# Second fundamental form proportional to the Hessian

Let $(M^n,g)$ be a Riemannian manifold and $f:M\to\mathbb{R}$ a smooth function. Then the graph $S=\{(p,f(p))\mid p\in M\}$ is a submanifold of $(M\times\mathbb{R},g+g_{\mathbb{R}})$ and carries the induced metric $\tilde{g}=g+\nabla f\otimes \nabla f$.

Now I want to know, how to calculate the second fundamental form $h$ of $S$. In my intuition at every point $h_{(p,f(p))}(df_p(\cdot),df_p(\cdot)):T_pM\times T_pM\to \mathbb{R}$ has to be proportional to the Hessian $\operatorname{hess}(f)_p(\cdot,\cdot):T_pM\times T_pM\to \mathbb{R}$. Does anyone know the exact relation between those?

EDIT: The idea is: Define $\tilde{f}:M\times\mathbb{R}\to \mathbb{R}; (p,r)\mapsto f(p)-r$. Then $\tilde{f}^{-1}(0)=S$ and following Sun Park Joe's comment the second fundamental form at $\tilde{p}=(p,r)$ is given by $h_{\tilde{p}}(v,w)=\frac{hess(\tilde{f})_{\tilde{p}}(v,w)}{\left|\nabla \tilde{f}_{\tilde{p}}\right|}=\frac{hess(f)_p(d(f^{-1})_\tilde{p}~v~~,~~d(f^{-1})_{\tilde{p}}~w)}{\sqrt{1+\left|\nabla f_p\right|}}$. I hope, I got everything right..

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I think the answer is yes, but you probably have to express your question in a more generalzied form. i.e. Let $M$ be a Riemannian manifold and $f\in C^{\infty}(M)$. Consider a smooth level set of $f$, i.e. say $\Sigma:=f^{-1}(0)$. Let $\nabla f$ be the gradient vector of $f$. Then the hypersurface $\Sigma$ has fundamental form $$II(v,w)=\langle \nabla_v \vec{n}, w \rangle$$ where $\vec{n}$ is the normal vector of $\Sigma$ which is exatctly $\nabla f$. And Hessian can be written as $$Hess(f)(v, w)=\langle \nabla_v \nabla f, w \rangle$$ where $v, w\in T_p(\Sigma)$ and $p\in \Sigma$. If you are not familer with the definition of Hessian or Second fundamental form, you can find them in any basic Riemannian Geometry book, for example Kobayashi-Nomizu's Fundamental of Diff. Geo. II

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Thank you for your answer, but my problem is the relation between the second fundamental form of the graph $graph(f)$ (not the preimage $f^{-1}(0)$) and the Hessian $hess(f)$. –  Untimed Nov 25 '12 at 22:01
Take the $M$ in my answer to be your $M\times \mathbb R$ and $\Sigma$ to be your graph of $f$. Now you can extend your function $f$ to a neighborhood of the graph, for example you can take the extension to be the signed distance function to the graphy. Then the normal vector to your graph is nothing by the grident of the extended $f$. –  Sun Nov 25 '12 at 22:12
Ok, thank you very much! I will edit my post. –  Untimed Nov 25 '12 at 22:19