6
$\begingroup$

In John Lee's book Riemannian Manifolds on page 33, Lee writes "Clearly a homogeneous Riemannian manifold that is isotropic at one point is isotropic at every point".

It seems that he means that he concluded this from the following argument: Suppose a Riemannian manifold $(M, g)$ is isotropic at $p \in M$ and homogeneous, and take $q \in M$, $v, w \in T_q M$ with $g_q(v, v) = g_q(w, w) = 1$. Because $M$ is homogeneous, there is a Lie group $G$ that acts smoothly and transitively on $M$ by isometries, so we can find $x \in G$ such that $x \cdot q = p$, hence $x_* v, x_*w \in T_qM$. Because $M$ is isotropic at $p$, there exists another Lie Group $H$ acting on $M$ smoothly by isometries such that the isotropy subgroup $H_p \subseteq H$ acts transitively on $T_pM$, so there exists $y \in H$ such that $y_*(x_* v) = x_* w$. We would then have $(x^{-1}yx)_*v = w$ as desired. However, Lee does not specify in the definitions of homogeneity and isotropy that $G, H$ have to be the same Lie group, so I am not sure why there is a Lie group acting smoothly by isometries on $M$ that contains all elements of the form $x^{-1}yx$, where $x \in G$, $y \in H$.

I do not want to use the fact that the isometry group of any Riemannian manifold can be made into a Lie group, since this fact is beyond the scope of this chapter, and is supposedly pretty difficult to prove. One solution I was thinking of was maybe using a coproduct (free product) of the Lie groups $G, H$, but I was not sure how to give this a topology or smooth structure. Is there a better way of finishing the above argument?

$\endgroup$
6
$\begingroup$

You're absolutely right. Without using the fact that the full isometry group is a Lie group, I think it would be very difficult to show that there's a Lie group containing all the elements of the form $x^{-1}yx$ for $x\in G$ and $y\in H$.

I think the Wikipedia definition pointed out by @MattSamuel has it right: There's no good reason to include the requirement that the action be by a Lie group in the definitions of homogeneous and isotropic. I've added a modification of my definition to my online correction list.

Thanks for pointing this out.

$\endgroup$
5
$\begingroup$

The definition of isotropic (which I found here) ensures only the existence of some isometries satisfying the conditions. Since $x$ and $y$ are isometries, so is $xyx^{-1}$. There is no requirement that the set of isometries that you "use" to satisfy these conditions (which is not necessarily unique) have any specific topology or even be closed under multiplication.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.