Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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
up vote 3 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.

share|cite|improve this answer
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
Yes. This is a standard abuse of notation, I'm afraid. – Zhen Lin Jun 7 '12 at 21:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.