- #1
jostpuur
- 2,116
- 19
I'm mainly hoping that somebody else might have done the same exercise earlier. In that case it could be possible to spot where I'm going wrong.
I'm supposed to prove that Lie algebras [tex]\mathfrak{o}(3)[/tex] and [tex]\mathfrak{sp}(2)[/tex] are isomorphic.
Let's see if I have the definitions right...
[tex]
\mathfrak{o}(3) = \{x_1e_1 + x_2e_2 + x_3e_3\;|\;x_1,x_2,x_3\in\mathbb{R}^3\},
[/tex]
where
[tex]
e_1 = \left(\begin{array}{rrr}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right), \quad
e_2 = \left(\begin{array}{rrr}
0 & 0 & -1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{array}\right),\quad
e_3 = \left(\begin{array}{rrr}
0 & 0 & 0\\
0 & 0 & 1\\
0 & -1 & 0 \\
\end{array}\right),
[/tex]
and
[tex]
\mathfrak{sp}(2n) = \{x_1a_1 + x_2a_2 + x_3a_3\;|\;x_1,x_2,x_3\in\mathbb{R}\}
[/tex]
where
[tex]
a_1 = \left(\begin{array}{rr}
0 & 1 \\
0 & 0 \\
\end{array}\right),\quad
a_2 = \left(\begin{array}{rr}
0 & 0 \\
1 & 0 \\
\end{array}\right),\quad
a_3 = \left(\begin{array}{rr}
1 & 0 \\
0 & -1 \\
\end{array}\right).
[/tex]
Lie brackets are
[tex]
[e_1,e_2] = e_3,\quad [e_2,e_3] = e_1,\quad [e_3,e_1] = e_2,
[/tex]
[tex]
[a_1,a_2] = a_3,\quad [a_2,a_3] = 2a_2,\quad [a_3,a_1] = 2a_1.
[/tex]
So I want an isomorphism [tex]\phi:\mathfrak{o}(3)\to\mathfrak{sp}(2)[/tex]. I tried to conclude that I could assume the isomorphism to be of form
[tex]
\phi(e_1) = \lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3
[/tex]
[tex]
\phi(e_2) = \pi_1 a_1 + \pi_2 a_2 + \pi_3 a_3
[/tex]
[tex]
\phi(e_3) = \alpha a_3,
[/tex]
because the cross-product structure on [tex]\mathbb{R}^3[/tex] is invariant under rotations. If there exists some isomorphism [tex]\bar{\phi}[/tex], then there would be a vector [tex]v\in\mathfrak{o}(3)[/tex] so that [tex]\bar{\phi}(v)= a_3[/tex]. Then I choose a new basis to [tex]\mathfrak{o}(3)[/tex] so that [tex]e_3\propto v[/tex].
Then it started to look like the task is impossible. The demand
[tex]
[\phi(e_2),\phi(e_3)] = \phi([e_2,e_3])
[/tex]
implies set of equations
[tex]
-2\pi_1\alpha = \lambda_1,\quad 2\pi_2\alpha = \lambda_2,\quad 0=\lambda_3,
[/tex]
and the demand
[tex]
[\phi(e_3),\phi(e_1)]=\phi([e_3,e_1])
[/tex]
implies set of equations
[tex]
2\alpha\lambda_1 =\pi_1,\quad -2\alpha\lambda_2 = \pi_2,\quad 0=\pi_3.
[/tex]
Then we get
[tex]
\lambda_1 = -2\pi_1\alpha = -4\alpha^2 \lambda_1,\quad\implies\quad \alpha=\pm\frac{i}{2},\;\textrm{or}\;\lambda_1 = 0
[/tex]
and both possibilities are unacceptable.
Homework Statement
I'm supposed to prove that Lie algebras [tex]\mathfrak{o}(3)[/tex] and [tex]\mathfrak{sp}(2)[/tex] are isomorphic.
Homework Equations
Let's see if I have the definitions right...
[tex]
\mathfrak{o}(3) = \{x_1e_1 + x_2e_2 + x_3e_3\;|\;x_1,x_2,x_3\in\mathbb{R}^3\},
[/tex]
where
[tex]
e_1 = \left(\begin{array}{rrr}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}\right), \quad
e_2 = \left(\begin{array}{rrr}
0 & 0 & -1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{array}\right),\quad
e_3 = \left(\begin{array}{rrr}
0 & 0 & 0\\
0 & 0 & 1\\
0 & -1 & 0 \\
\end{array}\right),
[/tex]
and
[tex]
\mathfrak{sp}(2n) = \{x_1a_1 + x_2a_2 + x_3a_3\;|\;x_1,x_2,x_3\in\mathbb{R}\}
[/tex]
where
[tex]
a_1 = \left(\begin{array}{rr}
0 & 1 \\
0 & 0 \\
\end{array}\right),\quad
a_2 = \left(\begin{array}{rr}
0 & 0 \\
1 & 0 \\
\end{array}\right),\quad
a_3 = \left(\begin{array}{rr}
1 & 0 \\
0 & -1 \\
\end{array}\right).
[/tex]
Lie brackets are
[tex]
[e_1,e_2] = e_3,\quad [e_2,e_3] = e_1,\quad [e_3,e_1] = e_2,
[/tex]
[tex]
[a_1,a_2] = a_3,\quad [a_2,a_3] = 2a_2,\quad [a_3,a_1] = 2a_1.
[/tex]
The Attempt at a Solution
So I want an isomorphism [tex]\phi:\mathfrak{o}(3)\to\mathfrak{sp}(2)[/tex]. I tried to conclude that I could assume the isomorphism to be of form
[tex]
\phi(e_1) = \lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3
[/tex]
[tex]
\phi(e_2) = \pi_1 a_1 + \pi_2 a_2 + \pi_3 a_3
[/tex]
[tex]
\phi(e_3) = \alpha a_3,
[/tex]
because the cross-product structure on [tex]\mathbb{R}^3[/tex] is invariant under rotations. If there exists some isomorphism [tex]\bar{\phi}[/tex], then there would be a vector [tex]v\in\mathfrak{o}(3)[/tex] so that [tex]\bar{\phi}(v)= a_3[/tex]. Then I choose a new basis to [tex]\mathfrak{o}(3)[/tex] so that [tex]e_3\propto v[/tex].
Then it started to look like the task is impossible. The demand
[tex]
[\phi(e_2),\phi(e_3)] = \phi([e_2,e_3])
[/tex]
implies set of equations
[tex]
-2\pi_1\alpha = \lambda_1,\quad 2\pi_2\alpha = \lambda_2,\quad 0=\lambda_3,
[/tex]
and the demand
[tex]
[\phi(e_3),\phi(e_1)]=\phi([e_3,e_1])
[/tex]
implies set of equations
[tex]
2\alpha\lambda_1 =\pi_1,\quad -2\alpha\lambda_2 = \pi_2,\quad 0=\pi_3.
[/tex]
Then we get
[tex]
\lambda_1 = -2\pi_1\alpha = -4\alpha^2 \lambda_1,\quad\implies\quad \alpha=\pm\frac{i}{2},\;\textrm{or}\;\lambda_1 = 0
[/tex]
and both possibilities are unacceptable.