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Given any two connections $\nabla_1, \nabla_2: \Omega^0 (V) \to \Omega^1 (V)$ on a vector bundle $V \to M$, their difference $\nabla_1 - \nabla_2$ is a $C^\infty (M)$-linear map $\Omega^0 (V) \to \Omega^1 (V)$.

Question: I have difficulties swallowing the implication that $\nabla_1 - \nabla_2 \in \Omega^1 (\text{End } V)$.

Of course, $\Omega^1 (\text{End } V) = \Gamma (T^\ast M \otimes \text{End } V)$, so this is saying that $\nabla_1 - \nabla_2$ is an endomorphism-valued 1-form. Also, given any section $s \in \Omega^0 (V)$, the difference $(\nabla_1 - \nabla_2) s$ at any point $m \in M$ is completely determined by the value $s(m)$, i.e. the operator $(\nabla_1 - \nabla_2) |_m$ is an endomorphism of the fiber $V|_m$, but I don't see how this is relevant, yet...

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    $\begingroup$ Your "question" is a statement. What do you want to know? It is a general fact that when you have a $C^\infty (M)$-linear map of global sections of vector bundles then you have a homomorphism of vector bundles. $\endgroup$
    – Zhen Lin
    Jun 7, 2012 at 16:57
  • $\begingroup$ That is it. I am not familiar with this general fact. Reference or explanation, please? $\endgroup$
    – Rick
    Jun 7, 2012 at 16:58
  • $\begingroup$ Also, everywhere I'm reading about this affine business of connections, they just say "it is a $C^\infty$-linear operator, so it follows that"... I'm missing that crucial link... $\endgroup$
    – Rick
    Jun 7, 2012 at 17:02

2 Answers 2

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Let $E$ and $F$ be vector bundles over a manifold $M$, and suppose I have a $C^\infty (M)$-linear map of global sections $\alpha : \Gamma (M, E) \to \Gamma (M, F)$. I claim this $\alpha$ is induced by a unique homomorphism of vector bundles $A : E \to F$.

Indeed, let $\vec{v}$ be a vector in the fibre $E_p$. By taking a local trivialisation and then multiplying by a bump function, I can get a global section $X \in \Gamma (M, E)$ such that $X |_p = \vec{v}$. Define $A (\vec{v}) = \alpha(X) |_p$. This is independent of the choice of $X$: if $Y$ is any other global section of $E$ with $Y |_p = \vec{v}$, then $(X - Y) |_p = \vec{0}$, so there is a smooth function $f : M \to \mathbb{R}$ and a global section $Z$ such that $f(p) = 0$ and $f Z = X - Y$. But then $C^\infty (M)$-linearity implies $$\alpha(X) = \alpha(X - Y) + \alpha(Y) = \alpha(f Z) + \alpha(Y) = f \alpha(Z) + \alpha(Y)$$ so by evaluating at $p$ we get $\alpha(X) |_p = \alpha(Y) |_p$, as claimed. Verifying that $A$ is indeed a vector bundle homomorphism is straightforward, and uniqueness is obvious.

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    $\begingroup$ So, am I understanding this right: In my case, $\nabla_1 - \nabla_2 : \Omega^0 (V) \to \Omega^1 (V)$ is then induced by some unique $A \in \text{Hom} (V, T^\ast M \otimes V)$, i.e. $A$ is a section of $T^\ast M \otimes (V^\ast \otimes V) = T^\ast M \otimes \text{End } V$, so then precisely $A \in \Gamma (M, T^\ast M \otimes \text{End } V) = \Omega^1 (\text{End } V)$. Now, there is some kind of identification of $A$ with $\nabla_1 - \nabla_2$, to say that $\nabla_1 - \nabla_2 \in \Omega^1 (\text{End } V)$? $\endgroup$
    – Rick
    Jun 7, 2012 at 21:04
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    $\begingroup$ Yes. This is a standard abuse of notation, I'm afraid. $\endgroup$
    – Zhen Lin
    Jun 7, 2012 at 21:45
  • $\begingroup$ @ZhenLin I think your argument only works for the case of line bundles, but the idea is the same for general vector bundles where instead of $Z$ you take a frame (previously showing that $\alpha$ is a local operator). $\endgroup$
    – inquisitor
    Jan 26, 2017 at 19:11
  • $\begingroup$ @ZhenLin I had a quick question. If $A \in \text{Hom} (V, T^\ast M \otimes V)$ then I know that one can identify, $\text{Hom} (V, T^\ast M \otimes V) \congruent V^\ast \otimes T^\ast M \otimes V,$ but then why is $A \in \Gamma(V^\ast \otimes T^\ast M \otimes V)$ and not in $V^\ast \otimes T^\ast M \otimes V$? $\endgroup$
    – Student
    Dec 22, 2020 at 16:24
  • $\begingroup$ It depends on whether your $\textrm{Hom}(-, -)$ is the internal or external hom. $\endgroup$
    – Zhen Lin
    Dec 22, 2020 at 22:06
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According to the Connection_(vector_bundle) article of Wikipedia, connecions are $R$-linear map from $\Gamma(E)$ to $\Gamma (T^\ast M \otimes E)$ (both are $C^\infty(M)$-modules), but not $C^\infty(M)$-linear map since:

$$\nabla(fσ)=f\nabla σ+df\otimes σ$$

Here $f \in C^\infty(M)$, the second term make it not $C^\infty (M)$-linear.

But the subtraction of two connection eliminated the second term making it a $C^\infty (M)$-linear map between $C^\infty(M)$-modules $\Gamma(E)$ and $\Gamma (T^\ast M \otimes E)$. So $$\nabla_1 - \nabla_2 \in Hom_{C^\infty(M)}(\Gamma(E),\Gamma (T^\ast M \otimes E))$$

Also

$$ Hom_{C^\infty(M)}(\Gamma(E),\Gamma(T^\ast M \otimes E)) \\\cong \Gamma(E)^*\otimes \Gamma(T^\ast M \otimes E) \\\cong \Gamma(E^*)\otimes \Gamma(T^\ast M \otimes E) \\\cong \Gamma(T^\ast M\otimes (E\otimes E^*)) \\\cong \Gamma(T^\ast M\otimes \text{End }E) $$

Then

$$\nabla_1 - \nabla_2 \in \Gamma(T^\ast M\otimes \text{End }E)=:\Omega^1 (M,\text{End } E)$$

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