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For the definition of germ, please see below.

I am having some difficulty internalizing the concept of germ due to an inability to think of concrete examples, which led to me having the following questions:

1. Are the germs of holomorphic functions (at a point $p$) simply single-element equivalence classes, because the Identity theorem implies that any two holomorphic functions which agree identically on a neighborhood are identical on their entire domains of definition?

2. Can one describe the members of the equivalence class of a smooth germ explicitly for simple enough examples? E.g. Let $M=\mathbb{R}$, and let $f(x)=x$, then is the germ of $x$ at $0$ just $$[x]_p=\{g\in C^{\infty}: g(0)=0, g'(0)=1,g^{(r)}(0)=0\ \forall\ r \ge 1 \}? $$

3. Do similar results hold for the members of the equivalence class of a $C^k$ germ, e.g. $$[x]_p = \{ g \in C^k: g(0)=0, g'(0)=1, g^{(r)}(0)=0\ \forall\ 1 \le r \le k \}? $$

4. Two distinct (real or complex) analytic functions cannot coincide on a neighborhood of a point -- is this the smallest class of functions for which this holds? (I.e. are germs non-trivial precisely for classes of functions which do not always coincide with their Taylor series, and are the equivalence classes simply the functions with the same Taylor series up to a certain order?)

My conjectures are motivated by the idea of derivatives being "infinitesimal" or local approximations of functions, as well as the fact that the standard example of a non-analytic smooth function is the only function I can think of which belongs to the germ of another function, but other than this intuition I have no reason to think that the germs of analytic, smooth, or differentiable functions can be described in the manner above.

That and this comment in Lee (on p.72):

The germ definition has a number of advantages. One of the most significant is that it makes the local nature of the tangent space clearer, without requiring the use of bump functions. Because there do not exist analytic bump functions, the germ definition of tangent vectors is the only one available on real-analytic or complex-analytic manifolds.

The only reason I might suspect that these are false is that they would lead to a much simpler definition (in my opinion) then the one given in the book. On the other hand, a definition of germs based on equality of Taylor coefficients would not generalize very well to classes of continuous functions and the like, which is perhaps the intent.


Definition: (taken from p.71 of Introduction to Smooth Manifolds by John Lee):

A smooth function element on [a smooth manifold] $M$ is an ordered pair $(f,U)$, where $U$ is an open subset of $M$ and $f: U \to \mathbb{R}$ is a smooth function. Given a point $p \in M$, let us define an equivalence relation on the set of all smooth function elements whose domains contain $p$ by setting $(f,U) \sim (g,V)$ if $f \equiv g$ on some neighborhood of $p$. The equivalence class of a function element $(f,U)$ is called the germ of $f$ at $p$. The set of all germs of smooth functions at $p$ is denoted by $C_p^{\infty}(M)$... Let us denote the germ at $p$ of the function element $(f,U)$ simply by $[f]_p$; there is no need to include the domain of $U$ in the notation, because the same germ is represented by the restriction of $f$ to any neighborhood of $p$. To say that two germs $[f]_p$ and $[g]_p$ are equal is simply to say that $f \equiv g$ on some neighborhood of $p$, however small.


Related questions (in which I could not find the answer): (1) (2) (3) (4) (5) (6) (7) (8) (9)

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    $\begingroup$ For 2, no you can't rely only on the derivatives at one point for comparing two $C^\infty$ functions. The counter-example is $f(x) = e^{-1/(1-x^2)}1_{|x|< 1}$ which is $C^\infty$ and whose derivatives of every order at $x=1$ vanish. Now if you assume analyticity, then yes comparing the derivatives at one point is enough, by the Taylor series and the identity theorem. $\endgroup$
    – reuns
    Commented Nov 1, 2016 at 20:34
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    $\begingroup$ If you're looking for the analog of Taylor polynomials on a manifold, these are called jets. $\endgroup$
    – ಠ_ಠ
    Commented Nov 1, 2016 at 20:43

1 Answer 1

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  1. The equivalence class of a function element $(f,U)$ around $p \in M$ will never consist of a single element unless the topology of $M$ is such that you can find an open neighborhood $V$ of $p$ which is "minimal" in the sense that any other open neighborhood $V$ of $p$ must contain it. Otherwise, you can take some open neighborhood $p \in V \subsetneq U$ and consider the different (but equivalent) function element $(f|_{V}, V)$.

    The identity theorem implies that if $(f, U)$ and $(g,V)$ are holomorphic equivalent function elements around $p$ and if $U \cap V$ is connected, then $f|_{U \cap V} = g|_{U \cap V}$ which won't necessarily be the case if you will work with smooth or continuous functions. However, a different version of the identity theorem states that if $f \colon U \rightarrow \mathbb{C}$ and $g \colon V \rightarrow \mathbb{C}$ are holomorphic functions with $p \in U \cap V$ and $f^{(i)}(p) = g^{(i)}(p)$ for all $i \geq 0$ then $f$ and $g$ will agree on an open neighborhood of $p$ and so they will define the same germ. Hence, you can describe a germ $[f]$ of holomorphic functions at $p$ by providing a list of all derivatives of $f$ (or any other $g$ with $[g] = [f]$) at $p$. In fact, it is more convenient to provide the list $(a_0, \dots, a_n, \dots)$ of the coefficients of the local power series expansion $f(z) = \sum_{n=0}^{\infty} a_n (z - p)^n$ which are given by $a_n = \frac{f^{(n)}(p)}{n!}$. This way, you can see that not all possible sequences arise as a germ of some function elements - only those sequences for which the power series defined by $\sum_{n=0}^{\infty} a_n (z - p)^n$ converges in a neighborhood of $p$. Thus, the set of germs of a holomorphic function at $p$ is in bijection with the set $$ \left \{ (a_0, \dots, a_n, \dots) \, \big| \, a_i \in \mathbb{C}, \limsup_{n \to \infty} |a_n|^{\frac{1}{n}} < \infty \right \}. $$

  2. You can't describe nicely the germ of a smooth/$C^k$/continuous function around $p \in M$ unless $M$ is zero dimensional. If $f \colon \mathbb{R} \rightarrow \mathbb{R}$ and $g \colon \mathbb{R} \rightarrow \mathbb{R}$ are smooth and define the same germ around $x = 0$ then $f^{(i)}(x) = g^{(i)}(x)$ for all $i \geq 0$ but this list is not enough to characterize explicitly the germ. For example, the zero function and the function $e^{-\frac{1}{x^2}}$ (a standard example for a function whose taylor series converges, but not to the function) share the derivatives of all orders at $x = 0$ but do not define the same germ. Characterizing the germ of a function in some category is the same as characterizing all possible local behaviors of functions and this is hopeless unless the category is rigid in some sense (for example, if you work with holomorphic/analytic/polynomial functions).

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  • $\begingroup$ Oh ok, so $e^{-\frac{1}{x^2}}$ doesn't define the same germ as $0$ at $x=0$ because $e^{-\frac{1}{x^2}}>0$ for all $x>0$ but $0=0$ for all $x>0$ so they can't possibly agree on any neighborhood of $0$? If so, that makes sense. $\endgroup$ Commented Nov 2, 2016 at 13:33
  • $\begingroup$ @William: Yeah, that's correct. $\endgroup$
    – levap
    Commented Nov 2, 2016 at 14:53
  • $\begingroup$ Can you take a look at my questions here and here? They’re related to your answer here. (What I’m trying to do with the second question is find a way to determine if a critical point is a local maximum or minimum of a function if the higher derivative test fails in a way that only depends on the germ of the function.) $\endgroup$ Commented Aug 5, 2018 at 5:42
  • $\begingroup$ @KeshavSrinivasan: I don't know the answers to your questions, but I don't think you can expect to get something reasonable. That is, I doubt you can get "a list" of invariants (definitely not a countable one) that characterizes a germ completely nor some germ based test which isn't trivial to determine whether a point is a local minimum or a maximum. $\endgroup$
    – levap
    Commented Aug 6, 2018 at 12:18
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    $\begingroup$ Why definitely not a countable one? Isn’t a smooth function, or any continuous function for that matter, uniquely determined by its values on the set of rational numbers? $\endgroup$ Commented Aug 6, 2018 at 14:04

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