Is the Wikipedia formula for the Gauss equation correct?

Question

I've been doing some computations and they don't seem to add up:

Consider an $n$-sphere immersed in $\mathbb{R}^{n+1}$.

Its Riemann tensor should be of the form $R_{abcd} = p *(g_{ac}g_{bd}-g_{ad}g_{bc})$, $p>0$.

Its second fundamental form should be of the form $\Pi_{ab}= k*g_{ab}$. https://en.wikipedia.org/wiki/Gauss%E2%80%93Codazzi_equations#Formal_statement

Wikipedia states that for $M\subset P$, the Gauss equation is with $X,Y,Z,W\in TM$ .

Now, if I apply the Gauss equation for $P=\mathbb{R}^{n+1}, M= \mathbb{S}^n$, then $R'$ vanishes leaving $R_{abcd} = (-k^{2}) (g_{ac}g_{bd}-g_{ad}g_{bc})$.

However $-k^2$ should be positive which is why I suspect the Wikipedia formula has $R$ and $R'$ switched around.

This is not the only place I've seen the formula like this. I'm really confused. Can you help me figure out where the error is?

Edit: Knowing $R_{bd} = {R^{c}}_{bcd}$ and R = ${R^{a}}_a$, (see here), a curvature tensor $R_{abcd} = (-k^{2}) (g_{ac}g_{bd}-g_{ad}g_{bc})$ would result in a scalar curvature of

$R= -k^2(n(n-1)) < 0$ for the sphere. The scalar curvature of the sphere should be positive.

Answer 1

After Jason DeVito's advice that there are two conventions, long discussions with Seub, and checking the computations of various formulas on Wikipedia pages, I've come up with the following heuristic for Wikipedia formulas:

1)Equations written using linear operator notation use one convention: $$\langle R(X,Y)Z, W \rangle = \langle \Pi (Y, Z), \Pi(X, W)\rangle - \langle \Pi (X, Z), \Pi(Y, W)\rangle $$ $$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z$$ (see Gauss Codazzi Equations, Second Fundamental Form,Ricci Curvature)

2)Equations written using Ricci calculus/Tensor notation use the opposite convention: $$ R_{xyzw} = \Pi_{xz} \Pi_{yw} - \Pi_{yz} \Pi_{xw}$$ $$A_{v;pq}-A_{v;qp}= A_{\beta}{R^{\beta}}_{vpq}$$ (see Riemann_curvature_tensor,List of formulas in Riemannian geometry)

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The idea is to swap the sign of the Riemann curvature tensor when transcribing between tensor notation and operator notation. I haven't seen any formula in Wikipedia that doesn't agree with this heuristic.

I myself prefer tensor notation, and am used to said convention, tough many do Riemannian geometry using operators and may be using the opposite convention.

Edit: The convention does appear written out explicitly: $$R^\rho{}_{\sigma\mu\nu} = dx^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})$$ It's weird, but all the Wiki's formulas seem to follow it.

Answer 2

First, as Jason deVito points out, there are indeed two different conventions for the Riemann curvature tensor, differing by a minus sign. But you and Wikipedia both use the more common one. Not only that, as an independent matter, there are also two different conventions for the second fundamental form, also differing by a minus sign. Here I believe that is that the second fundamental for on the sphere should be $$\Pi = - \sqrt{k} \, g = - \frac{1}{r} g$$ instead of $\Pi = \sqrt{k} \, g$. This is if we follow Wikipedia's sign convention (the more common one I believe) that $\Pi(u, v) = \langle \nabla_u v, n\rangle = \langle -\nabla_u n, v\rangle$ where $n$ is the unit normal. Note that the sign of the second fundamental depends crucially on the choice of the unit normal $n$ for your hypersurface. The unit normal is determined by choosing an orientation of both the manifold $P$ and the submanifold $M$. Here if you take the usual orientation of $\mathbb{R}^{n+1}$ and of $S^n$, you must choose the outward pointing normal for $n$.

That being said, even if you get the sign wrong on the second fundamental form, you get the same Gauss equation (following Wikipedia's formula), namely $$\langle R(X,Y)Z, W \rangle = k (\langle Y, Z\rangle \langle X, W\rangle - \langle X, Z\rangle \langle Y, W\rangle)$$ or following your notations $$R_{abcd} = k (g_{ad}g_{bc} - g_{ac}g_{bd}) = -k (g_{ac}g_{bd} - g_{ad}g_{bc})$$ But contrary to what you seem to think, there is no sign problem here, this is indeed what you expect for a Riemannian manifold of constant sectional curvature $k$.

EDIT: To answer your new question about Ricci and scalar curvature: I agree this is confusing. If we take your definition of the Ricci tensor, which agrees with Wikipedia's, namely \begin{equation} \operatorname{Ric}(X,Y) = \operatorname{tr}(V \mapsto R(X,V)Y) \end{equation} or, as you put it, $R_{bd} = {R^c}_{bcd}$, then we find

\begin{align} R_{bd} &= {R^c}_{bcd} = k({g^c}_d g_{bc} - {g^c}_c g_{bd} )\\ &= -k(n-1)g_{bd} \end{align} in other words $\operatorname{Ric} = -k(n-1)g$, where $n$ is the dimension of $M$. I agree with you that this is not what we want usually. Therefore the conclusion is that we must take the opposite sign convention in the definition of the Ricci tensor. I think the confusion comes from the fact that some authors take the opposite sign convention for the Riemann curvature tensor. By the way, here is the convention I use, I have not written it yet: $$R(X,Y) = \nabla_{X,Y}^2 - \nabla_{Y,X}^2$$ in other words $$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z$$

This definition agrees with that of Wikipedia in the pages Riemann curvature tensor and Ricci curvature. But the conclusion is that I disagree with their definition of the Ricci tensor by a minus sign. I also disagree with their sign in the paragraphs "Ricci Curvature" and "Special Cases" in the page Riemann curvature tensor. As for the page Ricci decomposition, I also disagree with the sign of the Ricci tensor, unless they take the opposite sign convention for the Riemann curvature tensor (it's not defined explicitly in that page). If you research, you'll see that quite a few authors choose the opposite sign convention for the Riemann curvature tensor (e.g. Gallot-Hulin-Lafontaine), which is probably the source of confusion.

EDIT2: More fighting about signs: see comments under OP's answer.