# Mean curvature operator in S^n

Consider the sphere $S^n$. By using the stereographic projection we can identify $S^n \setminus N$ with $\mathbb{R}^n$, where $N$ is the North pole of $S^n$. The metric then is given by $\frac{dx^2}{(1+x^2)^2}$. Now we consider the graph of a smooth function $f$ on an open subset of $\mathbb{R}^{n-1}$ as a subset of $\mathbb{R}^n$ with the above metric. Can you tell me what the mean curvature of this graph is in terms of $f$, or can you at least name me a book, where I can look this up? Every help would be appreciated.

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I'm a little confused here. Usually the mean curvature of the graph of a function in $\mathbb R^n$ is the trace of the derivative of the normal vector (i.e. the trace of the second fundamental form). What does this have to do with stereographic projection? –  treble Dec 17 '12 at 5:02
@treble. He does not consider the flat metric on $\mathbb R^{n - 1}$ but the one induced by stereographic projection. This clearly makes a difference. –  wspin Dec 17 '12 at 20:31
@wspin I agree that the OP seems to want to equip the graph of $f$ with the pullback metric (via $f^{-1}$) of the metric on $\mathbb R^{n-1}$ coming from stereographic projection. Thus the metric on the graph is not the same as the one induced from $\mathbb R^n$. This is precisely the problem. How do you want to define mean curvature in this case? –  treble Dec 19 '12 at 22:59
It should be the sum of the principal curvatures , i.e. the eigenvalues of the shape operator, over the dimension. –  wspin Dec 20 '12 at 7:29
The shape operator is defined using the connection on the ambient manifold, which in this case is just the ordinary directional derivative. The tangent planes (and thus the normal vector) of the graph of $f$ are independent of any metric on the graph of $f$. Thus it not clear (to me) how the shape operator ($X \to D_X n$) changes when the metric on the graph of $f$ changes. –  treble Dec 21 '12 at 1:50