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The ordinary definition of a tangent space uses the differentiable structure of differentiable manifolds and is hence not applicable to topological manifolds.

However for locally ringed spaces one can define the tangent space as the dual of the vector space obtained as a quotient by its maximal ideal, i.e. $\mathfrak{m}/\mathfrak{m}^2$.

Why is this latter construction not applicable to topological manifolds?

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  • $\begingroup$ Maybe because not every manifold is a variety (i.e. definable as the zero set of a system of multivariate polynomials). It's just a guess. $\endgroup$ – stressed out Aug 10 '18 at 15:34
  • $\begingroup$ But isn't every topological manifold a locally ringed space? I thought the stalks of the structure sheaf (of continuous functions) where local rings. Or are they not local? $\endgroup$ – NDewolf Aug 10 '18 at 15:40
  • $\begingroup$ Maybe you can do this but you don't get anything resembling a useful/traditional "tangent space"? I can't even fathom what that would mean outside the smooth context. $\endgroup$ – Randall Aug 10 '18 at 15:50
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If you look at the ring of germs of functions of a topological manifold at a point (you might as well take the point to be the origin in the manifold $\Bbb R^n$), you'll get a local ring, but a very non-Noetherian one. If the ring is $A$ and the maximal ideal $m$, then $A/m\cong\Bbb R$, as it should, bur $m/m^2$ will be infinite-dimensional over $\Bbb R$, so it's difficult to use it as a cotangent space at that point.

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  • $\begingroup$ So the construction is possible but just not very useful? $\endgroup$ – NDewolf Aug 10 '18 at 15:59
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    $\begingroup$ I am a bit confused by this answer: isn't it true that any square root of a (germ of) continuous function at a point is again continuous and therefore $m/m^2$ is trivial (like its dual: i.e. the space of all derivations at a point)? I'd rather say that this procedure gives a trivial tangent space than an infinite-dimensional one. $\endgroup$ – N. Ciccoli Aug 22 '18 at 6:30

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