Let $f$ be a smooth map from 3-dim manifold $M$ to $M'$ as submanifold of $R^4$, i.e. $f: M \rightarrow M' \subset R^4$. Suppose we have a positive and strongly convex norm function $\Phi$ on $R^4$ which induce a positive and strongly convex norm function $\Phi$ on the tangent space of $M'$ and smooth on the tangent bundle $TM'$. (Just like Euclidean metric induce an inner product on the submanifold of $R^3$)
Here positive and strongly convex norm mean that (1)$\Phi(v) \geq 0$, (2)$\Phi(cv) = c\Phi(v)$ for $c \geq 0 $ and (3) Hessian $\frac{\partial^2 \Phi^2}{\partial x_i \partial x_j}(v)$ is positive definite for every $v \in R^4$, and $x_i, x_j$ are coordinate charts in $R^4$. (which induce that Hessian $\frac{\partial^2 \Phi^2}{\partial v_i \partial v_j}(v)$ is positive definite for every $v \in T_{f(p)} M'$, and $v_i, v_j$ are coordinate charts in $T_{f(p)} M'$. And we can also deduce that unit sphere of $\Phi$ is convex).
Besides, we have following conditions:
- $d_{\tilde{v}} \Phi(d_p^2 f(v, w)) = 0$ for every $v, w \in T_p M$.
- $d_{\tilde{v}} \Phi(d_p^3 f(v, w, w_1)) + d_{\tilde{v}}^2 \Phi(d_p^2 f(v, w), d_p^2 f(v, w_1)) = 0 $ for every $v, w, w_1 \in T_p M$.
Here $\tilde{v} = d_p f(v)$
The question is: Can we use these conditions to prove that if for every pair $v, w \in T_p M$ which satisfies that $d_p^2 f(v, w)$ is in the tangent space $T_p M$, we have $d_p^2 f(v, w) = 0$?
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This question has two backgrounds:
- If it is 2-dim manifold, then from $d_{\tilde{v}} \Phi(d_p^2 f(v, w)) = 0$ and $d_{\tilde{w}} \Phi(d_p^2 f(v, w)) = 0$, we can manage to prove this.
- If it is Riemannian metric rather than this norm, then this is equivalent to the condition that first and second derivative of Riemannian metric are zero. So Christopher symbols $\Gamma_{ij}^k = 0$ and we can deduce that $d_p^2 f(v, w) = L_{ij} N$ with $L_{ij}$ as second fundamental form and get that $d_p^2 f(v, w) = 0$ if it is in $T_p M$