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The "Fundamental Theorem of Space Curves" (Wikipedia link; MathWorld link) states that there is a unique (up to congruence) curve in space that simultaneously realizes given continuous curvature $\kappa(s)$ and torsion $\tau(s)$ functions.
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Q. What is the closest equivalent generalization for surfaces in $\mathbb{R}^3$, and more generally for Riemannian manifolds in $\mathbb{R}^d$?

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  • $\begingroup$ Actually, I now remember that I asked a version of this question earlier: Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature. $\endgroup$ Commented Jan 26, 2014 at 12:37
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    $\begingroup$ I wouldn't call this question a 'version' of that question; rather, that one and this one are each special cases of a general class of questions concerning how much information about the invariants under rigid motion of an immersion one needs in order to determine that immersion up to rigid motion (and how freely such partial invariant information can be specified). Because many different subsets of the invariants suffice to specify an immersion up to rigid motion (and each is subject to its own set of constraints), there are many different questions and many different possible answers. $\endgroup$ Commented Jan 26, 2014 at 16:04
  • $\begingroup$ @RobertBryant: Thanks, Robert, you are right. I should have said "a related question" rather than "a version," but---alas!---comments cannot be edited after 5min. Your high-level view of these questions is immensely clarifying---Thanks again! $\endgroup$ Commented Jan 26, 2014 at 22:22

4 Answers 4

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I've added a few sentences to my answer to clarify something that some readers may be wondering about, which is why there isn't as simple an answer for surfaces in $3$-space as there is for curves in $3$-space.

First, a word of caution about the so-called 'Fundamental Theorem of Space Curves': You need to assume that the curvature $\kappa(s)$ is nowhere vanishing in order to get $\tau$ well-defined. The theorem is really about space curves endowed with a Frenet frame, and a Frenet frame for a given smooth curve may not be unique (or even be continuous) if $\kappa$ is allowed to vanish.

Second (still about curves), if one looks carefully at the FTSC, one sees that one actually has to specify three pieces of data along an interval $I\subset \mathbb{R}$ in order to determine a curve $c:I\to\mathbb{R}^3$ up to rigid motion: The functions $\kappa>0$, $\tau$, and the element of arc, $\mathrm{d}s$, which is a $1$-form. When $\kappa' := \mathrm{d}\kappa/\mathrm{d}s$ is nonzero, one could, alternatively, specify the triple of functions $(\kappa,\tau,\kappa')$ along $I$, for, then, one could recover $\mathrm{d}s$ as $(\mathrm{d}\kappa)/\kappa'$. This is satisfying because one has exactly three pieces of data along the domain of the curve to specify the three components of the curve up to rigid motion.

Now, for surfaces, one could hope that it would be possible to determine an immersed surface $f:S\to\mathbb{R}^3$ (which depends on a choice of three 'arbitrary' functions on $S$) uniquely up to rigid motion by specifying some three functions that are constructed out of $f$ by differentiation operations and algebra and that are unchanged when $f$ is replaced by $Rf$, where $R$ is a rigid motion of $\mathbb{R}^3$, i.e, what are often called 'Euclidean differential invariants' in the differential geometry literature. There certainly are plenty of such 'Euclidean differential invariants; for example, the Gauss curvature $K$ and the mean curvature $H$ (or, if one is worried about orientation effects, say $H^2$) are second order invariants, and higher order invariants are easily defined (such as, for example, $|\mathrm{d}H|^2$, etc.). However, it is a (nontrivial) theorem that no such triple of differential invariants exists that will uniquely characterize each immersion up to rigid motion. (Even stronger, it turns out that there is no differential invariant 'tensor' of rank $3$ that can do this, which includes data such as the first fundamental form, etc.) It turns out that the best one can do is specify 'excess data' to get the 'unique up to rigid motion characterization' property and then analyze what conditions one needs to put on the 'excess' in order to determine when the specified data actually does come from an immersion $f$. Bonnet's Theorem (discussed below) is one such example, and probably the most well-known (because we teach it in every beginning differential geometry course).

Now, for surfaces in $\mathbb{R}^3$, there is a different (and more traditional) generalization from the result that Anton mentions that may be what you want. It is usually known as Bonnet's Theorem, and it goes like this:

For any immersion of an oriented surface $f:S\to\mathbb{R}^3$, one defines two quadratic forms: The first fundamental form, $\mathrm{I}_f = \mathrm{d} f\cdot \mathrm{d} f>0$, and the second fundamental form, $\mathrm{I\!I}_f = - \mathrm{d} f\cdot \mathrm{d} n$, where $n:S\to \mathbb{R}^3$ is the oriented unit normal to the immersion.

Then the uniqueness theorem says that $f$ is uniquely determined up to oriented isometry by the pair $(\mathrm{I}_f,\mathrm{I\!I}_f)$, and the existence theorem says that, if $(\mathrm{I},\mathrm{I\!I})$ are given quadratic forms on a connected, oriented, simply connected surface $S$ that satisfy the Gauss equation and the Codazzi equations (see below), then there exists an immersion $f:S\to\mathbb{R}^3$ such that $(\mathrm{I},\mathrm{I\!I}) = (\mathrm{I}_f,\mathrm{I\!I}_f)$.

The Gauss equation is the condition $\det\mathrm{I\!I} = K(\mathrm{I})\ \det\mathrm{I}$, where $K(\mathrm{I})$ is the Gauss curvature of the (positive definite) quadratic form $\mathrm{I}$ and $\det$ means the operation on quadratic forms that, in local coordinates, is given by $$ \det( e\, dx^2 + 2f\, dx\ dy + g\, dy^2) = (eg-f^2) (dx\wedge dy)^{\otimes 2} $$ The Gauss equation is a single second order partial differential equation on the pair $(\mathrm{I},\mathrm{I\!I})$.

The Codazzi equations are $\delta_{\mathrm{I}}(\mathrm{I\!I})=0$, where $\delta_{\mathrm{I}}:C^\infty\bigl(S^2(T^*\!S)\bigr)\to C^\infty\bigl(T^*\!S\bigr)$ is a linear, first-order differential operator that is the $\mathrm{I}$-covariant derivative $\nabla_\mathrm{I}:C^\infty\bigl(S^2(T^*\!S)\bigr)\to C^\infty\bigl(T^*\!S\otimes S^2(T^*\!S)\bigr)$ followed by the appropriate contraction operation $c:T^*\!S\otimes S^2(T^*\!S)\to T^*\!S$. They constitute $2$ first-order partial differential equations on the pair $(\mathrm{I},\mathrm{I\!I})$.

Note that this uniqueness and existence theorem makes sense: A choice of $f:S\to\mathbb{R}^3$ is a choice of $3$ 'arbitrary' functions on the surface $S$ (subject only to the 'open' condition that they define an immersion of $S$ into $\mathbb{R}^3$) while a choice of a pair $(\mathrm{I},\mathrm{I\!I})$ represents a choice of $6$ 'arbitrary' functions on $S$ (the coefficients of the two quadratic forms in local coordinates, subject only to the condition that $\mathrm{I}$ be positive definite). Subjecting those $6$ functions to $3$ partial differential equations (i.e., the Gauss and Codazzi equations) restores the 'balance of arbitrariness' to $3$ arbitrary functions on the surface.

In higher dimensions, for submanifolds of $\mathbb{R}^d$, there is a similar theorem, but now $\mathrm{I\!I}$ gets replaced by a quadratic form taking values in a 'normal bundle' $N$, which is a vector bundle endowed with an Euclidean connection. The Gauss equation and the Codazzi equations generalize in a straightforward way, but when the 'normal bundle' has dimension greater than $1$, one also has a further set of equations relating the curvature of the connection on the normal bundle with the quadratic form $\mathrm{I\!I}$. The exact statement can be found in any good book on submanifold theory. If the ambient space is not $\mathbb{R}^d$ or, more generally, a space of constant sectional curvature, then the situation is more complicated, and I refer you to a book on submanifold theory for the ideas and details.

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  • $\begingroup$ Do you know a reference for the fact that one cannot use differential invariants to characterize an immersion? Preferably something accessible (I'm teaching a differential geometry for surfaces class, and if it's not too complex would try to cover it in some way or another). $\endgroup$ Commented Sep 27, 2023 at 14:08
  • $\begingroup$ @TobiasDiez: I'm not sure what you mean by 'one cannot use differential invariants to characterize an immersion'. Are you referring to the fact that one can have two non-congruent, smooth unit speed space curves with torsion $\tau\equiv0$ and equal curvatures $\kappa$? I gave a simple example of such in an answer to an earlier question (mathoverflow.net/q/428800). $\endgroup$ Commented Sep 27, 2023 at 18:29
  • $\begingroup$ Sorry for the confusion. I was referring to "There certainly are plenty of such 'Euclidean differential invariants. .... However, it is a (nontrivial) theorem that no such triple of differential invariants exists that will uniquely characterize each immersion up to rigid motion. " $\endgroup$ Commented Sep 28, 2023 at 15:31
  • $\begingroup$ @TobiasDiez: I see what you mean now. The proof that I had in mind looks at the geometry of the $k$-jet space $J^k(M,\mathbb{R}^3)$, where $M$ is a surface, with its natural action of the $6$-dimensional isometry group $G$ of Euclidean $3$-space. What we are asking for is a $G$-invariant function $\Phi:J^k(M,\mathbb{R}^3)\to\mathbb{R}^3$ such that two immersions $f_1,f_2:M\to\mathbb{R}^3$ are congruent under $G$ if and only if $\Phi(j^k(f_1))=\Phi(j^k(f_2))$. The proof that such a $\Phi$ does not exist involves some knowledge of exterior differential systems. $\endgroup$ Commented Sep 29, 2023 at 8:54
  • $\begingroup$ Okay, this makes sense. Thanks! $\endgroup$ Commented Oct 2, 2023 at 4:35
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The answer is rewritten.


First you need to find the analogs of natural parametrization, curvature and torsion for surfaces; the same can be done for all submanifolds, say of dimesion $m$.

Let us denote by $S^k(\mathbb R^m)$ the space of homogenious symmetric polynomials of degree $k$ on $\mathbb R^m$.

Assume that the submanifold is given locally by a chart $f\colon \mathbb R^m \to \mathbb R^d$. Then one can define a sequence of forms $G_n\colon \mathbb R^m\to S^{2{\cdot} n}(\mathbb R^m)$. Here $G_1$ is metric tensor $G_2(X)=\langle s(X,X),s(X,X)\rangle$, where $s\colon S^2(T)\to N$ is the second fundamental form. You can continue: in general, set $T^k_x$ to be the subspace of $\mathbb R^d$ spanned by the partial derivatives up to order $k$ at $x\in\mathbb R^m$. In this case $$G_{k+1}(X)=\left|(X^{k+1}f)_k^\bot\right|^2,$$ where $V\mapsto (V)^\bot_k$ denotes the orthogonal projection to the normal $T^k_x$.

So, $G_1$ is your analog of natural parametrization; the tensor field $G_2$ is the analog of curvature; $G_3$ is the analog of torsion.

The tensors $G_1,\dots,G_n$ describes $f$ at the point (say $x$) up order $n$ and a motion of the space. In particular, you can say if the surface is $n$-regular; i.e., $T^n_x$ has maximal possible dimension (I think Gromov use term "free" for $2$-regular).

Nondegeneracy up to $n$ plus and equality of the tensors up to $n+1$ imply congruence of submanifolds if dimension is small enough (say it is just enough to have $(n+1)$-regular submanifold). The tensors $G_i$ have to satisfy some equations (see the other answers).

You also get existence assuming dimension is big and $G_i$ are big. It follows from Gromov's analog of Nash--Kuiper theorem; it says that given embedding one can keep $G_1,\dots,G_n$ and add to $G_{n+1}$ arbitrary positive tensor; positive means sum of $\ell^{2\cdot (n+1)})$, where $\ell$ is a linear function. (Nash--Kuiper theorem states the same for $G_1$.)

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    $\begingroup$ For the generic surface in $\mathbb{R}^9$, $\dim T^1_x = 2$, $\dim T^2_x = 5$, and $\dim T^3_x = 9$. The induced forms $G_1$, $G_2$, and $G_3$ have a total of $3 + 5 + 7 = 15$ coefficient functions, which is more than $9$, the total number of functions needed to specify $f$ locally. Thus, the generic triple $(G_1,G_2,G_3)$ on a surface will not (even locally) come from an immersion $f$; there will have to be analogues of Gauss and Codazzi equations. Maybe $d(2,3) = 15$, while $d(2,2) = 8$, and you need only $(G_1,G_2)$ to determine the generic surface in $\mathbb{R}^8$ up to rigid motion? $\endgroup$ Commented Jan 26, 2014 at 14:01
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    $\begingroup$ One more comment: Note there cannot be an integer $d(2,1)$ with the properties that the quadratic form $G_1$ on a surface determines $f:S\to\mathbb{R}^{d(2,1)}$ up to rigid motion and that the generic $G_1$ works (at least locally). The only possibilities would be $d(2,1)=2$ (which doesn't work because the generic quadratic form is not flat) or $d(2,1)=3$ (which doesn't work because the first fundamental form of a surface in $3$-space does not determine it up to rigid motion). Thus, it appears that $d(m,n)$ does not always exist. Are we sure that $d(2,2)$ and $d(2,3)$ exist? $\endgroup$ Commented Jan 26, 2014 at 15:47
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    $\begingroup$ @RobertBryant, thank you for corrections, you are totally right, I took it from my memory without critical thinking. (As an excuse: I did not use the result.) $\endgroup$ Commented Jan 26, 2014 at 17:04
  • $\begingroup$ Thanks for the update, which is interesting. I feel compelled to add a note of caution, though, about your statement "The tensors $G_1,\ldots,G_n$ describes $f$ at the point (say $x$) up to order $n$ and a motion of the space." While this is true for $n=1$, it is not generally true for $n\ge 2$. To see this, note that $G_2$ has only $5$ coefficients, while the matrix of inner products of the three coefficients of $dx^2$, $dx\ dy$, and $dy^2$ in the second fundamental form $I\!I$ will have $6$ independent coefficients. Thus, one cannot recover $I\!I$ pointwise from $(G_1,G_2)$. $\endgroup$ Commented Jan 27, 2014 at 12:40
  • $\begingroup$ @RobertBryant, I wanted to say $G_i$ at all the points, in particular to recover second fundamental form you need the curvature tensor which is can found by taking derivatives of $G_1$. $\endgroup$ Commented Jan 27, 2014 at 21:22
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The fundamental equations you look for, are known as Gauss-Codazzi equations. I'm not an expert, but the Wiki article

http://en.wikipedia.org/wiki/Gauss%E2%80%93Codazzi_equations

gives an overview

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I think that the subject Submanifold Geometry can satisfy what you need. And you can find the generalization in the book: Geometry 1: Basic Idea and Concepts of Differential Geometry; Chapter 3, which can be found on the website http://en.bookfi.org/book/444417.

Now I want to give some my thoughts about it. The most generalization to talk about the question is to look at the question by considering the immersion $r:M^n\rightarrow N^{n+k}$. And we will have the Fundamental Theorem about: $r:M^n\rightarrow\mathbb R^{n+k}$ and you will find it on the book I provide.

However, there are two special things we should pay attention:

First is about curve: $r:I\rightarrow\mathbb R^n$. The curve's Fundemental Theorem does only rely on curvature and its uniqueness is under the natural parameter.

Second is about hypersurface $r:M^n\rightarrow\mathbb R^{n+1}$, like surface in $\mathbb R^3$. Because it is without torsion, then we can drop Ricci Equation. So the Fundemental Theorem is only depend on The Second Fundamental Form.

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