# The Diagonalization of a Riemannian Metric

Let $(M^n, ds^2)$ denote an $n$-dimensional Riemannian manifold. The Riemannian metric $ds^2$, in general, is given, in terms of local coordinates $x’_1,\cdots,x’_n$, by $$ds^2=\sum_{i,j}h_{ij}(x’_1,\cdots,x’_n)dx’_idx’_j$$ Since the metric tensor $(h_{ij})$ is symmetric, by the finite dimensional spectral theorem, there exists an orthogonal matrix $O$ such that $O^t(h_{ij})O=(g_{ii})$. Let $$\begin{pmatrix}dx_1\\\vdots\\dx_n\end{pmatrix}=O^t\begin{pmatrix}dx’_1\\\vdots\\dx’_n\end{pmatrix}$$ Then \begin{align*}ds^2&=\sum_{i,j}h_{ij}(x’_1,\cdots,x’_n)dx’_idx’_j\\&=\begin{pmatrix}dx’_1\\\vdots\\dx’_n\end{pmatrix}^t(h_{ij})\begin{pmatrix}dx’_1\\\vdots\\dx’_n\end{pmatrix}\\&=\left[O\begin{pmatrix}dx_1\\\vdots\\dx_n\end{pmatrix}\right]^tO(g_{ii})O^t\left[O\begin{pmatrix}dx_1\\\vdots\\dx_n\end{pmatrix}\right]\\&=\begin{pmatrix}dx_1\\\vdots\\dx_n\end{pmatrix}^t(g_{ii})\begin{pmatrix}dx_1\\\vdots\\dx_n\end{pmatrix}\\&=\sum_{i}g_{ii}dx_i^2\end{align*} Therefore, without loss of generality, we may assume that a Riemannian metric is of the diagonal form $$ds^2=\sum_{i}g_{ii}(x_1,\cdots,x_n)dx_i^2$$