Assignment-daixieTM为您提供波士顿学院Boston College MATH 8831 Geometry/Topology III 几何学/拓扑学代写代考和辅导服务！
Instructions:
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.
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 $\left(\mathcal{J}_A, \mathcal{J}_B\right)$ ?
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.
Let $\mathcal{J}$ be a generalized complex structure. Show that $e^{\theta \mathcal{J}}\left(T^*\right)$ is a Dirac structure for all $\theta$.
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 $e^{\theta\mathcal{J}}(T^*)$.
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 $H^3\left(S^1 \times S^2, \mathbb{Z}\right)$ ?
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
