3

I am curious about isometry group of new Riemannian manifolds out of old Riemannian manifolds. for example what we can say about isometry group of $N$ if we know the isometry group of $M$ and $f:M\to N$ a smooth Riemannian quotient map or smooth Riemannian submersion or a Riemannian covering map? Can one compute the isometry group of $N$ exactly? (for simplicity consider all manifolds are closed.)

Any (Book as) reference?

C.F.G
  • 8,789
  • Even in the case of Riemannian covering maps there is no simple procedure for such a computation. Suppose I told you that $M$ is a hyperbolic surface, hence, $N$ is also a hyperbolic surface, and $f$ is an irregular covering map. In general, isometries of $N$ will not lift to isometries of $M$ and vice-versa. – Moishe Kohan Feb 04 '21 at 23:00
  • No, not in general. One can get some information, for instance, if $N\to M$ is the universal covering with the deck-transformation group $G$, then $Isom(M)$ is isomorphic to the quotient of the normalizer of $G$ in $Isom(N)$ by $G$. But computing the normalizer is not easy either. – Moishe Kohan Feb 05 '21 at 16:39
  • OK, later...... – Moishe Kohan Feb 05 '21 at 22:31

1 Answers1

5

Even in the case of Riemannian covering maps, there is no simple description. One can get some information, for instance, if $X\to Y$ is the universal covering with the deck-transformation group $G$: Then the isometry group $Isom(Y)$ of $Y$ is isomorphic to the quotient of the normalizer of $G$ in $Isom(X)$ by $G$: $$ N_{Isom(X)}(G)/G\cong Isom(Y). $$ But computing the normalizer is not easy either. Sometimes, one can glean information by looking at symmetries of a suitably chosen fundamental domain of $G$ in $X$. But frequently it happens that $Y$ has more symmetries than that.

Moishe Kohan
  • 111,854
  • Thanks. This problem is hard enough (or impossible) to deal with or it is an ongoing research problem? – C.F.G Feb 06 '21 at 10:22
  • @C.F.G: I can find some references (computations in special cases) but, afaik, it is not the subject of current research. It's hard to say anything interesting in general about the question. – Moishe Kohan Feb 06 '21 at 10:56
  • Do you know of a counterexample with Riemannian manifolds $ X, Y $ where $ G $ is a discrete group of isometries of $ X $, and $ Y \cong X/G $ but it fails to be the case that $ N_{Isom(X)}(G)/G \cong Isom(Y) $? – Ian Gershon Teixeira Jan 26 '22 at 21:27
  • @IanGershonTeixeira: Yes. For instance, take $Y$ a hyperbolic surface of genus $2$, obtained from two tori $T_1, T_2$ with single hole by gluing along their boundary geodesics. Assume that $Y$ is sufficiently symmetric and the symmetry swaps $T_1, T_2$. Now take a homomorphism $h: \pi_1(Y)\to Z_2$ which sends $\pi_1(T_2)$ trivially and sends $\pi_1(T_1)$ nontrivially. Now, take the (normal) 2-fold covering of $Y$ corresponding to $ker(h)$. You can find similar examples even in the case when $Y$ is the "square" flat 2D torus and $X\to Y$ is a suitable 2-fold covering. – Moishe Kohan Jan 26 '22 at 21:57
  • Hmm I see is the situation any better if we ask that $ G $ is a group of constant displacement isometries (so called "Clifford-Wolf isometries")? – Ian Gershon Teixeira Jan 26 '22 at 22:01
  • @IanGershonTeixeira: No, this is exactly the case of the flat torus example. The issue is essentially algebraic: In these examples, the subgroup of $\pi_1(Y)$ defining the covering is normal (since the covering is regular) but is not characteristic (more precisely, is not stable under the action of the isometry group of $Y$). – Moishe Kohan Jan 26 '22 at 22:07
  • I see so I can only get away with this when $ X $ is simply connected or $ X $ is not simply connected but $ Y \cong X/\Gamma $ where $ \Gamma $ is a characteristic subgroup of of the isometry group of $ X $. I tried doing this illegally here https://math.stackexchange.com/questions/4327101/isometry-groups-of-flat-surfaces with covering of the flat cylinder over the flat Moebius strip and the flat Klein bottle, is that a case where it is acceptable? – Ian Gershon Teixeira Jan 26 '22 at 22:11
  • @IanGershonTeixeira: Yes for the Moebius strip but it can fail in the Klein bottle case (it depends on the precise choices: I did not read the linked question). – Moishe Kohan Jan 26 '22 at 22:14
  • ok fair enough would you happen to know what the isometry group of flat Klein bottle is? Is it always just $ O_2 $ the isometries of the fiber? And I suppose you are telling me that the isometry group of flat moebius strip is always $ (\mathbb{R} \rtimes O_1) \times O_1 $ (just isometries of the fiber direct product a cyclic two group). Or perhaps the Klein bottle isometry group also includes an extra cyclic two factor? – Ian Gershon Teixeira Jan 26 '22 at 22:18
  • Please can you give any reference for the formula ? – jhn142143 Feb 22 '23 at 10:49
  • @jhn142143 It's an exercise to prove, I do not have a reference. – Moishe Kohan Feb 22 '23 at 12:32