# Orthonormal basis for a tangent plane

Given a manifold $M$ described by the graph of an arbitrary smooth function $f:U \subset \mathbb{R}^2 \to \mathbb{R}$, I would like to construct an orthonormal basis for its tangent plane $T_pM$ at any point $p \in M$. We have that

$$M = \Gamma(f) := \{(x,y,z) \in \mathbb{R}^3 : (x,y) \in U, \; z = f(x,y)\}.$$

Partial differentiation yields

\begin{align} \Gamma_x &= (1,0,f_x), \\ \Gamma_y &= (0,1,f_y). \end{align}

Thus, we may take

\begin{align} \mathbf{e}_1 &:= \frac{\Gamma_x}{|\Gamma_x|} = \frac{1}{\sqrt{1+f_x^2}} (1,0,f_x), \\ \mathbf{e}_2 &:= \frac{\Gamma_y}{|\Gamma_y|} = \frac{1}{\sqrt{1+f_y^2}} (0,1,f_y), \end{align} as a basis $\mathbf{e}$ for $T_pM$. However, $\mathbf{e}$ is an orthonormal basis if and only if $\langle\Gamma_x,\Gamma_y\rangle = 0$, which is only possible when $f_xf_y = 0$.

My question, as dumb as it may sound, is: Is this always the case that $f_xf_y = 0$? Because it certainly doesn't look like it. If not, how can one easily construct an orthonormal basis for $T_pM$?

• Once you have any basis at all, you can apply Gram-Schmidt to it to get an orthonormal basis. – Qiaochu Yuan Jul 30 '14 at 0:08
• That is exactly what I thought, but for some reason I couldn't convince myself that $f_x$ and $f_y$ where not orthogonal to each other. – Hubble Jul 30 '14 at 0:10
• Have you considered any examples? – Qiaochu Yuan Jul 30 '14 at 0:11
• I guess I haven't. – Hubble Jul 30 '14 at 0:14

The problem is that you are forcing $\mathbf{e}_1$ to lie in the $xz$ plane, and forcing $e_2$ to lie in the $yz$ plane.
I think a nicer choice for an orthonormal basis is the basis with one vector along the level curves of $f$ and another vector along the direction of greatest change (gradient). Precisely, $$\mathbf{v}_1 = c_1 (-f_y, f_x,0) \\ \mathbf{v}_2 = c_2 ( f_x, f_y, f_x^2 + f_y^2)$$ where $c_1$ and $c_2$ are some normalization constants you can calculate. Also you can check fairly easily that $v_1, v_2$ are in the span of $\Gamma_1$ and $\Gamma_2$.