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Let us consider a standard metric and differentiable structure on $T^2$.

We first consider the diffeomorphism group and isometry group of $T^2$ on itself.

1. What is the diffeomorphism group of $T^2$?

According to Wiki, the diffeomorphism group of the torus has the homotopy-type of its linear automorphisms $S^1 \times S^1 \times GL(2, \mathbb{Z}).$ So is that true: $$\mathrm{Diff}(T^2)= S^1 \times S^1 \times GL(2, \mathbb{Z})?$$

2. What is the isometry group of $T^2$?

From isometry group of the riemannian manifold $\mathbb{T}^2$?, it seems that there is a ambiguity whether we have $$\mathrm{Isometry}(T^2)= D_8\rtimes_\varphi T^2?$$ where $\varphi\colon D_8\to \mathrm{Aut}(T^2)$ with $D_8$ the order 8 dihedral group.

3. Why are $\mathrm{Diff}(T^2)$ and $\mathrm{Isometry}(T^2)$ different?

Peter
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annie marie cœur
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    Are you interested in the homotopy types of these groups, or the groups themselves? – Peter May 22 '21 at 17:19
  • The groups themselves. But could you clarify what are the roles of the homotopy types of these groups? – annie marie cœur May 22 '21 at 17:22
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    Every topological space has a homotopy type. Two topological spaces can be different, but have the same homotopy type. For example, the circle and the annulus have the same homotopy type, but are different (not even homeorphic). – Peter May 22 '21 at 17:24
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    The groups $\mathrm{Diff}(T^2)$ and $\mathrm{Isometry}(T^2)$ are different, there are diffeomorphism which are not isometries. I don't know if these groups have the same or different homotopy types. – Peter May 22 '21 at 17:27
  • The isometry group of a flat torus depends on the lattice $\Lambda$ used to construct the torus: An isometry of a flat torus lifts to a Euclidean plane motion mapping $\Lambda$ to itself. Generically, the only symmetries of a lattice are translations and multiplication by $-1$. – Andrew D. Hwang May 22 '21 at 17:32
  • @Andrew D. Hwang - How is that related to the answer I wrote: $\mathrm{Isometry}(T^2)= D_8\rtimes_\varphi T^2?$ – annie marie cœur May 22 '21 at 17:42
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    The isometry group you assert is presumably for a flat square torus. – Andrew D. Hwang May 22 '21 at 18:22

1 Answers1

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Question 1 is answered in the comments. And for question 2, the formula that you copied from the other post is correct, presuming that by the "standard" metric you mean the product metric coming from the product formula $T^2 = S^1 \times S^1$ (as alluded to in the comments, there are many other interesting metrics on $T^2$, even many flat metrics, and their isometry groups can vary).

Regarding question 3, the groups $\text{Diff}(T^2)$ and $\text{Isometry}(T^2)$ are different in many ways, for many different meanings. Perhaps the simplest reason that the diffeomorphism group is never equal to the isometry group is because a diffeomorphism can "stretch" tangent vectors in a way that an isometry cannot.

In fact, on any Riemannian manifold $M$, for any point $x \in M$, and for any linear isomorphism $L : T_x M \to T_x M$, there exists a diffeomorphism $f : M \to M$ such that $f(x)=x$ and $Df_x = L : T_x M \to T_x M$ (this is an exercise in differential topology).

If you then pick a basis for $T_x M$ which is orthonormal with respect to the inner product that is specified by the given Riemannian metric on $M$, for any isometry $f : M \to M$ such that $f(x)=x$, its derivative $D_x f : T_x M \to T_x M$ is expressed in that basis as an element of the orthogonal group $\text{O}(n,\mathbb R)$, whereas the derivative at $x$ of a diffeomorphism that fixes $x$ is expressed as an element of the general linear group $\text{GL}(n,\mathbb R)$.

So all you have to do to get a diffeomorphism of $M$ that is not an isometry is to pick $L \in \text{GL}(n,\mathbb R) - \text{O}(n,\mathbb R)$.

For an independent, and somewhat different, but very interesting reason why $\text{Diff}(T^2)$ and $\text{Isometry}(T^2)$ differ: the group $\text{Diff}(T^2)$ has infinitely many path components, but no matter what Riemannian metric you pick on $T^2$ (not just the "standard" one), the image of the inclusion $\text{Isometry}(T^2) \hookrightarrow \text{Diff}(T^2)$ is contained in a union of only finitely many path components of $\text{Diff}(T^2)$. This can be expressed more clearly in algebraic topology language using the action of $\text{Diff}(T^2)$ on $H_1(T^2;\mathbb Z) \approx \mathbb Z^2$, but that's beginning to get rather far beyond the basic point of Question 3 so I'll stop now.

Lee Mosher
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  • +1 Thanks so much for this "Perhaps the simplest reason that the diffeomorphism group is never equal to the isometry group is because a diffeomorphism can "stretch" tangent vectors in a way that an isometry cannot." – annie marie cœur May 22 '21 at 18:32
  • Can you also answer this? https://math.stackexchange.com/q/4147754/141334 which Diff(M) = Isometry(M) likely? – annie marie cœur May 22 '21 at 18:33
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    Well, my answer here applies there as well: What I have shown here is that for any Riemannian manifold whatsoever, $\text{Diff(M)} \ne \text{Isometry}(M)$. – Lee Mosher May 22 '21 at 19:00
  • @LeeMosher Hi professor Mosher, could I ask your assistance here, please? I am researching answer to a question I have been stuck with for few days: I edited the question more times and now it is left only one step to solve the problem. Excuse me for the bother, but I am autodidact and so I thought you can help me. – Antonio Maria Di Mauro May 23 '21 at 07:36