This answer serves mainly to satisfy my own curiosity, especially since at this stage in the book Besse already introduces connections in vector bundles and exterior covariant derivatives (you should probably read/familiarize yourself with the stuff in this answer first). The following setup is motivated by Ivo Terek’s notes (you can also find a similar setup in an appendix to Taylor’s PDE volume 2). But I’ll second @Ted Shifrin’s comment that you really should consult a standard differential geometry text for a more down-to-earth presentation.
Tldr; for Ricci’s equation specifically, if you don’t find it in a textbook, then simply look at how Gauss’ equation is proved. The proof can be mimicked almost line by line, because these really are the same thing once you interchange the roles of the tangent and normal bundles.
The second fundamental form/shape tensor in Riemannian geometry captures the extrinsic properties of a submanifold. But more strictly speaking, we’re looking at “how” the tangent bundle of the submanifold sits inside of the tangent bundle of the ambient manifold. So, we can actually consider the setup at the general level of vector bundles. We’ll have a big bundle, and a direct subbundle decomposition. We’ll also have a connection on the big bundle, which induces connections on the subbundles, and we can then ask to compare these different connections. The difference is the second fundamental form (it measures how much the connection in the ambient bundle deviates from the induced ones on the subbundles… i.e extrinsic information). The key question to answer is how are the curvatures of these various connections related, and these are precisely the famous Gauss-Codazzi equations (the Ricci equation as you call it is essentially a ‘dual’ version of the Gauss’ equation). Since we have to understand various bundles, connections and curvatures simultaneously, it’s good to record some simple lemmas.
Lemma 1 (Curvature of direct-sum is direct sum of curvatures)
Let $(E,\pi,M)$ be a vector bundle over $M$, and $E=\bigoplus_{i=1}^nE_i$ a smooth subbundle decomposition. Suppose each $E_i$ has a connection $\nabla^{E_i}$, and equip $E$ with the induced direct sum connection $\nabla^E$. Then, letting $R,R_i$ be the curvatures of the connections $\nabla^E$ and $\nabla^{E_j}$ respectively, we have the diagonal block decomposition
\begin{align}
R=\text{diag}(R_1,\dots, R_n).
\end{align}
(Recall $R$ is an $\text{End}(E)\cong\bigoplus_{i,j=1}^n\text{Hom}(E_j,E_i)$-valued $2$-form on $M$, so it has an $n\times n$ block decomposition, with the $(i,j)$ term being a $\text{Hom}(E_j,E_i)$-valued $2$-form on $M$; the claim is that only the diagonal terms survive, and these are the curvatures $R_i$).
To prove this is very simple. Let $\psi$ be any $E$-valued $k$-form on $M$. We can decompose this as $\psi=\sum_{i=1}^n\psi_i$ where each $\psi_i$ is the $E_i$-valued portion of $\psi$. Now, by definition of $\nabla^E$ being the direct sum of the connections $\nabla^{E_i}$, and the definition of curvature, it easily follows that
\begin{align}
R\wedge_{\text{ev}}\psi&=d_{\nabla^E}(d_{\nabla^E}\psi)=\sum_{i=1}^nd_{\nabla^{E_i}}\bigg((d_{\nabla^E}\psi)_i\bigg)=\ \sum_{i=1}^nd_{\nabla^{E_i}}\bigg(d_{\nabla^{E_i}}(\psi_i)\bigg)=\sum_{i=1}^nR_i\wedge_{\text{ev}}\psi_i.
\end{align}
Since $\psi$ is arbitrary, it follows $R=\text{diag}(R_1,\dots, R_n)$.
Lemma 2 (Curvatures of different connections).
Let $(E,\pi,M)$ be a vector bundle with two connections $D,\nabla$, and let $A=D-\nabla$ be the difference. Then, the curvatures $R_D,R_{\nabla}$ are related by the equation
\begin{align}
R_D&=R_{\nabla}+d_{\nabla}A+A\wedge_{\circ}A.
\end{align}
For a proof, see this answer of mine.
Induced connections on subbundles and second fundamental form.
Let $(E,\pi,M)$ be a smooth vector bundle, and suppose we have a smooth direct sum decomposition $E=L\oplus L’$; let $P:E\to L$ and $P’:E\to L’$ be the induced projections. Suppose $E$ has a connection $D$. We can ‘project’ this to get a connection $\nabla$ on $L$ and $\nabla’$ on $L’$ as follows: for any section $\psi$ of $L$ and $\psi’$ of $L’$ and vector field $X$ on $M$,
\begin{align}
\nabla_X\psi:=P(D_X\psi),\quad\text{and}\quad \nabla’_X(\psi’):=P’(D_X(\psi’)).
\end{align}
This definition (modulo notation change) is exactly in Besse page 24, (1.10). It is easy to check these really are connections. Ok, now we have connections on the subbundles, so we can put them back together to get a connection $\nabla^{\oplus}=\nabla\oplus\nabla’$ on $E=L\oplus L’$. We now have two connections on $E$: our original one $D$, and our humpty-dumpty reconstructed $\nabla^{\oplus}$. There is no reason to expect these to be equal. Let us define $A:=D-\nabla^{\oplus}$; this is of course an $\text{End}(E)$-valued 1-form on $M$. We shall refer to this as the second fundamental form/shape tensor of the connection $D$ relative to the decomposition $E=L\oplus L’$. Now, the ‘ultimate’ question is to relate the curvatures of these various connections.
If we think of $D$ as being given first, then we usually think of it as an ‘intrinsic description of the geometry of $E$’ (in particular, the curvature of the connection $D$ is an intrinsic description of how the various fibers fit together). Likewise, $\nabla$ and $\nabla’$ are ‘intrinsic’ information to the subbundles $L,L’$. The fact that $A$ is by definition the difference $D=\nabla^{\oplus}=D-(\nabla\oplus\nabla’)$ means that $A$ measures the difference in the ambient geometry versus the two intrinsic geometries of $L,L’$, i.e it tells us the extrinsic information of how $L,L’$ sit inside $E$. Note I’m using the terms ‘intrinsic’ and ‘extrinsic’ in a vague sense, but hopefully this intuitively makes sense as to why the shape tensor captures information about ‘extrinsic geometry’. Slightly more precisely
Ok so notice we have direct sums, subbundles, connections, differences of connections, i.e we’re in a prime position to apply lemmas 1 and 2. But before we do so, we should record the following important property of the shape tensor.
Lemma 3 (Block decomposition of the shape tensor)
Keep notation as above. Then, $A$ has the block decomposition $\begin{pmatrix}0&\beta\\\alpha&0\end{pmatrix}$, where $\alpha,\beta$ are respectively $\text{Hom}(L,L’)$ and $\text{Hom}(L’,L)$-valued $1$-forms on $M$.
Proving this is very simple. Let $\psi$ be any section of $E$ and let subscripts like $\psi_L,\psi_{L’}$ denote the respective projections. Then,
\begin{align}
A\cdot\psi&:=D\psi-\nabla^{\oplus}\psi\\
&=D(\psi_L+\psi_{L’}) - [\nabla (\psi_L)+ \nabla’(\psi_{L’})]\\
&=D(\psi_L)+D(\psi_{L’}) - \left[\bigg(D(\psi_L)\bigg)_L+ \bigg(D(\psi_{L’})\bigg)_{L’}\right]\\ &=\underbrace{\bigg(D(\psi_L)\bigg)_{L’}}_{:= \alpha\cdot (\psi_L)} + \underbrace{\bigg(D(\psi_{L’})\bigg)_L}_{:=\beta\cdot (\psi_{L’})},
\end{align}
where I have combined the first and third terms, and the second and fourth. In other words, if $\psi$ is purely $L$-valued (i.e $\psi=\psi_L$, or equivalently $\psi_{L’}=0$), then the above equation shows that $A\cdot \psi=(D\psi)_{L’}$, i.e $A$ takes something $L$-valued and sends it to something $L’$-valued. This is exactly why the $\alpha$ appears in the lower left slot. Similarly, why $\beta$ appears in the top right slot.
Now, we can apply lemmas 1,2,3 to relate the curvatures of $D$ and $\nabla,\nabla’$. But just to clarify the notation, and all the connections involved:
- $D$ is a connection on $E$.
- $\nabla$ is a connection on $L$, $\nabla’$ a connection on $L’$
- $\nabla^{\oplus}:=\nabla\oplus\nabla’$ is the direct sum connection on $E$
- let $\text{End}(\nabla^{\oplus})$ denote the connection on $\text{End}(E)$ induced by $\nabla^{\oplus}$.
- let $\text{Hom}(\nabla,\nabla’)$ be the connection on $\text{Hom}(L,L’)$ induced by the connection $\nabla,\nabla’$ on $L,L’$ respectively, and likewise let $\text{Hom}(\nabla’,\nabla)$ denote the induced connection on $\text{Hom}(L’,L)$.
With all this out of the way, we have
\begin{align}
R_D&=R_{\nabla^{\oplus}}+d_{\text{End}(\nabla^{\oplus})}A+A\wedge_{\circ}A\tag{lemma 2}\\
&=
\begin{pmatrix}
R_{\nabla}&0\\
0&R_{\nabla’}
\end{pmatrix}
+
d_{\text{End}(\nabla^{\oplus})}
\begin{pmatrix}
0&\beta\\
\alpha& 0
\end{pmatrix}
+
\begin{pmatrix}
0&\beta\\
\alpha& 0
\end{pmatrix}
\wedge_{\circ}
\begin{pmatrix}
0&\beta\\
\alpha& 0
\end{pmatrix}\tag{lemmas 1,3}\\
&=
\begin{pmatrix}
R_{\nabla}&0\\
0&R_{\nabla’}
\end{pmatrix}
+
\begin{pmatrix}
0&d_{\text{Hom}(\nabla’,\nabla)}\beta\\
d_{\text{Hom}(\nabla,\nabla’)}\alpha& 0
\end{pmatrix}
+
\begin{pmatrix}
\beta\wedge_{\circ}\alpha&0\\
0&\alpha\wedge_{\circ}\beta
\end{pmatrix}\\
&=
\begin{pmatrix}
R_{\nabla}+\beta\wedge_{\circ}\alpha& d_{\text{Hom}(\nabla’,\nabla)}\beta\\
d_{\text{Hom}(\nabla,\nabla’)}\alpha & R_{\nabla’}+ \alpha\wedge_{\circ}\beta
\end{pmatrix}.
\end{align}
Rearranging slightly, we get
\begin{align}
\begin{pmatrix}
R_{\nabla}& (R_D)_{L’,L}\\
(R_D)_{L,L’}& R_{\nabla’}
\end{pmatrix}
&=
\begin{pmatrix}
(R_D)_{L,L}-\beta\wedge_{\circ}\alpha
&
d_{\text{Hom}(\nabla’,\nabla)}\beta\\
d_{\text{Hom}(\nabla,\nabla’)}\alpha
& (R_D)_{L’,L’}-\alpha\wedge_{\circ}\beta
\end{pmatrix}.\tag{$*$}
\end{align}
The diagonal equations express the curvature of the subbundles in terms of the ambient curvature and the shape tensor/second-fundamental form. These are the Gauss’ equation for $L,L’$ respectively. The off-diagonal equations express the mixed-components of the ambient curvature in terms of derivatives of the shape tensor; these are the Codazzi equations for $(L,L’)$, $(L’,L)$ respectively.
Specializing to the metric case
Let $(E,\pi,M)$ be a vector with a semi-Riemannian bundle metric $g$, and suppose $L\subset E$ is a non-degenerate subbundle, meaning that the $g$-orthogonal complement $L^{\perp}$ (which is a smooth subbundle) is complementary: $E=L\oplus L^{\perp}$. This is always the case if $g$ is Riemannian. So, the point is that with a bundle metric, we have a canonical choice of complement. Let $D$ now be a metric-compatible connection on $E$, and let $\nabla,\nabla^{\perp},\nabla^{\oplus}=(\nabla\oplus\nabla^{\perp})$ be the induced connections on $L,L^{\perp}$ and $E=L\oplus L^{\perp}$. Finally, let $A=D-\nabla^{\oplus}=D-(\nabla\oplus \nabla^{\perp})$ be the shape tensor.
Lemma 4 (metric-compatible connections, and subbundles)
With notation as just given, we have that
$\nabla$ and $\nabla’$ are metric-compatible
$A$, as an $\text{End}(E)$-valued object is skew-adjoint relative to $g$, i.e $A$ is a 1-form on $M$ with values in the space of skew-adjoint endomorphisms. In particular, $\beta=-\alpha^*$, where $*$ refer to the adjoint relative to $g$.
the curvature $R_D$ is also skew-adjoint-valued.
The first two bullet points are proved in Ivo’s notes linked above (page 26). The third bullet is a simple enough exercise (start with two sections $\psi,\phi$ of $E$, and consider the function $f=g(\psi,\phi)$. On the one hand, the usual second exterior derivative vanishes $d^2f=0$. On the other hand, use metric-compatibility to fully evaluate this… a few lines of algebra will give the desired skew-adjointness); in fact this is well-known from basic Riemannian geometry, particularly the skew-symmetry $R_{abcd}=-R_{bacd}$ (there’s also another skew-symmetry $R_{abcd}=-R_{abdc}$, but that is due to the curvature being a 2-form).
Let us now apply lemma (4) to the Gauss-Codazzi equations $(*)$:
\begin{align}
\begin{pmatrix}
R_{\nabla}& (R_D)_{\perp,\parallel}\\
(R_D)_{\parallel,\perp}& R_{\nabla^{\perp}}
\end{pmatrix}
&=
\begin{pmatrix}
(R_D)_{\parallel,\parallel}+\alpha^*\wedge_{\circ}\alpha
&
d_{\text{Hom}(\nabla^{\perp},\nabla)}\beta\\
d_{\text{Hom}(\nabla,\nabla^{\perp})}\alpha
& (R_D)_{\perp,\perp}+\beta^*\wedge_{\circ}\beta
\end{pmatrix}.\tag{$**$}
\end{align}
- the top left equation is called Gauss’ equation (for $L$)
- the bottom left equation is called the Codazzi equation (or Codazzi-Mainardi equation).
- the top right equation is just the negative adjoint of the bottom left equation, so it is redundant, which is why you’ll never see anyone mention it.
- the bottom right equation was what I called above Gauss’ equation for $L^{\perp}$. But it is more commonly known as Ricci’s equation.
These equations are very concise, because they’re homomorphism-valued differential 2-forms. Meaning that to fully understand them, you should first apply them to two vector fields on $M$ to get of the $2$-form nature, and then evaluate on either a section of $L$ or $L^{\perp}$. Since you asked about Ricci’s equation, let me now elaborate on that case specifically (the others are similar (in fact the Gauss’ equation is identical upon the interchange $L\leftrightarrow L^{\perp}$)).
Let $X,Y$ be vector fields on $M$, and let $\phi$ be a section of $L^{\perp}$. Then, fully evaluating Ricci’s equation, we get
\begin{align}
R_{\nabla^{\perp}}(X,Y)\cdot \phi&=(R_D)_{\perp,\perp}(X,Y)\cdot\phi+ (\beta^*\wedge_{\circ}\beta)(X,Y)\cdot \phi\\
&=[R_D(X,Y)\cdot\phi]_{\perp}+ [\beta(X)^*\circ \beta(Y)-\beta(Y)^*\circ \beta(X)]\cdot \phi.
\end{align}
So, if you now take the inner product with $\psi$, and use the definition of the adjoint, then you get
\begin{align}
\langle R_{\nabla^{\perp}}(X,Y)\cdot \phi,\psi\rangle_g&=
\langle R_D(X,Y)\cdot\phi,\psi\rangle_g
-
\langle\beta(X)\phi,\beta(Y)\psi\rangle_g
+
\langle\beta(Y)\phi,\beta(X)\psi\rangle_g,
\end{align}
which is precisely what you have (modulo notation).
Extreme special case: tangent bundle
A very important special case is of course when you start with a semi-Riemannian immersion $f:(M,g)\hookrightarrow (N,h)$. Now, consider the bundle $E=f^*(TN)$, and pullback the Levi-Civita connection of $N$ to $E$; call this $D$. Let $L=TM$, and $L^{\perp}$ the orthogonal complement of $L$ in $E$, and let $\nabla,\nabla^{\perp}$ be the induced connections.
What I called $\alpha$ above is often called the shape tensor (rather than the full $A$ above which I called the shape tensor). One other good thing to keep in mind is that $\alpha$ being a $\text{Hom}(TM,(TM)^{\perp})$-valued $1$-form on $M$ means it is by definition a bundle morphism $\alpha:TM\to \text{Hom}(TM,(TM)^{\perp})$, which is clearly equivalent to a bilinear bundle morphism $\mathrm{II}:TM\oplus TM\to (TM)^{\perp}$. It turns out that since the Levi-Civita connection is torsion-free, $\mathrm{II}$ is actually symmetric. As a result, for each $p\in M$ and normal vector $\nu\in (T_pM)^{\perp}$, the shape operator $S_{\nu}: T_pM\to T_pM$ defined by $\langle S_{\nu}(x),y\rangle:=\langle\mathrm{II}(x,y),\nu\rangle$ is actually a self-adjoint operator (which in the Riemannian case you can orthogonally diagonalize, by the spectral theorem, giving you the principal curvatures and directions (eigenvalues and eigenvectors) etc). Finally, your $B$ is essentially my $\beta$.
