# Why Riemannian metrics have to be smooth?

Why do Riemannian metrics have to be smooth? Can you give an example of a smooth curve with a none smooth metric and show me what possibly will go wrong if our metric is not smooth?

• One can do quite a bit of differential geometry with relaxed smoothness assumptions, I think. Often, “smooth” (in the sense of $C^\infty$) just means “enough smoothness for these proofs to go through”. As for the metric, you begin to struggle a bit if you don't at least have enough smoothness so that the differential equations defining the geodesics have unique solutions. Sep 13, 2014 at 7:49
• By paracompactness you can always create a smooth metric using partition of unity (or, in a more fancy way, you can find a deformation retract of the general linear group to the special orthogonal). Maybe you want to know an example of a not smooth but continuous metric… If you know about vector bundles I think you can come with an example. Anyway, at least continuity is necessary otherwise you could have an orthonormal basis on the tangent bundle going totally crazy, but since you want to differentiate, smoothness would be nice too (because of the determinant). Sep 13, 2014 at 8:11

To begin with, nothing has to be smooth. There are deep theories concerning manifolds which are only $C^k$ for some $k$, or topological manifolds, which don't even have to be differentiable.
It is the same with the metric tensor. This tensor actually gives an isomorphism between the tangent and cotangent spaces at every point on the manifold. When it is also smooth, it yields an isomorphism of vector bundles between the tangent and cotangent bundles, i.e. it transforms smooth $1$-forms into smooth vector fields, and vice versa. Just one example, out of many - Let $f:M\to\mathbb{R}$ be a smooth function, so the derivative $df$ is a smooth $1$-form, and the metric induces the corresponding vector field $\nabla f$, the gradient. If the metric is smooth, then so is $\nabla f$, thus it has a smooth flow function and so on. If the metric is not smooth, then neither is the obtained gradient, and it just means that this vector field does not fit into the smooth category.
• @user1 A Riemannian metric is indeed a tensor field. In particular, it is a map $g:M\to T^*M\otimes T^*M,$ and as such, it may be or not be $C^k$ for some $k\in\mathbb{N}$. Jun 11, 2019 at 14:05