Return to Answer

Setting aside the assumption that $\phi$ be a polynomial mapping for the moment, if one makes the 'nondegeneracy' assumptions

I don't know, offhand, whether there are any polynomial solutions to these conditions, but I think that shouldn't be hard to answer. (I suspect that there are such polynomial solutions, but I haven't had time to check this.)

The following analysis is a more-or-less standard approach using the so-called 'moving frame'. (I'm sure that the result itself is classical in some sense, though I don't know offhand where to look in the literature to find it.)

The conditions

1. $\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t)\bigr) =2 $,

2. $\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) \bigr) = 3$ for all $(s,t)\in[0,1]^2$, and

3. the subspace $W(s,t) =\mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) \bigr)\subset\mathbb{R}^4$ is not constant, in the sense that $W:[0,1]^2\to \mathrm{Gr}_3(\mathbb{R}^4)\simeq\mathbb{RP}^3$ has nonvanishing differential,

are actually independent of the choice of $st$-coordinates on the surface in $\mathbb{R}^4$ and so can be regarded as conditions on a surface $S\subset\mathbb{R}^4$ that, for local analysis purposes, can be taken to be smoothly embedded.

Now, let us pull back these functions and $1$-forms to $B_0(S)$ (but, as is customary, not notate the pullback). The definition of $B_0(S)$ implies that $$ \mathrm{d}x\wedge e_1\wedge e_2 = \mathrm{d}e_1\wedge e_1\wedge e_2\wedge e_3 = \mathrm{d}e_2\wedge e_1\wedge e_2\wedge e_3 = 0 $$ while the two expressions $$ \bigl(\,\mathrm{d}e_1\wedge e_1\wedge e_2,\ \mathrm{d}e_2\wedge e_1\wedge e_2\bigr) \qquad\text{and}\qquad \mathrm{d}e_3\wedge e_1\wedge e_2\wedge e_3 $$ are nowhere vanishing. Using the above structure equations, these are equivalent to the relations $$ \omega^3 = \omega^4 = \theta^4_1 = \theta^4_2 = 0,\tag3 $$ while $\omega^1\wedge\omega^2$ is nonvanishing, $\theta^4_3$ is nonvanishing, and the pair $(\theta^3_1,\theta^3_2)$ do not simultaneously vanish.

Meanwhile, using the structure equations, we have $$ 0 = \mathrm{d}\omega^3 = -\theta^3_1\wedge\omega^1 -\theta^3_2\wedge\omega^2, $$ so there must exist functions $h_{ij}=h_{ji}$ for $1\le i,j,\le 2$, not all simultaneously vanishing, such that $\theta^3_i = h_{ij}\omega^j$. The quadratic form $h = h_{ij}\,\omega^i\omega^j$ is then nonvanishing and well-defined up to multiples on the surface $S$. Moreover, for $1\le i\le 2$, we have $$ 0 = \mathrm{d}\theta^4_i = -\theta^4_k\wedge\theta^k_i = -\theta^4_3\wedge\theta^3_i\,. $$ Thus, since $\theta^4_3$ is nonvanishing, it follows that $\theta^3_1$ and $\theta^3_2$ are multiples of $\theta^4_3$. In particular, $\theta^3_1\wedge\theta^3_2$ vanishes identically, so $h_{11}h_{22}-{h_{12}}^2$ vanishes identically. Thus, the quadratic form $h$ has rank $1$.

Let $B_1(S)\subset B_2(S)$ denote the submanifold defined by $h_{11} = h_{12}=0$. It is a smooth submanifold of $B_0(S)$ of dimension $11$, and when all the forms and functions are pulled back to $B_1(S)$, we have $\theta^3_1 = 0$ while $\theta^3_2 = h_{22}\,\omega^2$. In particular, it now follows that $\theta^4_3$ is also a multiple of $\omega^2$, say $\theta^4_3 = f\,\omega^2$ for some $f$ (which is nonvanishing).

Further, we have $$ 0 = \mathrm{d}\theta^3_1 = -\theta^3_k\wedge\theta^k_1 = -\theta^3_2\wedge \theta^2_1 = -h_{22}\omega^2\wedge\theta^2_1, $$ so it follows (since $h_{22}$ is nonvanishing) that $\theta^2_1 = g\,\omega^2$ for some function $g$ on $B_1(S)$.

Now, we can compute that $$ \mathrm{d}\omega^2 = -\theta^2_j\wedge\omega^j = -g\,\omega^2\wedge\omega^1 - \theta^2_2\wedge\omega^2 = -(\theta^2_2 - g\,\omega^1)\wedge\omega^2. $$ Thus, $\omega^2$ is an integrable $1$-form, and, because it is semi-basic for the submersion $x:B_1(S)\to S\subset\mathbb{R}^4$, it follows that $\omega^2$ is a multiple of the $x$-pullback of a (nonvanishing) $1$-form on $S$. Thus, $S$ is foliated by (connected) curves whose $x$-preimages in $B_1(S)$ are codimension $1$ integral submanifolds of $S$.

I claim that these curves in $S$ are, in fact lines in $\mathbb{R}^4$. To see this, note that, when one pulls back to a leaf of $\omega^2=0$ in $B_1(S)$, one has $\theta^2_1 = \theta^3_1=\theta^4_1= 0$ as well, so one has $$ \mathrm{d}x = e_1\,\omega^1\qquad\text{and}\qquad \mathrm{d}e_1 = e_1\,\theta^1_1\,. $$ In particular, the direction of $e_1$ is fixed on this leaf and it is the tangent direction of the mapping $x$ restricted to this leaf. Hence the $x$-image of this leaf is an open interval in a line in $\mathbb{R}^4$. Thus, the surface is ruled, as claimed.

In particular $\mathcal{I}$ is the pullback of a well-defined Pfaffian system of rank $5$ on $F$, whose annihilator is the $4$-plane field $D\subset TF$. Moreover, it is easy to show that the three rank $6$ Pfaffian systems got by adjoining any one of $\omega^2$, $\theta^3_2$, or $\theta^4_3$ to $\mathcal{I}$ are themselves pullbacks of rank $6$ Pfaffian systems on $F$ whose annihilators in $TF$ are $3$-plane subbundles of $D$.

By its very construction, the projection of $B_1(S)$ into $F$ is a curve that is tangent to $D$ and not tangent to any of these three $3$-plane subbundles of $D$. Conversely, if $\gamma\subset F$ is any integral curve of $D$ that is not tangent to any of the three $3$-plane subbundles of $D$, its preimage in $\mathbb{R}^4\times\mathrm{GL}(4,\mathbb{R})$ is a submanifold of the form $B_1(S)$ for a surface $S$ satisfying our conditions, in fact, the surface swept out by the union of the lines represented by $\lambda(\gamma)$, where $\lambda:F\to \Lambda$ is the obvious map to the lines.

Setting aside the assumption that $\phi$ be a polynomial mapping for the moment (however, see below for a construction of a large family of polynomial solutions), if one makes the 'nondegeneracy' assumptions

It is not hard to write down polynomial solutions: For example, if $f:\mathbb{R}\to\mathbb{R}^4$ is a polynomial curve that satisfies $f'(s)\wedge f''(s)\wedge f'''(s) \wedge f''''(s)\not = 0$, then the mapping $\phi:\mathbb{R}^2\to\mathbb{R}^4$ given by $$ \phi(s,t) = f(s) + f'(s)\, t $$ (which parametrizes the 'tangential development' of the curve $f$) satisfies these conditions when $t\not=0$. (By replacing $t$ by $t{+}1$, say, one could arrange that $\phi$ be an immersion on all of $[0,1]^2$.)

The following analysis is a more-or-less standard approach to verifying the above description using the so-called 'moving frame'. (I'm sure that the result itself is classical in some sense, though I don't know offhand where to look in the literature to find it.)

The three conditions listed above are actually independent of the choice of $st$-coordinates on the surface in $\mathbb{R}^4$ and so can be regarded as conditions on a surface $S\subset\mathbb{R}^4$ that, for local analysis purposes, can be taken to be smoothly embedded.

Now, pull back these functions and $1$-forms to $B_0(S)$ (but, as is customary, not notate the pullback). The definition of $B_0(S)$ and the assumptions on $S$ imply that $$ \mathrm{d}x\wedge e_1\wedge e_2 = \mathrm{d}e_1\wedge e_1\wedge e_2\wedge e_3 = \mathrm{d}e_2\wedge e_1\wedge e_2\wedge e_3 = 0 $$ while the two expressions $$ \bigl(\,\mathrm{d}e_1\wedge e_1\wedge e_2,\ \mathrm{d}e_2\wedge e_1\wedge e_2\bigr) \qquad\text{and}\qquad \mathrm{d}e_3\wedge e_1\wedge e_2\wedge e_3 $$ are nowhere vanishing. Using the above structure equations, these imply the relations $$ \omega^3 = \omega^4 = \theta^4_1 = \theta^4_2 = 0,\tag3 $$ while, because of the three assumptions, $\omega^1\wedge\omega^2$ is nonvanishing, the pair $(\theta^3_1,\theta^3_2)$ do not simultaneously vanish, and $\theta^4_3$ is nonvanishing.

Meanwhile, the structure equations yield $$ 0 = \mathrm{d}\omega^3 = -\theta^3_1\wedge\omega^1 -\theta^3_2\wedge\omega^2, $$ so there must exist functions $h_{ij}=h_{ji}$ for $1\le i,j,\le 2$, not all simultaneously vanishing, such that $\theta^3_i = h_{ij}\omega^j$. The quadratic form $h = h_{ij}\,\omega^i\omega^j$ is then nonvanishing and well-defined up to multiples on the surface $S$. Moreover, for $i = 1$ or $2$, $$ 0 = \mathrm{d}\theta^4_i = -\theta^4_k\wedge\theta^k_i = -\theta^4_3\wedge\theta^3_i\,. $$ Thus, since $\theta^4_3$ is nonvanishing, it follows that $\theta^3_1$ and $\theta^3_2$ are multiples of $\theta^4_3$. In particular, $\theta^3_1\wedge\theta^3_2$ vanishes identically, so $h_{11}h_{22}-{h_{12}}^2$ vanishes identically. Thus, the quadratic form $h$ has rank $1$.

Let $B_1(S)\subset B_2(S)$ denote the submanifold defined by $h_{11} = h_{12}=0$. It is a smooth submanifold of $B_0(S)$ of dimension $11$, and when all the forms and functions are pulled back to $B_1(S)$, we have $\theta^3_1 = 0$ while $\theta^3_2 = h_{22}\,\omega^2$. In particular, it now follows that $\theta^4_3$ is also a multiple of $\omega^2$, say $\theta^4_3 = f\,\omega^2$ for some $f$ (which is nonvanishing).

Moreover, $$ 0 = \mathrm{d}\theta^3_1 = -\theta^3_k\wedge\theta^k_1 = -\theta^3_2\wedge \theta^2_1 = -h_{22}\,\omega^2\wedge\theta^2_1, $$ so it follows (since $h_{22}$ is nonvanishing) that $\theta^2_1 = g\,\omega^2$ for some function $g$ on $B_1(S)$.

Now, $$ \mathrm{d}\omega^2 = -\theta^2_j\wedge\omega^j = -g\,\omega^2\wedge\omega^1 - \theta^2_2\wedge\omega^2 = -(\theta^2_2 - g\,\omega^1)\wedge\omega^2. $$ Thus, $\omega^2$ is an integrable $1$-form, and, because it is semi-basic for the submersion $x:B_1(S)\to S\subset\mathbb{R}^4$, it follows that $\omega^2$ is a multiple of the $x$-pullback of a (nonvanishing) $1$-form on $S$. Thus, $S$ is foliated by (connected) curves whose $x$-preimages in $B_1(S)$ are codimension $1$ integral submanifolds of $\omega^2$.

I claim that these curves in $S$ are, in fact, lines in $\mathbb{R}^4$. To see this, note that, when one pulls back to a leaf of $\omega^2=0$ in $B_1(S)$, one has $\theta^2_1 = \theta^3_1=\theta^4_1= 0$ as well, so one has $$ \mathrm{d}x = e_1\,\omega^1\qquad\text{and}\qquad \mathrm{d}e_1 = e_1\,\theta^1_1\,. $$ In particular, the direction of $e_1$ is fixed on this leaf, and this is the tangent direction of the mapping $x$ restricted to this leaf. Hence the $x$-image of this leaf is an open interval in a line in $\mathbb{R}^4$. Thus, the surface is ruled, as claimed.

In particular $\mathcal{I}$ is the pullback of a well-defined Pfaffian system of rank $5$ on $F$, whose annihilator is the $4$-plane field $D\subset TF$. Moreover, it is easy to show that the three rank $6$ Pfaffian systems generated by adjoining any one of $\omega^2$, $\theta^3_2$, or $\theta^4_3$ to $\mathcal{I}$ are themselves pullbacks of rank $6$ Pfaffian systems on $F$ whose annihilators in $TF$ are each $3$-plane subbundles of $D$.

By its very construction, the projection of $B_1(S)$ into $F$ is a curve that is tangent to $D$ and not tangent to any of these three $3$-plane subbundles of $D$.

Conversely, if $\gamma\subset F$ is any integral curve of $D$ that is not tangent to any of the three $3$-plane subbundles of $D$, its preimage in $\mathbb{R}^4\times\mathrm{GL}(4,\mathbb{R})$ is a submanifold of the form $B_1(S)$ for a surface $S$ satisfying our conditions, in fact, the surface swept out by the union of the lines represented by $\lambda(\gamma)$, where $\lambda:F\to \Lambda$ is the obvious map to the lines.

Now, let us pull back these functions and $1$-forms to $B_0(S)$ (but, as is customary, not notate the pullback). The definition of $B_0(S)$ implies that $$ \mathrm{d}x\wedge e_1\wedge e_2 = \mathrm{d}e_1\wedge e_1\wedge e_2\wedge e_3 = \mathrm{d}e_2\wedge e_1\wedge e_2\wedge e_3 = 0 $$ while the two expressions $$ \mathrm{d}(e_1\wedge e_2)\wedge e_1\wedge e_2 \qquad\text{and}\qquad \mathrm{d}(e_1\wedge e_2\wedge e_3)\wedge e_1\wedge e_2\wedge e_3 $$ are nowhere vanishing. Using the above structure equations, these are equivalent to the relations $$ \omega^3 = \omega^4 = \theta^4_1 = \theta^4_2 = 0,\tag3 $$ while $\omega^1\wedge\omega^2$ is nonvanishing, $\theta^4_3$ is nonvanishing, and the pair $(\theta^3_1,\theta^3_2)$ do not simultaneously vanish.

I claim that these curves in $S$ are, in fact lines in $\mathbb{R}^4$. To see this, note that, when one pulls back to a leaf of $\omega^2=0$ in $B_1(S)$, one has $\theta^2_1 = \theta^3_1=\theta^4_1= 0$ as well, so one has $$ \mathrm{d}x = e_1\,\omega^1\qquad\text{and}\qquad \mathrm{d}e_1 = e_1\,\theta^1_1\,. $$ In particular, the direction of $e_1$ is fixed on this leaf and it is the tangent direction of the mapping $x$ restricted to this leaf. Hence the $x$-image of this leaf is a(n open interval in) a line in $\mathbb{R}^4$. Thus, the surface is ruled, as claimed.

Now, let us pull back these functions and $1$-forms to $B_0(S)$ (but, as is customary, not notate the pullback). The definition of $B_0(S)$ implies that $$ \mathrm{d}x\wedge e_1\wedge e_2 = \mathrm{d}e_1\wedge e_1\wedge e_2\wedge e_3 = \mathrm{d}e_2\wedge e_1\wedge e_2\wedge e_3 = 0 $$ while the two expressions $$ \bigl(\,\mathrm{d}e_1\wedge e_1\wedge e_2,\ \mathrm{d}e_2\wedge e_1\wedge e_2\bigr) \qquad\text{and}\qquad \mathrm{d}e_3\wedge e_1\wedge e_2\wedge e_3 $$ are nowhere vanishing. Using the above structure equations, these are equivalent to the relations $$ \omega^3 = \omega^4 = \theta^4_1 = \theta^4_2 = 0,\tag3 $$ while $\omega^1\wedge\omega^2$ is nonvanishing, $\theta^4_3$ is nonvanishing, and the pair $(\theta^3_1,\theta^3_2)$ do not simultaneously vanish.

I claim that these curves in $S$ are, in fact lines in $\mathbb{R}^4$. To see this, note that, when one pulls back to a leaf of $\omega^2=0$ in $B_1(S)$, one has $\theta^2_1 = \theta^3_1=\theta^4_1= 0$ as well, so one has $$ \mathrm{d}x = e_1\,\omega^1\qquad\text{and}\qquad \mathrm{d}e_1 = e_1\,\theta^1_1\,. $$ In particular, the direction of $e_1$ is fixed on this leaf and it is the tangent direction of the mapping $x$ restricted to this leaf. Hence the $x$-image of this leaf is an open interval in a line in $\mathbb{R}^4$. Thus, the surface is ruled, as claimed.

If you are interested, I can put in the analysis that yields these results.

The following analysis is a more-or-less standard approach using the so-called 'moving frame'. (I'm sure that the result itself is classical in some sense, though I don't know offhand where to look in the literature to find it.)

The conditions

1. $\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t)\bigr) =2 $,

2. $\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) \bigr) = 3$ for all $(s,t)\in[0,1]^2$, and

3. the subspace $W(s,t) =\mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) \bigr)\subset\mathbb{R}^4$ is not constant, in the sense that $W:[0,1]^2\to \mathrm{Gr}_3(\mathbb{R}^4)\simeq\mathbb{RP}^3$ has nonvanishing differential,

are actually independent of the choice of $st$-coordinates on the surface in $\mathbb{R}^4$ and so can be regarded as conditions on a surface $S\subset\mathbb{R}^4$ that, for local analysis purposes, can be taken to be smoothly embedded.

Let $B_0(S)$ denote the space of quintuples $(p;v_1,v_2,v_3,v_4)$ where $p$ lies in $S$ and the quadruple $(v_1,v_2,v_3,v_4)$ is a basis of $\mathbb{R}^4$ such that $(v_1,v_2)$ is a basis of $T_pS$ while $(v_1,v_2,v_3)$ is a basis of the $3$-dimensional subspace $W_pS\subset \mathbb{R}^4$. Then $B_0$ is a smooth submanifold of the product $\mathbb{R}^4\times\mathrm{GL}(4,\mathbb{R})$ that has dimension $2 + 4 + 3 + 4 = 13$.

It is useful to define $\mathbb{R}^4$-valued functions $x,e_i: \mathbb{R}^4\times\mathrm{GL}(4,\mathbb{R})\to\mathbb{R}^4$ such that $x(p;v_1,v_2,v_3,v_4) = p$ while $e_i(p;v_1,v_2,v_3,v_4) = v_i$. Then there exist unique (linearly independent) $1$-forms $\omega^i$ and $\theta^i_j$ on $\mathbb{R}^4\times\mathrm{GL}(4,\mathbb{R})$ such that (assuming the usual summation convention on repeated indices) the following *structure equations* hold: $$ \mathrm{d}x = e_i \omega^i \qquad\text{and}\qquad \mathrm{d}e_i = e_j\,\theta^j_i\,.\tag1 $$

$$ \mathrm{d}\omega^i = -\theta^i_j\wedge\omega^j \qquad\text{and}\qquad \mathrm{d}\theta^i_j = -\theta^i_k\wedge\theta^k_j\,.\tag2 $$

Now, let us pull back these functions and $1$-forms to $B_0(S)$ (but, as is customary, not notate the pullback). The definition of $B_0(S)$ implies that $$ \mathrm{d}x\wedge e_1\wedge e_2 = \mathrm{d}e_1\wedge e_1\wedge e_2\wedge e_3 = \mathrm{d}e_2\wedge e_1\wedge e_2\wedge e_3 = 0 $$ while the two expressions $$ \mathrm{d}(e_1\wedge e_2)\wedge e_1\wedge e_2 \qquad\text{and}\qquad \mathrm{d}(e_1\wedge e_2\wedge e_3)\wedge e_1\wedge e_2\wedge e_3 $$ are nowhere vanishing. Using the above structure equations, these are equivalent to the relations $$ \omega^3 = \omega^4 = \theta^4_1 = \theta^4_2 = 0,\tag3 $$ while $\omega^1\wedge\omega^2$ is nonvanishing, $\theta^4_3$ is nonvanishing, and the pair $(\theta^3_1,\theta^3_2)$ do not simultaneously vanish.

Meanwhile, using the structure equations, we have $$ 0 = \mathrm{d}\omega^3 = -\theta^3_1\wedge\omega^1 -\theta^3_2\wedge\omega^2, $$ so there must exist functions $h_{ij}=h_{ji}$ for $1\le i,j,\le 2$, not all simultaneously vanishing, such that $\theta^3_i = h_{ij}\omega^j$. The quadratic form $h = h_{ij}\,\omega^i\omega^j$ is then nonvanishing and well-defined up to multiples on the surface $S$. Moreover, for $1\le i\le 2$, we have $$ 0 = \mathrm{d}\theta^4_i = -\theta^4_k\wedge\theta^k_i = -\theta^4_3\wedge\theta^3_i\,. $$ Thus, since $\theta^4_3$ is nonvanishing, it follows that $\theta^3_1$ and $\theta^3_2$ are multiples of $\theta^4_3$. In particular, $\theta^3_1\wedge\theta^3_2$ vanishes identically, so $h_{11}h_{22}-{h_{12}}^2$ vanishes identically. Thus, the quadratic form $h$ has rank $1$.

Let $B_1(S)\subset B_2(S)$ denote the submanifold defined by $h_{11} = h_{12}=0$. It is a smooth submanifold of $B_0(S)$ of dimension $11$, and when all the forms and functions are pulled back to $B_1(S)$, we have $\theta^3_1 = 0$ while $\theta^3_2 = h_{22}\,\omega^2$. In particular, it now follows that $\theta^4_3$ is also a multiple of $\omega^2$, say $\theta^4_3 = f\,\omega^2$ for some $f$ (which is nonvanishing).

Further, we have $$ 0 = \mathrm{d}\theta^3_1 = -\theta^3_k\wedge\theta^k_1 = -\theta^3_2\wedge \theta^2_1 = -h_{22}\omega^2\wedge\theta^2_1, $$ so it follows (since $h_{22}$ is nonvanishing) that $\theta^2_1 = g\,\omega^2$ for some function $g$ on $B_1(S)$.

Now, we can compute that $$ \mathrm{d}\omega^2 = -\theta^2_j\wedge\omega^j = -g\,\omega^2\wedge\omega^1 - \theta^2_2\wedge\omega^2 = -(\theta^2_2 - g\,\omega^1)\wedge\omega^2. $$ Thus, $\omega^2$ is an integrable $1$-form, and, because it is semi-basic for the submersion $x:B_1(S)\to S\subset\mathbb{R}^4$, it follows that $\omega^2$ is a multiple of the $x$-pullback of a (nonvanishing) $1$-form on $S$. Thus, $S$ is foliated by (connected) curves whose $x$-preimages in $B_1(S)$ are codimension $1$ integral submanifolds of $S$.

There still remains the question of how one could 'generate' these surfaces, at least locally. I claim that, since these surfaces are, in fact, $1$-parameter families of lines, one should think of them as curves in the space of lines that satisfy some differential equations. Here is how one

Let $\mathcal{I}$ denote the Pfaffian system of rank $5$ that is generated by the five linearly independent $1$-forms $$ \omega^3,\ \omega^4,\ \theta^4_1\,,\ \theta^4_2\,,\ \theta^3_1 $$ (these are the $1$-forms that vanish when pulled back to $B_1(S)$ when $S\subset\mathbb{R}^4$ is a surface satisfying our hypotheses). By the structure equations, $$ \left. \begin{aligned} \mathrm{d}\omega^4 &\equiv 0\\ \mathrm{d}\theta^4_1 &\equiv 0\\ \mathrm{d}\omega^3 &\equiv -\theta^3_2\wedge\omega^2\\ \mathrm{d}\theta^3_1 &\equiv -\theta^3_2\wedge\theta^2_1\\ \mathrm{d}\theta^4_2 &\equiv \phantom{-}\theta^3_2\wedge\theta^4_3\\ \end{aligned}\ \right\} \mathrm{modulo}\ \mathcal{I} $$ It follows that $\mathcal{J}$, the Cartan system of $\mathcal{I}$, has rank $9$ and is spanned by the nine $1$-forms $$ \omega^3,\ \omega^4,\ \theta^4_1\,,\ \theta^4_2\,,\ \theta^3_1\,,\ \omega^2,\ \theta^2_1\,,\ \theta^3_2\,,\ \theta^4_3 $$ In fact, $\mathcal{J}$ is easily seen to be the bundle of $1$-forms on $\mathbb{R}^4\times\mathrm{GL}(4,\mathbb{R})$ that is semibasic for the projection to the $9$-manifold $F$, where the projection sends $(p; v_i)$ to the element of $F$ described by $$ \bigl (\lambda(p,v_1),\ [v_1\wedge v_2], [v_1\wedge v_2\wedge v_3]\ \bigr) $$ and where $\lambda(p,v_1)$ is the line through $p$

In particular $\mathcal{I}$ is the pullback of a well-defined Pfaffian system of rank $5$ on $F$, whose annihilator is the $4$-plane field $D\subset TF$. Moreover, it is easy to show that the three rank $6$ Pfaffian systems got by adjoining any one of $\omega^2$, $\theta^3_2$, or $\theta^4_3$ to $\mathcal{I}$ are themselves pullbacks of rank $6$ Pfaffian systems on $F$ whose annihilators in $TF$ are $3$-plane subbundles of $D$.

By its very construction, the projection of $B_1(S)$ into $F$ is a curve that is tangent to $D$ and not tangent to any of

It is a standard fact that the curves tangent to a 4-plane field in a smooth manifold are locally described (up to reparametrization) by prescribing $3$ functions of one variable.