# Sections form a vector space

Let $$p:E \to X$$ be a vector bundle and denote by $$\Gamma(E)$$ the set of all sections of $$E$$. I want to show that $$\Gamma(E)$$ is a real vector space and will show what I tried for the sum operation. For sections $$s_1,s_2 \in \Gamma(E)$$ and $$x \in X$$ we can define $$s_1+s_2(x):=s_1(x)+s_2(x)$$ which will give a well defined function $$s_1+s_2:X \to E$$ with the property that for every $$x \in X$$ we get that $$p((s_1+s_2)(x))=x$$ since the sum $$s_1(x)+s_2(x)$$ is the sum in $$p^{-1}(x)$$. So the only thing that one needs to show is that the sum is continuous.

We know that for every $$x \in X$$ there is an open $$U$$ around $$x$$ and an isomorphism of vector bundles $$f:p^{-1}(U) \to U \times \mathbb{R}^n$$ for some $$n$$. Thus we get a function $$U \to p^{-1}(U) \to U \times \mathbb{R}^n, u \mapsto s(u) \mapsto f(s(u)).$$ Note that on ever fiber $$p^{-1}(u)$$ with $$u \in U$$ there is a unique homomorphism $$g_u:p^{-1}(u) \to \mathbb{R}^n$$ with $$f(v)=(u,g_u(v))$$ where $$v \in p^{-1}(u)$$. So the composition above is given by $$u \mapsto (u,g_u(s(u))$$. In particular we get that the function for $$s_1+s_2$$ is given by: $$u \mapsto (u,g_u(s_1(u)+s_2(u))).$$ Defining the function $$g:p^{-1}(U), v \mapsto g_u(v)$$, $$p(v)=u$$ we get that $$g$$ is continuous because $$g=\pi_2 \circ f$$ where $$\pi_2$$ is the projection onto the second coordinate. We also know that the sum is continuous, and so the function $$u \mapsto (u,g_u(s_1(u)+s_2(u)))$$ is continuous. Composing this with the homeomorphism $$f^{-1}$$ we get the function $$u \mapsto s_1(u)+s_2(u)$$. So using that composition of continuous functions remains continuous we get that the sum $$s_1+s_2$$ is continuous on $$U$$. Therefore we can conclude that $$s_1+s_2$$ is continuous on $$X$$.

1. Is my approach correct? I think that I made at least one mistake namely when saying that the sum is continuous. While it is continuous on $$p^{-1}(u)$$ I think its not necessarily continuous on $$p^{-1}(U)$$ (when switching between fibers) which is what I need for the composition. I think this can be fixed by noting that $$g_u$$ is linear and so we have that the function $$u \mapsto (u,g_u(s_1(u)+s_2(u))$$ is equal to $$u \mapsto (u,g_u(s_1(u))+g_u(s_2(u)))$$ and so we get addition in $$\mathbb{R}^n$$ which is continuous.

2. I have also read this question where the answer does a similar approach I think. However, it is claimed that it is necessary to show the independence of the trivialization $$f$$. I don't see why this is the case. What I have shown is that using the existence of some trivialization we get that the sum is continuous, I don't see why one would need to show the independence here.

Your approach is correct. For any vector bundle, you correctly defined the sum of two sections, but concerning continuity of the sum you have indeed a gap.

I suggest to proceed as as follows.

1. Consider a trivial bundle $$\pi : Y \times \mathbb R^n \to Y$$. Then we have a bijection
$$b : C(Y,\mathbb R) \to \Gamma(Y \times \mathbb R^n), b(f)(y) = (y, f(y)).$$ Here $$C(Y,\mathbb R)$$ denotes the vector space of continuous real-valued functions on $$Y$$. Using this bijection we easily see that the sum $$s_1 + s_2$$ of sections is again a (continuous) section: We have $$b^{-1} (s_1) + b^{-1}(s_2) \in C(Y,\mathbb R)$$, thus $$b(b^{-1} (s_1) + b^{-1}(s_2)) \in \Gamma(Y \times \mathbb R^n)$$. But clearly $$b(b^{-1} (s_1) + b^{-1}(s_2)) = s_1 + s_2$$. Similarly a scalar multiple of a section is a section.

2. Now consider an arbitrary vector bundle $$p : E \to X$$. Since continuity is a local property, it suffices to show that $$(s_1 + s_2) \mid_U : U \to E$$ is continuous for each open $$U \subset X$$ such that the bundle is trivial over $$U$$ which means that there exists an isomorphism of vector bundles $$f :p^{-1}(U) \to U \times \mathbb R^n$$. Thus it suffices to show that $$f \circ (s_1 + s_2) \mid_U$$ is continuous. We have $$f \circ (s_1 + s_2) \mid_U = f \circ s_1 \mid_U + f \circ s_2 \mid_U$$, where the $$f \circ s_i \mid_U$$ are sections of the trivial bundle $$U \times \mathbb R^n$$. Now apply 1.

Note that the above proof clearly shows that it suffices to consider a single trivialization around each point of $$X$$. No transition functuons between different trivializations are involved.

• Thanks, but isn't what I stated in 1. a fix to the continuity issue? Also, would you mind commenting on 2? I think you indirectly did it by your 2. since you don't show independence of the trivialization either, but I'd like to get a comment on it. Commented Mar 16, 2023 at 11:08
• @user3118 See my update. Perhaps it was not fair to say that you have a gap, but your proof is a bit knotty. Commented Mar 16, 2023 at 11:56
• I would still like to know if my idea in 1. works. What I did was note that the function $u \mapsto (u,g_u(s_1(u)+s_2(u)))=(u,g_u(s_1(u))+g_u(s_2(u)))$ is continuous which gives continuity of the sum of the sections when composing with the homeomorphism $f^{-1}$. Here the linearity is used to translate the fiber sum to the sum on $\mathbb{R}^n$ which we know is continuous. Commented Mar 16, 2023 at 12:02
• I think you edited your comment and write that its knotty. Does that mean it is correct but overly complicated or incorrect and overly complicated? I would like to get verification whether my idea works so that I can learn from it. Commented Mar 16, 2023 at 18:39
• @user3118 It is correct, but you did two steps at the the same time (which I treated separately in my answer). This is what I wanted to say with "knotty". Commented Mar 16, 2023 at 21:19