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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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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. –  Zhen Lin Jun 7 '12 at 16:57
    
That is it. I am not familiar with this general fact. Reference or explanation, please? –  Rick Jun 7 '12 at 16:58
    
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... –  Rick Jun 7 '12 at 17:02
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up vote 2 down vote accepted

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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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 $\nabla_1 - \nabla_2 \in \Omega^1 (\text{End } V)$? –  Rick Jun 7 '12 at 21:04
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Yes. This is a standard abuse of notation, I'm afraid. –  Zhen Lin Jun 7 '12 at 21:45
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