So I thought of an example. The set is G={x}. It contains 1 element. the vector space considered is $S={f_1(x)=x+1,f_2(x)=x+2,...f_m(x)=x^2,...f_n=\sqrt{x}....}$
as an counter example
and I gave
So I thought of an example. The set is G={x}. It contains 1 element. the vector space considered is $S={f_1(x)=x+1,f_2(x)=x+2,...f_m(x)=x^2,...f_n=\sqrt{x}....}$
it is finite-dimensional. Consider the function $I(x):=1$; it is by construction an element of that space. For all functions $f$ from that space it holds that $f(x)=c_f I(x)$ for some real number $c_f$, so $f=c_f I$, showing that the set containing only $I$ is spanning. This set is also trivially linearly independent, and thus a basis. Hence, the sought vector space is one-dimensional
you should be able to generalize this if the set under consideration has $n$ elements
the point is similar to say that $\mathbb R$ is a one-dimensional vector space (over itself). there, you can write every element $x$ as a linear combination of $1$, namely $x=x 1$.
A fundamental principle of modern mathematics is that the way to understand a space M, given as some set of points, is to look at F(M), the set of functions on this space. This “linearizes” the problem, since the function space is a vector space, no matter what the geometrical structure of the original set is. If the set has a finite number of elements, the function space will be a finite dimensional vector space.
Has the following implication: If you consider some arbitrary space or set, with an arbitrary finite nr. of elements, the corresponding vector space, whose elements are functions, is a finite vector space.
So the statement implies a direct relationship between the NUMBER of elements of the set and the vector space
Not of the nr. of basis elements
Not on mapping of values
but on number of elements of the set and nr. of elements of the vector space
would it be correct to say that the lagrange multipliers are chosen so that whatever independent coords you choose, the remaining coords are determined automatically by the constraints?
I have one question about group action and homomoprhism. In wikipedia I read: "In mathematics, a group action, of a group G on a set S is a group homomorphism from from G to some group (under function composition) of functions from S to itself. It is said that G acts on S"
Group homomorphism is between two groups. While in the above statement we have a group and a set and we talk about group homomorphism. Is the group homomorphism between the group G and a group H, which is the representation of G, and it retains the group structure of G
@imbAF it is sort of a representatuon but it need not be a vector space representation cuz S need not be a vector space. so u can think of it as a generalisation of the notion of representations
when we have a vector space representation, the $g\in G$ is mapped to matrices which are functions $f: V--->V$. And the group operation becomes matrix composition
@imbAF V can be a vector space on which u want to define the group action (which is a special case of a set S on which u may want to define a group action)
I understand that, you have a group G, and via group homomorphism, you map every element of the group to some other group, call it H, which is the group of transformations of G. Then, H acts on a set S and gives an element which is part of S. The transformation from G to H, needs to retain the group structure of G. Is my understanding accurate until now?
And because there's a 1 to 1 mapping from G to H and the group structure is retained, when H acts on a set S, is the same as G acting on S, hence why we say "Group action of G on S"
i will describe it best. let F be the set of functions $f:S\to S$. Take a group G. Define a function $R: G --->F$ such that $R(g\cdot h)= R(g)oR(h)$. Then $R$ defines the action of group $G$ on set $S$
so it is a group homomorphism from G to the group of functions $F$
@imbAF ur description is pretty much the same, but I'm just writing my description to avoid ambiguities
@imbAF note that the mapping need not be 1 to 1 for it to be a group homomorphism
@RyderRude I did not mean that. I simply was saying that by us sticking to our role of me asking and you answering and not derailing, we can can fast to a conclusion
@RyderRude In the first box, we are given the definition of action of a group G on set G. Here you say something different than what is said in the first red box
@RyderRude Yes, a choice group homormorphism between $G$ and the group of bijective functions on $M$, $S(M)$ is equivalent to specifying some group action. That is a basic result. What the second red box however does is, as far as I understand, to consider the set of all functions $F(M)$ and define a group action as indicated in the text.
That is what we have discussed yesterday, to no avail I am afraid
But the idea is that the set of functions $f:M\to V$, where $V$ is a vector space, forms a vector space itself under natural definitions of the vector space operations.
but OK. Idk. the text seems, from my naive perspective, sloppy.
I really think they just want to say that: given a (left) group action on $M$, you get a group action on $F(M)$ by the definition they show. (which is left as an exercise to verify).
Just to at least confirm my understanding up to this point, ignoring for a moment that fact that the set of f is a vector space, which would require some conditions to be met and what not, is it safe to say that the set of f, is "gained" via a mapping from G to call the set of f F. And this mapping is a homomorphism
it is the usual problem. instead of reading a proper text(book) and starting from the basic notions, one tries to deal with more or less very specific statements in some random (?) text without understanding first the basic concepts.
The book was recommended to me by one of the members here. The other one was An Elementary Introduction to Groups and Representations by Brian C. Hall
What book should I read to understand group theory, representation and whatever else went down in this conversation. I would very gladly read that book
since, groups etc have been a thorn in my everyday life in physics
But is weird, whenever I ask about a recommendation for group theory etc, people never actually have any to give. Strangely considering that nearly everyone here is familiar with such concepts
@TobiasFünke But I didn't though did I? I made that assessment once you, someone of higher understanding on the matter than me, made that assessment. And I took your word for it, since you have more knowledge in the matter than me. I wasn't aware that mentioning the vector space notation, was the key to everything being accurate. Because I don't have fundamental understanding of the subject. Painfully unnecessary to have to explain this
And I read the definition too, and since I ask and you don't, it is an indicator that regardless the fact that we both read the same thing, we do not understand it in the same way....
and I never claimed that "mentioning the vector space notation, was the key to everything being accurate". I honestly did not understand all of your questions, except the one from yesterday; to me they appear very confusing and mixed up concepts (I don't claim that it is false, just that I do not get what you are (trying to) say). But there I could not convince you, since you seem to have a lack of understanding for basic concepts of vector spaces. Perhaps I just misunderstood you, though.
when I said: "the text seems, from my naive perspective, sloppy" I meant that I cannot make sense of what they mean with the information they give (you show us).
If you really ask for an advice: Start from the basics. Linear algebra, vector spaces and so on. Then basic group theory: axioms, definition of concepts, some examples
for all of that you do not have to read full books
but once you understand the basics, you can progress with some more advanced/special topics
@imbAF well just from what I read/hear. Of course you need some group theory in physics, but thus far I've never had to read a whole book or many chapters
@imbAF i was trying to prove that the action in the second box is a group action. we require $g\cdot ( h \cdot f)= gh \cdot f$. but $g\cdot (h\cdot f)= f(g^{-1}\cdot h^{-1}\cdot x)$, while $gh\cdot f= f(h^{-1}\cdot g^{-1} \cdot x)$
I have a simple question about this following action:
Let $L(G)$ be the vector space of all functions from $G$ to $\mathbb{C}$. Define an action of $G$ on $L(G)$ by
$$(\sigma f)(\tau) = f(\sigma ^{-1}\tau)$$
for all $\tau \in G$.
To show its a group action we need to show $ef=f$ and $(\sigma _...
Suppose that you have some group $G$ acting on a set $S$ under the action $\cdot: G \curvearrowright S$, and then you have a set of all functions $T = \{ f: S \to \mathbb{C} \}$. I want to define an action $*: G \curvearrowright T$.
Let $\omega_g: s \in S \mapsto g \cdot s$, then I want to defin...
Funnily enough, I think this whole discussion was completely pointless. They've asked the question on MathSE yesterday, and the comments there are exactly what we've pointed out. :d
@imbAF They are saying that the set of all functions $F(M)$ over a suitably nice space $M$ is a vector space. Then, you can study representations $\pi: G \to F(M)$ of a group $G$ over that vector space to glean properties of the original space. I don't think the text is ambiguous here.
@SillyGoose the text, at least the passage we've been shown, is ambiguous, because it is not clear why $F(M)$ would be a vector space
this depends on the range of the functions in $F(M)$, no? If the range is a vector space, then indeed $F(M)$ can be endowed with a vector space structure
separately, does fermi liquid theory solely deal with degenerate systems of fermions? that is, one is always concerned with a low temperature situation and can take things like $\mu = \epsilon_F$ and so on.
