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# 数学代写|随机分析代写Stochastic Calculus代考|In Metric Form

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## 数学代写|随机分析代写Stochastic Calculus代考|In Metric Form

Note that $\sigma^{\mathcal{A}}$ gives rise to a positive definite bilinear form on $T^* M$ :
$$\langle\phi, \psi\rangle_x=\phi(x)\left(\sigma_x^{\mathcal{A}}(\psi(x))\right)$$

and this induces an inner product on $E_x$ :
$$\langle u, v\rangle_x=\left(\sigma_x^{\mathcal{A}}\right)^{-1}(u)(v)$$
For an orthonormal basis $\left{e_i\right}$ of $E_x$, let $e_i^=\left(\sigma_x^{\mathcal{A}}\right)^{-1}\left(e_i\right)$. Then, $e_j^ \sigma^{\mathcal{A}}\left(e_i^\right)=$ $\left(\sigma_x^{\mathcal{A}}\right)^{-1}\left(e_j\right)\left(e_i\right)=\left\langle e_j, e_i\right\rangle$ and hence $$\langle\phi, \psi\rangle_x=\sum_i\left\langle e_j, e_i\right\rangle \phi\left(e_i\right) \psi\left(e_j\right)=\sum_i \phi\left(e_i\right) \psi\left(e_i\right) .$$ Likewise the symbol $\sigma^{\mathcal{A}^H}$ induces an inner product on $T^ N$ with the property that $\langle\phi \circ T p, \psi \circ T p\rangle=\langle\phi, \psi\rangle$ and a metric on $H \subset T N$ which is the same as that induced by $\mathrm{h}$ from $T M$. Note that $\sigma^{\mathcal{B}}=\sigma^{\mathcal{A}^H}+\sigma^{\mathcal{B}^V}$, where $\mathcal{B}^V$ is the vertical part of $\mathcal{B}$, and $\operatorname{Im}\left[\sigma^{\mathcal{B}^V}\right] \cap H={0}$. Let $\mu$ be an invariant measure for $\mathcal{A}^H$ and $\mu_M=p_*(\mu)$ the pushed forward measure which is an invariant measure for $\mathcal{A}$.
If $\mathcal{A}$ is symmetric,
\begin{aligned} \int_M\langle\mathrm{~d} f, \mathrm{~d} g\rangle \mu_M(\mathrm{~d} x) & =\int \sigma^{\mathcal{A}}(\mathrm{d} f, \mathrm{~d} g) \mu_M(\mathrm{~d} x) \ & =\frac{1}{2} \int[\mathcal{A}(f g)-f(\mathcal{A} g)-g(\mathcal{A} f)] \mu_M(\mathrm{~d} x) \ & =-\int_M f \mathcal{A} g \mathrm{~d} \mu_M(x) \end{aligned}

## 数学代写|随机分析代写Stochastic Calculus代考|On the Heisenberg Group

A Lie group is a group $G$ with a manifold structure such that the group multiplication $G \times G \rightarrow G$ and taking inverse are smooth. Its tangent space at the identity $g$ can be identified with left invariant vector fields on $G, X(a)=T L_a X(e)$ and we denote $A^*$ the left invariant vector field with value $A$ at the identity. The tangent space $T_a G$ at $a$ can be identified with $\mathrm{g}$ by the derivative $T L_a$ of the left translation map. Let $\alpha_t=\exp (t A)$ be the solution flow to the left invariant vector field $T L_a A$ whose value at 0 is the identity then it is also the flow for the corresponding right invariant vector field: $\dot{\alpha}s=\left.\frac{\mathrm{d}}{\mathrm{d} t}\right|{t=s} \exp ^{(t-s) A} \exp ^{s A}=T R_{\alpha_s} A$. Then $u_t=a \exp (t A)$ is the solution flow through $a$.

Consider the Heisenberg group $G$ whose elements are $(x, y, z) \in \mathbf{R}^3$ with group product
$$\left(x_1, y_1, z_1\right)\left(x_2, y_2, z_2\right)=\left(x_1+x_2, y_1+y_2, z_1+z_2+\frac{1}{2}\left(x_1 y_2-x_2 y_1\right)\right) .$$
The Lie bracket operation is $\left[(a, b, c),\left(a^{\prime}, b^{\prime}, c^{\prime}\right)\right]=\left(0,0, a b^{\prime}-a^{\prime} b\right)$. Note that for $X, Y \in \mathrm{g}, \mathrm{e}^X \mathrm{e}^Y=\mathrm{e}^{X+Y+\frac{1}{2}[X, Y]}$. If $A=(a, b, c)$, then $A^*=\left(a, b, c+\frac{1}{2}(x b-\right.$ $y a)$ ). Consider the projection $\pi: G \rightarrow \mathbf{R}^2$ where $\pi(x, y, z)=(x, y)$. Let
\begin{aligned} & X_1(x, y, z)=\left(1,0,-\frac{1}{2} y\right), \quad X_2(x, y, z)=\left(0,1, \frac{1}{2} x\right) \ & X_3(x, y, z)=(0,0,-1) \end{aligned}
be the left invariant vector fields corresponding to the standard basis of $\mathrm{g}$. The vector spaces $H_{(x, y, z)}=\operatorname{span}\left{X_1, X_2\right}=\left{\left(a, b, \frac{1}{2}(x b-y a)\right)\right}$ are of rank 2. They are the horizontal tangent spaces associated to the Laplacian $\mathcal{A}=\frac{1}{2}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)$ on $\mathbf{R}^2$ and the left invariant Laplacian $\mathcal{B}:=\frac{1}{2} \sum_{i=1}^3 L_{X_i} L_{X_i}$ on $G$. The vertical tangent space is ${(0,0, c)}$, and there is a horizontal lifting map from $T_{(x, y)} \mathbf{R}^2$ :
$$h_{(x, y, z)}(a, b)=\left(a, b, \frac{1}{2}(x b-y a)\right)$$

## 数学代写|随机分析代写Stochastic Calculus代考|In Metric Form

$$\langle\phi, \psi\rangle_x=\phi(x)\left(\sigma_x^{\mathcal{A}}(\psi(x))\right)$$

$$\langle u, v\rangle_x=\left(\sigma_x^{\mathcal{A}}\right)^{-1}(u)(v)$$

\begin{aligned} \int_M\langle\mathrm{~d} f, \mathrm{~d} g\rangle \mu_M(\mathrm{~d} x) & =\int \sigma^{\mathcal{A}}(\mathrm{d} f, \mathrm{~d} g) \mu_M(\mathrm{~d} x) \ & =\frac{1}{2} \int[\mathcal{A}(f g)-f(\mathcal{A} g)-g(\mathcal{A} f)] \mu_M(\mathrm{~d} x) \ & =-\int_M f \mathcal{A} g \mathrm{~d} \mu_M(x) \end{aligned}

## 数学代写|随机分析代写Stochastic Calculus代考|On the Heisenberg Group

$$\left(x_1, y_1, z_1\right)\left(x_2, y_2, z_2\right)=\left(x_1+x_2, y_1+y_2, z_1+z_2+\frac{1}{2}\left(x_1 y_2-x_2 y_1\right)\right) .$$

## MATLAB代写

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