Let $f:M^{n}\rightarrow \tilde{M}^{m}$ be an isometric immersion, that is, $\langle\cdot, \cdot\rangle_{f}=\langle\cdot, \cdot \rangle$, where $\langle \cdot, \cdot \rangle$ is the metric with respect to $M$. Also denote $\langle \cdot,\cdot\rangle_{\tilde{M}}$ by the metric associeted with $\tilde{M}$.
Denote $f^{*}T\tilde{M}$ the induced bundle over $M$ whose fiber at $x \in M$ is $T_{f(x)}\tilde{M}$. The orthogonal complement of $f_{*}T_{x}M$ in $T_{f(x)}\tilde{M}$ is called the normal space of $f$ at $x$ and is denoted by $N_{f}M(x)$. The normal bundle $N_{f}M$ of $f$ is the vector subbundle of $f^{*}T\tilde{M}$ whose fiber at a point $x\in M$ is $N_{f}M(x)$. Then we have the orthogonal decomposition $$ f^{*}T\tilde{M}=f_{*}TM\oplus N_{f}M.$$ Given $X, Y$ vector fields in $M$, decompose $$ \tilde{\nabla}_{X}f_{*}Y=(\tilde{\nabla}_{X}f_{*}Y)^{T}+(\tilde{\nabla}_{X}f_{*}Y)^{\perp}.$$ As $$ \nabla_{X}Y:=f_{*}^{-1}(\tilde{\nabla}_{X}f_{*}Y)^{T}$$ defines a compatible torsion-free connection on $TM$ we have by uniqueness that $\nabla_{X}Y$ is the Levi-Civita connection of $M^{n}$. The map $\alpha:\Gamma(TM) \times \Gamma(TM) \rightarrow \Gamma(N_{f}M$) defined by $$ \alpha^{f}(X,Y)=(\tilde{\nabla}_{X}f_{*}Y)^{\perp}$$ is called the second fundamental form of $f$. The shape operator $A_{\xi}$ of $f$ at $x \in M$ with repect to $\xi \in N_{f}M(x)$ is defined by $$ \langle A_{\xi}X,Y\rangle=\langle \alpha(X,Y), \xi \rangle_{\tilde{M}}.$$
Definition: Let $f, g:M^{n}\rightarrow \mathbb{R}^{m}$ be immersions. We say that $f$ and $g$ are conformal, if there exists $\varphi \in C^{\infty}(M)$ such that the metrics induced by $f$ and $g$ are related by $$ \langle \cdot , \cdot\rangle_{g}=\varphi^{2}\langle \cdot, \cdot\rangle_{f}$$
Problem: I would like to know if someone could show me the relations between the second fundamental forms of $f$ and $g$.
I thought of using the relation $$ \nabla_{X}^{g}{Y}=\nabla_{X}^{f}Y+\frac{1}{\varphi}(Y(\varphi)X+X(\varphi)Y-\langle X,Y \rangle_{f}\mbox{grad}_{f}\varphi)$$ where $\mbox{grad}_{f}\varphi$ is defined by $$ X(\varphi)=\langle \mbox{grad}_{f}\varphi, X \rangle_{f}.$$
Best Answer
There is no relation. Take the simplest case when $n=1$ and $m=2$. That is, $f, g$ are two immersed curves in $\mathbb R^2$. One can give both $f, g$ the arc-length parametrization, which in your notations means $\langle \cdot, \cdot\rangle_g = \langle \cdot, \cdot\rangle_f$, or $\varphi = 1$. But there is no restrictions on $f$ and $g$, and so no restrictions on the curvature $\kappa$.
On the other extreme, let $f$ be any isometric immersion and let $g = f\circ h$, where $h : M \to M$ is any conformal immersion. They have the same second fundamental forms, and it has nothing to do with the conformal factor.