The Affine Property of Connections on Vector Bundles

Question

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...

Answer 1

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.

Answer 2

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

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