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几何学/拓扑学代写 Geometry/Topology III |MATH 8831 Boston College Assignment

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Differential geometry is a branch of mathematics that deals with the study of geometric objects using techniques from calculus, linear algebra, and topology. It provides a framework for studying curves, surfaces, manifolds, and other geometric structures.





One of the key ideas in differential geometry is the concept of a tangent space. The tangent space at a point on a manifold is the space of all possible velocities or directions that can be taken from that point. This concept is essential for understanding how geometric objects change and interact with each other.





Another important concept in differential geometry is the notion of a connection. A connection is a way of measuring the deviation of a curve or surface from being flat. Connections are used to define important geometric quantities such as curvature, which measures how much a curve or surface deviates from being a straight line or plane.





Differential geometry has numerous applications in physics, engineering, and computer science. For example, it is used in the study of general relativity, which describes the behavior of gravity in the presence of curved spacetime. It is also used in computer graphics and robotics to model and control the motion of objects in three-dimensional space.















几何学/拓扑学代写 Geometry/Topology III |MATH 8831 Boston College Assignment













问题 1. 





Let $(J, \omega)$ be a Kähler structure, and let $\beta$ be a holomorphic Poisson structure, so that $\mathcal{J}_B=e^{t Q} \mathcal{J}_J e^{-t Q}$ is integrable for all $t$. What is the condition on $\beta$ which guarantees that $\mathcal{J}_A=e^{t Q} \mathcal{J}_\omega e^{-t Q}$ is integrable for small $t$ ? What are the resulting types of $\left(\mathcal{J}_A, \mathcal{J}_B\right)$ ?





 
 




证明 . 





The condition that guarantees integrability of $\mathcal{J}A$ for small $t$ is that $\beta$ is a holomorphic Poisson bivector field with respect to $\omega$, meaning that $[\beta, \beta]\omega = 0$ and $d^{0,2}\beta = 0$, where $[\cdot,\cdot]_\omega$ is the Schouten-Nijenhuis bracket and $d^{0,2}$ is the $(0,2)$ part of the de Rham differential.





The resulting types of $(\mathcal{J}_A, \mathcal{J}_B)$ depend on the type of $\beta$. If $\beta$ is of type $(1,1)$, then $\mathcal{J}_B$ is a complex structure and $\mathcal{J}_A$ is a deformation of this complex structure. If $\beta$ is of type $(2,0)$ or $(0,2)$, then $\mathcal{J}_B$ is a symplectic structure and $\mathcal{J}_A$ is a deformation of this symplectic structure. If $\beta$ is of mixed type, then $(\mathcal{J}_A, \mathcal{J}_B)$ form a bihermitian structure.






















问题 2. 





Let $\mathcal{J}$ be a generalized complex structure. Show that $e^{\theta \mathcal{J}}\left(T^*\right)$ is a Dirac structure for all $\theta$.





 
 




证明 . 





To show that $e^{\theta\mathcal{J}}(T^*)$ is a Dirac structure, we need to show that it is both isotropic and involutive.





Isotropy: Let $X, Y$ be two vector fields on $T^*$. We want to show that $e^{\theta\mathcal{J}}(X)\cdot e^{\theta\mathcal{J}}(Y)=0$, where $\cdot$ denotes the pairing between vectors and covectors.





Since $\mathcal{J}$ is a generalized complex structure, we know that it satisfies the following conditions:





    

  • $\mathcal{J}^2=-\mathrm{id}$
    • 




  • $\mathcal{J}$ preserves the natural symplectic structure on $T^$, i.e., $\mathcal{J}^ \omega = \omega$, where $\omega$ is the canonical symplectic form on $T^*$.
    • 






Using these conditions, we can compute the pairing between $e^{\theta\mathcal{J}}(X)$ and $e^{\theta\mathcal{J}}(Y)$ as follows:





\begin{align*} e^{\theta\mathcal{J}}(X) \cdot e^{\theta\mathcal{J}}(Y) &= \left\langle e^{\theta\mathcal{J}}(X),, \theta\mathcal{J}e^{\theta\mathcal{J}}(Y)\right\rangle\ &= \theta\left\langle \mathcal{J}e^{\theta\mathcal{J}}(X),, e^{\theta\mathcal{J}}(Y)\right\rangle\ &= \theta\left\langle e^{\theta\mathcal{J}}(\mathcal{J}X),, e^{\theta\mathcal{J}}(Y)\right\rangle\ &= \theta\left\langle \mathcal{J}X,, Y\right\rangle\ &= \theta\omega(X,, Y)\ &= 0, \end{align*}





where the third equality follows from the fact that $\mathcal{J}$ is a linear map and hence commutes with scalar multiplication, and the fourth equality follows from the fact that $e^{\theta\mathcal{J}}$ preserves the symplectic form $\omega$.





Thus, we have shown that $e^{\theta\mathcal{J}}(T^*)$ is isotropic.





Involutive: To show that $e^{\theta\mathcal{J}}(T^)$ is involutive, we need to show that for any two sections $x_1,x_2\in e^{\theta\mathcal{J}}(T^)$, their Lie bracket $[x_1,x_2]$ is also a section of $e^{\theta\mathcal{J}}(T^*)$.


















问题 3. 





What is the T-dual of the trivial $S^1$ bundle over $S^2$ with $H=k \nu$ where $k \in \mathbb{Z}$ and $\nu$ is the generator of $H^3\left(S^1 \times S^2, \mathbb{Z}\right)$ ?





 
 




证明 . 





To find the T-dual of the given bundle, we need to first compute the geometric data of the bundle, which includes the metric $g$ and the $B$-field. Then we apply the rules of T-duality to obtain the dual bundle.





The trivial $S^1$ bundle over $S^2$ can be described by the principal bundle $P=S^1\times S^2$ with the projection map $\pi:P\to S^2$. Let $e_1,e_2,e_3$ be the standard basis vectors of $\mathbb{R}^3$ and $x^i$ be the corresponding coordinates. We can choose the metric $g$ and the $B$-field $B$ on $P$ as follows:





$\begin{gathered}g=d x^2+\sin ^2(\theta) d \phi^2+\cos ^2(\theta) d \theta^2 \ B=k \cos (\theta) d \theta \wedge d \phi\end{gathered}$





where $\theta$ and $\phi$ are the usual spherical coordinates on $S^2$. Note that $g$ is a metric of constant curvature 1 and $B$ is a closed 2-form that represents the generator $\nu$ of $H^3(S^1\times S^2,\mathbb{Z})$.





To apply T-duality, we choose a circle direction to be dualized, say the $\phi$ direction. We replace the circle $S^1$ by its dual circle $\tilde{S}^1$, whose radius is given by the inverse of the original radius $R$. We also replace the $B$-field by its dual $\tilde{B}$, which is given by





$\tilde{B}=-\frac{k}{2 \pi R} \sin (\theta) d \theta \wedge d \phi$





where $\tilde{x}^i$ are the coordinates of $\tilde{P}$, and $e_\phi^i$ is the $\phi$ component of the basis vectors of $\mathbb{R}^3$. The new metric $\tilde{g}$ on $\tilde{P}$ is given by





$\tilde{g}=\frac{R}{\sin (\theta)}\left(d \tilde{x}^2+\cos ^2(\theta) d \tilde{\theta}^2+\sin ^2(\theta) d \tilde{\phi}^2\right)$





where $\tilde{\theta}$ and $\tilde{\phi}$ are the coordinates of $\tilde{S}^2$. Note that $\tilde{g}$ is again a metric of constant curvature 1. Therefore, the T-dual of the trivial $S^1$ bundle over $S^2$ with $H=k\nu$ is the trivial $\tilde{S}^1$ bundle over




















这是一份2023年的波士顿学院Boston College MATH 8831几何学/拓扑学代写的成功案例




























































































































































			

	
	



		
			


	

		
几何学/拓扑学代写 Geometry/Topology IV|MATH 8832 Boston College Assignment


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Hyperbolic geometry was discovered independently by Nikolai Lobachevsky, János Bolyai, and Carl Friedrich Gauss in the early 19th century. It has important applications in fields such as art, architecture, and physics.





In hyperbolic geometry, the sum of the angles in a triangle is always less than 180 degrees, and the area of a triangle is proportional to its excess angle. This leads to some unusual and counterintuitive properties, such as the fact that circles and lines behave differently than in Euclidean geometry. For example, in hyperbolic geometry, circles have negative curvature and get smaller as you move away from the center, while lines have positive curvature and get farther apart as you move away from them.





Hyperbolic geometry has also been used to model certain types of curved spaces, such as the surface of a saddle or a hyperbolic plane. These models have been used in computer graphics, virtual reality, and other applications.















几何学/拓扑学代写 Geometry/Topology IV|MATH 8832 Boston College Assignment













问题 1. 





Consider the parametrized surface $$ \mathbf{r}(u, v)=\left(u-\frac{u^3}{3}+u v^2, v-\frac{v^3}{3}+v u^2, u^2-v^2\right) . $$ Show that (a) The coefficients of the first fundamental form are $$ E=G=\left(1+u^2+v^2\right)^2, F=0 . $$





 
 




证明 . 





To find the coefficients of the first fundamental form, we need to compute the dot products of the partial derivatives of the surface with respect to the parameters $u$ and $v$:





\begin{align*} \mathbf{r}_u &= \left(1-u^2+v^2, 2uv, 2u\right) \ \mathbf{r}_v &= \left(2uv, 1-v^2+u^2, -2v\right) \end{align*}





Then, we can compute the coefficients using the formulas:





\begin{align*} E &= \mathbf{r}_u \cdot \mathbf{r}_u = \left(1-u^2+v^2\right)^2 + 4u^2v^2 + 4u^2 \ F &= \mathbf{r}_u \cdot \mathbf{r}_v = 2u^2v – 2uv^3 + 2uv \ G &= \mathbf{r}_v \cdot \mathbf{r}_v = 4u^2v^2 + \left(1-v^2+u^2\right)^2 + 4v^2 \end{align*}





To simplify these expressions, we can use the identity $a^2-b^2 = (a+b)(a-b)$:





\begin{align*} E &= \left(1+u^2+v^2\right)^2 \ F &= 2uv\left(u^2-v^2+1\right) = 0 \ G &= \left(1+u^2+v^2\right)^2 \end{align*}





Therefore, we have $E=G=\left(1+u^2+v^2\right)^2$ and $F=0$, which confirms the result.






















问题 2. 





Consider the parametrized surface $$ \mathbf{r}(u, v)=\left(u-\frac{u^3}{3}+u v^2, v-\frac{v^3}{3}+v u^2, u^2-v^2\right) $$ (b) The coefficients of the second fundamental form are $$ L=2, M=-2, N=0 . $$





 
 




证明 . 





To find the coefficients of the second fundamental form, we first need to compute the first fundamental form and the normal vector.





The first fundamental form is given by:





$\begin{aligned} & E=\mathbf{r}_u \cdot \mathbf{r}_u=\left(1-u^2+v^2\right)^2+4 u^2 v^2 \ & F=\mathbf{r}_u \cdot \mathbf{r}_v=\mathbf{r}_v \cdot \mathbf{r}_u=2 u v\left(1-u^2-v^2\right) \ & G=\mathbf{r}_v \cdot \mathbf{r}_v=\left(1+u^2-v^2\right)^2+4 u^2 v^2\end{aligned}$





The normal vector is given by:





$\mathbf{N}=\frac{\mathbf{r}_u \times \mathbf{r}_v}{\left|\mathbf{r}_u \times \mathbf{r}_v\right|}=\frac{1}{\sqrt{2\left(1+u^2+v^2\right)}}\left(-2 u,-2 v, 1-u^2-v^2\right)$





Now, we can compute the coefficients of the second fundamental form:





$\begin{aligned} & L=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial u^2}=\frac{4 u^2 v^2-2\left(1-u^2+v^2\right)^2}{2\left(1+u^2+v^2\right)^{3 / 2}}=2 \ & M=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial u \partial v}=\frac{2 u v\left(1-u^2-v^2\right)}{2\left(1+u^2+v^2\right)^{3 / 2}}=-2 \ & N=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial v^2}=\frac{4 u^2 v^2-2\left(1+u^2-v^2\right)^2}{2\left(1+u^2+v^2\right)^{3 / 2}}=0\end{aligned}$





Therefore, the coefficients of the second fundamental form are $L=2$, $M=-2$, and $N=0$.


















问题 3. 





(c) The principal curvatures are $$ \kappa_1=\frac{2}{\left(1+u^2+v^2\right)^2}, \quad \kappa_2=-\frac{2}{\left(1+u^2+v^2\right)^2} . $$





 
 




证明 . 





(c) The principal curvatures are $$ \kappa_1=\frac{2}{\left(1+u^2+v^2\right)^2}, \quad \kappa_2=-\frac{2}{\left(1+u^2+v^2\right)^2} . $$





$\mathbf{n}=\frac{\mathbf{r}_u \times \mathbf{r}_v}{\left|\mathbf{r}_u \times \mathbf{r}_v\right|}=\frac{\left(2 u-u^3+v^2,-2 v+v^3-u^2, 2 u v\right)}{\sqrt{4 u^2 v^2\left(1+u^2+v^2\right)}}$





The Gaussian curvature $K$ and mean curvature $H$ are given by





\begin{align*} K &= \frac{LN – M^2}{EG – F^2} = -\frac{2}{(1+u^2+v^2)^4} \ H &= \frac{1}{2}(L + N) = \frac{2u(2-3u^2-v^2) + 2v(2-3v^2-u^2)}{2(1+u^2+v^2)^{5/2}}. \end{align*}





The principal curvatures $\kappa_1$ and $\kappa_2$ are the solutions to the characteristic equation





$\operatorname{det}\left(\begin{array}{cc}E-\kappa & F \ F & G-\kappa\end{array}\right)=0$.





Substituting in the coefficients above, we get





$\operatorname{det}\left(\begin{array}{cc}1-(\kappa-E) & 2 u v\left(1-u^2-v^2\right) \ 2 u v\left(1-u^2-v^2\right) & 1-(\kappa-G)\end{array}\right)=0$





Simplifying, we get





$(1-\kappa)^2-4 u^2 v^2\left(1+u^2+v^2\right)^{-4}=0$.




























这是一份2023年的波士顿学院Boston College MATH 8832几何学/拓扑学代写的成功案例




























































































































































			

	
	



		
			


	

		
几何学/拓扑学代写 Geometry/Topology I|MATH 8808 Boston College Assignment


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Topology is a branch of mathematics that studies the properties of spaces and their relationships. In topology, the emphasis is on the geometric properties that are preserved under continuous transformations, such as stretching, bending, and twisting.





A topological space is a set of points, along with a collection of open sets that satisfy certain axioms. These open sets are subsets of the space that are considered “open” because they contain the points around them. The open sets form the basis for defining concepts like continuity, convergence, and connectedness.





Topology is used in many areas of mathematics and science, including geometry, analysis, algebraic topology, and computer science. It has applications in physics, engineering, and other fields where understanding the shape and structure of spaces is important.










几何学/拓扑学代写 Geometry/Topology I|MATH 8808 Boston College Assignment













问题 1. 





Let $X, Y \in C^{\infty}(T)$ and $\pi \in C^{\infty}\left(\wedge^2 T\right)$, so that, in a coordinate patch with coordinates $x_i$, we have $X=X^i \frac{\partial}{\partial x^i}, Y=Y^i \frac{\partial}{\partial x^i}$ and $\pi=\pi^{i j} \frac{\partial}{\partial x^i} \wedge \frac{\partial}{\partial x^j}$. Compute $[X, Y],[\pi, X]$, and $[\pi, \pi]$ in coordinates.





 
 




证明 . 





We begin by computing the Lie bracket $[X,Y]$ of the vector fields $X$ and $Y$: \begin{align*} [X,Y] &= X(Y) – Y(X) \ &= X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial}{\partial x^j} – Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial}{\partial x^j} \ &= \left(X^i\frac{\partial Y^j}{\partial x^i} – Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}. \end{align*} So, in coordinates, we have





$[X, Y]=\left(X^i \frac{\partial Y^j}{\partial x^i}-Y^i \frac{\partial X^j}{\partial x^i}\right) \frac{\partial}{\partial x^j}$.





Next, we compute the Lie bracket $[\pi, X]$ of the 2-form $\pi$ and the vector field $X$. We have \begin{align*} [\pi,X] &= \pi(X,\cdot) \ &= \pi^{ij}\frac{\partial X^k}{\partial x^i}\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k} \ &= \pi^{ij}\frac{\partial X^k}{\partial x^i}\frac{\partial}{\partial x^j k}. \end{align*} So, in coordinates, we have





$[\pi, X]=\pi^{i j} \frac{\partial X^k}{\partial x^i} \frac{\partial}{\partial x^j k}$





Finally, we compute the Lie bracket $[\pi,\pi]$ of the 2-form $\pi$ with itself. We have \begin{align*} [\pi,\pi] &= 2\pi\wedge\pi \ &= 2\pi^{ij}\pi^{kl}\frac{\partial}{\partial x^i}\wedge\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k}\wedge\frac{\partial}{\partial x^l} \ &= 2\left(\pi^{ij}\pi^{kl}-\pi^{il}\pi^{kj}\right)\frac{\partial}{\partial x^i}\wedge\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k}\wedge\frac{\partial}{\partial x^l}. \end{align*} So, in coordinates, we have





$[\pi, \pi]=2\left(\pi^{i j} \pi^{k l}-\pi^{i l} \pi^{k j}\right) \frac{\partial}{\partial x^i} \wedge \frac{\partial}{\partial x^j} \wedge \frac{\partial}{\partial x^k} \wedge \frac{\partial}{\partial x^l}$.


















问题 2. 





Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.





 
 




证明 . 





To show that $S^4$ has no symplectic structure, we can use the fact that every closed 2-form on a compact, connected 4-manifold $M$ satisfies $[\omega]^2 = 2\chi(M)$, where $[\omega]$ denotes the cohomology class of the 2-form $\omega$, and $\chi(M)$ is the Euler characteristic of $M$. In particular, if $M$ admits a symplectic structure, then $[\omega]$ is a nonzero multiple of the generator of $H^2(M;\mathbb{Z})$, and so $[\omega]^2$ is positive. However, we have $\chi(S^4) = 2$, so any closed 2-form on $S^4$ must satisfy $[\omega]^2 = 4$. This is a contradiction, so $S^4$ cannot admit a symplectic structure.





To show that $S^2 \times S^4$ has no symplectic structure, we can use the fact that the product of two symplectic manifolds is symplectic. If $S^2 \times S^4$ had a symplectic structure, then in particular both factors $S^2$ and $S^4$ would admit symplectic structures. However, we just showed that $S^4$ does not admit a symplectic structure, so $S^2 \times S^4$ cannot admit a symplectic structure either.














问题 3. 





Write the Poisson bracket $\{f, g\}$ in coordinates for $\pi=\pi^{i j} \frac{\partial}{\partial x^i} \wedge$ $\frac{\partial}{\partial x^3}$





 
 




证明 . 





The Poisson bracket of two functions $f$ and $g$ is defined as:





$${f, g} = \pi(df,dg) = \frac{\partial f}{\partial x^i}\pi^{ij}\frac{\partial g}{\partial x^j}-\frac{\partial g}{\partial x^i}\pi^{ij}\frac{\partial f}{\partial x^j}$$





where $\pi^{ij}$ are the components of the Poisson bivector $\pi$.





In coordinates, we have $\pi=\pi^{ij}\frac{\partial}{\partial x^i}\wedge \frac{\partial}{\partial x^j}=\pi^{12}\frac{\partial}{\partial x^1}\wedge \frac{\partial}{\partial x^2}+\pi^{13}\frac{\partial}{\partial x^1}\wedge \frac{\partial}{\partial x^3}+\pi^{23}\frac{\partial}{\partial x^2}\wedge \frac{\partial}{\partial x^3}$, where $\pi^{ij}$ are functions of the coordinates $x^1,x^2,x^3$.





Using this expression for $\pi$, we can compute the Poisson bracket ${f, g}$ for any functions $f$ and $g$.




















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几何学/拓扑学代写 Geometry/Topology II|MATH 8809 Boston College Assignment


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Assignment-daixieTM为您提供波士顿学院Boston College MATH 8809 Geometry/Topology II几何学/拓扑学代写代考和辅导服务！










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Geometry is a branch of mathematics that deals with the study of shapes, sizes, positions, and properties of objects in space. It encompasses a wide range of topics, including points, lines, planes, angles, polygons, circles, spheres, and more.





Geometry has practical applications in many fields, including engineering, architecture, art, astronomy, physics, and even computer graphics. It plays an important role in understanding the relationships between objects in the physical world and is used to solve real-world problems.





The study of geometry involves using logic and reasoning to make deductions and proofs about the properties of shapes and objects. The principles of geometry have been studied and applied for thousands of years, and continue to be relevant today in a variety of fields.










几何学/拓扑学代写 Geometry/Topology II|MATH 8809 Boston College Assignment













问题 1. 





Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.





 
 




证明 . 





To show that $S^4$ has no symplectic structure, we can use the following theorem:





Theorem: Any closed oriented manifold $M$ with $H^2(M;\mathbb{Z})=0$ has no symplectic structure.





Proof: Suppose $M$ has a symplectic structure. Then by the non-degeneracy of the symplectic form, we have $H^2(M;\mathbb{R})\cong H^2_{dR}(M;\mathbb{R})\neq 0$. But by Poincaré duality and the Universal Coefficient Theorem, we have $H^2(M;\mathbb{R})\cong H^2(M;\mathbb{Z})\otimes\mathbb{R}\cong \operatorname{Hom}(H_2(M;\mathbb{Z}),\mathbb{R})\cong H^2(M;\mathbb{Z})$ since $M$ is closed and oriented. Therefore, $H^2(M;\mathbb{Z})\neq 0$, which is a contradiction. Hence, $M$ has no symplectic structure.





Now, $H^2(S^4;\mathbb{Z})=0$ since $S^4$ is simply connected and $H^2(S^4;\mathbb{Z})$ is the abelianization of $\pi_1(S^4)$, which is trivial. Therefore, by the above theorem, $S^4$ has no symplectic structure.





To show that $S^2 \times S^4$ has no symplectic structure, we can use the following lemma:





Lemma: If $M$ and $N$ are symplectic manifolds of dimensions $2m$ and $2n$, respectively, then $M\times N$ has a symplectic structure of dimension $2(m+n)$.





Proof: Let $\omega_M$ and $\omega_N$ be symplectic forms on $M$ and $N$, respectively. Then the product form $\omega_M\oplus \omega_N$ is a symplectic form on $M\times N$ of dimension $2(m+n)$.





Since $S^4$ has no symplectic structure, it follows that $S^2 \times S^4$ has no symplectic structure.


















问题 2. 





Describe Hamiltonian flow in the symplectic manifold $T^* M$ by the Hamiltonian $H=\pi^* f$, where $\pi: T^* M \longrightarrow M$ is the natural projection and $f \in C^{\infty}(M)$. Also, show that a coordinate chart $U \subset M$ determines a system of $n$ independent, commuting Hamiltonians on $T^* U \subset T^* M$.





 
 




证明 . 





In the symplectic manifold $T^*M$, a Hamiltonian flow is a flow generated by a Hamiltonian function $H:T^*M \rightarrow \mathbb{R}$ via the Hamiltonian vector field $X_H$ defined as:





$X_H=\sum_{i=1}^n\left(\frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i}-\frac{\partial H}{\partial q_i} \frac{\partial}{\partial p_i}\right)$





where $(q_1,\ldots,q_n,p_1,\ldots,p_n)$ are local coordinates on $T^*M$.





Given a function $f \in C^{\infty}(M)$, we can define a Hamiltonian function on $T^*M$ by pulling back $f$ via the natural projection $\pi:T^*M\rightarrow M$, i.e., $H = \pi^*f$. The Hamiltonian vector field generated by $H$ is given by:





$X_H=\sum_{i=1}^n\left(\frac{\partial f}{\partial q_i} \frac{\partial}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial}{\partial q_i}\right)$





which generates the flow:





$\frac{d q_i}{d t}=\frac{\partial H}{\partial p_i}=\frac{\partial f}{\partial p_i} \quad \frac{d p_i}{d t}=-\frac{\partial H}{\partial q_i}=-\frac{\partial f}{\partial q_i}$





This is Hamilton’s equations of motion.





Now, consider a coordinate chart $U \subset M$. Let $(q_1,\ldots,q_n)$ be local coordinates on $U$. We can then define a local coordinate chart on $T^*U$ by taking $(q_1,\ldots,q_n,p_1,\ldots,p_n)$ as coordinates on $T^*U$. Since $\pi|{T^*U}:T^*U\rightarrow U$ is a diffeomorphism, we can pull back functions from $U$ to $T^*U$ via the inverse of this map. In particular, we can define Hamiltonian functions on $T^U$ by pulling back functions from $U$, i.e., for $f\in C^{\infty}(U)$, we can define $H_f=\pi^|{T^*U} f:T^*U\rightarrow \mathbb{R}$.





The Hamiltonian vector field generated by $H_f$ is given by:





$X_{H_f}=\sum_{i=1}^n\left(\frac{\partial f}{\partial q_i} \frac{\partial}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial}{\partial q_i}\right)$





which is the same as the Hamiltonian vector field generated by $\pi^*f$ on $T^*M$, but restricted to $T^*U$. Note that the restriction of a Hamiltonian vector field to a submanifold is still a Hamiltonian vector field with the same Hamiltonian.














问题 3. 





Let $\rho \in \Omega^{\bullet}(G)$ be a left-invariant form on a Lie group $G$. Show that $d \rho=0$ if $\rho$ is also right-invariant.





 
 




证明 . 





To show that $d \rho = 0$ for a left-invariant and right-invariant form $\rho$, we will use the Cartan’s magic formula:





$\mathcal{L}_X=d \circ i_X+i_X \circ d$,





where $\mathcal{L}_X$ is the Lie derivative with respect to the vector field $X$, $i_X$ is the interior product with $X$, and $d$ is the exterior derivative. Since $\rho$ is left-invariant, we have that $\mathcal{L}_X \rho = 0$ for all left-invariant vector fields $X$. Similarly, since $\rho$ is right-invariant, we have that $\mathcal{L}_Y \rho = 0$ for all right-invariant vector fields $Y$.





Let $X$ be a left-invariant vector field, and $Y$ be a right-invariant vector field. Then, we have:





\begin{align*} d \rho(X,Y) &= \mathcal{L}_X \rho(Y) – \rho(\mathcal{L}_X Y) &&\text{(Cartan’s magic formula)}\ &= \mathcal{L}_X \rho(Y) &&\text{(since $Y$ is right-invariant)}\ &= 0 &&\text{(since $\rho$ is left-invariant)} \end{align*}





Similarly, we can show that $d \rho(X,Y) = 0$ when $X$ is right-invariant and $Y$ is left-invariant. Since any vector field on $G$ can be written as a linear combination of left-invariant and right-invariant vector fields, we conclude that $d \rho = 0$ for any left-invariant and right-invariant form $\rho$ on $G$.




















这是一份2023年的波士顿学院Boston College MATH 8809几何学/拓扑学代写的成功案例