# Curvature formula

Suppose I have a homogeneous cubic polynomial $f(w,x,y,z)$ and I let $X$ be the set of points in $\mathbb{R}^4$ where $f(w,x,y,z)=0$ and $w^2+x^2+y^2+z^2=1$. Suppose that this is a smooth surface, and I give it the metric inherited from $\mathbb{R}^4$. Does anyone know a nice formula for the curvature of $X$ in terms of $w,x,y,z$ and the coefficients of $f$? Doubtless this could be worked out by going through the whole business of local coordinates and Christoffel symbols but I imagine that that would be quite lengthy and the final answer is probably not too complicated.

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## 1 Answer

There is basic formula in Riemannian geometry that, used twice, gives fairly direct answer. Let $M$ be a Riemannian manifold, let $N$ be a codimension $1$ submanifold of $M$, and let $f$ be a function defined on $M$. Call $f$ also the restriction of $f$ to $N$. Then the Hessian of $f$ on $N$ and the restriction to $TN$ of the Hessian of $f$ on $M$ are related by $$Hess_N(f) = Hess_M(f)+df(n) II~,$$ where $n$ is the oriented unit normal of $N$ in $M$ and $II$ is the second fundamental form of $N$ in $M$.

You can apply this first for $S^3$ considered as a hypersurface in $R^4$, and obtain that $$Hess_{S^3}(f) = Hess_{R^4}(f)+(\partial_rf) g~,$$ where $g$ is induced metric on $S^3$. Then you can apply it to $X$ considered as a hypersurface of $S^3$ to obtain that $$0 = Hess_X(f) = Hess_{S^3}(f) + df(n) II~,$$ where $n$ is the unit normal to $X$ in $S^3$ and $II$ its second fundamental form.

Putting all this together yields a formula $$II = -(Hess_{R^4}(f) + (\partial_rf) g)/df(n)~.$$ The curvature of $X$ the follows from the Gauss formula, $K=1+det(II)$.

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Just one comment: since $X$ is a level set of $f$ in $\mathbb{S}^3$, $df(n)$ is nothing more than the norm of the projection of $df$ to $T^*S^3$. – Willie Wong Jul 9 '11 at 14:38