I have an isometri $\psi$ of a riemannien manifold that coincides with identity on a nonempty open set. How do you show that it must be the identity?

I thought of proving that it's closed.

$\{x\in M\ \vert\ \psi(x) = x\} = \{x\in M\ \vert\ (\psi-\text{Id})(x) = 0\}$ which is the inverse of a closed set and therefore closed.

Is this good?


You're not there yet. First, if $M$ is just a Riemannian manifold, then $(\psi-\text{Id})(x)$ makes no sense -- you can't subtract points in an abstract manifold. But you could use the fact that every smooth manifold can be embedded in some Euclidean space, and perform the subtraction there. With that interpretation, your proof does show that the set of points where $\psi(x)=x$ is a closed subset of $M$.

But that's not enough. In fact, it is the case that if $f$ and $g$ are any two continuous maps from a topological space to a Hausdorff space, then the set of points $x$ where $f(x)=g(x)$ is a closed subset of the domain. (It's a good exercise in point-set topology to prove this without relying on an embedding in $\mathbb R^n$.)

In fact, the statement you're trying to prove is not true. For example, if $M$ is the union of two disjoint unit circles in the plane, and $f$ acts as the identity on one circle and as a nontrivial rotation on the other, then $f$ satisfies the hypothesis but not the conclusion. You have to assume $M$ is connected.

With $M$ connected, it would be sufficient to prove that the set of points $x$ such that $f(x)=g(x)$ is both open and closed. (Do you see why?)

In your case, you know there's a nonempty open set of points where $f(x)=g(x)$, but that does not imply that the set of all points where $f(x)=g(x)$ is open. You're going to have to use some nontrivial properties of isometries, such as how they interact with the exponential map, to prove this.

In fact, you probably won't be able to prove that the set of points where $f(x)=g(x)$ is open -- for example, a reflection of the circle is an isometry whose fixed point set is closed but not open. Instead, you should look at the set of points where more is true about $f$ and $g$ -- I'll let you figure out what a good set is, but the key is that you have to assume enough of a match between $f$ and $g$ at each point to ensure that they play well with the exponential map.

  • $\begingroup$ Indeed, what I did seemed to easy and didn't use the isometric condition. I'll post an answer once I find it. $\endgroup$ – tomak Jun 20 '17 at 9:22

So the result is based on the following result:

If $\psi^1$ and $\psi^2$ are two isometries from $(M,g)$ to $(N,h)$ such that there exists $x\in M$ s.t. $\psi^1(x) = \psi^2(x)$ and $\psi^1_{*x} = \psi^2_{*x}$ then $\psi^1 = \psi^2$

We show that the set respecting these two conditions is closed and open. We know that it's closed so it remains to show that it's open.

take $x\in A:=\{x\in M\ \vert\ \psi^1(x) = \psi^2(x),\psi^1_{*x} = \psi^2_{*x}\}$.

There exists $W$ open s.t. for every $y\in W$ there is a unique geodesic from $x$ to $y$. Then $\psi^1\circ \gamma$ and $\psi^2\circ \gamma$ are geodesics from $\psi^1(x)$ to $\psi^1(y)$ and from $\psi^2(x)$ to $\psi^2(y)$ respectively. But $\psi^1(x) = \psi^2(x)$ and $\psi^1_{*x}(X_x) = \psi^2_{*x}(X_x)$ for all $X_x$ so by unisicty of geodesics, $\psi^1(y) = \psi^2(y)$. So on $W$, $\psi^1 = \psi^2$.

$\psi^1_{*x}Y_y(f) = Y_y(f\circ \psi^1) = Y_y(f\circ \psi^2) = \psi^2_{*x}Y_y(f)$ since $Y_y$ is local. So $\psi^1_{*x} = \psi^2_{*x}$

So $W\subset A$

Now all we need to see is that our initial hypothesis of the question satisfies the property and we are finished.

  • 1
    $\begingroup$ Note that you must assume $M$ and $N$ are connected. $\endgroup$ – Neal Jun 20 '17 at 20:05

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