I want to find the orthonormal frame associated to the following Riemann metric on $\Bbb R^5$:
$$g=\begin{pmatrix} 1+y^2+t^2 & yw &t&0&-y\\yw& 1+w^2+t^2&0&t&-w\\ t&0&1&0&0\\ 0&t&0&1&0\\-y&-w&0&0&1 \end{pmatrix},$$ where $(x,y,z,w,t)$ is local coordinate chart. It is clear to me that $e_5=\frac{\partial}{\partial t},e_4=\frac{\partial}{\partial w},e_3=\frac{\partial}{\partial z}$ but I don't know how to compute the $e_1$ and $e_2$?
I don't know how to do it directly. Probably using Gram-Schmidt Algorithm ?. In my experience, you may write down the metric as $ds^2 = g_{ij} dx^i dx^j$. Rearrange the terms gives \begin{equation} ds^2 = dx^2 + dy^2 + (wdy+ydx-dt)^2 +(dw+tdy)^2 + (dz+tdx)^2 \end{equation} take the coframe field $$\theta^1 = dx, \quad \theta^2=dy,\quad \theta^3=wdy+ydx-dt, \quad \theta^4=dw+tdy, \quad \theta^5 = dz+tdx$$ Then use transformation law for the basis vector to get the orthonormal frame. If $\theta^{\mu}=e^{\mu}_i dx^i$ and let $[E^{i}_{\mu}]$ be the inverse of $[e^{\mu}_i]$, then the orthonormal frame is $e_{\mu} = E_{\mu}^i \partial_i$.
$\textbf{EDIT :}$
I kind a feel uneasy by finding the orthonormal frame without Gram-Schmidt. So here i present the result by using that algorithm. Because $\{\partial_x,\partial_y, \partial_z, \partial_w, \partial_t\}$ is a local frame, the GS method gives an orthonormal frame $\{E_i\}_{i=1,\dots, 5}$ as $$ E_i = \frac{\partial_i - \sum_{j=1}^{i-1} (\partial_i \cdot E_j) E_j}{|\partial_i - \sum_{j=1}^{i-1} (\partial_i \cdot E_j) E_j|} $$
$$ E_1 = \partial_t, \quad E_2=\partial_w,\quad E_3 = \partial_z, \quad E_4 = \partial_y+w\partial_t-t \partial_w, \quad E_5 = \partial_x-t\partial_z +y \partial_t $$ By the way, in this case actually Gram-Schmidt is faster than my construction before.