math214:02-19

# 2020-02-19, Wednesday

## Vector Bundle

Definition : A vector bundle is a quadruple $(E, \pi, M ,F)$, such that

• $E, M$ are smooth manifolds
• $\pi: E \to M$ is a surjective submersion. For each $U \subset M$, we set $E|_U := \pi^{-1}(U)$.
• $F$ is a vector space of rank $n$ over $\R$.
• there exists a trivializing cover, ie. an open cover $\gdef\cU{\mathcal U} \cU$ of $M$, and for every $U \in \cU$ a diffeomorphism $$\Psi_U: E|_U \to U \times F, \quad v \mapsto (p=\pi(v), \Phi^U_p(v))$$
• $\Phi^U_p: E_p \to F$ is a diffeomorphism for any $p \in U$.
• If $U, V \in \cU$ are two trivializing neighborhoods with non-empty overlap $U \cap V$, then, the map $\Phi_{VU}(p): \Phi^V_p \circ (\Phi^U_p)^{-1}: F \to F$ is a linear isomorphism. And moreover, the map $$p \mapsto \Phi_{VU}(p) \in GL(n, \R)$$ is smooth (i.e. each entry of hte matrix is a smooth function of $p$)

Given a smooth manifold $M$, and fiber $F = \R^n$, how to specify the data of a vector bundle? We just need to specify a cover $\cU$ and some gluing data: $$g_{\alpha\beta} = g_{\alpha \gets \beta}: U_\alpha \cap U_\beta \to Aut(F)=GL(\R,n)$$ for any $U_\alpha ,U_\beta \in \cU$, that satisfies the tricycle condition

1. $g_{\alpha \beta} \circ g_{\beta \alpha}=1_F$ over $U_\alpha \cap U_\beta$
2. $g_{\alpha \beta} \circ g_{\beta\gamma} \circ g_{\gamma \alpha} = 1_F$ over $U_\alpha \cap U_\beta \cap U_\gamma$.

### Example of Vector Bundles

$E(n,k) \to Gr(n, k)$ tautological bundle over the Grassmannian.

$O(-1) \to \C\P(1)$.

### Operations on Vector Bundles

Direct Sum, Tensor Product, Hom.

## Cotangent Bundle

Some linear algebra first. Let $V$ be a finite dimensional vector space over $\R$

Dual vector space of $V$ is the vector space of linear function on $V$, denoted as $V^*$.

If $V$ is a basis $\{E_1, \cdots, E_n\}$, then there is a basis basis $\{ \epsilon^i\}$ of $V^*$, called dual basis to $\{E_i\}$, satisfying $\epsilon^i( E_j) = \delta_j^i$.

Given a basis for $V$ and corresponding dual basis as above, then a vector $v \in V$ and a covector $\omega \in V^*$ can be written as $$v = \sum_i v^i E_i , \quad \omega = \sum_j \omega_j \epsilon^j$$ The canonical pairing $(\omega, v) = \omega(v)$ can be written as $$(\omega, v) = \sum_i \omega_i v^i.$$

If $f: V \to W$ is a linear map, then there is a dual map $f^*: V^* \to W^*$, given by $$f(\varphi)(w) = \varphi(f(w)), \quad w \in W, \varphi \in V^*.$$

Now, let $M$ be a smooth manifold, $p \in M$, $T_p M$ its tangent space over $p$. We define the cotangent space at $p$ to be $$T^*_p M: = (T_p M)^*.$$

Given a coordinate system near $p$, $(U, (x^1, \cdots, x^n))$, we have basis for $T_p M$ as $\{ \d_i|_p = \frac{\d}{\d x^i}\vert_p$. The dual basis of $\{\d_i\}|_p$ is denoted as $\{dx^i\}|_p$. Hence, a covector $\omega$ at $p$ can be written as $\omega = \sum_i \omega_i dx^i|_p$

Exercise: figure out the transformation rule for the coefficients $\omega_I$ if we change coordinates.

### Covector Fields (Differential 1-form)

Recall how we define smooth vector field over $M$: it is an assignment $p \mapsto X(p) \in T_p M$ for $p \in M$, such that if we write it in coordinate patch $(U, (x_i))$, we have $$X(p) = \sum_i X(p)^i \d_i$$ where $X(p)^i:U \to \R$ are smooth functions.

Simiarly, we define smooth covector fields over $M$ as an assignment $p \mapsto \omega(p) \in T^*_p M$, such that in coordinates we have $$\omega(p) = \sum_i \omega(p)_i dx^i$$ where $\omega(p)_i$ are smooth.

Prop : The cotangent bundle $T^*M = \sqcup_p T_p^*M$ is a vector bundle over $M$.

## Poincaré Lemma For 1-form

Given a differential 1-form, we can do line integral.

A differential 1-form is closed, if the value of the line integral is invariant under isotopy that fixes the end-points.

A differential 1-form $\lambda$ is exact, if there exists a function $f$, such that $df = \lambda$. We say $f$ is a primitive of $\lambda$.

Over a unit ball (or more generally, connected space with trivial $\pi_1$), any closed 1-form is exact. We can find a primitive of $\lambda$, by fixing a point $p_0$ in the space, and define $f(p) = \int_{p_0}^p \lambda$. 