# Isometry of $\mathbb{R}^4$ Torus

I am working on an exercise and I am stumped. The exercise is to prove that $$X(u,v)=(\sin u,\cos u, \sin v, \cos v)$$ is an isometry. This is in the chapter of Theorema Egregium so I assume they mean "Locally isometric" when they say that. I am also aware of that they are isometric if and only if they have the same first fundamental form.

I easily got $$X_u=(\cos u,-\sin u, \sin v,\cos v)$$ $$X_v=(\sin u,\cos u, \cos v,-\sin v)$$ and from there $$X_u\cdot X_u = 2$$ $$X_u\cdot X_v = 0$$ $$X_v\cdot X_v = 2$$

So far so good, for $\mathbb{R}^2$ I figured our $X$ would be $X(u,v)=(u,v)$ as a suitable one as it is fairly trivial and naturally get $$X_u\cdot X_u = 1$$ $$X_u\cdot X_v = 0$$ $$X_v\cdot X_v = 1$$ and poof it dies. Where am I going wrong? Is this a bad tactic to take it on? Is a more direct approch better or?

• There's an (obvious) mistake in your calculation...... "I easily got..." – user99914 Sep 23 '15 at 8:29
• I have been staring it for so long I am currently blind, care to point it out so I can feel stupid for a brief moment and then happy to find the solution? – Zelos Malum Sep 23 '15 at 8:30
• $\frac{\partial}{\partial u} \sin u \neq -\sin u$ and some others, recalculate your derivatives – Christoph Sep 23 '15 at 8:31
• Actually I just realized all eight terms in $X_u, X_v$ are calculated incorrectly.. You have to take partial derivative term by term. – user99914 Sep 23 '15 at 8:32
• OH YEAH! That is it, I forgot to make them zeros due to it. You are perfectly correct, wow thank you. Yes that WAS obvious now that it is pointed out, thank you! – Zelos Malum Sep 23 '15 at 8:34