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I'm trying to understand a calculation in this paper (equation (3.8)). With some details removed, the setup is as follows.

Let $G$ be a Lie group, and $g(t)$ a curve in $G$ satisfying the SDE

$$dg(t) = T_e L_{g(t)}H_1 dW_t + T_e L_{g(t)}H_2 dt + T_e L_{g(t)}u(t) dt,$$ where $H_1,H_2$ are elements of the Lie algebra, $u(t)$ is a curve in the Lie algebra, and $W_t$ is a real-valued brownian motion. The SDE / stochastic integrals are in the Stratonovich sense. My knowledge of processes and SDE's on manifolds are from Elton Hsu's book. According to the construction in this book, a solution to an equation such as the above is defined by being a solution to the following euclidean SDE, for every smooth real-valued function f on G,

$$df(g(t)) = (T_e L_{g(t)}H_1)f\hspace{1mm} dW_t + (T_e L_{g(t)}H_2)f\hspace{1mm}dt + (T_e L_{g(t)}u(t))f \hspace{1mm} dt.$$

In the paper, they then consider a sort of perturbation of the curve g; $g(t)e_{\epsilon}(t)$, where e_{\epsilon}(t) is another curve in G. I don't think it is important to my question, but $e_\epsilon (t)$ is defined by a set of ODE's, equation (3.2) in the paper.

They then state that 'by Itö's formula' $g_\epsilon(t) := g(t) e_\epsilon(t)$ fulfills the SDE

$$dg_\epsilon(t) = T_e L_{g_\epsilon(t)}Ad_{e_{\epsilon}^{-1}(t)}(H_1) dW_t + T_e L_{g_\epsilon(t)}Ad_{e_{\epsilon}^{-1}(t)}H_2 dt + T_e L_{g_\epsilon(t)}Ad_{e_{\epsilon}^{-1}(t)}u(t) dt + T_e L_{g_\epsilon(t)} (T_{e_\epsilon (t)} L_{e^{-1}_\epsilon(t)}\dot{e}_\epsilon(t) dt $$

which (my calculation) is equivalent to

$$d(g(t) e_\epsilon (t)) = T_e (L_{g(t)}\circ R_{e_\epsilon (t)})H_1\hspace{1mm} dW_t + T_e (L_{g(t)}\circ R_{e_\epsilon (t)})H_2\hspace{1mm}dt + T_e (L_{g(t)}\circ R_{e_\epsilon (t)})u(t) \hspace{1mm} dt + T_{e_\epsilon (t)} L_{g(t)} \dot{e}_\epsilon(t) dt\hspace{1mm}$$

I haven't been able to verify this, or find hints elsewhere. I have a feeling that I'm missing (or lack knowledge of) something very simple. Any hints would be greatly appreciated. Even if only for the deterministic case ($H_1 = 0$).

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1 Answer 1

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To understand (3.8), I think it helps to write down (2.4) and (3.2) from that paper. From (3.2), note that $e_{\epsilon}(t)$ satisfies the ODE $$ \dot e_{\epsilon}(t) = \epsilon T L_{e_{\epsilon}(t)} \dot v(t) \;. \tag{1} $$ By the Euclidean Itô-Stratonovich conversion formula applied in the Lie algebra, note from (2.4) that $g(t)$ satisfies the Itô SDE $$ d g(t) = T L_{g(t)} \left( \sum_i H_i dW^i_t + u(t) dt \right) \;. \tag{2} $$

Note also that the product rule for Itô processes applied to $g_{\epsilon}(t) = g(t) e_{\epsilon}(t)$ reduces to the standard product rule because $e_{\epsilon}(t)$ has finite variation. Hence, from (1), (2), and the product rule, \begin{align} d g_{\epsilon}(t) &= T R_{e_{\epsilon}(t)} d g(t) + T L_{g(t)} \dot e_{\epsilon}(t) \\ &= T L_{g_{\epsilon}(t)} \left( \sum_{i} \operatorname{Ad}_{e_{\epsilon}(t)^{-1}} H_i dW^i_t + \operatorname{Ad}_{e_{\epsilon}(t)^{-1}} u(t) dt \right) + \epsilon T L_{g_{\epsilon}(t)} \dot v \tag{3} \end{align} where we used the identities \begin{align} T R_{e_{\epsilon}(t)} \; T L_{g(t)} = T L_{g(t)} \; T R_{\epsilon(t)} = T L_{g_{\epsilon}(t)} \; \operatorname{Ad}_{e_{\epsilon}(t)^{-1}} \;, \qquad T L_{g(t)} \dot e_{\epsilon}(t) = \epsilon T L_{g_{\epsilon}(t)} \dot v(t) \;. \end{align} Up to an Itô-Stratonovich correction term, (3) is exactly (3.8) in that paper.

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  • $\begingroup$ Thanks a lot for your answer. However, I am new to these processes on Lie groups, so there are some things I need to understand before I can accept your answer. I will write my questions in comments below. $\endgroup$ Commented Jul 17, 2019 at 8:45
  • $\begingroup$ 1. I don't understand the notation $dg(t) = TL_{g(t)}(H_i dW_t + u(t) dt)$ (where $H_i$ is an element, and $u(t)$ a curve, in the Lie algebra), which is also used in the paper. Should it be understood as an equation for the differential of $g(t)$, as in $\frac{d}{ds}|_{s=t}\hspace{1mm} g(s) = RHS$? From reading the rest of the paper, I got the impression that it should be read as $g(t) = TL_{g(t)}(H_i) dW_t + TL_{g(t)}(u(t)) dt)$ (i.e. as integrals of the left-invariant vector fields on the manifold). Maybe these are equivalent viewpoints? $\endgroup$ Commented Jul 17, 2019 at 9:16
  • $\begingroup$ I realize now that all of my questions disappear if the interpretation of the LHS of (2) as $\dot{g}(t) = \frac{d}{ds}|_{s=t}g(s)$ is correct. I haven't seen this notation before (identifying stochastic differentials with actual differentials), do you have a reference for this? Or maybe it is specific to the type of equation above. $\endgroup$ Commented Jul 17, 2019 at 9:34
  • $\begingroup$ Lastly, a minor question: you write "Note also that the product rule for Itô processes applied to $g_{\epsilon}(t) = g(t) e_{\epsilon}(t)$ reduces to the standard product rule because $e_{\epsilon}(t)$ has finite variation.". But the Itô product rule is not actually used in your calculation, is it? Only the product rule for the differential in a Lie group, as I understand your calculations. The Itô product rule, and the notion of finite variation of $e_\epsilon (t)$, isn't applicable for $g_\epsilon (t)$ and $e_\epsilon (t)$, since they are processes in the Lie group, right? $\endgroup$ Commented Jul 17, 2019 at 9:39
  • $\begingroup$ Regarding 1, please note that $T L_g$ denotes left multiplication by $g$ on elements of the tangent space of the Lie group $G$. Hence, $T L_{g(t)}(H_i(g(t)) dW_t^i + u(t) dt)$ essentially "maps" $H_i(g(t))$ and $u(t)$ belonging to the Lie algebra, i.e., $T_e G$, into the tangent space of the Lie group at $g(t)$, i.e., $T_{g(t)} G$. The point of doing that is to obtain an SDE on the Lie group by taking a standard one on the Lie algebra and left translating it to obtain one on $G$. The interpretation of the LHS of (2) is the integral version of (2) — exactly as in the Euclidean case. $\endgroup$ Commented Jul 17, 2019 at 10:17

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