Suppose one has a smooth non-degenerate curve $\gamma: [0,1] \to {\bf R}^n$ into Euclidean space (thus $\gamma'$ never vanishes), with the property that the velocity $\gamma'(t)$ and acceleration $\gamma''(t)$ are always parallel, thus $$ \mathrm{dim}\ \mathrm{span}( \gamma'(t), \gamma''(t) ) \leq 1$$ for all $t \in [0,1]$. To avoid technicalities let us assume that $\gamma$ is a polynomial. Then it is easy to see that $\gamma$ actually traverses a straight line (there is no curvature).

More generally, if we have $$ \mathrm{dim}\ \mathrm{span}( \gamma'(t), \dots, \gamma^{(k)}(t) ) \leq k-1$$ for some fixed $k$ and all $t$, then one can show that the curve $\gamma([0,1])$ is trapped inside some (affine) $k-1$-dimensional subspace of ${\bf R}^n$. (For instance, the $k=2$ case corresponds to the case when the curve has no torsion and is thus a plane curve.) One can prove this for instance by showing that the $k$-form $\gamma'(t) \wedge \dots \wedge \gamma^{(k-1)}(t)$ has a direction that is fixed in $t$ (assuming it does not degenerate to zero, in which case one can instead induct).

A variant of this is the Monge-Cayley-Salmon theorem: if $\phi: [0,1]^2 \to {\bf R}^3$ is a polynomial immersed surface in three dimensions (thus $D\phi$ has maximal rank everywhere), and for each point $(s,t) \in (0,1)^2$ there is a smooth non-degenerate curve $\gamma = \gamma_{s,t}: [0,1] \to \phi([0,1]^2)$ on the surface passing through $\phi(s,t)$ at time zero, thus $\gamma(0) = \phi(s,t)$ and $\gamma'(0) \neq 0$, which is linear to third order, in that $$ \mathrm{dim}\ \mathrm{span}( \gamma'(0), \gamma''(0), \gamma'''(0) ) \leq 1$$ (or equivalently, $\phi(s,t)$ is a flecnode of the surface, in that there is a tangent line that matches the surface to third order). Then the Monge-Cayley-Salmon theorem asserts that $\phi([0,1]^2)$ is a ruled surface. (I discuss this theorem in this blog post; the previous results about curves are used in the proof of the Monge-Cayley-Salmon theorem.)

In all of the above results, the hypothesis is about the local geometry of the surface (a condition on some finite Taylor expansion of $\phi$, or equivalently some finite jet of the surface), but the conclusion constrains the global geometry of the surface (in particular locating linear spaces that globally relate to the surface).

My (somewhat ill-posed) question concerns the following variant of the Monge-Cayley-Salmon situation: suppose that $\phi: [0,1]^2 \to {\bf R}^4$ is a polynomial immersed surface, and suppose that one has the dimension condition $$ \mathrm{dim}\ \mathrm{span}( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) ) \leq 3$$ on the first and second partial derivatives of $\phi$ for all $(s,t) \in [0,1]^2$ (thus the Taylor expansion to second order of $\phi([0,1]^2)$ around any point is always at most three-dimensional rather than four). Does this place a strong geometric constraint on the surface $\phi([0,1]^2)$, such as being ruled, or being trapped in a three-dimensional subspace of ${\bf R}^4$? I am a bit vague on what type of conclusion I want here, but it should somehow control the "global" geometry of the surface in a manner similar to the previous examples. The requirement that $\phi$ be a polynomial might be unnecessary, but that is what actually occurs for the application I have in mind.

One can also pose this question for higher dimensional varieties in higher dimensional Euclidean spaces, but the above situation of two-dimensional surfaces in ${\bf R}^4$ seems to be the first non-trivial case that is not directly treated by the previous assertions about curves.

  • 3
    $\begingroup$ I think what most people call "the" Cayley-Salmon theorem is the one asserting that there are 27 lines on a smooth (or at least, general) cubic surface. I'm probably not the only one confused by your terminology, so maybe you should clarify. (It's particularly confusing that in your blog post you say it goes back to "at least 1915", when both Cayley and Salmon had been dead for quite some time!) $\endgroup$ – Gro-Tsen Jan 26 '17 at 20:36
  • $\begingroup$ seems related to the classical Hesse-Gordan-Noether problem as, e.g., in the two articles arxiv.org/abs/1506.06387 and arxiv.org/abs/1312.1618 $\endgroup$ – Abdelmalek Abdesselam Jan 26 '17 at 22:17
  • 1
    $\begingroup$ Fair enough. The result was first discovered by Monge anyway, so I'll change the attribution to avoid confusion. $\endgroup$ – Terry Tao Jan 27 '17 at 3:18

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

  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,

then one can show that the surface $\phi\bigl([0,1]^2\bigr)\subset\mathbb{R}^4$ is ruled (and does not lie in an affine $3$-space).

Such surfaces locally depend on three arbitrary functions of one variable in Cartan's sense. One way of describing them is this: Let $\Lambda$ be the space of (affine) lines in $\mathbb{R}^4$, a manifold of dimension $6$. Consider the ($9$-dimensional) bundle $\pi:F\to\Lambda$ whose fiber over $\lambda\in\Lambda$ is the flag variety of the $3$-dimensional vector space $\mathbb{R}^4/\lambda'$, where $\lambda'\subset\mathbb{R}^4$ is the linear subspace parallel to $\lambda$. Then there exists a smooth $4$-plane field $D\subset TF$ such that, if $\gamma\subset F$ is a generic curve tangent to $D$, then regarding $\gamma$ as a $1$-parameter family of affine lines via $\pi(\gamma)\subset \Lambda$, the union of these lines is a surface satisfying the above assumptions. (Here 'generic' means that the tangents to $\gamma$ do not lie in triple of hyperplanes in $D$.)

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.

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, 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.

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 can think of this system of equations:

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 the direction $v_1$.

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.

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.

  • 2
    $\begingroup$ I would be interested in seeing more details, thanks. Right now I don't see how the bundle $F$ is interacting with the subspaces $W$. $\endgroup$ – Terry Tao Jan 27 '17 at 3:22
  • 2
    $\begingroup$ @TerryTao: Ok, I've put it in. If you have questions, or something needs clarification, please let me know. $\endgroup$ – Robert Bryant Jan 27 '17 at 14:23
  • 1
    $\begingroup$ Very nice application of exterior differential systems. $\endgroup$ – Deane Yang Jan 27 '17 at 15:17
  • 2
    $\begingroup$ Thanks for this! I am slowly going through the calculations. Is there a Lie group interpretation of the 1-forms $\omega^i$ and $\theta^j_i$? It looks like they should somehow be associated to the affine group ${\bf R}^4 \rtimes GL(4,{\bf R})$, but I don't see the precise relation yet. $\endgroup$ – Terry Tao Jan 28 '17 at 18:26
  • 2
    $\begingroup$ @TerryTao: Yes, they are the left-invariant forms of the affine group. You can think of this group as embedded in $\mathrm{GL}(5,\mathbb{R})$ as the matrix $$g=\begin{pmatrix}1&0&0&0&0\\ x&e_1&e_2&e_3&e_4\end{pmatrix}=\begin{pmatrix}1&0\\x&e\end{pmatrix}.$$ Then the equation $$\mathrm{d}g = g\begin{pmatrix}0&0\\ \omega&\theta\end{pmatrix} = g\,\gamma$$ is the matrix form of the first structure equation, ie., $\gamma = g^{-1}\,\mathrm{d}g$. The second equation is then just $$\mathrm{d}\gamma=-\gamma\wedge\gamma.$$ $\endgroup$ – Robert Bryant Jan 28 '17 at 19:01

Just some suggestions:

I am concerned you are talking about parameterization (as would be the natural thing in an engineering problem) rather than the intrinsic geometry of the curve itself (as a geometer should).

  • in your first case $v'(t) \propto v(t)$

  • in your second case $ \sum a_k(t)v^{(k)}(t) = 0$ for some time-dependent functions $a_k(t)$.

  • in your third case, it seems too much to have a third-order tangent everywhere but I'm not sure. who knows? You have certainly ruled it out.

I don't know how the everywhere existence of a particular catastrophe leads to a global restriction. There is a Gauss map from curves to projective space, or from surfaces to Grassmanian $\mathrm{Gr}(2,n)$ which encodes how the tangent plane changes as you move around the surface.

Whenever you write down an equation or do an approximations, polynomials have been the choice for describing these kind of relations. Therefore, algebraic geometry -- and algebraic varieties are always the main source of examples.

e.g. a Lie group is a variety

For me it's the fact that your theorem comes from a textbook, suggests the main geometers of the time knew your result under a different name.

I am researching the name of the map $C \to SO(k)$ from your curve to the first $k$ derivative at that point. There must be an analogue of the Gauss map there.

Just a bit from Wikipedia:

In algebraic geometry, ruled surfaces were originally defined as projective surfaces in projective space containing a straight line through any given point. This immediately implies that there is a projective line on the surface through any given point, and this condition is now often used as the definition of a ruled surface: ruled surfaces are defined to be abstract projective surfaces satisfying this condition that there is a projective line through any point. This is equivalent to saying that they are birational to the product of a curve and a projective line. Sometimes a ruled surface is defined to be one satisfying the stronger condition that it has a fibration over a curve with fibers that are projective lines. This excludes the projective plane, which has a projective line though every point but cannot be written as such a fibration.

Instance of Monge-Cayley-Salmon


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.