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So I just noticed that the set of functions with a fixed point $$f(x_0)=x_0,$$ are closed under composition $$(f\circ g)(x):=g(f(x)),$$ and with $e(x)=x$, the inverible functions even seem to form a (non-commutative) group.

Then if one chooses another point $x_1$ and restricts the set to the functions which also have $x_1$ as a fixed point, then it is again closed and so on.

If I have one parameterized point (i.e. a curve, or even a couple of those), then solving $f_t(x(t))=x(t)$ for the families $f_t$ should give me morphisms between the functions for different values of $t$.

Are there general considerations regarding this?

And is this somehow related to the characterization of points of a manifold via the ideal of functions which evaluate to $0$ that point?

Edit 1: Might be just a general property of homeomorphisms or something, although I don't associating picking out isolated fixed points with these kind of things.

Edit 2: I now see that this might relate a translation/transformation of points in the manifold to a transformation of the function algebra over that manifold. This has some features: If you take two points $y_1$ and $y_2$ and transformations along the curves $Y_1(t),Y_2(t)$ with $Y_1(0)=y_1, Y_1(1)=y_2$ and $Y_2(0): =y_2, Y_2(1)=y_1$ (they move into each other), then the fuction set with both fixed points $Y_1(t),Y_2(t)$ makes a loop as $\{Y_1(0),Y_2(0)\}=\{Y_1(1),Y_2(1)\}=\{y_1,y_2\}$. The particular form of the curves have an impact on how the function set looks in between.

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+1 interesting thing... –  draks ... Aug 9 '12 at 8:46
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Picking a fixed point just corresponds to removing that point from the domain and range of your functions (when looking at the invertible functions that is). –  Tobias Kildetoft Aug 9 '12 at 8:49
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Congratulations: you have discovered the category of pointed sets/spaces/manifolds/etc. –  Zhen Lin Aug 9 '12 at 9:35
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@Tobias: But that's only true for in the category of sets, not in the category of topological space, for instance. Right? –  Rasmus Aug 9 '12 at 9:37
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You are looking at the stabilizer subgroup for a group action on a set. –  i. m. soloveichik Sep 13 '12 at 15:11
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1 Answer

I see that this question hasn't received a full answer and that may be due to that vast number of fields which study fixed points and the lack of specificity of which objects you're interested in (although the points you have made in fact were quite general and apply to many objects). However, I'll try to answer in the context of manifolds.

We begin with a manifold $M$ and the space of all orientation preserving homeomorphisms from $M$ to itself which we will denote by $Homeo(M)$ (if $M$ is smooth we can in fact restrict to diffeomorphisms of $M$). We may also restrict to homeomorphisms which fix a particular set $N\subset M$ and we will denote this subgroup of $Homeo(M)$ by $Homeo(M,N)$. It is a subgroup for the very reasons you mentioned in your post. Now, $Homeo(M)$ is intractable. Therefore, it is often useful to examine this group up to some sort of equivalence. In this case, we may choose to use isotopy equivalence (i.e. two homeomorphisms are equivalent if there exists an arc of homeomorphisms between them). We denote this group up to equivalence as $\pi_0(Homeo(M,N))$.

The reason why certain subgroups which fix a subset of $M$ are interesting can be demonstrated in the following example: consider a closed annulus $A$. Denote the boundary of $A$ as $\partial A\cong S^0\times S^1$. Also consider the subgroup of $Homeo(A)$ which fix $\partial A$, $Homeo(A,\partial A)$. It is easy to visualize why $\pi_0(Homeo(A))\cong 0$ whereas $\pi_0(Homeo(A,\partial A)\cong\mathbb{Z}$. Intuitively, this means that homeomorphisms of the annulus can be undone if we allow isotopies which do not fix the boundary (we can "untwist"); however, working with a fixed boundary gives us a non-trivial mapping class group ($\pi_0(Homeo(M,\partial M)$). This notion is discussed in detail in "A Primer on Mapping Class Groups" by Benson Farb and Dan Margalit. Hope this demonstrates one use of this idea that you discovered. There are many many others.

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I hope you don't mind but I formatted your answer into paragraphs so that it was easier to read. –  Daniel Rust Apr 20 at 20:41
    
Not at all! Its much easier to read, thank you! –  Joseph Zambrano Apr 21 at 2:06
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