Let $M$ be a submanifold of a Riemannian manifold $\widetilde{M}$. Let $A$ be the second fundamental form of $M$.

Suppose that, for all $p \in M$, the linear map $A(v, \cdot)\: \colon T_{p}M \to N_{p}M$ has rank $k$ for every nonzero $v \in T_{p}M$.

Does this condition define some well-known class of submanifolds (if any)?

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    $\begingroup$ I think you want to have your map be of rank $k$ for every nonzero $v\in T_pM$. Otherwise, taking $v=0$, you will have to have $k=0$, and then you'll only get the totally geodesic submanifolds. $\endgroup$ – Robert Bryant Sep 16 '18 at 11:53
  • $\begingroup$ Good point. I have now edited the question. $\endgroup$ – MK7 Sep 17 '18 at 8:07

Sometimes, this is an open condition, so that it does not impose any differential equations on the submanifold. For example, consider the case of a surface $\Sigma^2$ in a $4$-manifold $M^4$. The condition that the rank of $A(v,\cdot):T_p\Sigma\to N_p\Sigma$ be equal to $2$ for every nonzero $v\in T_p\Sigma$ is an open condition on the submanifold $\Sigma$, so once one has such a submanifold, any $C^2$-nearby submanifold will also have this property. Probably, someone has a name for this class of surfaces, but I don't think that it is well-known.

There are many examples of such surfaces in $\mathbb{R}^4$. For example, let $\Sigma\subset\mathbb{C}^2$ be a holomorphic curve without flexes regarded as a smooth surface in $\mathbb{R^4}=\mathbb{C}^2$.

  • $\begingroup$ Thanks a lot for your answer. So you are saying that such condition singles out a class which is distinct from the class of submanifolds having constant relative nullity index $n-k$, right? $\endgroup$ – MK7 Sep 17 '18 at 8:17
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    $\begingroup$ @MK7: I believe so. In my understanding, the index of relative nullity at a point $p\in M$ is the dimension of the kernel of the quadratic form $|\mathrm{I}\!\mathrm{I}|_p:T_pM\times T_pM\to N_pM$, which could be zero even when $A(v,\cdot)$ fails to have constant rank for nonzero $v$. $\endgroup$ – Robert Bryant Sep 17 '18 at 13:07
  • $\begingroup$ OK. My main concern is in fact the other direction, i.e., whether there exist submanifolds with index of relative nullity smaller than $n-k$ satisfying the given condition. $\endgroup$ – MK7 Sep 17 '18 at 20:24

This condition means that the 1-nullity $\nu_1=n-k$.

Let $U,\ V$ and $W$ be real vector spaces of finite dimension, and let $\beta:V\times U\to W$ be a bilinear form. The nullity subspace $\mathcal{N}\left(\beta\right)\subset U$ of $\beta$ is defined by

$$\mathcal{N}\left(\beta\right)=\left\{Y\in U:\beta\left(X,Y\right)=0\text{ for all }X\in V\right\},$$

and its image subspace $\mathcal{S}\left(\beta\right)\subset W$ by

$$\mathcal{S}\left(\beta\right)=\text{span}\left\{\beta\left(X,Y\right):X\in V\text{ and }Y\in U\right\}.$$

Assume that $W$ has a positive definite inner product and that $\beta:V\times V\to W$ is a symmetric bilinear form. For an $s$-dimensional subspace $U^s\subset W$, we denote by $\beta_{U^s}:V\times V\to U^s$ the map given by


where $\pi_{U^s}$ stands for the orthogonal projection $\pi_{U^s}:W\to U^s$. The $s$-nullity $\nu_s$ of the bilinear form $\beta$ is defined by

$$\nu_s=\max_{U^s\subset W}\left\{\dim\mathcal{N}\left(\beta_{U^s}\right)\right\}$$

for each integer $1\leq s\leq\dim W$.

For an isometric immersion $f:M^n\to\tilde M^m$, the $s$-nullity $\nu_s\left(x\right)$ at $x\in M^n$ is defined as the $s$-nullity of its second fundamental form $\alpha$ at $x$.

If the subspaces $U^s$ in the definition of $\nu_s\left(x\right)$ are restricted to subspaces of $N_1\left(x\right)$, then one obtains the $s$-nullity of $f$ on the first normal space, which we denote by $\nu_s^*\left(x\right)$. Notice that when $k=\dim N_1\left(x\right)$ then $\nu_k^*\left(x\right)$ is the usual index of relative nullity.

The concept of $s$-nullities plays a key role in the study of rigidity aspects of submanifolds. Roughly speaking, the $s$-nullities measure the degeneracy of the extrinsic geometry of a submanifold. As a general rule of thumb, the higher the $s$-nullities, the more deformable the submanifold. Indeed, do Carmo-Dajczer rigidity result states that in low codimension submanifolds with low $s$-nullities are rigid.

Theorem. An isometric immersion $f:M^n\to\mathbb{Q}_c^{n+p},\ p\leq5,$ with $s$-nullities satisfying $\nu_s\leq n-2s-1$ for all $1\leq s\leq p$ at any point is rigid.

Moreover, the above assumption on the $s$-nullities also implies infinitesimal rigidity.

Since rigid submanifolds tend not to use more codimension than strictly necessary to immerse their second fundamental form, it is not surprising that the notion of $s$-nullities plays also a role in the study of reduction of codimension. Indeed, the next result ensures the parallelism of the first normal bundle for submanifolds with low $s$-nullities in low codimension.

Proposition. Let $f:M^n\to\mathbb{Q}_c^m$ be a 1-regular isometric immersion such that $\text{rank}\,N_1=q\leq n-1$. If $\nu_s^*\left(x\right)<n-s$ for all $1\leq s\leq q$ at any point $x\in M^n$, then the first normal bundle $N_1$ is parallel.

Since a generic set of vectors is as linearly independent as possible, it follows that generic submanifolds will have as high $s$-nullities as possible. In the particular case of 1-nullity, if $\nu_1=n-k$ everywhere, then the index of relative nullity $\nu=n-k$. Thus every shape operator $A_\xi|_{\mathcal{N}^\perp}$ is an isomorphism. The maximal dimension of a subspace of symmetric isomorphisms of a $k$-dimensional real vector space is given by


where $\rho\left(k\right)$ is the Hurwitz-Radon number. Thus, in codimension $p\leq p\left(k\right)$, a generic $n$-dimensional submanifold with constant index of relative nullity $\nu=n-k$ will have $\nu_1=n-k$.

  • $\begingroup$ Thanks for the answer. How do you see that $\nu_{1}=n-k$ everywhere implies $\nu=n-k$? $\endgroup$ – MK7 Sep 18 '18 at 14:55
  • $\begingroup$ I am sorry, I have misread your notation and thought your linear map $A$ stood for the shape operator rather than the second fundamental form itself. Anyway, the same genericity principle of linear algebra that I mention above allows you to conclude that your condition is generic as far as the dimension and codimension of $M$ permit the existence of a family of linear maps $A\left(v,\cdot\right)$ satisfying your condition. $\endgroup$ – SubGeo Sep 19 '18 at 14:17
  • $\begingroup$ I am not sure I understand your comment completely, sorry. Is it possible to have a submanifold with $\nu_{1} = n-k$ everywhere, but $\nu < n-k$? $\endgroup$ – MK7 Sep 19 '18 at 15:06
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    $\begingroup$ @MK7: Yes, it is possible. But if in addition you assume that every shape operator $A_\xi$ for $\xi\neq0$ has constant rank $k$, then it is no longer possible. In this case, since the orthogonal complement $\Delta^\perp$ of the relative nullity subspace is spanned by the images of the $A_\xi$'s, if the $A_\xi$'s did not share the same kernel, then a suitable normal direction $\xi$ would yield you an $A_\xi$ of rank strictly greater than $k$, which is a contradiction. Thus, all $A_\xi$´s have the same kernel of dimension $k$, namely $\Delta$, whence $\nu=n-k$. $\endgroup$ – SubGeo Sep 19 '18 at 16:11

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