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kakarukeys
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Can isomorphism of matrix groups
[tex]\phi: G_1 \rightarrow G_2[/tex]
always be expressed by
[tex]\phi(M) = S M S^{-1}[/tex]?
[tex]\phi: G_1 \rightarrow G_2[/tex]
always be expressed by
[tex]\phi(M) = S M S^{-1}[/tex]?
Right, this thread is the closest to the topic I could find.
My question is more convention choice than anything else.
The finite Cyclic [tex]C_3[/tex] group is defined by:
[tex]{e,c,b(=c^2)}[/tex]
where [tex]e[/tex] is the identity, [tex]c[/tex] is rotation through [tex]\frac{2\pi}{3}[/tex] etc. I'm keeping formalities to a minimum here.
Clearly, we are rotating a triangle with directed sides in one plane through three angles, yeah?
Now, my question has to do with representations of this group.
Let me begin with how I am learning about groups.
The definition I'm working from (and taking the example from) is in line with H F Jones' "Groups, Representations and Physics Second Ed".
A representation of a Group [tex]G[/tex] is the pair [tex]{\iota, V}[/tex] where [tex]V[/tex] is the vector space which is also a group and
[tex]\iota : G \mapsto V[/tex]
is a homomorphism. With the usual condition that [tex]\iota[/tex] preserves the group structure.
So, some books understandably skip the generality of [tex]V[/tex] and claim that [tex]V=M_{n \times n}[/tex] the set of invertible nxn matrices.
And to represent [tex]C_3[/tex] we will use the finite subspace
[tex]R(\theta)= \begin{bmatrix}\cos \theta & -\sin \theta & {0} \\ \sin \theta & \cos \theta & {0} \\ {0} & {0} & {1}\end{bmatrix} [/tex]
where [tex]\theta=0, \frac{2 \pi}{3},\frac{4\pi}{3}[/tex]
A representation of [tex]C_3[/tex] is the pair [tex]{\iota, M}[/tex] where [tex]M[/tex] is the vector space with three elements
[tex]M=R(0),R(\frac{2 \pi}{3}),R(\frac{4\pi}{3})[/tex]
[tex]\iota : G \mapsto M[/tex]
Now, finally, I can ask my question.
According to the book [tex]R(\frac{2 \pi}{3})[/tex] would be the "representation" of [tex]c[/tex] in [tex]C_3[/tex]. I find this kind of confusing.
So what would ([tex]\textbf{x'},\textbf{x}[/tex] are just Cartesian column vectors)
[tex]\textbf{x'}=R(\theta)\textbf{x}[/tex]
be?
Can someone shed some light on how to view this definition as intuitive. I understand that a representation shouldn't demand a co-od system. I suppose, to me, it just seems like we're identifying an 'operator' ([tex]R(\theta)\in M_{n \times n}[/tex]) with a 'state' ([tex]e,c,b \in C_3[/tex]).
May I redefine in more familiar language(to my course).
A representation of C_3 is the pair {i,R^2}, where
i : C_3 -> R^2
is the function taking the following form:
i(g) = R(2pi/3)
where, g \in C_3 and R is the 2X2 rotation matrix.