# How to prove a function is an isometry

1. Mar 12, 2008

### Cincinnatus

I need to show a particular map f:M-->N is an isometry (globally). M,N are riemannian manifolds, p is a point on M.

That is, I need to show:

<u,v>_p = <df_p(u),df_p(v)>_f(p) (for all p and for all u,v in T_p(M)?)

I think my problem is that I don't understand what this statement really means. Are u,v elements of the tangent space T_p(M)? Then doesn't that mean I should identify them with partial derivatives? But this doesn't make sense unless I apply them to functions...

I see that in euclidean space we have:

<e_i,e_j>=g_ij where g_ij is the identity matrix of the appropriate size

I don't see how that statement connects to the other one. What is "f" in this context?

Thanks for you help...

Last edited: Mar 12, 2008
2. Mar 13, 2008

### mrandersdk

okay an isometry is a function that preserves the metric. So you have two riemannian manifolds M and N and hence given two metrics g_M and g_N.

A metric is a tensor that takes two tangent vectors as input. The way we map tangent vectors from the tangent bundle of M to the tangent bundle of N is by the differential df.

so you see that the metric is preserved if as you say:

<u,v>_p = <df_p(u),df_p(v)>_f(p) (for all p and for all u,v in T_p(M)?)

because:

<u,v>_p = g_M(u,v)(p) (i added a (p) because you have to evaluate at p in M)
<df_p(u),df_p(v)>_f(p) = g_N(df_p(u),df_p(v))(f(p))

so you need to calculate the differential of f at p in M, use it on two tangents then aply the metric in N, and see this gives the same as just aplying g in M to the two tangent vectors. This can be done in coordinates or without depending on the concrete exercise.

You are right that the metric in euclidean space just corresponds to the identity matrix, (in coordinates), but that does not give you that any function f is an isometry, you can easily see that in R^2, if we set f(x)=x^2 then whe don't have: ||x-y|| = ||f(x)-f(y)|| for all x,y in R^2, which like what you need to prove for a manifold, but because only the tangent space has a metric it is a bit more complicated.

hope this explain the general case.

Now for your more concrete example. Lets look at $$R^2$$ with the standard basis $$e_1,e_2$$, lets look at the linear function defined on the basis by

$$f(e_1)=e_2, f(e_2) = e_1$$ extend by linearity to all $$R^2$$.

This function is clearly an isometry because it only flip the coordinate system, I know that the metric actually works on the tangent space at a point in R^2, but because R^2 is so "nice" we can identify T_pM with R^2 (if you don't see this don't worry, maybe if you understand the rest you can ask about this i will explain) so thats why i can say it is an isometry, just by looking on what the function does on R^2, but in general for a manifold it depends on the differential of f.

But lets us try to show it in your notation of a riemannian manifold. So let M be the manifold $$R^2$$ and N the same namely $$R^2$$ (i know i said that that you need to check ||x-y||=||f(x)-f(y)||) in R^2, but because we look at R^2 as a manifold now, we need to be a bit carefull (as said above), the metric now works on the tangent space, so I will try to be careful and only use the metric on tangent. As i said above lets take two tangent vectors v,u to a point p in R^2=M. These are as you say just directional derivatives that is ve can write

$$v = v_1 \partial_1(p) + v_2 \partial_2(p)$$, $$\partial_i(p) = \frac{\partial}{\partial x_i}|_p$$
$$u = u_1 \partial_1(p) + u_2 \partial_2(p)$$

so we see that

$$g_M(v,u)_p = g(v_1 \partial_1(p) + v_2 \partial_2(p),u_1 \partial_1(p) + u_2 \partial_2(p))$$

remember we have (for R^2)

$$g_M = g_N = g_{ij} dx^i \otimes dx^j = \delta_{ij} dx^i \otimes dx^j$$ einstein summation assumed.

so we get:

$$g_M(v,u)_p = \delta_{ij} dx^i(v_1 \partial_1(p) + v_2 \partial_2(p)) \otimes dx^j(u_1 \partial_1(p) + u_2 \partial_2(p)) = v_1 u_1 + v_2 u_2$$

this is actually just the usual inner product, you see. Remeber that $$dx^i$$ is defined by $$dx^i(\partial_j) = \delta^i_j$$ (maybe I write to much but don't at what level you are, so just tell me if you know stuff I said, then I can leave it out in the next post, if another post is needed)

So know we need to calculate the differential df_p(v). To do this we need some curve $$\gamma(t)$$ such that $$\gamma(0) = p, \gamma'(0) = v$$. It is not hard to see that such a curve could be

$$\gamma(t) = (v_1 e_1 + v_2 e_2)t + p$$

I said not hard to see, hate when people do this so lets check. $$\gamma'(0)$$
works on differential functions from M to R (and thus need to be evaluated on such), so let h be such a function and p = (p_1,p_2)

$$\gamma'(0)h = \frac{\partial}{\partial t}(h \circ\gamma(t))|_{t=0} = \frac{\partial}{\partial t}(h(v_1 t+p_1, v_2 t+p_2 )) |_{t=0} = (\partial_1h)(\frac{\partial(v_1 t+p_1)}{\partial t}) + (\partial_2h)(\frac{\partial(v_2 t+p_2)}{\partial t})|_{t=0} =$$
$$(\partial_1h)(v_1)+ (\partial_2h)(v_2)|_{t=0} = (v_1\partial_1+ v_2\partial_2)h|_{t=0}= (v_1\partial_1+ v_2\partial_2)h$$

this is for all h so

$$\gamma'(0)= v_1\partial_1+ v_2\partial_2$$

so i was right :-). Now we got our curve, lets get to df_p(v), this is defined as

$$df_p(v) = df_p(\gamma'(0)) = \frac{\partial}{\partial t}(f \circ \gamma(t))|_{t=0} = \frac{\partial}{\partial t}(f((v_1 e_1 + v_2 e_2)t))|_{t=0}= \frac{\partial}{\partial t}((v_1 e_2 + v_2 e_1)t)|_{t=0} = v_1 e_2 + v_2 e_1$$

the same thing happens for u so

$$df_p(u) = u_1 e_2 + u_2 e_1$$ of cause to see this you need to find a simelar curve that fulfills $$\gamma(0) = p, \gamma'(0) = u$$

we are know ready to evaulate

$$g_N(df_p(v),df_p(u))_p = g_N(v_1 e_2 + v_2 e_1,u_1 e_2 + u_2 e_1)_p$$

like before this is just the usual inner product (now on N, but N=R^2) so

$$g_N(df_p(v),df_p(u))_p = v_1 u_1 + v_2 u_2 = v_2 u_2 + v_1 u_1 = g_M(v,u)_p$$

you see that in this case, what p i chose didn't play any role so this is for all p in M,
so we are done, I think i did it in a so general way that M and N could have been any manifolds, the onyl thing i used was that it was easy to find curves in R^2, you need to make a more general curve maybe only demanding $$\gamma(0) = p, \gamma'(0) = u$$, not even writing an explicit form of $$\gamma$$, this depend on your problem.

3. Mar 15, 2008

### Cincinnatus

Wow, thanks so much for the detailed answer.

One question (I'll probably have more as I digest this) is, do I always need to come up with curve in M? What do I do if it's not so clear how to write down a curve... Actually, I'm trying to show a function on the upper half plane is an isometry with respect to the metric g_ij=identity(2)*(1/y^2). So, I guess I could write down a curve...

Separately, I also want to show that metric is induced by the euclidean metric on R^2 since the upper half plane is a subspace of R^2...

4. Mar 16, 2008

### mrandersdk

if it imposible to write down the curve it can be a bit tricky, but sometimes all you need is to know that it satiesfy $$\gamma(0)=p,\gamma'(0)=v$$. There are also other equivalent deffinitions of the differetial, thats why it completely depends on the exercise what way you should do it. Sometimes on a general manifold you can construct the curves from something which is called the exponential function on a manifold (not the same as the normal exp), this is a very usefull construction because it makes it possible to calculate differentials of a general manifold when doing proofs.

By the way the upper half plane is only a subspace if it is over the postive real number, because if v is in that then -v is not, just a comment.

5. Mar 16, 2008

### Cincinnatus

Oops, I meant it was a submanifold, I didn't mean a (vector) space over only the positive real numbers. This wasn't really important though.