# Isometry group on a coset manifold

In ''Einstein Gravity in a Nutshell'' Zee says

''On a coset manifold $$G/H$$, the isometry group is evidently just $$G$$''

when discussing the relation between the Killing vector fields and Lie algebras (Ch IX 6 - App 3). Is there some intuition and/or physicist-level-of-rigor-proof for this ''evidently''? E.g. can we somehow know that Killing vectors for $$S^{d}$$ form Lie algebra $$so(d+1)$$ without calculating the commutators of Killing vectors?

## 2 Answers

In the generality as stated in your question this doesn't seem correct. I am assuming $$G$$ is a Lie group with a left-invariant metric.

There is an obvious mapping $$G\to Isom(G/H)$$ where $$g\in G$$ acts on $$G/H$$ by left multiplication. This does neither have to be injective not surjective however.

Consider the extreme case that $$G=H$$. In that case every element becomes trivial. The other extreme case, $$G=e$$ would require that all isometries are obtained as left-multiplications by a fixed element. This is not true.

As a concrete example, the sphere $$S^2$$ can be written as $$SO(3)/SO(2)$$, but its isometry group is $$O(3)$$. In fact, it can also be written $$O(3)/O(2)$$, and in general a representation of the form $$G/H$$ will not be unique, so that in this form the statement cannot hold.

You might be possibly swamped by a surfeit of unfamiliar terms and symbols, but look at a beachball, or our earth, for d=2, $$x^2+y^2+z^2=R^2$$

All three rigid rotations: $$L_x,L_y, L_z$$, so SO(3), preserve the above constraint and map the earth onto itself preserving all distances (isometry).

If you eliminate the redundant coordinate z in favor of the other two, $$z=\pm\sqrt{R^2-x^2-y^2}$$, you singled out $$L_z$$, manifestly an earth-spin isometry, leaving z invariant. (In physics, this is a "linearly realized", unbroken symmetry.)

But it is not the only one transiting on circles on the globe. Spinning around a spot in the Pacific near Genovesa Island (Galapagos), you might arc your way from Greenwich to the North pole, etc... The motions $$L_x,L_y$$ do not leave z invariant (in physicsese, they are nonlinearly realized, "spontaneously broken" symmetries).

The symmetries of the earth are still the three $$L_x,L_y, L_z$$. You might like to illustrate this in your formalism.