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Consider the following Riemannian metric: $$g_{ij}(x):= (1-\psi)\dfrac{(\delta_{ik}x^{k})(\delta_{jl}x^{l})}{|x|^{2}}+ \psi\delta_{ij},$$ where $$|x|:=\sqrt{\delta_{ij}x^{i}x^{j}}\qquad , \qquad \psi:= \Big[\frac{s_{\lambda}(|x|)}{|x|}\Big]^{2},$$ and $$ s_{\lambda}(|x|)= \begin{cases} \frac{\sin(\sqrt{\lambda}|x|)}{\sqrt{\lambda}}, &\lambda > 0 \\ |x|, &\lambda=0 \\ \frac{\sinh(\sqrt{-\lambda}|x|)}{\sqrt{-\lambda}}, &\lambda < 0 \\ \end{cases} $$ I have to check

  • For $\lambda \leq 0$, the matrix $(g_{ij})$ is positive-definite at every $x\in \mathbb{R}^{n}$.
  • For $\lambda > 0$, the matrix $(g_{ij})$ is positive-definite at every $x < \frac{\pi}{\sqrt{\lambda}}$.

How can i easily do this?

Help me please.

user
  • 1,342

1 Answers1

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Implicit in the definition of your Riemannian metric, whatever the value of $\lambda$, is that at $x=0$, you get $g_{ij}(0) = \delta_{ij}$, which simply recovers the Euclidean inner product; thus, you only need to worry about $x \neq 0$. Let $\langle \cdot, \cdot \rangle$ denote the Euclidean inner product, so that for $v$, $w \in \mathbb{R}^n$, $$ \langle v,w \rangle = \delta_{ij} v^i v^j, \quad |v| = \sqrt{\langle v,w \rangle}. $$ Fix $0 \neq x \in \mathbb{R}^n$. Then for any $v \in \mathbb{R}^n$, $$ g(x)(v,v) := g_{ij}(x)v^iv^j\\ = (1-\psi)\frac{(\delta_{ik}x^k)(\delta_{jl}x^l)}{|x|^2}v^iv^j + \psi \delta_{ij}v^iv^j\\ = (1-\psi)\frac{(\delta_{ik}v^ix^k)(\delta_{jl}v^jx^l)}{|x|^2} + \psi \delta_{ij}v^iv^j\\ = (1-\psi)\frac{\langle v,x \rangle^2}{|x|^2} + \psi |v|^2, $$ where, in particular, $$ \frac{\langle v,x \rangle^2}{|x|^2} \geq 0. $$ At this point, do you see how to check case by case when $g(x)$ is positive definite?

Branimir Ćaćić
  • 13,684