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I'm trying to find a definition for Gauss-Kronecker curvature of submanifolds of $\Bbb R^n$, but I'm only finding it for hypersurfaces. I would like to know if someone knows any text which works in $\Bbb R^n$, but not kicking it and going all out for manifolds. (I'm self-studying this). I also have some ideas, which I would like to know if are in the right way.

Let $M^k \subset \Bbb R^n$ be a oriented submanifold. Let $\{{\bf e}_1, \cdots, {\bf e}_k\}$ and $\{{\bf N}_1, \cdots, {\bf N}_{n-k}\}$ be orthonormal bases for the tangent and normal spaces, respectively.

  • The second fundamental form at the point ${\bf p} \in M, {\rm II}: T_{\bf p}M \times T_{\bf p}M \to (T_{\bf p}M)^\perp$ can be defined by: $${\rm II}({\bf v},{\bf w}) = \sum_{i = 1}^{n-k} -\langle {\rm d}{{\bf N}_i({\bf v}),{\bf w}}\rangle {\bf N}_i.$$

  • Since $\{{\bf e}_i\}_{i=1}^k$ is an orthonormal basis, we define the trace of ${\rm II}$ by $${\rm tr}({\rm II}) = \sum_{i=1}^k {\rm II}({\bf e}_i,{\bf e}_i).$$

  • So we define the mean curvature vector by: $${\bf H} = \frac{1}{n}{\rm tr(II)}.$$

Here my problem begins. I'm trying to base myself in this part of Kühnel's Differential Geometry: Curves - Surfaces - Manifolds, in page $118$:

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  • I'm tempted to define the Gauss-Kronecker curvature as: $$K = \frac{\det_{\langle \cdot,\cdot \rangle}({\rm II})}{\det({\rm I})}.$$ But this notion of $\det_{\langle \cdot, \cdot \rangle}$ does not seems precise enough for me. This bothers me, because if the dimension of the submanifold is greater than $2$, what would be taking the "product" of three vectors? (the "product" of two vector is simply their inner product).

But what is in favor of this definition is the following: in $\Bbb R^3$, we have one shape operator $\mathcal{S}_{\bf N} = -{\rm d}{\bf N}$. Here, with codimension $n-k$, we will have $n-k$ shape operators related to the initial fixed base: $$\mathcal{S}_{{\bf N}_1} = -{\rm d}{\bf N}_1, \ldots ,\mathcal{S}_{{\bf N}_{n-k}} = -{\rm d}{\bf N}_{n-k}.$$ I made a few examples in $\Bbb R^4$, which seems to indicate that: $$\sum_{i=1}^{n-k}\det(-{\rm d}{\bf N}_i) = \frac{\det_{\langle \cdot, \cdot \rangle}({\rm II})}{\det({\rm I})}.$$ Since in $\Bbb R^3$ we have $K = \det(\mathcal{S}_{\bf N})$, it would be natural to define $K$ as this sum, at least for $n=4$.

A final remark: I do not want a definition which uses principal curvatures. I want a definition that I can use safely in Lorentz-Minkowski space $\mathbb{R}^n_1$ (there are timelike surfaces with no principal directions).

I'm sorry for the long post, but I really need some pointers here, and I wanted to share my efforts with you. Thanks for the attention.

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  • $\begingroup$ This might already be equivalent to one of your definitions but for me a natural definition would be the following: Define the G-K curvature to be a function of unit normal vectors. For each unit normal vector $\nu$, the second fundamental form is a symmetric matrix $\nu\cdot \mathrm{II}$ with respect to an orthonormal basis of tangent vectors. Define the curvature at $\nu$ to be $K(\nu) = \det (\nu\cdot\mathrm{II})$. $\endgroup$
    – Deane Yang
    Feb 4, 2015 at 22:50
  • $\begingroup$ Hi Deane, thanks for the comment. I'm not familiar with the notation $\nu \cdot {\rm II}$, and I understood the following: If we write ${\rm II}({\bf e}_i,{\bf e}_j) = \sum_k h_{ij}^k {\bf N}_k$, if $\nu = {\bf N}_k$, the matrix $\nu \cdot {\rm II}$ would have as entries the $h_{ij}^k$? I quite expected the G-K curvature to be a number which didn't depended of the direction I took. $\endgroup$
    – Ivo Terek
    Feb 4, 2015 at 22:58
  • $\begingroup$ I don't see any way to define the Gauss-Kronecker curvature as a single number. Only as a function of the normal vector. $\endgroup$
    – Deane Yang
    Feb 5, 2015 at 0:26
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    $\begingroup$ You could take the average value over all unit vectors. $\endgroup$
    – Deane Yang
    Feb 5, 2015 at 0:28
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    $\begingroup$ Yes, you have it right. $\endgroup$
    – Deane Yang
    Feb 5, 2015 at 0:55

2 Answers 2

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One possible reason for you to be having trouble finding the 'right' definition of Gauss-Kronecker curvature is that you haven't really told us what properties you want this curvature to have. I gather that you want it to be a scalar that is a function of the second fundamental form somehow and that, in the hypersurface case, it should be the classical Gauss-Kronecker curvature, which, in the hypersurface case, is defined as the ratio of the $\mathbf{N}_1$-pullback of the volume form on $S^{n-1}$ (where $\mathbf{N}_1:M\to S^{n-1}$ is the (oriented) Gauss map) to the induced volume form on $M$.

The natural way to generalize this is to generalize the Gauss map. In other words, for an oriented manifold and frame field as you have defined it above, consider the mapping $$ ^\ast\gamma = \mathbf{N}_1\wedge \mathbf{N}_2\wedge\cdots \wedge\mathbf{N}_{n-k}: M\to \mathrm{Gr}_{n-k}(\mathbb{R}^n) $$ or, what, by duality, is essentially the same thing, the generalized Gauss mapping $$ \gamma = \mathbf{e}_1\wedge \mathbf{e}_2\wedge\cdots \wedge\mathbf{e}_{k}: M\to \mathrm{Gr}_{k}(\mathbb{R}^n), $$ where $\mathrm{Gr}_{p}(\mathbb{R}^n)$ means the Grassmannian of oriented $p$-planes in $\mathbb{R}^n$, which is a homogeneous space of $\mathrm{SO}(n)$ of dimension $p(n{-}p)$ and which carries a natural $\mathrm{SO}(n)$-invariant metric, $h$.

Then one natural definition of a scalar $S$ is to take the ratio of the volume form of $\gamma^*h$ to the induced volume form on $M$. Up to a sign, this reduces to the Gauss-Kronecker curvature when $k=n{-}1$ (i.e., the hypersurface case). The formula in terms of the coefficients of the second fundamental form is then $$ S = \sqrt{\det(H_{j l})} \qquad\text{where}\qquad H_{jl} = \sum_{\alpha, i} h_{\alpha i j}h_{\alpha i l} $$ where the index $\alpha$ satisfies $1\le \alpha\le n{-}k$ and the latin indices satisfy $1\le i,j\le k$.

Another natural scalar that is well-defined when $k=2$ or $k=n{-}2$ and $n$ is even is to note that $\mathrm{Gr}_{2}(\mathbb{R}^n)$ and $\mathrm{Gr}_{n-2}(\mathbb{R}^n)$ both carry a canonical $\mathrm{SO}(n)$-invariant $2$-form, $\Omega$. When $k=2$, you could just define a scalar $W$ to be the ratio of $\gamma^*\Omega$ to the induced area form on $M^2$. The formula is $$ W = \sum_{\alpha} h_{\alpha11}h_{\alpha22}-{h_{\alpha12}}^2 $$ This agrees with the Gauss-Kronecker curvature when $n=3$ and generalizes it when $n>3$. When $n=2m$ and $k=n{-}2=2m{-}2$, you could take $L$ to be the scalar that is the ratio of $(\gamma^*\Omega)^{m-1}/((m{-}1)!)$ to the volume form on $M$. Of course, $W$ and $L$ are the same when $(k,n) = (2,4)$.

Of course, $S$, $W$, and $L$ generally have different properties, even when they are all defined. The good thing is that these things don't depend on a positive definite metric. They'll work for any submanifold on which the first fundamental form is nondegenerate.

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    $\begingroup$ Isn't $W$, up to a constant factor, just the scalar curvature of $M$, at least in the case of Euclidean signature? Also, I am not entirely convinced by the last paragraph. Ideally the definition should be invariant under isometric transformations of the ambient space, but since your definitions use the $SO(n)$ invariant metric on the Grassmannian, this is in general not going to be preserved by $SO(1,n-1)$ actions on $\mathbb{R}^{n}_1$ for example. On the other hand, it seems natural to replace the Riemannian Grassmannian in your example directly by... $\endgroup$ Feb 13, 2015 at 13:23
  • $\begingroup$ ... the space of all linear subspaces of a fixed signature in $\mathbb{R}^n_\nu$. You get a (non-compact in general) homogeneous space of the Lorentz group (or the $SO$ group of whatever the ambient signature is). If I am doing the math correctly then the $W$ defined relative to this group would indeed be just the scalar curvature. Though I am not too sure what $L$ would be in this case. $\endgroup$ Feb 13, 2015 at 13:30
  • $\begingroup$ @WillieWong: Of course, in my final sentence, I did not mean that you should use the $\mathrm{SO}(n)$-invariant metric! I meant that there are analogs in the other signatures for the subsets of the Grassmannians that correspond to subspaces on which the inner product is nondegenerate. Perhaps I should have inserted a mutatis mutandis in that sentence, but I assumed that this would be clear to the reader. $\endgroup$ Feb 13, 2015 at 13:37
  • $\begingroup$ I had guessed that's what you meant. But considering the OP's background I think it would be worth making explicit. $\endgroup$ Feb 13, 2015 at 13:39
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    $\begingroup$ @WillieWong: You may be right about that. Of course, you are also right that $W$ is the Gauss curvature of the induced metric, but my intended point was that the form $\Omega$ is a natural object that can be used in conduction with the Gauss map to make a construction that works in all codimensions (for surfaces) and can be used in codimension $2$ (in the even dimensional case) to define a different scalar invariant. I just wanted to emphasize that there are several 'natural' constructions of scalar quantities from the second fundamental form. $\endgroup$ Feb 13, 2015 at 13:45
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Just a remark about the n=4 case: one could proceed by analogy with the n=3 in which the Gauss-Kronecker curvature K happens to be equal to the curvature of the induced metric on the surface (intrinsic curvature). This is Gauss' famous Teorema Egregium (observe also that this still holds true, maybe up to sign, in the Lorentzian case). In the case of surfaces in 4-space, one could look for an expression of the second fundamental form which happens to be equal, again, to the intrinsic curvature of the surface. An examination of Gauss equation $$ \langle R(X,Y)Z,W \rangle + \langle h(Y,Z) ,h(X,W)\rangle - \langle h(Y,W) ,h(X,Z) \rangle= 0$$ shows that this quantity will be $$ K(\nu_1) + K(\nu_2),$$ where $(\nu_1,\nu_2)$ is an orthonormal basis of the normal space and I follow Deane Yang's notation. In other words, the average on unit vectors, as suggested Deane.

I don't see any generalization in higher dimension.

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