Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How is that any other vector bundle over a Riemannian manifold can be turn into a Riemannian manifold as well?

I think that the question is of interest due to the presence in math.SE of several related entries.

Edit:

I would emphasize that the definition of vector bundle is the standard that one can look at, say, wikipedia.

I am reporting that the presence in of the word metric is a cause to my confusion.

One shouldn't use the word metric to define "Riemannian structure" on a manifold and not to confuse with the concept of "metric as a distance function".

share|improve this question
5  
Riemannian manifolds are smooth manifolds. Vector bundles over smooth manifolds are smooth manifolds. Any smooth manifold can be given a Riemannian metric. –  Neal Jan 12 at 4:21
2  
Yes, as long as your definition of manifold includes paracompact, you're done. –  Ted Shifrin Jan 12 at 4:27
2  
Math.se is not the appropriate place to make challenges. If you want to ask something, go ahead and ask it; if not, well... –  Mariano Suárez-Alvarez Jan 12 at 5:06
2  
This forum is at the appropriate level for lots of questions. I think the debate here is due to your issuing the problem/question as a challenge and, to make matters worse, making the completely vague statement "This question can be solved with ideas recently displayed here." What are we all supposed to make of that? Scour the past weeks of questions? ... And, yes, the "standard construction" to which I referred is partitions of unity (hence the importance of paracompactness in the definition of topological manifold). –  Ted Shifrin Jan 13 at 18:17
3  
There is a way to solve the problem without partitions of unity, using only elementary constructions. Basically the idea is to find a way to put a Riemann metric on the fibres of the new vector bundle, then take the product Riemann metric. –  Ryan Budney Jan 16 at 18:47

1 Answer 1

That a $M$ is a Riemannian manifold means that there is a $C^{\infty}$-section $$g:M\to TM\otimes TM,$$ which each $g_p$ is a bilinear, symmetric, positively-definite and non-degenerated map $g_p:T_pM\times T_pM\to{\Bbb{R}}$.

For the tangent bundle $TM$ can be induced a Riemannian structure assembling a $C^{\infty}$-section $$\bar{g}:TM\to TTM\otimes TTM,$$ where $TTM$ is tangent bundle of $TM$:

Remembering that there is a well defined $\pi_*:TTM\to TM$ called the differential of $\pi$, one of this sections is $\bar{g}$ is given by $$\bar{g}_{(V,p)}(X,Y)=g_p(\pi_*X,\pi_*Y)+g_p(D_{\alpha'}V,D_{\beta'}V),$$ where $V\in T_pM$, $\alpha:I\hookrightarrow TM$ and $\beta:J\hookrightarrow TM$ are curves that comply $\alpha(0)=\beta(0)=p$ as well as $$\alpha'(0)=X\quad\mbox{and}\quad \beta'(0)=Y.$$

Similarly for the cotangent bundle $TM^*$ a section $TM^*\to TM^*\otimes TM^*$ can be constructed.

First, consider that for a $f$ covector in $T_pM^*$ there exists a $F\in T_pM$ such that $f(x)=g_p(F,x)$ for all $x\in T_pM$. This $f$ is the Riesz's representative for $f$. It is standard that the assemble: $$g^*_p(h,k)=g_p(H,K)$$ with $H,K$ representing to $h,k$ respectively, gives a inner product $g_p^*:T_pM^*\times T_pM^*\to{\Bbb{R}}$.

So for the $TM^*$ we define $$\tilde{g}_{(f,p)}(h,k)=g_p(\pi_*H,\pi_*K)+g_p(D_{\mu'}F,D_{\nu'}F),$$ for $F\in T_pM$ representative for $f$ and for a pair of curves $\mu:I\hookrightarrow TM$ and $\nu:J\hookrightarrow TM$ such that $\mu(0)=\nu(0)=p$ and $\mu'(0)=H$ and $\nu'(0)=K$, both $H,K$ representatives for $h,k$ respectively.

Then, this two previous sections can be carried to any finite rank vector bundle $E\to M$ to a section $$E\to TE\otimes TE,$$ that I'll describe at posterior updates.

It is also pending to extend this partial answer to the case of infinite rank vector bundle over $M$.

share|improve this answer

Your Answer

 
discard

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.