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--- math214:04-01 [2020/03/31 22:25]
pzhou
+++ math214:04-01 [2020/04/03 14:20] (current)
pzhou
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+$$ \gdef\vect{\text{Vect}} \gdef\lcal{\mathcal L} \gdef\End{\text{End}} \gdef\Hom{\text{Hom}}$$
 ====== 2020-04-01, Wednesday ======
 Today and Friday, we will follow Nicolascu's note 3.3, and discuss connection on vector bundle.
@@ Line 20: / Line 23: @@
 ==== Connection and Covariant Derivative ====
-$$ \gdef\vect{\text{Vect}} \gdef\lcal{\mathcal L} $$
 Let $\pi: E \to M$ be a vector bundle. A connection should satisfies the following
   - Data: $\nabla: \vect(M) \times C^\infty(M, E) \to C^\infty(M, E)$, $(X, \sigma) \mapsto \nabla_X(\sigma)$.
@@ Line 33: / Line 36: @@
 where $T^*M \ot E$ is the tensor product of two vector bundles, and $C^\infty(M, T^* M \otimes E)$ is smooth section of it. Sometimes, we use the following notation: recall $\Omega^0(M)$ is 0-form, ie. smooth function, and $\Omega^1(M)$ is the space of 1-forms. We use $\Omega^0(M, E) = C^\infty(M, E)$, and $\Omega^1(M,E) = C^\infty(M, T^* M \otimes E)$, then
 $$\nabla: \Omega^0(M, E) \to \Omega^1(M, E). $$
+In general, we define
+$$ \Omega^k(M, E) = C^\infty( \wedge^k(T^*M) \ot E) $$
+as $k$-forms with coefficient in $E$.
+** Example 1 **: Trivial vector bundle with trivial connection. Suppose $E = M \times \R^k$, then we can define the trivial connection $\nabla$ on $E$ as
+$$ \nabla (f_1, \cdots, f_k) = (df_1, \cdots, df_k). $$
+The space of connection is not a linear space, but rather, an affine linear space. Suppose $\nabla^0, \nabla^1$ are both connections on $E$. Then, for any smooth function $f \in C^\infty(M)$, the linear combination
+$$ f\nabla^0 + (1-f) \nabla^1: \Omega^0(M, E) \to \Omega^1(M, E). $$
+is still a connection. Indeed, the Leibniz rule works.
+** Proposition ** Suppose $\nabla^0, \nabla^1$ are both connections on $E$, then $A = \nabla^1 - \nabla^0 \in \Omega^1(M, \End(E))$.
+Remark: In plain words, for any $X \in \vect(M)$, and $\sigma$ section of $E$, we have
+$A_X(\sigma)(p)$ only depends on $\sigma(p)$, ie., only the value of $\sigma$ at $p$, not the derivatives of $\sigma$ at $p$.
+** Proof: ** We have $A: \Omega^0(M, E) \to \Omega^1(M, E)$ an $\R$-linear map automatically. We need to show that $A$ is $\C^\infty(M)$ linear, that is, for any $f \in C^\infty(M)$, $\sigma \in C^\infty(M, E)$, we want
+$$ A (f \sigma) = f A(\sigma). $$
+Indeed, this is easy to check
+$$ A(f \sigma) = (\nabla^1 - \nabla^0) (f \sigma) = (df - df ) \otimes \sigma + f (\nabla^1 - \nabla^0) (\sigma) = f A (\sigma).$$
+** Prop **: the space of all possible connections on $E$, denoted as $\acal(E)$ is an affine vector space model on $\Omega^1(\End(E))$.
+Given the previous propostion, we only need to prove that
+  - this space $\acal(E)$ is not empty, and
+  - for any $\nabla \in \acal(E)$, $A \in \Omega^1(\End(E))$, $\nabla + A \in \acal(E)$.
+The first statement can be shown using partition of unity. The second statement is an easy check.
+==== Tensor, Hom and Dual ====
+Let $E_1, E_2$ be two vector bundles on $M$. Recall that we can define the following
+$$ E_1 \ot E_2, \quad \Hom(E_1, E_2) $$
+as vector bundles on $M$, where  the fiber satisfies $(E_1 \ot E_2)_p = (E_1)_p \ot (E_2)_p$ and $\Hom(E_1, E_2)_p = \Hom( (E_1)_p, (E_2)_p)$. We define the dual bundle of a vector bundle $E$ as $E^\vee := \Hom(E, \underline{\R})$ where
+$\underline{\R} = \R \times M$ is the trivial bundle.
+Suppose $(E_1, \nabla^1)$ and $(E_2, \nabla^2)$ are equipped with connections, we define connection on the tensor and hom bundle as
+$$ \nabla( \sigma_1 \ot \sigma_2) = \nabla ( \sigma_1) \ot \sigma_2 + \sigma_1 \ot \nabla (\sigma_2) $$
+where we omit the superscript on $\nabla$ and we use the identification
+$$\Omega^1(M, E_1 \ot E_2) = \Omega^1(M) \ot \Omega^0(M, E_1) \ot \Omega^0(M, E_2),$$
+where the tensor on the RHS is over $\C^\infty(M)$.
+Suppose $T \in \Hom(E_1, E_2)$, then we define
+$(\nabla T)(\sigma_1) = \nabla (T \sigma_1) - T (\nabla \sigma_1).$
+==== Moving Frame ====
+Let $M$ be a smooth manifold of dimension $n$, $E$ a rank $r$ vector bundle on $M$ over $\R$. Suppose $(U, (x_1,\cdots, x_n))$ is a coordinate chart on $M$, and furthermore $E|_U$ is trivializable, with a trivialization $(e_\alpha)_{\alpha=1,\cdots,r}$. Then we can write a local section $u \in \Gamma(U, E)$ as
+$$ u = \sum_\alpha u^\alpha e_\alpha = u^\alpha e_\alpha,$$
+using Einstein summation convention.
+And we can write
+$$ \nabla(u) =  \nabla(u^\alpha e_\alpha) = d(u^\alpha) \ot e_\alpha + u^\alpha \ot \nabla(e_\alpha). $$
+Hence, if we know how $\nabla$ acts on the frame $e_\alpha$, we know how it acts on any sections. We may write
+$$ \nabla(e_\alpha) = \Gamma^{\beta}_{i \alpha} dx^i \ot  e_\beta$$
+These coefficients $\Gamma^{\beta}_{i \alpha}$ encodes the data of the connection over $U$.
+More abstractly, we can say: given a trivilization of $E|_U = U \times \R^r$, we may consider the trivial connection $d_U$ on $U \times \R^r$, then over $U$, we have
+$$ \nabla|_U = d_U + A_U $$
+where $A_U \in \Omega^1(U, \End(E))$ is called the ** local ** conection 1-form.

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