# Is there a topological notion of the derivative?

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?

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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 7 at 17:47
This question may be of interest as well –  Ilya Jan 7 at 18:33
I agree with gnometorule that outside diff geometry there's likely not enough structure to discuss derivatives. Do you know about: 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 9 at 4:23

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).

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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 8 at 3:10
@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 8 at 7:54
Alright. Thanks for the input. –  Alex Petzke Jan 8 at 16:31
@Marc van Leeuwen: What about the book: Topology from the differentiable viewpoint? –  Adam Dec 2 at 7:42
@Adam: What indeed? I don't know the book. –  Marc van Leeuwen Dec 2 at 7:59

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!

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woops I guess this was actually from long ago... –  Elden Elmanto Nov 3 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 3 at 15:59
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 4 at 3:20
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 4 at 17:54