# commuting property of connections and bundle homomorphisms

I have the following situation: $E,F$ are (smooth) vector bundles over a smooth manifold $M$. Assume we are given connections $\nabla^E,\nabla^F$ on $E,F$ and a homomorphism of $M$-bundles $P:E\rightarrow F$.

Note that $\nabla^E$ induces a connection on the dual bundle $E^*$, which we denote by $\nabla^{E^*}$.

$\nabla^{E^*}, \nabla^{F}$ induce a connection on the tensor product bundle $E^*\otimes F \cong Hom(E,F)$, which we denote by $\nabla^{E^*\otimes F}$.

Here is the question:

Let $X\in \Gamma(E)$ . The homomorphism $P$ can be identifed with a $C^\infty(M)$-module homomorphism: $\hat P:\Gamma(E)\rightarrow \Gamma(F)$. We can also identify $P$ as a section of the Hom-bundle: $\bar P\in \Gamma(Hom(E,F))=\Gamma(E^*\otimes F)$.

Now take $V\in \Gamma(TM),X\in \Gamma(E)$. Is the following equality true?

$(\nabla_V^{E^*\otimes F} \bar P)(X)=\nabla_V^F (\hat P(X)) \in \Gamma(F)$

In other words, I am asking whether we can change the order of differentiation and action on sections.

At first glance there is something strange here, since the right side is independent of the connection on $E$.

In more detail:

1) $(\nabla_V^{E^*\otimes F} \bar P)(X).$

Here I think of ($\nabla_V^{E^*\otimes F} \bar P)$ as a section of the hom-bundle, and using the standard identification I can think of it as a bundle-homomorphism, and then (another identification) as a homomrphims of (modules of) sections $\Gamma(E)\rightarrow \Gamma(F)$.

2) $\nabla_V^F (\hat P(X)).$

Here I first use $\hat P$ to get a section $F$ and then I differentiate it covariantly along $V$.

As @mollyerin noted, your formula is false. The correct formula is $$\nabla_V^F \big(\hat P(X)\big) = \big(\nabla_V^{E^*\otimes F} \bar P\big)(X) + \hat P\big( \nabla_V^E (X)\big).$$ This follows essentially from the definition of the tensor product connection and the fact that covariant differentiation commutes with contraction, after noting that $$\hat P(X) = C(\bar P \otimes X),$$ where $C$ denotes contraction between the $E^*$-index of $\bar P$ and the $E$-index of $X$.

• Thanks for your answer. It is quite elegant. I have two questions: (1) I feel it is strange that in order to prove equality which involves only the connections on $E,F,E^* \otimes F$ we need to go through the connection on the "larger" bundle $E^* \otimes F \otimes E$. Do you see a way to avoid this "climbing"? (2) I only managed to see the validity $\hat P(X) = C(\bar P \otimes X)$ by using of components in local frames. (See the details I have added in my answer). Is there a coordinate-free way to see this? Thanks. Jul 2, 2015 at 20:58
• I have now found a way to show the equality without climbing to $E^* \otimes F \otimes E$. (you can see the alternative solution I have added). The essential reason of course still remains the commuting of differentiation with contraction as you mentioned. (Though the only contraction I use is on $E^* \otimes E$, not on $E^* \otimes F \otimes E$). Jul 6, 2015 at 14:00

This seems to be false.

As a counterexample, let $E = F$ be the trivial one-dimensional line bundle over $M = \mathbb{R}$, let $\nabla^E = \nabla^F$ be some flat connection realizing the trivialization, and let $P$ be the identity map. One checks that $\bar{P}$ is flat according to the connection $\nabla^{E^* \otimes E}$, so that for any $V \in \Gamma(TM)$ $$\nabla^{E^* \otimes E}_V \bar{P} = 0,$$ the zero endomorphism of $E$ over $M$; that is, the left side of your equation is zero regardless of $X$. But the right side reduces to (using $E = F$ and $P = 1$) $$\nabla_V^E X,$$ which certainly need not be zero.

(To check that $\bar{P}$ is flat, it suffices for instance to note that if $X$ is a flat global section of $E$ and $\theta$ the global dual (hence flat) section of $E^*$, then $\bar{P} = \theta \otimes X$ as a section of $E^* \otimes E$.)

Here is an alternative answer which does not involve the connection on $E^* \otimes F \otimes E$.

We work again in local components:

Step 1: Assume first that $\bar P = e^i \otimes f_j$. (We denote $X=x^ke_k$). Then $\hat P(X) = (e^i \otimes f_j)(x^ke_k)=x^ke^i(e_k)f_j=x^if_j$

(1) $\nabla_V^F \big ( \hat P(X) \big) = \nabla_V^F ( x^if_j) = (V\cdot x^i)f_j+x^i(\nabla_V^Ff_j)$.

On the other hand:

$\nabla^{E^* \otimes F}_V \bar{P} = \nabla^{E^* \otimes F}_V (e^i\otimes f_j) = (\nabla_V^{E^*}e^i) \otimes f_j + e^i \otimes (\nabla_V^Ff_j) \Rightarrow$

(2) $\widehat{ \big( \nabla_V^{E^*\otimes F} \bar P\big)}(X) = (\nabla^{E^* \otimes F}_V \bar{P} \big )(x^ke_k) = (\nabla_V^{E^*}e^i)(X) \cdot f_j + x^i (\nabla_V^Ff_j)$

and also,

(3) $\hat P(\nabla_V^EX) = (e^i \otimes f_j)(\nabla_V^EX)=e^i(\nabla_V^EX)f_j$

So we get:

$\widehat{ \big( \nabla_V^{E^*\otimes F} \bar P\big)}(X) + \hat P(\nabla_V^EX) \stackrel{(2),(3)}{=} \big ((\nabla_V^{E^*}e^i)(X) + e^i(\nabla_V^EX) \big )f_j + x^i (\nabla_V^Ff_j) =$

$[V \big (e^i(X) \big )]\cdot f_j + x^i (\nabla_V^Ff_j) = (V\cdot x^i)f_j+x^i(\nabla_V^Ff_j) \stackrel{(1)}{=} \nabla_V^F \big ( \hat P(X) \big)$

As required.

Step2: We show the case we handled in step (1) implies the general result.

By the $\mathbb{R}-\text{Linearity}$ of the connection it is enough to show the result for $Q=fP$ where $f\in C^\infty(M)$ and $\bar P = e^i \otimes f_j$ as before.

First note that

(a) $\nabla_V^F \big ( \hat Q(X) \big) = \nabla_V^F \big ( f \hat P(X) \big) = (V\cdot f) \hat P(X) + f\nabla_V^F \big ( \hat P(X) \big)$

Now:

$\widehat{ \big( \nabla_V^{E^*\otimes F} \bar Q\big)}(X) + \hat Q(\nabla_V^EX) = \widehat{ \big( \nabla_V^{E^*\otimes F} \overline {fP}\big)}(X) + \widehat {fP}(\nabla_V^EX) =$

$\widehat{ \big( \nabla_V^{E^*\otimes F} f \bar P\big)}(X) + f \hat P(\nabla_V^EX) = [\big ( \widehat {(V\cdot f)\bar P }\big) + f \big( \nabla_V^{E^*\otimes F} \bar P\big )](X) + f \hat P(\nabla_V^EX) = (V\cdot f) \hat P(X) + f \big[ \widehat{ \big( \nabla_V^{E^*\otimes F} \bar P\big)}(X) + \hat P(\nabla_V^EX) \big] \stackrel{step (1)}{=} (V\cdot f) \hat P(X) + f\nabla_V^F \big ( \hat P(X) \big) \stackrel{(a)}{=} \nabla_V^F \big ( \hat Q(X) \big)$

For completeness, I am writing here the details of the solution suggested by Jack Lee:

Lemma1: $\hat P(X) = C(\bar P \otimes X)$

Derivation of the formula based on Lemma1:

$\nabla_V^F \big(\hat P(X)\big) \stackrel{Lemma 1}{=} \nabla_V^F \big(C(\bar P \otimes X)\big) = C\big(\nabla_V^{(E^*\otimes F) \otimes E} (\bar P \otimes X)\big) = C\big( \nabla_V^{E^*\otimes F} \bar P \otimes X + \bar P \otimes \nabla_V^E X \big) = C\big( \nabla_V^{(E^*\otimes F)} \bar P \otimes X \big) + C\big( \bar P \otimes \nabla_V^E X \big) \stackrel{Lemma 1}{=} \widehat{ \big( \nabla_V^{E^*\otimes F} \bar P\big)}(X) + \hat P\big( \nabla_V^E (X)\big)$

(Where we used Lemma1 twice in the last equality).

Proof of Lemma1:

We use local frames. Take a small enough neighbourhood $U \subseteq M$ over which both $E,F$ (and hence $E*$) are trivial. Let $e_i,e^i,f_j$ be local frames for $E,E^*,F$ respectively.

Then we can write in our tensors in components w.r.t these local frames:

$\bar P = a_i^je^i \otimes f_j, X=x^ke_k \Rightarrow \bar P \otimes X = a_i^je^i \otimes f_j \otimes x^ke_k = a_i^j x^k e^i \otimes f_j \otimes e_k \Rightarrow$

$C(\bar P \otimes X)=a_k^jx^kf_j$ Lemma1: $\hat P(X) = C(\bar P \otimes X)$

Derivation of the formula based on Lemma1:

$\nabla_V^F \big(\hat P(X)\big) \stackrel{Lemma 1}{=} \nabla_V^F \big(C(\bar P \otimes X)\big) = C\big(\nabla_V^{(E^*\otimes F) \otimes E} (\bar P \otimes X)\big) = C\big( \nabla_V^{E^*\otimes F} \bar P \otimes X + \bar P \otimes \nabla_V^E X \big) = C\big( \nabla_V^{(E^*\otimes F)} \bar P \otimes X \big) + C\big( \bar P \otimes \nabla_V^E X \big) \stackrel{Lemma 1}{=} \widehat{ \big( \nabla_V^{E^*\otimes F} \bar P\big)}(X) + \hat P\big( \nabla_V^E (X)\big)$

(Where we used Lemma1 twice in the last equality).

Proof of Lemma1:

We use local frames. Take a small enough neighbourhood $U \subseteq M$ over which both $E,F$ (and hence $E*$) are trivial. Let $e_i,e^i,f_j$ be local frames for $E,E^*,F$ respectively.

Then we can write our tensors in components w.r.t these local frames:

$\bar P = a_i^je^i \otimes f_j, X=x^ke_k \Rightarrow \bar P \otimes X = a_i^je^i \otimes f_j \otimes x^ke_k = a_i^j x^k e^i \otimes f_j \otimes e_k \Rightarrow$

$C(\bar P \otimes X)=a_k^jx^kf_j$

Also, we get $\hat P(X) = (a_i^je^i \otimes f_j)(x^ke_k)=a_i^jx^ke^i(e_k)f_j=a_k^jx^kf_j$ Also, we get $\hat P(X) = (a_i^je^i \otimes f_j)(x^ke_k)=a_i^jx^ke^i(e_k)f_j=a_k^jx^kf_j$