<s>[INST] Gram-Schmidt process when metric is not an Euclidean Can we use a Gram-Schmidt process for finding an orthogonal vector if a metric on a space is not an Euclidean one?
These are the vectors and the metric.
If we have the next two vectors:
\begin{align}
\notag
A_{1}=c_{1}y_{1}^{2}y_{2}^{2}u(u+2t)(u+y_{2}^{2})\frac{\partial}{\partial x_{2}},
\end{align}
\begin{align}
\notag
A=-c_{11}\frac{\partial}{\partial x_{1}}-c_{12}\frac{\partial}{\partial x_{2}}-\frac{u(u+2t)}{u+y_{2}^{2}}(c_{8}(u+y_{1}^{2})+c_{1}y_{1}y_{2})\frac{\partial}{\partial y_{1}},
\end{align}
and the metric
\begin{align}
\notag
G=
\begin{pmatrix}
u+vy_{1}^{2} & vy_{1}y_{2} & 0& 0 \\
vy_{1}y_{2} & u+vy_{2}^{2} & 0& 0 \\
0 & 0 & \frac{u+vy_{2}^{2}}{u(u+2tv)}& -\frac{vy_{1}y_{2}}{u(u+2tv)}\\
0& 0& -\frac{vy_{1}y_{2}}{u(u+2tv)} & \frac{u+vy_{1}^{2}}{u(u+2tv)} ,
\end{pmatrix}
\end{align}
If we use Gram-Schmidt in this case, we get a vector which is orthogonal to $A_{1}$ but not to $A$.\
On the other hand, we can find vectors that span a normal bundle of a submanifold generated by $A_{1}$ and $A$, using an almost complex structure.\
We need to find a normal component of a given vector with respect to $A_{1}$ and $A$, that is why we need a vector that is orthogonal to $A_{1}$ and $A$. So, my other question would be is one of the vectors in the normal bundle enough for finding mentioned component?
Thank you!
 [/INST] In the exact same way, but instead of using the dot product, you use whatever inner product you're given.
Example: Consider $\mathbb{R}^2$ with the inner product 
$$\langle (x_1, x_2), (y_1, y_2)\rangle = 2x_1y_1 + x_1y_2 + x_2y_1 + 3x_2y_2.$$ 
Given the basis $\{(1, 0), (0, 1)\}$ of $\mathbb{R}^2$, what do we get when we apply the Gram-Schmidt process? As 
$$\|(1, 0)\|^2 = \langle (1, 0), (1, 0)\rangle = 2(1)(1) + (1)(0) + (0)(1) + 3(0)(0) = 2,$$ 
$u_1 = \frac{1}{\sqrt{2}}(1, 0)$. Now
\begin{align*}
u_2' &= (0, 1) - \left\langle (0, 1), \frac{1}{\sqrt{2}}(1, 0)\right\rangle\frac{1}{\sqrt{2}}(1, 0)\\ 
&= (0, 1) - \left(2(0)\left(\frac{1}{\sqrt{2}}\right)+(0)(0)+(1)\left(\frac{1}{\sqrt{2}}\right)+3(1)(0)\right)\frac{1}{\sqrt{2}}(1,0)\\ 
&= (0,1)-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}(1,0)\\
&= (0,1) - \frac{1}{2}(1, 0)\\
&= \left(-\frac{1}{2}, 1\right)
\end{align*}
and 
\begin{align*}
\|u_2'\|^2 &= \left\|\left(-\frac{1}{2},1\right)\right\|^2\\ 
&= \left\langle\left(-\frac{1}{2},1\right), \left(-\frac{1}{2},1\right)\right\rangle\\ 
&= 2\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right)(1) + (1)\left(-\frac{1}{2}\right)+3(1)(1)\\ 
&= \frac{5}{2}
\end{align*}
so $u_2 = \frac{\sqrt{2}}{\sqrt{5}}(-\frac{1}{2}, 1)$. So an orthonormal basis for $\mathbb{R}^2$, with the inner product above is 
$$\{u_1, u_2\} = \left\{\frac{1}{\sqrt{2}}(1, 0), \frac{\sqrt{2}}{\sqrt{5}}\left(-\frac{1}{2}, 1\right)\right\}.$$
</s>