I About Arnold's ODE Book Notation

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In Arnold's book, ordinary differential equations 3rd. WHY Arnold say Tg:M→M instead of Tg:G→S(M) for transformations Tfg=Tf Tg,
Tg^-1=(Tg)^-1.

Let M be a group and M a set. We say that an action of the group G on the set M is defined if to each element g of G there corresponds a transformation Tg : M→M of the set M, to the product and inverse elements corresponds Tfg=TfTg, Tg^-1=(Tg)^-1
 

fresh_42

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You know that you only address members who a) have this book and b) are willing to take it from the shelf?
 
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Yes, but Arnold is popular and this paragraph is being kill me.
 

fresh_42

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As you wish. I could probably answer your question if I only had the book or you had put a little effort in describing the situation. I prefer to read books written in my own language. So, good luck! I'm out.
 

George Jones

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My guess is that ##T : G \rightarrow S\left(M\right)## with ##g \in G \mapsto T_g \in S\left(M\right)##.
 
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My guess is that ##T : G \rightarrow S\left(M\right)## with ##g \in G \mapsto T_g \in S\left(M\right)##.
Yes I think the same, then Arnold is wrong with
Tg:M→M.
 

George Jones

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Yes I think the same, then Arnold is wrong with
Tg:M→M.
If what I wrote is correct, then Arnold is correct., i.e., I, in part, wrote ##T_g \in S\left(M\right)##. This means that ##T_g : M \rightarrow M##. This is fairly common notation.
 
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If what I wrote is correct, then Arnold is correct., i.e., I, in part, wrote ##T_g \in S\left(M\right)##. This means that ##T_g : M \rightarrow M##. This is fairly common notation.
Then it is a old fashioned notation, thanks
 

George Jones

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Then it is a old fashioned notation, thanks
It currently is bog-standard notation used in an almost uncountable number of books on my shelves.
 
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But if T takes m to m, how is the same which T takes g of G(or gh of G) to S(M)=group of all bijective transformations of M.
And say Arnold's books are ""pedagogic".
 

George Jones

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And say Arnold's books are ""pedagogic".
Put an end to your sarcastic comments, or I will put an end to helping you.

But if T takes m to m
This is not what I wrote. I wrote

##T : G \rightarrow S\left(M\right)##
In other words, ##T## takes ##G## to ##S\left(M\right)##. Consequently, using functional bracket notation, ##T\left(g\right)## is in ##S\left(M\right)##, i.e., ##T\left(g\right)## takes ##M## to ##M##. Here, bracket notation becomes too cluttered/confusing, so it is conventional to denote ##T\left(g\right)## as ##T_g##.
 
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Yes, I understand, but in the book Arnold first defines an action:
Let M be a group and M a set. We say that an action of the group G on the set M is defined if to each element g of G there corresponds a transformation Tg : M→M of the set M, to the product and inverse elements corresponds Tfg=TfTg, Tg^-1=(Tg)^-1

Why? It is correct(Arnold talk about homomorphims after)?
 

martinbn

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Yes, I understand, but in the book Arnold first defines an action:
Let M be a group and M a set. We say that an action of the group G on the set M is defined if to each element g of G there corresponds a transformation Tg : M→M of the set M, to the product and inverse elements corresponds Tfg=TfTg, Tg^-1=(Tg)^-1

Why? It is correct(Arnold talk about homomorphims after)?
It is all correct and standard. Every algebra book (any maths book really) that defines action of a group on a set uses these notation.

Probably what confuses you is that you have a map ##f:A\rightarrow B##, where the target set ##B## is a set of maps, say between the sets ##X## and ##Y##, so the image of any element ##a\in A## is map itself ##f(a) : X \rightarrow Y##.
 

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