The germ captures the local behaviour of a function at a point of a topological space. I'd like to know how it is different from a 1-form or its relationship with a 1-form. Pardon me if its a naive question in case its too simple please give a link adding a comment.

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    $\begingroup$ Do you know the definitions? If you do, what are your definitions, and where do you find them confusing? $\endgroup$
    – Zhen Lin
    Jul 20, 2011 at 14:38

3 Answers 3


The first point to make is that a differential $1$-form is something not unlike a function and has a domain in a meaningful sense. By contrast, a germ is only really defined at a point, so is very very local. Like a function, however, a differential $1$-form does have a ‘value’ at a point and so can be ‘evaluated’ at that point.

The second point to make is that a germ is a very general concept. It can be defined for arbitrary functions, not just continuous or differentiable or analytic ones! On the other hand, differential $1$-forms are built from differentiable functions (obviously). Two functions can have the same differential at a point but different germs; on the other hand if two functions have the same germ at a point, then their differentials must agree at that point.

Let's be precise.

Definition. Let $X$ be a topological space, $E$ any set, and $D_1, D_2$ any two open subsets of $X$. Let $f_1 : D_1 \to X$, $f_2 : D_2 \to X$ be any two functions, and $p \in D_1 \cap D_2$. We say $f_1$ and $f_2$ have the same germ at $p$ just if there is an open neighbourhood $U$, $p \in U \subseteq D_1 \cap D_2$, such that the restrictions of $f_1$ and $f_2$ to $U$ are equal, i.e. for every $p'$ in $U$, we have $f_1(p') = f_2(p')$. This induces an equivalence relation on the set of all functions defined in an open neighbourhood of $p$, and the germ of a function at $p$ is the equivalence class under the relation so defined.

Proposition. Let $X$ be a smooth manifold, and $f_1 : D_1 \to \mathbb{R}$, $f_2 : D_2 \to \mathbb{R}$ two functions defined on open subsets $D_1$, $D_2$. Suppose $D_1 \cap D_2$ is non-empty, $p \in D_1 \cap D_2$, and $f_1$ and $f_2$ have the same germ at $p$. Then,

  1. $f_1$ is continuous at $p$ if and only if $f_2$ is continuous at $p$.
  2. $f_1$ is differentiable at $p$ if and only if $f_2$ is differentiable at $p$, and moreover their derivatives are equal when they exist.
  3. Ditto, with $n$-times differentiable and the $n$-th derivative instead.
  4. Ditto, with analytic and the power series (under a fixed chart) about $p$ instead.

So we see that germs really do capture the ‘local’ data of a function near a point. (Or we can flip this on its head and say that a property is local if and only if the property can be defined in terms of germs.)

Definition. Let $X$ be a smooth manifold, and let $D$ be an open subset of $X$. Let $f_1, f_2 : D \to \mathbb{R}$ be two functions which are smooth on $D$. Let $p \in D$. Then, we say $f_1$ and $f_2$ have the same differential at $p$ if their derivatives agree at $p$. [This is the main difference from germs: We don't need them to agree on an open set!] If the derivatives also agree on an open neighbourhood of $p$ then we say $f_1$ and $f_2$ have the same differential germ at $p$. We write $\mathrm{d}f_p$ for the differential germ at $p$ of a smooth function $f : D \to \mathbb{R}$.

A differential $1$-form on $D$ is an assignment of a differential germ $\omega_p$ to each point $p$ of $D$ such that for each point $p$, there is an open neighbourhood $U$, $p \in U \subseteq D$, such that there is a smooth function $f : U \to \mathbb{R}$ for which $\omega_{p'} = \mathrm{d}f_{p'}$ for each $p'$ in $U$.

Notice that we only require $\omega_{\bullet}$ to agree with the differential germs of a smooth function locally; the function chosen may be different from point to point. In other words, there need not be a function $f : D \to \mathbb{R}$ such that $\omega_p = \mathrm{d}f_p$ for every $p$ in $D$. For example, take $D = \mathbb{R}^2 \setminus \{ (0, 0) \}$, the punctured plane. Then there is a differential $1$-form $$\omega = \frac{-y \mathrm{d}x + x \mathrm{d}y}{x^2 + y^2}$$ which is not of the form $\mathrm{d}f$ for any $f : D \to \mathbb{R}$. (The reason is simple enough: when you integrate $\omega$ over a loop going around the origin once, you get a non-zero result, but the fundamental theorem of calculus tells us that the integral of $\mathrm{d}f$ around any loop must be zero.)

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    $\begingroup$ In a Physics book it is given that a 1-form is nothing but a contangent vector ( defined here : math.stackexchange.com/questions/36591/… ) and the assignment of a 1-form at each point of the manifold is called a 1-form field. But in your answer a differential 1-form is defined as an assignment of differential germ at each point in a set, which leaves a doubt for that, a 1-form is different from an differential 1-form. also what is a 1-form field according to your definitions ? $\endgroup$
    – Rajesh D
    Jul 23, 2011 at 4:00
  • $\begingroup$ That's just terminology. I choose to use the word ‘differential 1-form’ for sections of the cotangent bundle (i.e. covector fields). $\endgroup$
    – Zhen Lin
    Jul 23, 2011 at 7:04

Given a point $p$ in a topological space $X$ you might be interested in functions $f:\ U\to{\mathbb R}$ defined in various neighborhoods $U$ of $p$. If you don't care about the behavior of such $f$ far away from $p$ you call two such functions $f_i:\ U_i\to{\mathbb R}$ equivalent if there is a smaller neighborhood $V\subset U_1\cap U_2$ on which $f_1$ and $f_2$ coincide. The equivalence classes of this relation are called ${\it germs}$ of functions at $p$. Note that we have not computed derivatives; it didn't even matter whether our functions were continuous.

If the space $X$ is actually a differentiable manifold then we can do calculus on $X$. In particular, "any" function $f$ defined on an open set $U\subset X$ has its differential $df:\ T_U\to{\mathbb R}$ which is the "multivariable and coordinate free" substitute for the derivative of a calculus 101 function $f:\ {\mathbb R}\to{\mathbb R}$. This differential is a function of two variables $p\in U$ and $X\in T_p$ and is linear in the second variable. The following are true:

(a) $df$ is a $1$-form on the domain $U$ of $f$.

(b) If $f_1$ and $f_2$ coincide in a neighborhood $V$ of $p$ then their differentials $df_1$, $df_2$ coincide there as well. It follows that each function germ at $p$ determines a unique "differential germ".

On the other hand it is not true that the germ of $df$ at $p$ determines the germ of $f$ at $p$: It determines $f$ only up to an additive constant. And more seriously: It is not true that any given $1$-form $\omega$ in a neighborhood $U$ of $p$ is the $df$ of some $f$; the reason being that there are integrability conditions.

  • $\begingroup$ Shouldn't the differential of an $\mathbb R$-valued function map to the tangent bundle of $\mathbb R$? $\endgroup$
    – Rasmus
    Jul 20, 2011 at 17:33
  • $\begingroup$ For your last statement, the Poincare lemma says that all you need to check is that $\omega$ is closed. (If $U\subset \mathbb{R}$, then this is automatic.) $\endgroup$ Jul 20, 2011 at 17:37
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    $\begingroup$ @Rasmus: A $1$-form, in particular the differential of a function $f$, is for each $p\in U$ a linear functional on the tangent space $T_p$, so it is a function $(p, X)\mapsto df(p).X\in{\mathbb R}$. $\endgroup$ Jul 20, 2011 at 17:49
  • $\begingroup$ @Aaron: That's what I meant when I spoke of "integrability conditions". $\endgroup$ Jul 20, 2011 at 17:49
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    $\begingroup$ @Rasmus If $\underline{\mathbf{R}} = X \times \mathbf{R}$ is the trivial bundle of rank $1$ on $X$, then I don't think you lose anything by writing a vector bundle morphism $f\colon TX \to \underline{\mathbf{R}}$ as $TX \to \mathbf{R}$. If $p\colon TX \to X$ is the standard projection, then $f$ is of the form $f(y) = (p(y), g(y))$ for some function $g\colon TX \to \mathbf{R}$ that is linear when restricted to each fibre $T_xX$, and that determines everything, no? $\endgroup$ Jul 21, 2011 at 0:14

In this post I am only talking about differentiable maps. Having the same germ seems to be much more stringent that having the same differential. For example, the functions $f,g:\mathbb{R}_{\geq 0}\rightarrow \mathbb{R}$ defined by $f(x)=x$ and $g(x)=log(x)$ have the same differential at $x=1$, but do not have the same germ. Also the functions $x$ and $x+3$ have the same differential everywhere, but nowhere have the same germ. Functions having the same germ, do have the same differential there.

I believe the following is true. If two functions have the same differential in some neighborhood of a point, and are equal in this point, then the functions have the same germ.

Edit: removed the remark of my uncertainty after the comments.

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    $\begingroup$ I agree with your final statement. Writing everything out in coordinates seems to make this obvious. $\endgroup$ Jul 20, 2011 at 14:48
  • $\begingroup$ Isn't that just the fundamental theorem of calculus? $\endgroup$ Jul 20, 2011 at 17:30

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