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Fix $n\in\mathbb{N}$. A vector bundle of rank $n$ is a smooth map $\pi:E\rightarrow B$ between manifolds such that $\forall p\in B: E_p := \pi^{-1}(p)$ is an $n$-dimensional vector space and $\forall p\in B$, there exists a neighborhood $U$ of $p$ and a diffeomorphism $\psi:E\mid_U:=\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ such that $\operatorname{pr}_1\circ\psi = \pi$ and $\psi\mid_{E_q}:E_q\rightarrow\{q\}\times\mathbb{R}^n$ is a vectorspace isomorphism for all $q\in U$.

Suppose $E$ is a vector bundle over a manifold $M$. (I suppose by this they mean there exists $\pi:E\rightarrow M$ as above). Prove that for all $x\in M$ one can construct an inner product (symmetric, positive definite, bilinear form): $g_x : E_x\times E_x\rightarrow \mathbb{R}$ which depends smoothly on $x$.

"$g$ depends smoothly on $x$" means: $g(v,w)$ is a smooth function on $M$ for all smooth sections $v,w$ of $E$.

I have tried to construct this norm by using the standard norm on $\mathbb{R}$. Take $x\in M$. There is an open neighborhood $U$ of $x$ for which there exists a diffeomorphism $\psi$ as above. Let us define $g_x((a,b)) = <\operatorname{pr}_2(\psi(a)),\operatorname{pr}_2(\psi(b))>_\mathbb{R}$. Since this is symmetric, $g_x$ will be symmetric. Since this is positive definite, $g_x$ will also be positive definite. Now for bilinearity. For $a_1,a_2\in E_x$, we have $\psi(a_1)+\psi(a_2) = \psi(a_1+a_2)$, since $\psi$ is an isomorphism. And since $\psi(a_i)\in\{x\}\times\mathbb{R}^n$. $\operatorname{pr}_2(\psi(a_1)+\psi(a_2))=\operatorname{pr}_2(\psi(a_1))+\operatorname{pr}_2(\psi(a_2))$. Since we have a vector space isomorphism we can also show this for scalar multiplication. I am unsure however what this field of scalars is exactly.

Then for the final part I need to show that $g$ depends smoothly on $x$. But here I am lost completely. How can we evaluate a section in $g$? As far as I know there is no correspondence between sections and elements of $E_x$. A hint I was given is that we can use partitions of unity, but I have no clue how this ties in with what I have constructed.

Jarne Renders
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1 Answers1

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Suppose your vector bundle is $\pi: E\rightarrow M$ and has rank $n$. Let $U_\alpha$ be a cover of $M$ such that $\varphi: \pi^{-1}(U_\alpha)\rightarrow U_\alpha\times \mathbb{R}^n$ are local trivialisations. To construct a smooth inner product on $\pi^{-1}(U_\alpha)$, iust take the inner product $g_\alpha$ on $U_\alpha\times \mathbb{R}^n$ which at $p\in U_\alpha$ is just the standard inner product on $\mathbb{R}^n$. It is not hard to show that this is smooth: just use local charts.

Now take a partition of unity $\{\lambda_\alpha\}_{\alpha\in A}$ subordinate to the cover $\{U_\alpha\}_{\alpha\in A}$. Then $$ g = \sum_{\alpha\in A}\lambda_\alpha g_\alpha\,, $$ is a smooth metric on $\pi: E\rightarrow M$. This metric is clearly smooth since for sections $s,t$ of $E$ we have that $g_\alpha(s,t)$ is smooth, and $\lambda_\alpha$ is smooth. I'll leave it to you to show that $g$ is indeed a metric on $E$.

Andre of Astora
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  • @EpsilonDelta I'm not sure what you mean by $g$ is defined on $M$. The way I have defined it $g$ is a map from $\Gamma(E)\times \Gamma(E)$ to $C^\infty(M) $. This is because $g_\alpha$ similarly maps pairs of sections of $U_\alpha\times \mathbb{R}^n$ to $C^\infty(U_\alpha) $. – Andre of Astora Nov 11 '19 at 11:17
  • @EpsilonDelta My apologies. I mean $\pi^{-1}(U_\alpha)$. – Andre of Astora Nov 11 '19 at 11:24
  • Yes regarding your first comment, I do mean to make $g$ a map from $\Gamma(E)\times\Gamma(E)$ to $C^\infty(M)$, where an element $\Gamma(E)$ is the space of smooth maps $s:M\rightarrow E$ such that $\pi\circ s = \text{id}$ - called a smooth section of $E$. A smooth metric on this vector bundle is a map $g:\Gamma(E)\times\Gamma(E)\rightarrow C^\infty(M)$. The standard way to check that a choice of inner product $g_p$ for $\pi^{-1}(p)$ at each $p\in M$ is indeed a metric is to check that $g(s,t)$ is smooth for all $s,t\in \Gamma(E)$. – Andre of Astora Nov 11 '19 at 11:33
  • Yes it is smooth however it doesn't give the object you're looking for. You want $(g_\alpha)p$ to be an inner product of $\pi^{-1}(p)$ for $p\in U\alpha$, but $\pi^{-1}(U_\alpha)\times\pi^{-1}(U_\alpha)$ contains points such at $((p,v),(q,w)$ where $p$ and $q$ are distinct points. Rather, you need a fibre product. See https://en.wikipedia.org/wiki/Bundle_metric – Andre of Astora Nov 11 '19 at 11:49
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    So I'm a little bit confused with the definition of $g$. If I understand it correctly $g_\alpha$ is a norm on $U_\alpha \times \mathbb{R}^n$ so it is a map from $(U_\alpha \times \mathbb{R}^n) \times (U_\alpha \times \mathbb{R}^n)$ to $\mathbb{R}$, while $\lambda_\alpha$ is a map form $M$ to $\mathbb{R}$. So how do we get that $g$ is a map from $\Gamma(E) \times \Gamma(E)$ to $C^\infty(M)$? – Mee98 Nov 11 '19 at 14:12
  • @VishnuM How can one construct an inner product $g_\alpha$ on $U_\alpha\times \mathbb{R}^n$ like that? What is $g_\alpha((p,x),(q,y))$? I do not see how this is well defined. – Jarne Renders Nov 11 '19 at 17:25
  • @JarneRenders when I say an inner product on $U_\alpha\times\mathbb{R}^n$ I mean an inner product on $\mathbb{R} ^n$ at each $p$. – Andre of Astora Nov 11 '19 at 20:21