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Homomorphism Group Theory

Hi. Let
Unparseable latex formula:

A=\left( \begin{array}{cc}[br]\alpha & \beta \\[br]-\bar{\beta} & \bar{\alpha} \end{array} \right)

where α,β \alpha, \beta are complex and αˉ,βˉ \bar{\alpha}, \bar{\beta} are their complex conjugates. We also have that det(A)=1 det(A)=1 . The matrix B is given as
Unparseable latex formula:

B=\left( \begin{array}{cc}[br]\gamma& \delta\\[br]-\bar{\delta} & \bar{\gamma} \end{array} \right)

where γ,δ \gamma, \delta are complex and γˉ,δˉ \bar{\gamma}, \bar{\delta} are their complex conjugates. We also have that det(B)=1 det(B)=1 . The map from 2×23×3 2 \times 2 \to 3 \times 3 for A is given by

Unparseable latex formula:

[br]A\left( \begin{array}{c}[br]i \\[br]-k \\[br]j \end{array} \right)A^{-1}=\left( \begin{array}{ccc} i & -k & j \end{array} \right) \mu(A)[br]


where μ(A) \mu(A) is a 3×3 3 \times 3 matrix.

How do I show that there exists a group homomorphism? i.e.[br]μ(AB)=μ(A)μ(B).[br]\mu(AB)=\mu(A)\mu(B). I don't know how to justify changing
[br]A(ikj)μ(B)A1=(ikj)μ(AB),[br][br]A\left( \begin{array}{ccc} i & -k & j \end{array} \right)\mu(B) A^{-1}=\left( \begin{array}{ccc} i & -k & j \end{array} \right) \mu(AB),[br]
to
Unparseable latex formula:

[br]A\left( \begin{array}{c}[br]i \\[br]-k \\[br]j \end{array} \right) A^{-1}\mu(B)=\left( \begin{array}{ccc} i & -k & j \end{array} \right) \mu(AB),[br]


in order to show that group homomorphism exists. Thank you for your help
(edited 9 years ago)
Reply 1
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Reply 2
What do you mean exactly? A group homomorphism between which groups?

Can you post the whole question?
Reply 3
Original post by james22
What do you mean exactly? A group homomorphism between which groups?

Can you post the whole question?


I am currently learning about lie groups and I am stuck on this question. I am looking for a homomorphism between μ(A)   and   μ(B) \mu(A)~~~ \text{and}~~~ \mu(B) . Thank you for your help
Reply 4
Original post by gardash
I am currently learning about lie groups and I am stuck on this question. I am looking for a homomorphism between μ(A)   and   μ(B) \mu(A)~~~ \text{and}~~~ \mu(B) . Thank you for your help


Oh right. I don't know anything about lie groups I'm afraid.
Reply 5
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Original post by gardash
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Interestingly, I get μ(AB)=μ(B)μ(A)\mu(A B) = \mu(B) \mu(A). I'll denote by vv the vector i,-k,j; then:
vμ(AB)=ABvB1A1=A(vμ(B))A1=vμ(B)μ(A)v \mu(AB) = AB v B^{-1}A^{-1} = A (v \mu(B)) A^{-1} = v \mu(B) \mu(A).

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