Recently I was learning the Riemannian manifold and when calculating the Christoffel symbol I was in some trouble:
In $\mathbb{R}^n$, let $g_{ij}=\delta_{ij}+\frac{x^ix^j}{K^2-|x|^2}$, where $K$ is a constant. To calculate the Christoffel symbol, we need to find the inverse matrix of $(g_{ij})$. However, I don't know how to do this.
Could anyone give me some hints? Thanks a lot! Any help will be appreciated!
Well, how do you calculate the inverse of an ordinary $n \times n$ matrix? I'd first write the $n \times n$ matrix for $g = (g_{ij})$ as $g = I + \lambda x^T x$ where $\lambda = \frac{1}{K^2 - \lvert x \rvert^2}$ and $x = (x^1, \ldots, x^n)^T$ is a column vector.
Then you can just read-off/check that $(I + \lambda x^T x)^{-1} = I - \frac{\lambda x^T x}{1 + \lambda x x^T}$, since we have \begin{align*} (I + \lambda x^T x)(I - \frac{\lambda x^T x}{1 + \lambda x x^T}) &= (I + \lambda x^T x) - (I + \lambda x^T x) \frac{\lambda x^T x}{1 + \lambda x x^T}\\ &= (I + \lambda x^T x) - \frac{\lambda x^T x}{1 + \lambda x x^T} - \frac{\lambda^2 x^T (x x^T) x}{1 + \lambda x x^T}\\ &= (I + \lambda x^T x) - \frac{\lambda x^T x}{1 + \lambda x x^T} - (\lambda x x^T) \frac{\lambda x^T x}{1 + \lambda x x^T} \\ &= (I + \lambda x^T x) - (1 + \lambda x x^T) \frac{\lambda x^T x}{1 + \lambda x x^T}\\ &= (I + \lambda x^T x) - \lambda x^T x\\ &= I. \end{align*}
So, the inverse of $g$ has entries $$ (g^{-1})_{ij} = \delta_{i j} - \frac{x^i x^j}{K^2 - \lvert x \rvert^2 + \lvert x \rvert^2} = \delta_{i j} - \frac{x^i x^j}{K^2}. $$