# Is it true that any Riemannian metric on cylinder can be deformed to product metric?

Is it true that any Riemannian metric on cylinder $$\Bbb S^1\times \Bbb R$$ can be deformed to standard product metric?

Are there some standard references about such classifications?

Yes, this is true. Here is a not-quite-full generalization. Unfortunately, I do not know of any references for such questions.

Proposition. Let $$(M, g)$$ be a Riemannian manifold of dimension $$m$$ such that $$M$$ admits a global orthonormal frame $$e_1, \dots, e_m$$. (This means that each $$e_i$$ is a vector field on $$M$$, and for every $$p \in M$$ the vectors $$e_i(p)$$ form an orthonormal basis for $$T_p M$$.) Then any other Riemannian metric on $$M$$ continuously deforms into $$g$$.

Remarks. The cylinder $$\mathbb{S}^1 \times \mathbb{R}$$ with the product metric is an example of such an $$M$$. I believe this proposition is true even when $$M$$ doesn't admit a global orthonormal frame; see my remarks at the bottom of the post.

Proof. If $$h$$ is some other Riemannian metric on $$M$$, then we get a map $$\phi_h: M \to GL_m \mathbb{R} / O_m\mathbb{R}$$ as follows.

Given $$p$$ in $$M$$ choose an $$h$$-orthonormal basis $$f_1, \dots, f_m$$ for $$T_pM$$. Writing $$e_i(p) = \sum_j a_{ij} f_j$$, let $$A$$ be the matrix whose $$(i,j)$$-th coefficient is $$a_{ij}$$.

If $$f_1', \dots, f_m'$$ is some other oriented $$h$$-orthonormal basis, then we write $$e_i(p) = \sum_j a'_{ij} f_j'$$.

Write $$f_i' = \sum_j b_{ij} f_j$$. Then the matrix $$B = (b_{ij})$$ is in $$O_m \mathbb{R}$$, and $$e_i(p) = \sum_j a'_{ij} f_j' = \sum_{j, k} a'_{ij} b_{jk} f_k,$$ or $$a_{ik} = \sum_j a'_{ij} b_{jk}$$, which shows that the matrix $$A$$ is well-defined up to multiplication on the right by an element of $$O_m \mathbb{R}$$.

The point is that (given our choice of global frame) deformation classes of metrics on $$M$$ are classified by homotopy classes of maps into $$GL_m \mathbb{R} / O_m \mathbb{R}$$. But the Gram-Schmidt process shows that $$GL_m \mathbb{R} / O_m \mathbb{R}$$ is homotopy equivalent to a point. $$\square$$

Remarks. More generally, one might wish to pay attention to orientation issues in the above; you could ask whether there is an orientation-preserving deformation, in which case you get a map into $$GL_m \mathbb{R} / SO_m \mathbb{R}$$, which is homotopy equivalent to two points, and the argument goes through as before (assuming the oriented metric you want to preserve has the same orientation as the one you want to deform it into).

Making the deformation above smooth should be straightforward; I just didn't want to think about it.

If $$M$$ is not globally trivializable, there's still a fiber bundle $$E$$ over $$M$$ whose fibers are isomorphic to $$GL_m\mathbb{R} / O_m \mathbb{R}$$ whose sections classify Riemannian metrics on $$M$$. I believe it is still true then that the space of global sections of $$E$$ is contractible (since the fibers $$GL_m\mathbb{R} / O_m \mathbb{R}$$ are contractible); see this Math:SE post, for instance, though I have not gone through the details myself. It should follow that the assumption that $$M$$ admits a global frame can be eliminated in the proposition above.