Problem: Find Basis for Tangent Space of Torus
Suppose I have a torus
$$
T^2 = \left\{(x,y,z)\in\mathbb{R}^3\, : \, \left(x^2 + y^2 + z^2 + R^2 – r^2\right)^2 – 4R^2(x^2+
y^2) = 0\right\}
$$
How do I find an orthonormal basis for its tangent space $T_x$ and for $T_x^{\perp}$?
My Attempt
In this paper that one should basically first compute $Q$, the transpose of the Jacobian and then compute its QR decomposition to find the basis, so I tried this approach. Since the constraint defining the torus is $q:\mathbb{R}^3\to\mathbb{R}$ we know that the transpose of its Jacobian will be $3\times 1$ and given by
$$
Q =
\begin{pmatrix}
\partial_x q(x,y,z) \\
\partial_y q(x,y,z) \\
\partial_z q(x,y,z)
\end{pmatrix} =
\begin{pmatrix}
4x(x^2 + y^2 + z^2 + R^2 – r^2) – 8xR^2 \\
4y(x^2 + y^2 + z^2 + R^2 – r^2) – 8yR^2 \\
4z(x^2 + y^2 + z^2 + R^2 – r^2)
\end{pmatrix} \in \mathbb{R}^{3\times 1}
$$
Now, however, I don't end up with a matrix, but with a colum vector and I should be taking its QR decomposition to find the basis?
Best Answer
I find the case of a sphere sphere $S^2 = \{(x, y, z) \in \mathbb R^3 : x^2 + y^2 + z^2 - r^2 = 0\}$ easier to illustrate. As explained here, for all $p \in S^2$ we have $$T_p S^2 = \ker df_p,$$ where $f(x, y, z) = x^2 + y^2 + z^2 - r^2$. Explicitly, for $p = (a, b, c)$ we compute that $$df_p = \begin{pmatrix} 2a & 2b & 2c \end{pmatrix},$$ so we see that the kernel is $$\ker df_p = \{v \in \mathbb R^3: \langle a, b, c \rangle^T \cdot v = 0 \}.$$ This set is no other than a plane with normal vector $\langle a, b, c \rangle$, which indeed is the tangent space to $S^2$ at $p$.
Edit: So, finding an ON basis for the tangent space amounts to finding an ON basis of the kernel of $df_p$. As the kernel is perpendicular to the row space of $df_p$, an ON basis of the normal space amounts to an ON basis of the row space of $df_p$. In this case $df_p$ has one row, so the vector $df_p^T$, once normalized, is your desired basis.