$\ker \alpha(X,.)$的可积性：黎曼子流形下的判定与条件

Hey there, let's break down this question about the integrability of $\ker \alpha(X, \cdot)$ for a 3-dimensional submanifold $N^3$ embedded in a 4-dimensional Riemannian manifold $M^4$:

First off, straight to the point: $\ker \alpha(X, \cdot)$ is NOT always integrable. Its integrability hinges on the geometry of the second fundamental form $\alpha$, the chosen non-vanishing tangent vector field $X$, and the ambient/manifold structures. Let's dive into the details:

Why It's Generally Non-Integrable #

To check if a smooth distribution is integrable, we rely on the Frobenius Theorem: a distribution $\mathcal{D}$ is integrable if and only if for any smooth vector fields $Y,Z \in \mathcal{D}$, their Lie bracket $[Y,Z]$ also lies in $\mathcal{D}$.

For $\mathcal{D} = \ker \alpha(X, \cdot)$, take any $Y,Z \in \mathcal{D}$ — meaning $\alpha(X,Y) = 0$ and $\alpha(X,Z) = 0$. We need to verify if $\alpha(X, [Y,Z]) = 0$.

Using the Weingarten and Gauss formulas, along with the symmetry of the second fundamental form ($\alpha(Y,Z) = \alpha(Z,Y)$), we can derive:
$$\alpha(X, [Y,Z]) = (\nabla_Y \alpha)(X,Z) - (\nabla_Z \alpha)(X,Y)$$
Here, $\nabla$ is the Riemannian connection on $TN$ induced from $M^4$. In most cases, this expression doesn't vanish, so $\mathcal{D}$ fails the Frobenius condition.

As a concrete example: consider the 3-sphere $S^3$ as a submanifold of $\mathbb{R}^4$. Pick $X$ as a unit tangent vector field on $S^3$ — the second fundamental form here is $\alpha(X,Y) = -g(X,Y)\nu$ (where $\nu$ is the unit normal vector). The kernel $\ker \alpha(X, \cdot)$ is the orthogonal complement of $X$, which is the horizontal distribution of the Hopf fibration. This distribution is non-integrable because the Lie bracket of two horizontal vectors lands in the vertical (fiber) direction, violating Frobenius' condition.

Sufficient Conditions for Integrability #

Here are key scenarios where $\ker \alpha(X, \cdot)$ is integrable:

• Case 1: $X$ is a principal curvature direction with zero principal curvature
  If $X$ is a principal direction, there exists a function $\lambda$ such that $\alpha(X,Y) = \lambda g(X,Y)\nu$ (since $N^3 \subset M^4$, the normal bundle is 1-dimensional, so $\nu$ is unique up to sign). If $\lambda = 0$, then $\alpha(X,Y) = 0$ for all $Y \in TN$, making $\ker \alpha(X, \cdot) = TN$ — the entire tangent bundle is trivially integrable.
• Case 2: The second fundamental form is "parallel" with respect to $X$
  Formally, if $(\nabla_Y \alpha)(X,Z) = (\nabla_Z \alpha)(X,Y)$ for all $Y,Z \in TN$, then from our earlier derivation, $\alpha(X, [Y,Z]) = 0$. This satisfies the Frobenius condition, so the distribution is integrable. A simpler special case is when $\nabla_X \alpha = 0$ (the second fundamental form is parallel along $X$).
• Case 3: $X$ is a Killing vector field with compatible Weingarten map
  If $X$ is a Killing vector field on $N$, and the Weingarten map $W_X$ (defined by $g(W_X Y, Z) = \alpha(X,Y,Z)$) has integrable eigendistributions, then $\ker W_X = \ker \alpha(X, \cdot)$ will be integrable.

Additionally, trivial cases like $N$ being a totally geodesic submanifold ($\alpha \equiv 0$, so $\ker \alpha(X, \cdot) = TN$) or $N$ being totally umbilical with $X$ corresponding to a zero principal curvature also guarantee integrability.

内容的提问来源于stack exchange，提问作者Renato Moreira