Complete self-shrinkers with constant norm of the second fundamental form (Q2070961)

Let \(X: M\rightarrow \mathbb{R}^{n+p}\) be an \(n\)-dimensional submanifold of the \((n+p)\)-dimensional Euclidean space. It is called a self-shrinker if it satisfies \(\vec{H} +X^{\perp}=0\) for the normal component \(X^{\perp}\) of the position vector \(X\) and the mean curvature vector \(\vec{H}\) of the submanifold. \newline In [Math. Z. 284, No. 1--2, 537--542 (2016; Zbl 1354.53074)], \textit{Q.-M. Cheng} and \textit{S. Ogata} proved that a 2-dimensional complete self shrinker in \(\mathbb{R}^3\) with constant squared norm of the second fundamental form is isometric to one of the three: \(\mathbb{R}^2\), \(S^1(1)\times \mathbb{R}\), or \(S^2(\sqrt{2})\). In this paper, the authors have develop further and investigate 3-dimensional complete self-shrinker \(M\) in Euclidean space \(\mathbb{R}^4\) with constant squared norm of the second fundamental form. \(S=\sum_{i,j}h_{ij}^2\) and \(f_{4}=\sum_{i,j,k,l}h_{ij}h_{jk}h_{kl}h_{li}\) are constant, then \(M\) is isometric to one of the four: \(\mathbb{R}^3\), \(S^1(1)\times \mathbb{R}^2\), or \(S^2(\sqrt{2})\times \mathbb{R}^1\) or \(S^3(\sqrt{3})\). They also provide that \(S\) must be either 0 or 1 and \(f_4\) must be 0,\(\frac{1}{3},\frac{1}{2}\) and 1.