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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.

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  • $\begingroup$ I'd check the (most recent) book by Dajczer & Tojero, on the chapter about submanifold theory. It's possible to relate the scalar curvature of a submanifold of a space form with the ambient curvature, the mean curvature of the submanifold, and the norm of the third fundamental form. $\endgroup$ Commented Aug 20, 2021 at 22:03
  • $\begingroup$ I can't seem to make sense of your edit: If $M=f^{-1}(1)\cap N$ for a linear map $f$, then there is a global frame (i.e. a set of global vector fields that form a basis at each point): Choose a basis for $\operatorname{ker}(f)$, and extend each vector to a constant vector fields on $N$. These fields can be restricted to $TM$, where they form a basis at each point. $\endgroup$ Commented Aug 21, 2021 at 18:17
  • $\begingroup$ You are right, I should specify the fact that the overcomplete basis I have is preferred in some sense, and the choice of a global basis you suggest makes computations more difficult, and that is why I would like to be able to use the overcomplete basis. I understand it may sound unclear, if you want, I can write down what is the explicit situation I am facing, but I do not want to bother you too much :) $\endgroup$ Commented Aug 21, 2021 at 18:41
  • $\begingroup$ Does $N$ have the flat metric induced by its embedding into $\mathbb{R}^n$ or a possibly different one? $\endgroup$ Commented Aug 21, 2021 at 19:14
  • $\begingroup$ I just re-edited the question to give as many details as possible, and I apologize if now seems too "narrow" and not interesting $\endgroup$ Commented Aug 21, 2021 at 20:14

2 Answers 2

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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 $M$.

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  • $\begingroup$ Thank you very much for the answer. Unfortunately, even if it explains what happens with Riemannian immersions, it does not quite answer what I had in mind. However, it was my fault for not being completely clear in writing my question. I thus edited it in order to explain better what I had in mind. $\endgroup$ Commented Aug 21, 2021 at 16:51
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    $\begingroup$ @F.M.C. I've added a bit about how to apply this to your situation. In the case where $M$ is a level curve, one can do the entire computation in $N$ using any coordinates/frame on $N$. $\endgroup$ Commented Aug 22, 2021 at 14:36
  • $\begingroup$ One point that simplifies things: When you compute the shape operator $S = \nabla n$, you get a $(1,1)$-tensor on $N$ such that $S^A{}_B\nabla^B f = 0$. As a consequence, you do not need to find a frame on $M$ itself or split the metric into normal and tangential parts. You just do all of the tensor calculations, including the trace, with respect to the frame you already have on $N$. The components normal to $M$ will automatically vanish. $\endgroup$ Commented Aug 22, 2021 at 17:32
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    $\begingroup$ @Deane I believe the identity should read $S=-(\nabla n)|_{TM}$ (at least in the sign convention I'm using). If one defines the extension to $TN$ by $\widetilde{S}=-\nabla n$, then the normal component $\widetilde{S}(n)=-\nabla_nn$ will not necessarily vanish. Instead one should project out $n$ to to ensure that $\widetilde{S}(n)=0$, yielding $\widetilde{S}{}^A{}_B=-(\nabla_Cn^A)P^C{}_B$. $\endgroup$ Commented Aug 22, 2021 at 17:57
  • $\begingroup$ Yes, you're right. $\endgroup$ Commented Aug 22, 2021 at 21:14
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$\newcommand{\SU}{\operatorname{SU}}$ 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:

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

  2. 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 $1$).

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  • $\begingroup$ Thank you very much for the help. However, I do not see how to use @Kajelad's formula because it requires a splitting of the ambient metric in terms of the normal and "tangential" components, and it requires contraction among tensors (which is essentially taking the trace), and I am not sure I can use an overcomplete basis to do all this (that is precisely the core of my question). $\endgroup$ Commented Aug 22, 2021 at 8:13

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