I read some geometry material which states about kernel space. For given manifold $(g, \mathcal{M})$ and distribution $A \equiv \mbox{span}(\{a_i\})$. We know two (tangent) vectors $u, v \in T_x \mathcal{X}$ can operate using the metric operator $g(u, v) = \langle u, v \rangle = g_{ij} u^i v^j$. Let us define distribution $B \equiv \mbox{span}(\{b_j\})$ such that null inner product $\langle a_i, b_j \rangle = 0$.
In Euclidean space $\mathbb{R}^n$, metric entry $g_{ij}$ is the Dirac delta $\delta_{ij}$, and inner product $\langle u, v \rangle$ becomes $u^\intercal v$, which means $v \in \mbox{ker}(u^\intercal)$. It is fairly easy to get an algebraic expression for v-belonging vector space, like $I - \frac{u u^\intercal}{u^\intercal u}$. For more than 1 vector, the kernel space is spanned by $I - (A^\intercal)^+ A^\intercal$.
We can accomplish something similar to curved space such that inner product $\langle a_i, b_j \rangle$ turns into $\langle a_i, e^k \rangle b^k_j$. Assembling all vectors $\{a_i\}$ leads to equality below
$$\underbrace{\begin{bmatrix} \lvert \\ \langle a_i, e^k \rangle \\ \lvert\end{bmatrix}}_{A^\intercal} b^k_j = 0$$
My question is: do you know any other way to proceed?
Example: An unicycle has constraints $\dot{x} \sin{\theta} - \dot{y} \cos{\theta} = 0$. In flat $\mathbb{R}^n$ space, it would be fairly easy: $b_1 = \begin{bmatrix} \cos{\theta} \\ \sin{\theta} \\ 0 \end{bmatrix}$ and $b_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ On a surface $\mathcal{S}$, we have $\dot{\gamma}^1 \sin{\theta} - \dot{\gamma}^2 \cos{\theta} = 0$, such that vector decomposition $\dot{\gamma}$ requires coordinate vector field $\dot{\gamma}^l X_l$, for vector $X_l = \varphi_{;l}$ from chart $(U_i, \varphi_i)$. We look for vector field $\{b_j\}$ such that $\left\langle \sin{\theta} X_i -\cos{\theta} X_2, b_j \right\rangle_{\mathbb{R}^n} = 0$.