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I read this:


Let $(M,g)$ be a compact Riemannian manifold and let $W$ be a vector bundle (rank $n$) over $M$ with $h_W$ a bundle metric of $W$ and $D$ a bundle connection of $W$.

I choose $W$ to be the trivial vector bundle $M \times \mathbb{R}$.

Let $V=T^0_0 M \otimes W = W$ be the vector bundle of $W-$valued (0,0)-tensors on $M$.

Endow $V$ with the bundle metric $$h := (\cdot, \cdot)^0_0 \otimes h_W$$ where $(\cdot, \cdot)^0_0 := g^{\otimes 0}\otimes g^{*\otimes 0}$ is the bundle metric on $T^0_0M$ induced by $g$. Equip $V$ with the metric connection $$\nabla = \nabla(\nabla_g, D)$$ induced by the Levi-civita connection of $M$ and connection $D$ of $W$.

It follows that $\nabla u = Du$ for $u \in \mathcal{T}^0_0(M, W)$.


  1. Am I right that in my case $h = h_W$? I am very confused about how the notation should be simplified for my simple trivial bundle case.
  2. Now the last sentence seems wrong to me. Don't we need $\nabla u = \nabla_g u$ where $\nabla_g$ is the Levi-Civita connection on $M$, whenever $u \in \mathcal{T}^0_0(M, W)$ (i.e., whenever $u$ is a smooth function on $M$)???

The purpose of all this stuff is to set up Sobolev spaces over vector or tensor bundles. I only need sobolev spaces over $M$, so I thought of taking the trivial vector bundle like I wrote above in order for the theory to simplify, but I have a lot of trouble understanding. My second point for example, I would expect $\nabla$ to be the ordinary gradient for a function on $M$ but it's not. It's a different connection $D$, and I don't know what I should choose $h_W$ and $D$ to be to make it work. Thanks.

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    $\begingroup$ What does it mean to give a bundle the Riemannian manifold's metric and connection when the bundle isn't $TM$? $\endgroup$
    – user98602
    Jan 15, 2015 at 15:55
  • $\begingroup$ Please see my updated question. $\endgroup$
    – aaa
    Jan 15, 2015 at 17:15

1 Answer 1

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This is a pretty subtle issue. For a section $s$ of $W$, $s=1\otimes s=s\otimes 1$, so you'd think you could write $\nabla(s)=\nabla_g s\otimes 1+s\otimes D 1=\nabla_g s\otimes 1+0= \nabla_g s$. This only works assumes $D 1=0$. If this holds, then indeed we have $D=\nabla_g$. And for a metric connection on a bundle isometric to the trivial line bundle, this is the only choice of $D$, because the metric uniquely determines the flat sections as those of constant norm.

On the other hand, it's worth mentioning that for non-metric connections we could take $d+\omega$ with $\omega$ an arbitrary 1-form for the connection on $W$, and then the constant sections would not be the flat sections. And this could easily come up if your $h_W$ isn't the standard metric: the flat sections for the metric connection will be those of constant length for $h_W$. But even in that case your $h_W$ will coincide with $h$, which is the usual metric multiplied by $h_W(1,1)$: $$h(f_1\otimes s_1,f_2\otimes s_2)=h(f_1s_1\otimes 1,f_2s_2\otimes 1)=(f_1s_1,f_2s_2)^0_0h_W(1,1)=h(1\otimes f_1s_1,1\otimes f_2s_1)=h_W(f_1s_1,f_2s_2)$$

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  • $\begingroup$ thanks for the reply. If I understood you, if we constrain $D$ to be a metric connection (which we have to, according to the text), then in the trivial vector bundle case we must have $D$ equalt to the manifold $M$'s connection. Is it correct? In this case I get what I expected and wanted (regarding the last paragraph in my OP). $\endgroup$
    – aaa
    Jan 18, 2015 at 10:49
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    $\begingroup$ Yep, that's right. $\endgroup$ Jan 18, 2015 at 14:56
  • $\begingroup$ Could you still write this answer today? (it's a nice one!) $\endgroup$
    – user98602
    Oct 6, 2018 at 13:25

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