# Induced Riemannian metric

I'm new to differential geometry and have only begun reading about Riemannian metrics. I have the following problem from Jost.

We equip $\mathbb{R}^{n+1}$ with the inner product $$\langle x,y \rangle = -x^0 y^0 + x^1y^1 + \cdots + x^ny^n$$ for $x = (x^0, x^1, ..., x^n)$ and $y= (y^0, y^1, ..., y^n)$. Set $$H^n = \{ x \in \mathbb{R}^{n+1} \ : \ \langle x,x \rangle = -1, \ x^0 > 0 \}.$$ I want to show that $\langle \cdot, \cdot \rangle$ induces a Riemannian metric on the tangent spaces $T_p H^n \subset T_p \mathbb{R}^{n+1}$ for $p \in H^n$.

If there are any surrounding comments that you would like to add for pedagogical purposes that would be greatly appreciated. I have a fair background in manifolds but mainly from the perspective of complex analysis.

## 1 Answer

The main point here is the fact that one can identify $T_{H,p}$ with $p^{\perp}$ (with respect to your metric!). Once we have done that, we can notice that the inner product, restricted to $p^{\perp}$, is positive definite (via Sylvester Theorem).

Now, why is it possible to have that identification?

Let $c(t)$ be a (smooth) curve in H such that $c(0)=p$. Then we have got that $\langle c(t),c(t)\rangle=-1$. Hence, differentiating we get $2\langle c'(0),p\rangle = 0$. Now H is n-dimensional, so the claim is valid.