In the previous post, we looked at invariant polynomials. The topic for this one is going to be something called connection $1$-forms and curvature $2$-forms.

For the remaining of this post let $M$ be a smooth manifold and $\nabla$ a connection on a smooth rank $n$ vector bundle $\pi : E \to M$. To begin with, if you have taken a class in Riemannian Geometry, you might know a way to describe connections locally using the Christoffel symbols $\Gamma^k_{ij}$. Instead of this, we are going to adapt to the Cartan Formalism and describe the connection locally using differential forms.

Let’s first remind ourselves that a connection $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ is $C^\infty(M)$-linear in the first argument, but not $C^\infty(M)$-linear in the second argument. However, it turns out that being $C^\infty(M)$-linear in the first argument and satisfying the Leibniz rule in the second argument is enough to imply that $\nabla$ is a Local Operator. The proof can be found in Lee’s book on Riemannian Geometry, but for the sake of completeness, I’ll prove it here.

Let $\nabla$ be a connection on a vector bundle $E$ over a manifold $M$, $X \in \Gamma(TM)$, and $s \in \Gamma(E)$. If either $X$ or $s$ vanishes identically on an open subset $U \subset M$, then $\nabla_X s \equiv 0$ on $U$.
Consider an open neighborhood $U \subset M$ and $p \in U$. Suppose that $X$ vanishes identically on $U$. Choose a bump function $\varphi \in C^\infty(M)$ with support in $U$ such that $\varphi(p) = 1$. Now $\varphi X \equiv 0$ on $M$ since $X$ vanishes on $U$ and $\varphi$ on $M \setminus U$. Thus $$ \nabla_{\varphi X}s = 0. $$ Since $\nabla$ is $C^\infty(M)$-linear in the first argument we have $$ 0 = \nabla_{\varphi X}s = \varphi\nabla_X s. $$ Evaluating at $p$ yields $$ 0 = \nabla_{\varphi X}s\vert_p = \varphi(p)\nabla_X s\vert_p = \nabla_X s\vert_p. $$ As $p$ is arbitary, $\nabla_X s \equiv 0$ on $U$. Suppose then that $s \equiv 0$ on $U$ and $p \in U$. Again, choose a bump function $\psi$ supported on $U$ such that $\psi(p) \equiv 1$ in a neighborhood of $p$. By our choice of $\psi$ the derivative $X_p\psi$ vanishes. Since $s$ vanishes on the support of $\psi$ we have that $\psi s \equiv 0$ on $M$. It follows that $\nabla_X(\psi s) \equiv 0$ on $M$. Evaluating at $p$ yields $$ 0 =\nabla_X(\psi s)\vert_p = (X_p\psi) s(p) + \psi(p)\nabla_X s\vert_p = \nabla_X s\vert_p $$ and the result follows.

Now since local operators can be restricted to any open subset, a connection on a vector bundle can be restricted to any open subset.

Suppose now that $U$ is a trivializing open set for $E$ and $(e_i)$ a frame for $E$ over $U$. Let $X \in \Gamma(U, TM)$. Since any local section $s \in \Gamma(U, E)$ is a linear combination $s = \sum s^je_j$, the section $\nabla_X s$ can be computed from $\nabla_X e_j$ by using linearity and the Leibniz rule. Indeed, we have that

\[\begin{align*} \nabla_X s &= \nabla_X \left(s^j e_j\right) \\ &= ds^j(X)e_j + s^j\nabla_X e_j. \end{align*}\]

Describing $\nabla_X e_j$ is merely linear algebra. Let $(\varepsilon^i)$ be the dual coframe, we have that

\[\nabla_X e_j = \sum_i \varepsilon^i(\nabla_X e_j)e_i.\]

We set $\omega^i_j(X) := \varepsilon^i(\nabla_X e_j)$ and call these the connection $1$-forms1. The matrix $[\omega^i_j]$ is called the connection matrix of the connection $\nabla$ relative to the local frame. Without fixing $X \in \Gamma(U, TM)$, we have

\[\nabla e_j = \sum \omega^i_j \otimes e_i\]

and

\[\begin{align*} \nabla s &= \nabla(s^je_j) \\ &= ds^j \otimes e_j + s^j\nabla e_j \\ &= ds^j \otimes e_j + s^j \omega^i_j \otimes e_i \\ &= ds^j \otimes e_j + \omega^i_js^j \otimes e_i \\ &= ds + As, \end{align*}\]

which, if you recall, is exactly in the form we described in Connections.

So a connection $\nabla$ on $E\vert_U$ determines a unique connection matrix $[\omega^i_j]$ relative to the local frame. Conversely, any matrix $[\omega^i_j]$ of $1$-forms determines a connection on $E\vert_U$ as follows. For $X, Y \in \Gamma(U,TM)$ set

\[\nabla_X e_j = \sum \omega^i_j(X)e_i,\]

and define $\nabla_X Y$ by applying the Leibniz rule to $Y = \sum_j Y^je_j$:

\[\begin{align*} \nabla_X Y &= \nabla_X\left(\sum_j Y^je_j\right) \\ &= (XY^j)e_j + Y^j\omega^i_j(X)e_i \\ &=(XY^j + Y^j\omega^i_j(X))e_i. \end{align*}\]

As you might notice the story here is pretty much the same as with the Christoffel symbols2. The reason we do it in terms of differential forms now is that we get a neat pathway to characteristic classes.

What about curvature? Similarly for $X,Y \in \Gamma(TM)$ and $s \in \Gamma(E)$ the section $R(X,Y)s \in \Gamma(E)$ can be computed from $R(X,Y)e_j$ when $(e_i)$ is a local frame over $U \subset M$. Again if $(\varepsilon^i)$ is the dual coframe

\[R(X,Y)e_j = \sum \varepsilon^i(R(X,Y)e_j)e_i.\]

Analogously as with the connection $1$-forms we define the curvature $2$-forms as $\Omega^i_j(X,Y) := \varepsilon^i(R(X,Y)e_j)$.

You might recall from your studies in Riemannian Geometry that the curvature endomorphism can be written in terms of local coordinates $(U,x^1,\dots,x^n)$ as

\[R = R^l_{ijk} dx^i \otimes dx^j \otimes dx^k \otimes \partial_l\]

where the coefficients $R^l_{ijk}$ are defined by

\[R^l_{ijk} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \Gamma^m_{jk}\Gamma^l_{im} - \Gamma^m_{ik}\Gamma^l_{jm}.\]

Well in terms of this so-called Cartan Formalism, we end up having $\Omega^l_k(\partial_i,\partial_j) = R^l_{ijk}$ and since $R^l_{ijk}$ is expressible by the Christoffel symbols it would be reasonable to expect that there is a relationship between the connection $1$-forms and curvature $2$-forms right? This turns out to be true and we state it as the following theorem.

Let $\nabla$ be a connection on a vector bundle $E \to M$ of rank $n$. Relative to a local frame $(e_i)$ over $U$, the curvature forms $\Omega^i_j$ are related to the connection forms $\omega^i_j$ by the second structural equation: $$ \Omega^i_j = d\omega^i_j + \sum_k \omega^i_k \wedge \omega^k_j. $$
Let $X, Y \in \Gamma(U, TM)$. We have that $$ \begin{align*} \nabla_X\nabla_Y e_j &= \nabla_X \sum_k\left(\omega^k_j(Y)e_k\right) \\ &= \sum_k X\omega^k_j(Y)e_k + \sum_k\omega^k_j(Y)\nabla_X e_k \\ &= \sum_i X\omega^i_j(Y)e_i + \sum_{i,k} \omega^k_j(Y)\omega^i_k(X)e_i. \end{align*} $$ Swapping $X$ and $Y$ yields $$ \begin{align*} \nabla_Y\nabla_Xe_j = \sum_i Y\omega^i_j(X)e_i + \sum_{i,k} \omega^k_j(X)\omega^i_k(Y)e_i. \end{align*} $$ Hence $$ \begin{align*} R(X,Y)e_j &= \nabla_X\nabla_Y e_j - \nabla_Y\nabla_X e_j - \nabla_{[X,Y]}e_j \\ &= (X\omega^i_j(Y) - Y\omega^i_j(X) - \omega^i_j([X,Y]))e_i + (\omega^i_k(X)\omega^k_j(Y)- \omega^i_k(Y)\omega^k_j(X))e_i \\ &= d\omega^i_j(X,Y)e_i + \omega^i_k\wedge \omega^k_j(X,Y)e_i \\ &= (d\omega^i_j + \omega^i_k\wedge \omega^k_j)(X,Y)e_i. \end{align*} $$ Comparing this with how we defined $\Omega^i_j(X,Y)$ we see that $$ \Omega^i_j = d\omega^i_j + \sum_k \omega^i_k \wedge \omega^k_j. $$

In matrix notation, the above can be written as $\Omega = d\omega + \omega \wedge \omega$.

Before we wrap this up, let’s consider how the connection $1$-forms and curvature $2$-forms transform under a change of frame. Recalling that the curvature endomorphism is a section of the bundle $\left(\bigotimes^2 T^\ast M \right) \otimes \operatorname{End}(E)$ i.e. an $\operatorname{End}(E)$-valued $(0,2)$-tensor field. These two observations will end up looking awfully similar to what we discussed in Invariant Polynomials, guiding us towards characteristic classes.

Suppose that $U \subset M$ is an open set and $(e_i)$ a local frame over $U$. Writing the frame as a row vector $e = \begin{bmatrix} e_1 & \cdots & e_n\end{bmatrix}$, then in matrix notation we can write $\nabla_X e_j = \sum \omega^i_j(X)e_i$ as

\[\nabla_X e = e\omega(X).\]

As a function of $X$ we have $\nabla e = e\omega$. Suppose now that $(\bar{e}_i)$ is another local frame over $U$. Let $\bar{\omega} = [\bar{\omega}^i_j]$ and $\bar{\Omega} = [\bar{\Omega}^i_j]$ be the connection and curvature matrices with respect to the frame $(\bar{e}_i)$. As sections over $U$ we can write

\[\bar{e}_j = \sum_i a^i_je_i\]

which yields a matrix of functions $a = [a^i_j]$ where each $a^i_j$ is a smooth function on $U$. It follows that $a : U \to \mathrm{GL}(n,\Bbb R)$. In matrix notation $\bar{e} = ea$.

Suppose that $(e_i)$ and $(\bar{e}_i)$ are two local frames over $U \subset M$ such that $\bar{e} = ea$ for some $a : U \to \mathrm{GL}(n,\Bbb R)$. If $\omega$ and $\bar{\omega}$ are the connection matrices and $\Omega$ and $\bar{\Omega}$ the curvature matrices of a connection $\nabla$ relative to the two frames, then
  1. $\bar{\omega} = a^{-1}\omega a + a^{-1}da$.
  2. $\bar{\Omega} = a^{-1}\Omega a$.
For the first item note that $$ \begin{align*} \nabla \bar{e} &= \nabla(ea) \\ &= (\nabla e)a + eda \\ &= (e\omega)a + eda \\ &= \bar{e}a^{-1}\omega a + \bar{e}a^{-1}da \\ &= \bar{e}(a^{-1}\omega a + a^{-1}da). \end{align*} $$ As $\nabla \bar{e} = \bar{e}\bar{\omega}$ it follows that $\bar{\omega} = a^{-1}\omega a + a^{-1}da$. For the second item remember that the curvature is $\operatorname{End}(E)$-valued. That is for a point $p$ and $X_p, Y_p \in T_pM$ we have that $R(X_p,Y_P) \in \operatorname{End}(E_p)$ with matrix $[\Omega^i_j(X_p,Y_p)]$ relative to the basis $(e_i)$ at $p$. A change of basis should thus lead to a conjugate matrix. Writing $R(X,Y)e_j = \sum \Omega^i_j(X,Y)e_i$ in matrix notation yields $$ R(X,Y)e = \begin{bmatrix} e_1 & \cdots & e_n\end{bmatrix}[\Omega^i_j(X,Y)] = e\Omega(X,Y) $$ and so $R(e) = e\Omega$. It follows that $$ \begin{align*} R(\bar{e}) &= R(ea) \\ &= R(e)a \\ &=e\Omega a \\ &= \bar{e}a\Omega a \end{align*} $$ which yields that $\bar{\Omega} = a^{-1}\Omega a$.

Ring a bell regarding those invariant polynomials and conjugation? That’s what we’ll discuss in the next post.

  1. These are indeed $1$-forms as $\omega^i_j : \Gamma(TM) \to C^\infty(M)$ is a $C^\infty(M)$-linear map. ↩

  2. Infact, note that in local coordinates $(U, x^1,\dots,x^n)$ we have $\omega^k_j(\partial_i) = \Gamma^k_{ij}$. ↩