# About Arnold's ODE Book Notation

• I
• Martin T
So what you are looking for is a function from the set of all maps in ##X## to the set of all maps in ##Y##, which is just what ##T_g## defines.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. That is correct. If ##T## takes ##g## to ##S\left(M\right)##, then ##T_g## is in ##S\left(M\right)##.f

#### 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

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.

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.

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

mathwonk
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.

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

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".

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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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##.

Martin T