# Non-compact form of an algebra.

1. Jul 2, 2013

### llorgos

I read that in string theory the Virasoro algebra contains an $SL(2,R)$ subalgebra that is generated by $L_{-1}, L_{0}, L_{1}$. I read that this is the non-compact form of the $SU(2)$ algebra. Also, that as $SU(2)$ and $SO(3)$ have the same Lie algebra, so do $SL(2,R)$ and $SO(2,1)$.

Can someone explain all the above statementes? I understand what a compact group is and I have seen that $SU(2)$ and $SO(3)$ have the same Lie algebra. But, what do the other statements mean?

Can you also give a more intuitive explanation?

Thank you very much.

2. Jul 2, 2013

### Ben Niehoff

What's intuitive to one person might not be to another.

To obtain a "non-compact form" of a Lie algebra, you multiply some of the generators by $i$. So if $X_1, X_2, X_3$ have the $SU(2)$ algebra

$$[X_1, X_2] = X_3, \qquad [X_2, X_3] = X_1, \qquad [X_3, X_1] = X_2,$$
then the new set of generators given by

$$\tilde X_1 \equiv i X_1, \qquad \tilde X_2 \equiv i X_2, \qquad \tilde X_3 \equiv X_3$$
will have the algebra

$$[\tilde X_1, \tilde X_2] = -\tilde X_3, \qquad [\tilde X_2, \tilde X_3] = \tilde X_1, \qquad [\tilde X_3, \tilde X_1] = \tilde X_2,$$

which is the algebra of $SL(2,\mathbb{R})$.

A Lie group is "compact" or "non-compact" depending on the eigenvalues of its Cartan-Killing form. If all the eigenvalues are negative, then the group is compact. If some of the eigenvalues are positive, then the group is non-compact. The Killing form of $SL(2, \mathbb{R})$ has signature $(+, +, -)$, and it has the topology $R^2 \times S^1$, so you see there are two noncompact directions, and one compact, matching the signs in the Killing form.

Both $SU(2)$ and $SL(2,\mathbb{R})$ algebras are subalgebras of $SL(2, \mathbb{C})$, which is simply

$$[X_i, X_j] = \varepsilon_{ijk} X_k,$$
except that we allow complex linear combinations of the generators.

3. Jul 2, 2013

### llorgos

would it be possible to talk a bit more about the second part, the compactness or non-compactness of a Lie group?

In any case thank you very much.

4. Jul 2, 2013

### dextercioby

Ben nicely touched the compactness of a Lie group from the compactness of its Lie algebra. But a group is a topological space on its own, so its compactness is defined in terms of open sets, open covers and subcovers.