- #1

- 2,112

- 18

## 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.