Scalar curvature of Riemannian immersion I apologize in advance if this is a silly question, or if it is well-known to people studying Riemannian geometry.
In the context of Riemannian submersions, it is possible to relate the scalar curvature of  the two manifolds (in particular, I refer to corollary 9.37 of Besse's book "Einstein manifolds").
I would like to know if something similar also holds for Riemannian immersions. For instance, is it possible to relate the scalar curvature of the 2-sphere with the (vanishing) scalar curvature of the ambient space (for instance using something like the second fundamental form or something similar)?
EDIT
I realize it is better if a give some other details.
The case I am interested in is the following: $N=P\times SU(n)$ where $P=\{\mathbf{x}\in\mathbb{R}^{n}:x^{1}>x^{2}>....>x^{n}>0\}$ with $(x^{1},...,x^{n})$ is the natural Cartesian coordinate system in $\mathbb{R}^{n}$, and $M=f^{-1}(1)\cap N$ where $f(\mathbf{x},U)=\sum_{j}x^{j}$.
On the manifold $N$ there is the Riemannian metric
$$
g=\sum_{j}\frac{\mathrm{d}x^{j}\otimes\mathrm{d}x^{j}}{x^{j}} + \sum_{k}G_{k}(\mathbf{x})\theta^{k}\otimes\theta^{k},
$$
where $\{\theta^{k}\}$ is a preferred system of left-invariant one forms on $SU(n)$ (or better, their pullback to $N$ with respect to the obvious projection map), and we endow $M$ with the Riemannian metric $h$ which is the pull-back of $g$ with respect to the canonical immersion $i: M\rightarrow N$.
There is a basis of globally-defined vector fields $\{\partial_{x^{j}},X_{k}\}$ on $N$, where the $X_{k}$'s are tangent to $SU(n)$ and are dual to the $\theta^{k}$'s,  and there is a basis of globally-defined differential forms $\{\mathrm{d}x^{j},\theta^{k}\}$.
Clearly, using the pullback with respect to the inclusion map, we can still use $\{\mathrm{d}x^{j},\theta^{k}\}$ as an overcomplete basis of differential forms on $M$, and it is clear that $h$ "looks" exactly the same using this basis.
Regarding vector fields, by setting $Y_{j}=\partial_{x^{j}} - \sum_{l}x^{l}\partial_{x^{l}}$, we obtain an overcomplete basis $\{Y_{j},X_{k}\}$ of vector fields.
The situation is similar to that of the unit 2-sphere in $\mathbb{R}^{3}$ where the overcomplete basis of globally-defined vector fields is obtained by considering the restriction to the 2-sphere of the fundamental vector fields of the standard action of $SO(3)$ on $\mathbb{R}^{3}$ having the unit 2-sphere as an orbit, while the overcomplete basis of globally-defined differential forms is obtained considering the pull-back of the $\mathrm{d}x^{j}$ to the sphere, with $(x^{1},x^{2},x^{3})$ a Cartesian coordinate system on $\mathbb{R}^{3}$.
I want to compute the scalar curvature of $(M,h)$. It can be computed in terms of an orthogonal basis of tangent vectors at each point of $M$, but I am not able to do it because I was not able to quickly find such a basis. Therefore, I thought I could do it using the overcomplete basis introduced before, but up to now I failed. I realized that the difficulty I face boils down, essentially, to the fact that for computing the Ricci tensor and the scalar curvature I need to take the trace of something, and taking a trace requires choosing a basis, and can not be done using an overcomplete basis.
This made me wonder if, for a Riemannian immersion as the one described above, it is possible to define the Ricci tensor and the scalar curvature using an overcomplete basis of vector fields defined in terms of vector fields on the ambient manifold.
If you have read up to now, I thank you for the patience, and I welcome any hint.
 A: It seems what you're asking form is a straightforward application of the Gauss equation. Let $\iota:M\to N$ be an isometric immersion, $R$ be the Riemann tensor of $M$, $\widetilde{R}$ be the pullback of the Riemann tensor of $N$, and let $\operatorname{II}$ be the second fundamental form. These are related by the Gauss equation, which we can write in abstract index notation, using roman indices for $TM$ and greek indices for $NM$:
$$
R_{abcd}=\widetilde{R}_{abcd}+g_{\alpha\beta}\operatorname{II}^\alpha{}_{ac}\operatorname{II}^\beta{}_{bd}-g_{\alpha\beta}\operatorname{II}^\alpha{}_{ad}\operatorname{II}^\beta{}_{bc}
$$
Contracting all of the indices to obtain the scalar curvature gives
$$
R=\widetilde{R}+g_{\alpha\beta}g^{ac}g^{bd}(\operatorname{II}^\alpha{}_{ac}\operatorname{II}^\beta{}_{bd}-\operatorname{II}^\alpha{}_{ad}\operatorname{II}^\beta{}_{bc})
$$
Edit:
For the case given in the edited version of the question, we can simplify this expression: If $M=f^{-1}(1)$ where $1$ is a regular value of a smooth function $f$, we can define a global unit normal vector $n=\operatorname{grad}f/\|\operatorname{grad}f\|$. This allows us to define the shape operator $S$ by $\operatorname{II}(u,v)=\langle u,S(v)\rangle n$, and express the scalar curvature in terms of it:
$$
R=\widetilde{R}+S^a{}_aS^b{}_b-S^a{}_bS^b{}_a
$$
Let capital roman indices denote tensors in $N$. If you want to carry out this computation in $N$, you can define a $(1,1)$ tensor $\widetilde{S}$ on $N$ such that $\widetilde{S}|_{TM}=S$ and $\widetilde{S}(n)=0$. This then gives
$$
R=\widetilde{R}+\widetilde{S}{}^A{}_A\widetilde{S}{}^B{}_B-\widetilde{S}{}^A{}_B\widetilde{S}{}^B{}_A
$$
provided we evaluate the expression at a point in $M$. Starting with the identity $S(u)=-\nabla_un$, one can show that $\widetilde{S}{}^A{}_B=-P^A{}_CH^C{}_DP^D{}_B$ where $P^A{}_B=\delta^A_B-n^An^Cg_{CB}$ is the orthogonal projection onto $n^\perp$ and $H^A{}_B=g^{AC}\nabla_C\nabla_Bf$ is the Hessian of $f$ with an index raised. These can all be explicitly computed using your global frame on $M$.
A: $\newcommand{\SU}{\operatorname{SU}}$
Here are some comments and a sketch on what to do.
The first observation is that you can simplify the metric $g$
by using coordinates
$$
u^j = 2\sqrt{x^j}\ \implies du^j = \frac{dx^j}{\sqrt{x^j}}.
$$
It follows that $g$ restricted to $P$ is
$$
\sum_j\frac{dx^j\otimes dx^j}{x^j} = \sum_{j} (du^j)^2.
$$
In particular, this is a flat metric, and
$$
g = \sum_j(du^j)^2 + \sum_k G_k\left(\frac{(u^1)^2}{2}, \dots, \frac{(u^n)^2}{2}\right)\theta^k\otimes\theta^k.
$$
Depending on what the $G_k$ are, these coordinates might be easier to work with.
The second is that @Kajelad's answer is still the right one. You just need to do the following:

*

*Calculate the scalar curvature of $N$. You can use the frame $(du^1, \dots, du^n, \theta^1, \dots)$ to do this.


*Find the formula for the second fundamental form of $M$. This can be done by calculating the covariant derivative of the Gauss map. Since $\nabla f$ is normal to $M$, the Gauss map is $\nabla f/|\nabla f|$.  You can use your formulas to calculate this.
This all can now be substituted into @Kajelad's formula (which can be simplified a bit, since the codimension of $M$ is $1$).
