math214:04-01

$$ \gdef\vect{\text{Vect}} \gdef\lcal{\mathcal L} \gdef\End{\text{End}} \gdef\Hom{\text{Hom}}$$

Today and Friday, we will follow Nicolascu's note 3.3, and discuss connection on vector bundle.

Suppose we are given a smooth manifold $M$. Let $f$ be a smooth function on $M$. We know two things:

- We know what it means for $f$ to be constant.
- Given a tangent vector $v_p \in T_p M$, we know how to take directional derivative $v_p(f)$.

So, why things are different when we talk about a vector bundle $\pi: E \to M$? Let's think about the following thing:

- Given a local section $\sigma \in \Gamma(U, E)$ for an open set $U \In M$, what does it mean when we say $\sigma$ is a 'constant section'? Or can we say it? (recall a section means a smooth map $\sigma: U \to E$ such that $\pi \circ \sigma = id:U \to U$.)
- Given a point $p \in M$, $\sigma$ a section of $E$ defined in a neighborhood of $p$, and a tangent vector $v_p \in T_p M$, can we take 'directional derivative' of $\sigma$ along $v_p$?

Of course, “yes, we can!”, you may say, since we can find a local trivialization of $E$ over $U$ $$ \pi^{-1}(U) \cong U \times \R^k $$ then a section $\sigma$ over $U$ can be described by an $k$-tuple of functions $(\sigma_1, \cdots, \sigma_k): U \to \R^k$, and we all know how to take derivative of functions. Any objection?

**The above approach is problematic.** How do you know the answer you get does not depends on the local trivialization? Giving a local trivialiation is like giving a local frame $(e_1, \cdots, e_k)$ of the vector bundle. If we use the above prescription to define 'taking derivatives' of a section, then, we are declaring these local sections $e_1, \cdots, e_p \in \Gamma(U, E)$ are 'constant'. That is a huge bias, a huge choice to make. In other words, the notion of 'constant section' would depends on the choice of the trivialization. Hence, the approach in the previous paragraph is not right (in general).

You may remember that we have encountered the same problem, when we try to take derivative of a tangent vector field $X$, and we found a solution called 'Lie derivative'. Here, we introduce the notion of *connection* and *covariant derivatives*.

Let $\pi: E \to M$ be a vector bundle. A connection should satisfies the following

- Data: $\nabla: \vect(M) \times C^\infty(M, E) \to C^\infty(M, E)$, $(X, \sigma) \mapsto \nabla_X(\sigma)$.
- $\C^\infty(M)$-linear: suppose $f \in C^\infty(M)$, we want $\nabla_{fX}(\sigma) = f \nabla_X(\sigma)$. Note that Lie derivative does not satisfy this property: $\lcal_{fX} Y = [fX, Y] = f[X,Y] + [f, Y]X = f \lcal_X Y - Y(f) X$
- Leiniz rule: suppose $f \in C^\infty(M)$, we want

$$ \nabla_X(f \sigma) = \nabla_X(f) \sigma + f \nabla_X(\sigma) = X(f) \sigma + f \nabla_X(\sigma).$$

One can concisely reformulate the above conditions, using differential one-form. A connection is the following data $$ \nabla: C^\infty(M, E) \to C^\infty(M, T^* M \otimes E) $$ such that $$ \nabla(f \sigma) = df \ot \sigma + f \nabla(\sigma), \quad \forall f \in C^\infty(M) $$ 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.

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 $(\nabla T)(\sigma_1) = \nabla (T \sigma_1) - T (\nabla \sigma_1).$

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.

math214/04-01.txt · Last modified: 2020/04/03 14:20 by pzhou