# I About Arnold's ODE Book Notation

#### Martin T

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?

#### Martin T

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

#### Martin T

Here the Page, the Last paragragh, thanks.

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#### Martin T

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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Science Advisor
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.

#### Martin T

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
Science Advisor
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.

#### Martin T

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
Science Advisor
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 $T_g$.

#### Martin T

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

Science Advisor
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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