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# 数学代写|随机分析代写Stochastic Calculus代考|The Diffeomorphism Group Example

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## 数学代写|随机分析代写Stochastic Calculus代考|The Diffeomorphism Group Example

If $M$ is a compact smooth manifold and $X$ is smooth we may consider an equation on the space of smooth diffeomorphisms $\operatorname{Diff}(M)$. Define $\tilde{X}(f)(x)=X(f(x))$ and $\tilde{X}0(f)(x)=X_0(f(x))$ and consider the $\operatorname{SDE}$ on $\operatorname{Diff}(M)$ : $$\mathrm{d} f_t=\tilde{X}\left(f_t\right) \circ \mathrm{d} B_t+\tilde{X}_0\left(f_t\right) \mathrm{d} t$$ with $f_0(x)=x$. Then, $f_t(x)$ is solution to $\mathrm{d} x_t=X\left(x_t\right) \circ \mathrm{d} B_t$ with initial point $x$. Fix $x_0 \in M$, we have a map $\theta: \operatorname{Diff}(M) \rightarrow M$ given by $\theta(f)=f\left(x_0\right)$. Let $\mathcal{B}=\frac{1}{2} L{\tilde{X}i} L{\tilde{X}i}$ and $\mathcal{A}=\frac{1}{2} L{X_i} L_{X_i}$. Then,
$$h_f(v)(x)=\tilde{X}(f)\left(Y\left(f\left(x_0\right)\right) v\right)(x)=X(f(x))\left(Y\left(f\left(x_0\right)\right) v\right) .$$

Consider the polar coordinates in $\mathbf{R}^n$, with the origin removed. Consider the conditional expectation of a Brownian motion $W_t$ on $\mathbf{R}^n$ on $\left|W_t\right|$ where $\left|W_t\right|$, and $n$-dimensional Bessel Process, $n>1$, lives in $\mathbf{R}_{+}$. For $n=2$, we are in the situation that $p: \mathbf{R}^2 \rightarrow \mathbf{R}$ given by $p:(r, \theta) \mapsto r$. The $\mathcal{B}$ and $\mathcal{A}$ diffusion are the Laplacians, $\mathcal{A}^H=\frac{\partial^2}{\partial r^2}$. The map $p(r, \theta)=r^2$ would result the lifting map $v \frac{\partial}{\partial x} \mapsto\left(\frac{v}{2 r}, 0\right)=\frac{v}{2 r} \frac{\partial}{\partial r}$

At this stage, we note that if $B_t$ is a one dimensional Brownian motion, $l_t$ the local time at 0 of $B_t$ and $Y_t=\left|B_t\right|+\ell_t$, a 3-dimensional Bessel process starting from 0 . There is the following beautiful result of Pitman:
$$E\left{f\left(\left|B_t\right|\right) \mid \sigma\left(Y_s: s \leq t\right)\right}=\int_0^1 f\left(x Y_t\right) \mathrm{d} x=V f\left(Y_t\right)$$
where $V$ is the Markov kernel: $V(x, \mathrm{~d} z)=\frac{\mathbf{1}_{0 \leq z \leq x}}{x} \mathrm{~d} z[2,21]$.
A second example, [11], which demonstrates the twist effect is on the product space of the circle. Let $p: S^1 \times S^1 \rightarrow S^1$ be the projection on the first factor. For $0<\alpha<\frac{\pi}{4}$, define the diffusion operator on $S^1 \times S^1$ :
$$\mathcal{B}=\frac{1}{2}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+\tan \alpha \frac{\partial^2}{\partial x \partial y}$$
and the diffusion operator $\mathcal{A}=\frac{1}{2} \frac{\partial^2}{\partial x^2}$ on $S^1$. Then,
\begin{aligned} \mathcal{B}^V & =\frac{1}{2}\left(1-(\tan \alpha)^2\right) \frac{\partial^2}{\partial y^2} \ \mathcal{A}^H & =\frac{1}{2}\left(\frac{\partial^2}{\partial x^2}+(\tan \alpha)^2 \frac{\partial^2}{\partial y^2}\right)+\tan \alpha \frac{\partial^2}{\partial x \partial y} \end{aligned}

## 数学代写|随机分析代写Stochastic Calculus代考|Parallel Translation

The horizontal lift map $u_t$ can also be thought of solutions to:
$$\mathrm{d} u_t=\sum H\left(e_i\right)\left(u_t\right) \circ \mathrm{d} \sigma_t$$
In fact, if $\dot{v}t$ is the horizontal lift of $\dot{\sigma}_t, \dot{v}_t=\sum{i=1}^n\left\langle\dot{\sigma}t, e_i\right\rangle H\left(e_i\right)\left(\tilde{\sigma}_t\right)$. Note that, $/ / t(\sigma)$ is not a solution to a Markovian equation, the pair $\left(/ / t(\sigma), u_t\right)$ is. In local coordinates for $v_t^i$ the ith component of $/ / t(\sigma)(v), v \in T{\sigma_0} M$,
$$\mathrm{d} v_t^k=-\Gamma_{i, j}^k\left(\sigma_t\right) v_t^j \circ \mathrm{d} \sigma_t^i .$$
If $\sigma_t$ is the solution of the $\operatorname{SDE~} \mathrm{d} x_t^k=X_i^k\left(x_t\right) \circ \mathrm{d} B_t^i+X_0^k\left(x_t\right) \mathrm{d} t$, then
$$\mathrm{d} v_t^k=-\Gamma_{i, j}^k\left(x_t\right) v_t^j X_i^k\left(x_t\right) \circ \mathrm{d} B_t^i-\Gamma_{i, j}^k\left(x_t\right) v_t^j X_0^k\left(x_t\right) \mathrm{d} t$$

## 数学代写|随机分析代写Stochastic Calculus代考|The Diffeomorphism Group Example

$$h_f(v)(x)=\tilde{X}(f)\left(Y\left(f\left(x_0\right)\right) v\right)(x)=X(f(x))\left(Y\left(f\left(x_0\right)\right) v\right) .$$

$$E\left{f\left(\left|B_t\right|\right) \mid \sigma\left(Y_s: s \leq t\right)\right}=\int_0^1 f\left(x Y_t\right) \mathrm{d} x=V f\left(Y_t\right)$$

$$\mathcal{B}=\frac{1}{2}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+\tan \alpha \frac{\partial^2}{\partial x \partial y}$$

\begin{aligned} \mathcal{B}^V & =\frac{1}{2}\left(1-(\tan \alpha)^2\right) \frac{\partial^2}{\partial y^2} \ \mathcal{A}^H & =\frac{1}{2}\left(\frac{\partial^2}{\partial x^2}+(\tan \alpha)^2 \frac{\partial^2}{\partial y^2}\right)+\tan \alpha \frac{\partial^2}{\partial x \partial y} \end{aligned}

## 数学代写|随机分析代写Stochastic Calculus代考|Parallel Translation

$$\mathrm{d} u_t=\sum H\left(e_i\right)\left(u_t\right) \circ \mathrm{d} \sigma_t$$

$$\mathrm{d} v_t^k=-\Gamma_{i, j}^k\left(x_t\right) v_t^j X_i^k\left(x_t\right) \circ \mathrm{d} B_t^i-\Gamma_{i, j}^k\left(x_t\right) v_t^j X_0^k\left(x_t\right) \mathrm{d} t$$

## MATLAB代写

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