# About Arnold's ODE Book Notation

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## Main Question or Discussion Point

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

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fresh_42
Mentor
2018 Award
You know that you only address members who a) have this book and b) are willing to take it from the shelf?

Yes, but Arnold is popular and this paragraph is being kill me.

fresh_42
Mentor
2018 Award
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
Staff Emeritus
Gold Member
My guess is that $T : G \rightarrow S\left(M\right)$ with $g \in G \mapsto T_g \in S\left(M\right)$.

Here the Page, the Last paragragh, thanks.

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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
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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 $T_g$.

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)?

Saw

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martinbn
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 $f(a) : X \rightarrow Y$.