# Is the projection from $\mathbb R^n$ locally isometric to the flat torus?

I’m reading a problem asking to introduce a Riemannian metric on $$T^n$$ in such a way that the projection $$\pi(x_1,...,x_n)=(\exp(ix_1),...,\exp(ix_n))$$ from $$\mathbb R^n$$ to $$T^n$$ is a local isometry. Does the flat metric apply here?

By definition, flat torus is defined as the product of n copies of the Riemannian manifold $$S^1$$ with its metric induced by $$\mathbb R$$.

• Maybe you ought to clarify what the projection $\pi$ you refer to is? – T_M Dec 13 '18 at 15:16
• I editted my question @T_M – user555729 Dec 13 '18 at 15:19
• isn't the flat metric exactly defined via this property? related to this: math.stackexchange.com/questions/450110/… – Balou Dec 13 '18 at 15:19
• I’m reading DoCarmo’s book and the definition there is different @Balou – user555729 Dec 13 '18 at 15:23
• maybe you should state then how "the flat metric" is defined in DoCarmo – Balou Dec 13 '18 at 15:29

Notice that the natural projection written above maps a point of $$R^n$$ to a point in $$T^n$$. This may seem odd at first but notice how the coordinate curves of $$R^n$$ get mapped to $$T^n$$. Because $$e^{ix}$$ is a point in a complex circle, the projection essentially wraps each axis of $$R^n$$ into a circle of the torus. This maps coordinate curves into coordinate curves. Since the projection is evidently the cartesian product of a number of complex circles (which are the same as $$S^1$$) you can introduce the product metric the same as if you were working with real numbers exclusively. Since you can take the vectors to be coordinate curves locally, the product metric is a local isometry.