I am reading Bailin and Love's review on Kaluza Klein theories. On section 4.1 they start talking about infinitesimal isometries generated "with a particular generator $t_a$ of the isometry group".

$I+id\sigma{}t_a$ $y^n\to{}y'^n=y^n+d\sigma{}\xi_a^n(y)$

where the $\xi_a^n(y)$ are just killing vectors.

Immediately thereafter it says

For any representation of the non-Abelian isometry group, the eigenvalues of a diagonal generator $t_a$ will be integral multiples of some lowest (positive) eigenvalue $g_{min}$

Why is that so?

And also, I have a suspicion that where it says representation of the non-Abelian (Lie) group it really means representation of the Lie algebra associated. Is my suspicion right?


The eigenvalue of the generator $t_a$ are integer multiples of $g_{min}$ because $t_a$ is a generator of a (cyclic) $U(1)$ group and $$ \exp(2\pi i t_a/ g_{min}) = 1 $$ holds as an operator equation. This equation says that the exponentiation of the generator with some imaginary coefficient that I parameterized as $2\pi i / g_{min}$ is equal to the identity. The rotation of a sphere by $2\pi$ is an example. When the equation above acts on an eigenstate, $t_a$ is replaced by its eigenvalue, but because $\exp(2\pi i n)=1$ only for $n\in{\mathbb Z}$, it follows that the eigenvalue of $t_a/g_{min}$ is integer.

The representation of a Lie algebra and the representation of the corresponding Lie group is the same thing. One may be used for the other.

  • $\begingroup$ So, do all the generators of isometry groups always generate a $U(1)$ group? $\endgroup$ – Yossarian Mar 10 '14 at 17:40
  • 1
    $\begingroup$ Not really. If you pick an "irrational" combination of two natural generators, $t_a+K t_b$ where $K$ is irrational, the group that this combined generator generates is isomorphic to $R^+$ and not $U(1)$ because you will never return quite to the same place - the periodicity disappears. But it's always possible to choose a Cartan (maximal commuting) subalgebra of commuting and compact generators isomorphic to $U(1)^\ell$. Any generator of a compact group may be written as a combination of some generators in some $U(1)^\ell$ subalgebra. $\endgroup$ – Luboš Motl Mar 11 '14 at 7:15

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