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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?

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  • $\begingroup$ 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. $\endgroup$ Sep 13, 2014 at 7:49
  • $\begingroup$ 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). $\endgroup$
    – user40276
    Sep 13, 2014 at 8:11

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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.

But when considering the category of smooth manifolds and smooth maps, certain constructions have to be smooth in order for the resulting objects to fit into this category. For example, the inverse image of a regular value under a smooth map is again a smooth manifold. If we omit the requirement that the initial map is smooth, the resulting manifold will not be smooth either (actually, sometimes it will not be a manifold at all). Alternatively, consider smooth vector fields. We know that such a vector field has a flow function which induces diffeomorphisms from the given manifold to itself. If the vector field is not smooth, the resulting automorphisms will not be smooth either, thus will not preserve the basic structure of any object in this category.

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.

There are many other examples for things that "go wrong" if the metric is not smooth. The "bad" outcome is always that some obtained map or manifold or any other object will not be smooth either. It is worth mentioning that this "bad" outcome is not always so bad. As written above, there is a lot to do with non-smooth constructions. However, there are some results which require smoothness or at least the existence of many derivatives, and these results may fail to hold if one uses a non-smooth metric.

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  • $\begingroup$ How does one define smoothness of a metric? In my text it is a tensorfields on the product of two modules, I dont see how one can define smoothness from that. $\endgroup$
    – user123124
    Jun 10, 2019 at 13:43
  • $\begingroup$ @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 $k\in\mathbb{N}$. $\endgroup$ Jun 11, 2019 at 14:05

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