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My topology book says that "A function $f:U \to \mathbb{R}^m$ from an open set $U$ in $\mathbb{R}^n$ into $\mathbb{R}^m$ is smooth provided that $f$ has continuous partial derivatives of all orders. A function $f:A \to \mathbb{R}^m$ from an arbitrary subset $A$ of $\mathbb{R}^n$ to $\mathbb{R}^m$ is smooth provided that for each $x$ in $A$ there is an open set $U$ containing $x$ and a smooth function $F:U \to \mathbb{R}^m$ such that $F$ agrees with $f$ on $U \cap A$."

It really just leaves things at that, assuming knowledge of calculus including partial derivatives (which I do have). What I'm curious about is...

Is there a topological notion of the derivative? If there is not, is there a generalization of the derivative designed to allow the notion to make sense in a purely topological context?

I have never seen any references to such an idea. There is a topological notion of limit (see here), but can this be used to define a topological definition of the derivative?

Alex Petzke
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  • General and algebraic topology, not really to my knowledge. Differential topology though studies diffeomorphisms (smooth maps) - what you define above. As we talk locally euclidian then, we naturally can rely on knowledge of such spaces (such as partial derivatives). This doesn't answer your question, but you probably won't find a 'topological derivative' outside differential topology. – gnometorule Jan 07 '13 at 17:47
  • This question may be of interest as well – SBF Jan 07 '13 at 18:33
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    I agree with gnometorule that outside diff geometry there's likely not enough structure to discuss derivatives. Do you know about: http://en.wikipedia.org/wiki/Tangent_bundle ? For example differential equations evolving on manifolds are generated by vector fields that are contained in the tangent bundles. – alancalvitti Jan 09 '13 at 04:23
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    Similar: https://math.stackexchange.com/questions/1267268/why-cant-differentiability-be-generalized-as-nicely-as-continuity – Akiva Weinberger Jan 20 '20 at 19:44
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    See https://math.stackexchange.com/a/3966531/76284. – user76284 Dec 30 '20 at 19:25

4 Answers4

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There can be no purely topological definition of deriviative, because neither is the notion of differentiability preserved under homeomorphisms, nor (in cases where it happens to be preserved) does the derivative transform well under homeomorphisms (for instance the derivative could be nonzero before, and zero after application of a homeomorphism). General topology simply does not deal with notions of differentiation; you need a different category than topological spaces for that (for instance that of differentiable manifolds).

Marc van Leeuwen
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  • Do you mean that if we have two spaces, $X$ and $Y$, a function $f:X \to X$, and a homeomorphism $h:X \to Y$, then it's not necessarily true that for $x \in X$ we have $f'(x) = f'(h(x))$? I don't know if that necessarily makes sense (does $f$ 'act the same' on $X$ and $Y$ since they're homeomorphic?), I just need some clarification on what exactly you're saying. I do get the general idea though. – Alex Petzke Jan 08 '13 at 03:10
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    @AlexPetzke: In the situation you describe one would in fact be interested in the derivative at $h(x)$ of the "conjugate" $g=h\circ f\circ h^{-1}$. If $h$ were a diffeomorphism that derivative would be related to $f'(x)$ by $g'(h(x))=h'(f(x))\circ f'(x)\circ h'(x)^{-1}$, but if $h$ is just a homeomorphism this makes no sense, and there is no clear relation at all between $f'(x)$ and $g'(h(x))$. – Marc van Leeuwen Jan 08 '13 at 07:54
  • Alright. Thanks for the input. – Alex Petzke Jan 08 '13 at 16:31
  • @Marc van Leeuwen: What about the book: Topology from the differentiable viewpoint? – Adam Dec 02 '13 at 07:42
  • @Adam: What indeed? I don't know the book. – Marc van Leeuwen Dec 02 '13 at 07:59
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I am not sure if you count this as topological $X$ and I must admit that this might be unnecessarily "high-brow" but if your topological space is a locally ringed space (it's not that bad - just think of attaching a ring to every open set in some coherent way and when you "zoom" into a point $x$ you get a local ring $O_{X,x}$ - a ring with only one maximal ideal. Think manifolds or Euclidean space with rings of continuous $\mathbb{R}$-valued functions at each point), then one may define the (co)tangent space at $x$ as the vector space $m_x/m_x^2$ over the base field $O_{X,x}/m_x$ where $m_x$ is the unique maximal ideal of $O_{X,x}$.

The motivation of (the purely algebraic process of) quotienting out by the 2nd power of the ideal is exactly capturing the intuition of a derivative - you want to linearize everything in sight. This is what's done in algebraic geometry where the intuitive notion of smoothness is trickier and sometimes absent, but you still want to somehow have them anyway.

Hope that was at least fun!

Elden Elmanto
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  • woops I guess this was actually from long ago... – Elden Elmanto Nov 03 '13 at 14:08
  • Not a problem at all! I have sort of read what you said in your first paragraph, but your second paragraph I really got something from. I never had any idea why the quotient by the square of the ideal is taken. That's very interesting. I do understand the relationship between linearity and the derivative, but could you say something more about what it means to linearize and how quotienting by the square of the ideal accomplishes this? – Alex Petzke Nov 03 '13 at 15:59
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    Dear @AlexPetzke, of course! The simplest example is to look at the following case (technically the ring is not local but it's okay) - if you look at the polynomial ring $k[x]/(x^2)$ we see that all expressions here are of the form $ax + b$. You cannot have any higher powers - this is the way in which you are restricting yourself to only "first order" or "linear information" when you look at the ring. – Elden Elmanto Nov 04 '13 at 03:20
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    Could you give an example for a non-polynomial ring? I can't see as intuitively how it works in that case. Maybe I'll even make this a question of its own. – Alex Petzke Nov 04 '13 at 17:54
  • You can define the cotangent space this way, but isn't it trivial in the topological category? Meaning $m_x^2=m_x$, since for any function $f$ vanishing at $x$ we may write $f=\sqrt{|f|}\cdot\sqrt{|f|}$ hence the cotangent space is trivial. This doesn't work in the smooth or algebraic category, because $\sqrt{f}$ is not smooth or algebraic. – ziggurism Dec 21 '18 at 17:21
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Usually to define a derivative you need a smooth structure, because as Marc van Leeuwen points out in his answer, critical points are otherwise not preserved under homeomorphism.

I am submitting this answer just to point out the existence of a tangent microbundle in the category of topological manifolds. This may get closer to a "purely topological" notion of derivative.

ziggurism
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I’m sorry .My last answer was not transferred properly .How can I send the mathematics symbols? However I am sending that again. I asked a similar question in www.artofproblemsolving.com, with topic name topological derivation definition. But I didn’t receive any answer. I’d like to explain more about this Idea. Indeed it seems that derivative and integral are concepts which are independent of coordinate system and maybe even independent of function concept. A geometric view of derivative is tangent. A tangent line of a curve at a point is a line which is very close to the curve at close points .This closeness can be interpreted by topological tools. On the other hand the integral of a function ∫abf(x)dx apparently depends on coordinate system, but by more inspection we see that coordinate system is only the boundary of integration .

Let’s denote a curve in a plane by the set cl(E) –E which E is open and is the greatest possible set in this respect. The following definition can be suggested for derivative: C= cl(E)-E , E is the greatest possible set. C is the curve. cl(NWEWV) -(NWEWV) = [cl(N)W(cl(V)-V)] [cl(V)W(cl(N)-N)] =( cl(N)W(cl(V))-( VWN) E,V and N are open .VT (a family of open sets) The important point in this way is the selection of type of V. If V be the open areas between two crossed line we get to ordinary derivative but we may select another sets. H= Vi (on all solutions of the above equation)  S1, S2 / cl(V)-V= S1 S2 , S1WS2 ={x} x N D= cl(H)-H , H is the greatest possible set. D is the derivative. But the above definition is useless and can’t be used to combination of function that is very important.

christ
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