Connection on the cotangent bundle

connectionsdifferential-formsdifferential-geometry

I'm reading "Differential forms and connections" by R. Darling and I must have made a mistake in problem 4 in section 9.4. It states:

Prove that $\nabla_X \omega := \iota_X\, d\omega$ is not a connection on the cotangent bundle.

But it's additive in both arguments and $\nabla_{hX} \omega = \iota_{hX}\, d\omega = h\iota_{X}\, d\omega=h\nabla_X \omega$. The only problem can be the Leibniz rule:

$$\nabla_X (h\omega) = \iota_X\, d(h\omega)=\iota_X\, (dh\, \wedge\, \omega + h\,d\omega)=X(h)\, \omega + h\, \iota_x\,d\omega = h\nabla_X\omega + X(h)\omega,$$

so it's seems to work as well. Where did I make a mistake?

Best Answer

There's a missing term in the penultimate expression in your display equation.

Since $\iota_X$ is an antiderivation and $dh$ is a $1$-from, expanding the quantity $$\iota_X (\color{#df0000}{dh \wedge \omega} + \color{#0000df}{h \,d\omega})$$ gives $$\color{#df0000}{(\iota_X dh) \wedge \omega - dh \wedge \iota_X d\omega} + \color{#0000df}{h \iota_X d\omega}.$$ By definition the first term is $X(h) \omega$ and the last is $h \nabla_X \omega$.

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It seems to me that people usually include the $\mathbb{R}$-linearity as part of the definition (correct me if I'm wrong); i.e. define $\nabla$ on the simple tensors by the product rule above, and then extend it linearly over $\mathbb{R}$ to the whole space $C^{\infty}(E\otimes F)$.

Actually, due to the fact that $C^{\infty}(E\otimes F)\simeq C^{\infty}(E)\otimes_{C^{\infty}(M)}C^{\infty}(F)$ which OP has mentioned in the comment above, it follows that additivity (i.e. \begin{align} \nabla_X(s\otimes t+s'\otimes t')=\nabla_X(s\otimes t)+\nabla_X(s'\otimes t') \end{align} which is weaker than $\mathbb{R}$-linearity) is sufficient.

With this in mind, here is an alternative proof, which is essentially a direct computation, though perhaps it may look less elegant.

Let $A\in C^{\infty}(E\otimes F)$. Such $A$ can be expressed as a finite sum of simple tensors, though not uniquely in general. So we want to show that if it can be written in the following two ways \begin{align} A=\sum_is_i\otimes t_i=\sum_j\tilde{s}_j\otimes\tilde{t}_j \end{align} (the number of summands in these two expressions may be different in general), then \begin{align} \nabla_X\left(\sum_is_i\otimes t_i\right) =\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) & & (*) \end{align} Let $(e_{\alpha})$ and $(\epsilon_{\beta})$ be smooth local frames for $E$ and $F$ respectively. Then one can write \begin{align} s_i=s_i^{\alpha}e_{\alpha},\qquad t_j=t_j^{\beta}\epsilon_{\beta},\qquad \tilde{s}_i=\tilde{s}_i^{\alpha}e_{\alpha},\qquad \tilde{t}_j=\tilde{t}_j^{\beta}\epsilon_{\beta} \end{align} (Einstein summation convention is assumed.) Then we will have \begin{align} \sum_is_i\otimes t_i =\left(\sum_is_i^{\alpha}t_i^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta}, \qquad \sum_j\tilde{s}_j\otimes\tilde{t}_j =\left(\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta} \end{align} Since $\{e_{\alpha}\otimes\epsilon_{\beta}\}$ is a smooth local frame for $E\otimes F$, by uniqueness of local components we must have \begin{align} \sum_is_i^{\alpha}t_i^{\beta}=\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta} & & (1) \end{align} Taking exterior derivative we then have \begin{align} \sum_i\left(t_i^{\beta}ds_i^{\alpha}+s_i^{\alpha}dt_i^{\beta}\right) =\sum_j\left(\tilde{t}_j^{\beta}d\tilde{s}_j^{\alpha} +\tilde{s}_j^{\alpha}d\tilde{t}_j^{\beta}\right) & & (2) \end{align} Now let $\omega_{\alpha}^{\beta}$ and $\theta_{\alpha}^{\beta}$ be the connection 1-forms of $\nabla^E$ and $\nabla^F$ respectively, so that e.g. \begin{align} \nabla_X s_i=\left[Xs_i^{\alpha}+s_i^{\beta}\omega_{\beta}^{\alpha} (X)\right]e_{\alpha} & & (3) \end{align} and similar identities hold for $\nabla_Xt_i$, $\nabla_X\tilde{s}_j$ and $\nabla_X\tilde{t}_j$. Then we can compute \begin{align} &\nabla_X\left(\sum_is_i\otimes t_i\right) \\ %%% &=\sum_i\left(\nabla_Xs_i\otimes t_i+s_i\otimes\nabla_Xt_i\right) \\ %%% &=\underbrace{\left(\sum_i\left((Xs_i^{\alpha})t_i^{\beta} +s_i^{\alpha}(Xt_i^{\beta})\right)\right)}_{=:I}e_{\alpha}\otimes\epsilon_{\beta} +\underbrace{\left(\sum_is_i^{\gamma}t_i^{\beta}\right)}_{=:II} \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta} +\underbrace{\left(\sum_is_i^{\alpha}t_i^{\gamma}\right)}_{=:III} \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta} \end{align} where the first step follows by additivity and product rule, while the second step is obtained by substitution of (3) and rearranging the terms.

Now $\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)$ will have the same expression, except that all of the $s_i,t_i$ are replaced by $\tilde{s}_j,\tilde{t}_j$; says \begin{align} \nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) =\tilde{I}\cdot e_{\alpha}\otimes\epsilon_{\beta} +\tilde{II}\cdot \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta} +\tilde{III}\cdot \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta} \end{align} Then:

By (2), we have $I=\tilde{I}$.
By (1), we have $II=\tilde{II}$ and $III=\tilde{III}$.

Hence, (*) holds as desired.

Different notions of connection

The first definition sounds much more like the exterior covariant derivative associated to a linear connection (as in the second definition); another common notation for it is $d^{\nabla}$ or $d_{\nabla}$. Also, in the second definition, you write

assigning to a vector field $X$ and a section $Y$ of $V$ a vector field $\nabla_XY$...

the second instance of vector field (emphasis mine) is a typo, it should say "a section $\nabla_XY$ of $V$".

Also, in the first definition, $D^2s$, the second covariant exterior differential of a smooth section $E$-valued $q$ form $s$ on $M$, equals the 'wedge' of the curvature, $R$, of the connection (which is a smooth $\text{End}(E)$-valued $2$-form on $M$, i.e $R\in \mathcal{C}^{\infty}_2(M,\text{End}(E))$) and $s$: \begin{align} D^2s&=R\wedge_{\epsilon}s. \end{align} Here, $\wedge_{\epsilon}$ denotes the wedge product with respect to evaluation of endomorphisms on vectors: see Ivo Terek's answer here for details. Its value on a pair of vector fields $X,Y$ is \begin{align} (D^2s)(X,Y)&=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s. \end{align}

As far as terminology is concerned, I find this usage confusing, but the operators are essentially the same when acting on sections of $E$, i.e $D:\mathcal{C}^{\infty}_0(M,E)\to \mathcal{C}^{\infty}_1(M,E)$ and $\nabla:C^{\infty}(TM)\otimes C^{\infty}(E)\to C^{\infty}(E)$ are related as: \begin{align} (Ds)(X)&=\nabla_Xs. \end{align} Or, $Ds=\nabla s$, where it is understood that there is an open slot to be filled in by a vector field $(Ds)(\cdot)=\nabla_{(\cdot)}s$; this open slot for a vector field is what gives the 1-form nature.

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Different notions of connection

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