Pg.248 of "Textbook in Tensor Calculus and Differential Geometry" by Prasun Nayak.




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Let us suppose that $\lambda\_{h|}^i$
is not a unit vector and therefore, the mean curvature $M\_h$ in
this case is given by


$$M\_h=-\frac{R\_{ij}\lambda\_{h|}^i\lambda\_{h|}^j}{g\_{ij}\lambda^i\_{h|}\lambda^j\_{h|}} \tag{1}$$




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A proof was not given so I am trying to derive this.


For when $\lambda\_{h|}^i$ are components of an orthornormal set of vectors
$$M\_h=\sum\_{k}K\_{hk}=\sum\_k\frac{\lambda\_{h|}^l\lambda\_{k|}^i\lambda\_{h|}^j\lambda\_{k|}^rR\_{lijr}}{\lambda\_{h|}^l\lambda\_{k|}^i\lambda\_{h|}^j\lambda\_{k|}^r(g\_{lj}g\_{ir}-g\_{lr}g\_{ij})}=\sum\_k\frac{\lambda\_{h|}^l\lambda\_{k|}^i\lambda\_{h|}^j\lambda\_{k|}^rR\_{lijr}}{(1)\cdot(1)-0}$$
$$=\lambda\_{h|}^l\lambda^j\_{h|}R\_{lijr}g^{jr}=-\lambda\_{h|}^l\lambda^j\_{h|}R\_{lj}$$


where $\sum\_k\lambda^i\_{k|}\lambda^r\_{k|}=g^{ir}$.


Now for when $\lambda\_{h|}^i$ are not components of an orthonormal set of vectors, succumbing to the same approach gives


$$M\_h=\sum\_k K\_{hk}=\frac{\lambda\_{h|}^l\lambda\_{h|}^j R\_{lijr}}{(g\_{lj}\lambda\_{h|}^l\lambda\_{h|}^j)g\_{pq}}\sum\_k\frac{\lambda\_{k|}^i\lambda\_{k|}^r}{\lambda\_{k|}^p\lambda\_{k|}^q} \tag{2}$$
It is not obvious to me how $(2) \to (1)$. Modifying the definition of $M\_h$
$$M\_h=\frac{\sum\_k\lambda\_{h|}^l\lambda\_{k|}^i\lambda\_{h|}^j\lambda\_{k|}^rR\_{lijr}}{(g\_{lj}\lambda\_{h|}^l\lambda\_{h|}^j)\sum\_k g\_{pq}\lambda^p\_{k|}\lambda\_{k|}^q}=-\frac{R\_{ij}\lambda\_{h|}^i\lambda\_{h|}^j}{g\_{ij}\lambda^i\_{h|}\lambda^j\_{h|}N} \tag{3}$$


which is the nearest I can get to $(1)$, but with an extra factor $\frac{1}{N}$.




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What's gone wrong? Is there a way to derive $(1)$ or is $(1)$ a definition itself?



Mean Gaussian curvature from non-unit vector
It's really quite simple to obtain your equation (1) for a non-unit vector from the unit-vector result (your second, unnumbered equation).


Start from your second equation which gives the mean curvature (Ricci curvature) in terms of the Ricci tensor $\mathbf{R}$ and a unit vector ${\lambda}\_{h|}$:
$$M\_h=-\sum\_{i,j}\lambda\_{h|}^i\lambda^j\_{h|}R\_{ij}.$$
Keep in mind that the index $h$ labeling the direction of the vector is fixed, only the component indices $i,j$ are summed over. Now define a new vector $\tilde{\lambda}\_{h|}$ in the same direction $h$, but with a non-unit length, so the unit vector ${\lambda}\_{h|}$ is obtained by dividing $\tilde{\lambda}\_{h|}$ by its length, obtained from the metric tensor $\mathbf{g}$:
$${\lambda}\_{h|}=\frac{\tilde{\lambda}\_{h|}}{\sqrt{\sum\_{k,l}g\_{kl}\lambda\_{h|}^k\lambda\_{h|}^l}}.$$
Combination of these two equations gives the desired expression,
$$M\_h=-\frac{\sum\_{i,j}\tilde{\lambda}\_{h|}^i\tilde{\lambda}^j\_{h|}R\_{ij}}{\sum\_{k,l}g\_{kl}\tilde{\lambda}\_{h|}^{k}\tilde{\lambda}\_{h|}^{l}}.$$




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Notice that there is no need to consider a non-orthogonal set, it's only the length of the single vector $\lambda\_{h|}$ that is changed. So in your equation (2) you can still assume that the vectors $\lambda\_{k|}$ with $k\neq h$ form an orthonormal set, that seems to be the origin of the difficulty you encountered.


