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another seemingly innocent problem in Loring Tu's Introduction to Manifolds is 2.2 (Algebra structure in $C_p^\infty$) that says

Define carefully addition, multiplication, and scalar multiplication in $C_p^\infty$. Prove that addition in $C_p^\infty$ is commutative.

That "carefully" is somehow scaring me as I found pretty obvious (probably too much) where some pages before the author said "the addition and multiplication of functions induce corresponding operations on $C_p^\infty$ making it into an algebra over $\mathbb{R}$".

In particular, if two real valued functions $f_1$ and $f_2$ have the same values in the same neighborhood $U$ of $p$ and $g_1$ and $g_2$ have the same values (different from $f$) in the same neighborhood I can pick $f_1+g_1$ or $f_2+g_1$ or $g_2+f_2$ etc. as all valid representatives of the sum of the two germs $f$ and $g$ at $p$, where commutativity and smoothness of the sum would derive from the properties of the real valued sum and of the derivation respectively ... or am I missing something here?

Thanks for any hint!

latelrn
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  • You seem to be missing something because I don't understand how your paragraph defines the sum of two germs – Ben Grossmann Apr 02 '20 at 22:50
  • I guess a more "careful" answer might begin something like: For $f_{1},f_{2}$ elements of the equivalence class $f$ we have $f_{1}=f_{2}$ on some neighbourhood $U$, while for $g_{1},g_{2}$ elements of the equivalence class $g$ we have $g_{1}=g_{2}$ on some neighbourhood $V$, so there is a neighbourhood $U \cap V$ on which $f_{1}=f_{2}$ and $g_{1}=g_{2}$ ... – Ali Apr 02 '20 at 22:53
  • See also https://math.stackexchange.com/q/4446882/349785. – Paul Frost Feb 03 '25 at 17:03

1 Answers1

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The definition of a germ is given as follows:

Consider the set of all pairs $(f,U)$,where $U$ is a neighborhood of p and $f:U\to\Bbb R$ is a $C^\infty$ function. We say that $(f,U)$ is equivalent to $(g,V)$ if there is an open set $W \subset U \cap V$ containing $p$ such that $f = g$ when restricted to $W$. This is clearly an equivalence relation because it is reflexive, symmetric, and transitive. The equivalence class of $(f,U)$ is called the $germ$ of $f$ at $p$.

I will use $[(f,U)]$ to denote the equivalence class of $(f,U)$. Note that ultimately, we need to be able to define $[(f,U)] + [(g,V)]$ given any $C^\infty$ functions $f,g$ and neighborhoods $U,V$ of $p$.

In other words, given an $[(f,U)]$ and $[(g,V)]$, your definition should be able to let us generate a function $h$ and neighborhood $W$ for which $[(f,U)] + [(g,V)] = [(h,W)]$.

In order for this operation to be "well defined", it needs to be defined in such a way that if $(f_1,U_1)\sim(f_2,U_2)$ and $(g_1,V_1) \sim (g_2,V_2)$, then the $h,W$ corresponding to $[(f_1,U_1)] + [(g_1,V_1)]$ matches the equivalence class of $[(f_2,U_2)] + [(g_2,V_2)]$.

One valid definition is as follows:

Given $(f,U)$ and $(g,V)$, we define $[(f,U)] + [(g,V)] = [(h,W)]$, where $W = U \cap V$ and $h:W \to \Bbb R$ is defined by $h(x) = f(x) + g(x)$.

Ben Grossmann
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  • thanks Omnomnomnom and @Ali , so I believe the whole point here lies in the definition of $W$. but, in other terms, any representatives of $(f,U)$ and $(g,V)$ will define a $(h,W)$ that belongs to the sum of the germs, provided $W=U \cap V$. But what about being commutative? What does Tu want me to prove explicitely? I guess that the sum of the two germs is commutative because real sum and intersection between sets are commutative... no? – latelrn Apr 03 '20 at 18:47
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    @l4teLearner Yes, that's all you need for commutativity. The only tricky part about this question is the point about the set $W$ – Ben Grossmann Apr 03 '20 at 18:52