-\ell)^3 \biggr]. $$

(B.11)

Similarly, we have

$$ \varXi_2^{h+\frac{1}{2},0}- \varXi_2^{h,1}= \biggl(h+\frac{1}{4} \biggr) \biggl[ \biggl(\sum_\ell a_{2\ell+1}(h-\ell) \biggr)^2-\sum_\ell a_{2\ell +1}(h- \ell)^2 \biggr]. $$

(B.12)

Substituting (B.8)–(B.10) into (B.11) and (B.12), yields the two identities

(B.13)

Hence

$$\varXi_1^{h+\frac{1}{2},0}-\varXi_1^{h,1}= \varXi_2^{h+\frac{1}{2},0}-\varXi_2^{h,1} $$

and (B.1) is proved.

Proof of (B.2)

By definition,

Hence,

Therefore,

$$\varXi_3^{h+\frac{1}{2},0}-\varXi_3^{h,1}= \frac{1}{2} \biggl[ \biggl(\sum_\ell a_{2\ell+1}(h-\ell) \biggr)^3-\sum _\ell a_{2\ell+1}(h-\ell)^3 \biggr] . $$

Substituting (B.8) and (B.10) into the above equality, we have

$$\varXi_3^{h+\frac{1}{2},0}-\varXi_3^{h,1}= \frac{3}{2} \biggl(h+\frac {1}{4} \biggr) \biggl(\frac{1}{16}- \sum_\ell a_{2\ell+1}\ell^2 \biggr) . $$

Combining with (B.13) yields

$$\varXi_3^{h+\frac{1}{2},0}-\varXi_3^{h,1}= \frac{3}{2} \bigl(\varXi_1^{h+\frac{1}{2},0}-\varXi_1^{h,1} \bigr) . $$

which proves (B.2).

Proof of (B.3)

By definition of ϒ _i with k=4,

Hence,

Therefore,

$$ \begin{aligned}[b] \varUpsilon_1^{h+\frac{1}{2},0}- \varUpsilon_1^{h,1}&=\frac{1}{4} \biggl[\sum _\ell a_{2\ell+1}(h-\ell)\sum _\ell a_{2\ell+1}(h-\ell) (h-\ell -1) \\ &\quad {}-\sum _\ell a_{2\ell+1}(h-\ell)^2(h-\ell-1) \biggr]. \end{aligned}$$

(B.14)

Similarly, we have

$$ \varUpsilon_2^{h+\frac{1}{2},0}- \varUpsilon_2^{h,1}=\frac{1}{2} \biggl(h- \frac{1}{4} \biggr) \biggl[ \biggl(\sum_\ell a_{2\ell+1}(h-\ell) \biggr)^2-\sum _\ell a_{2\ell+1}(h-\ell)^2 \biggr]. $$

(B.15)

Substituting (B.8)–(B.10) into (B.14) and (B.15) yields

$$\varUpsilon_1^{h+\frac{1}{2},0}-\varUpsilon_1^{h,1}= \frac{1}{2} \biggl(h-\frac{1}{4} \biggr) \biggl(\frac{1}{16}- \sum_\ell a_{2\ell+1}\ell^2 \biggr), $$

$$\varUpsilon_2^{h+\frac{1}{2},0}-\varUpsilon_2^{h,1}= \frac{1}{2} \biggl(h-\frac{1}{4} \biggr) \biggl(\frac{1}{16}- \sum_\ell a_{2\ell+1}\ell^2 \biggr). $$

Hence,

$$\varUpsilon_1^{h+\frac{1}{2},0}-\varUpsilon_1^{h,1}= \varUpsilon_2^{h+\frac {1}{2},0}-\varUpsilon_2^{h,1}, $$

and (B.3) is proved.