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What is the isometry group of the riemannian manifold $\mathbb{T}\times \mathbb{T}$ where $\mathbb{T}=\{z\in \mathbb{C}\ :\ |z|=1\}$ is the classical torus?

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Is this a homework problem? What have you tried so far? Do you have thoughts on what the answer should be? –  Aaron Jan 25 '12 at 17:53
    
I know the isometry group of $\mathbb{T}$. More generally, how to compute the isometry group of a product manifold? I does not think that it is the product of isometry groups in the general case. –  Zouba Jan 25 '12 at 17:58
    
To be clear, do you mean to put the product Riemannian metric on $\mathbb{T}^2$? You are right that the isometry group of $M\times N$ isn't always the product of the isometry groups - consider $\mathbb{R}^2 = \mathbb{R}\times\mathbb{R}$ (everything with standard metrics)$. –  Jason DeVito Jan 25 '12 at 18:05
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Your definition of $\mathbb T$ is actually of a circle. –  Grumpy Parsnip Jan 25 '12 at 18:12
    
Speaking of $\mathbb R^2$, it is worthwhile to note that $\mathbb R^2$ is a covering space of $\mathbb T^2$, and we can lift the Riemannian metric on $\mathbb T^2$ to $\mathbb R^2$. It might be easier to think about isometries of $\mathbb R^2$ and then asking which ones induce isometries of $\mathbb T^2$. –  Aaron Jan 25 '12 at 18:18

2 Answers 2

up vote 3 down vote accepted

I'll just work out some of the pieces and let you work out the details.

First, the isometry group contains a canonical copy of $\mathbb{T}^2$ obtained by multiplication by elements in $\mathbb{T}^2$ (recalling that $\mathbb{T}^2$ is a Lie group). This is already enough to see that every point of $\mathbb{T}^2$ can be moved to any other by an isometry, i.e., $\mathbb{T}^2$ is homogeneous.

This means that we really only need to figure out what the isometries are which fix a point.

Thinking of $\mathbb{T}^2$ as a square with sides identified, I'm thinking of the point as being the bottom left (=top left = top right = bottom right) corner. Focus on the vertical and horizontal lines emanating from this point. I claim each of those are geodesics (if parametrized correctly) and that they have the same length, and that they are shorter than any other closed geodesics starting at the point we chose.

This shows that there are at most 8 isometries which fix this point: The vertical line can go to either the horizontal line or vertical traversed one way or the other (4 choices) and the horizontal line must go to the other line, but we can still choose which way to traverse it. This is a group isomorphic to $\mathbb{Z}/2\mathbb{Z}^3$ generated by 3 elements: Change direction vertically, change direction horizontally, swap vertical an horizontal.

It follows from this that the isometry group is generated by $\mathbb{T}^2$ and these other 8 isometries. I'll leave it to you to prove that $\mathbb{T}^2$ is in the center of the isometry group (which is easy after noting that you're just checking conjugation by these 8 extra isometries, and that you can reduce this to checking only 2 cases).

This should be enough for you to write down the whole isometry group.

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Thank you very much. First question: is there a description of this group using the semi-direct product of this group (of $\mathbb{T}^2$ by $\matrhbb{Z_2}^3$? –  Zouba Jan 25 '12 at 19:24
    
A last question which seems more difficult: Do you know necessary and sufficient conditions on the manifolds $M$ and $N$ in order to have $Isom(M\times N)=Isom(M)\times Isom (N)$? –  Zouba Jan 25 '12 at 19:29
    
For the first, I think you can show, using this, that it's just a direct product. For the second, I'd guess that whenever $M$ iss NOT isometric to $N$, then $Iso(M\times N) = Iso(M)\times Iso(N)$, espeically in the case of $M$ and $N$ compact. But I'm not sure. –  Jason DeVito Jan 25 '12 at 20:56
    
I posted a supplemental answer. –  Grumpy Parsnip Jan 26 '12 at 2:46
    
As Jim points out in his answer, I got (at least) 2 things wrong: First, the isotropy group of order 8 is not $\mathbb{Z}/2\mathbb{Z}^3$ as I said, but rather $D_8$ as he says. Likewise, the answer is a semidirect product, not a direct product as I guessed. I'm not going to edit my answer because he already supplies the details on fixing it. –  Jason DeVito Jan 26 '12 at 2:55

I'd like to complement Jason's excellent and insightful answer by pointing out that the stabilizer of a point that he calculates is the dihedral group $D_8$ and the full isometry group is not $D_8\times \mathbb T^2$ as he guessed, but actually the semidirect product $D_8\rtimes_\varphi \mathbb T^2$ where $\varphi\colon D_8\to Aut(\mathbb T^2)$ is the obvious map. You can see that it's not a direct product by calculating on a simple example. Say take $\rho\in D_8$ to be reflection in the $y$-axis and $T$ to be the isometry moving things to the right by $.25$ units. Then $T\rho(0)= 0.25\neq \rho T(0)=.75$.

It shouldn't be too hard to write an explicit isomorphism from $Isom(\mathbb T^2)$ to $D_8\rtimes_\varphi \mathbb T^2$.

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I can't believe I thought it was a direct product. I also got the $\mathbb{Z}/2\mathbb{Z}^3$ part wrong which is obvious since "swap horizontal and vertical" and "reverse direction vertically" actually generate everything. Thanks! –  Jason DeVito Jan 26 '12 at 2:54

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