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The Gauss, Codazzi-Mainardi and Ricci equations are the three fundamental equations of a Riemannian immersion.

The Gauss equation inputs 4 tangent vectors, the Codazzi-Mainardi inputs 3 tangent vectors and 1 normal vector and the Ricci equation inputs 2 tangent vectors and 2 normal vectors.

Are there analogues for these equations which input 3 normal vectors and 1 tangent vector or 4 normal vectors?

AlexInorbit
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1 Answers1

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Our goal is to understand the ambient curvature in terms of the curvatures of submanifolds. You always need 2 tangent vectors (in the correct slots); this is because curvature is an endomorphism-valued $2$-form. The other two slots can be filled in with tangent/normal vectors, and this basically corresponds to a block decomposition (relative to the tangent, normal direct sum decomposition) of the curvature endomorphism. In order to fully do this, one has to of course introduce the shape tensor/ second fundamental form.

Conceptually the different roles of the various tangent/normal vectors can be clarified in the vector bundle setting, where you start with a vector bundle $(E,\pi,M)$ together with a connection $\nabla$. Fix a direct sum decomposition $E=L\oplus L’$. The most natural questions then are:

  • How does the connection $\nabla$ on $E$ induce connections on $L,L’$ respectively? (Answer: just differentiate with $\nabla$, then project onto the subbundle)
  • How do these two connections differ from the ambient connection $\nabla$? (Answer: the shape tensor)
  • How are the curvatures of these various connections related, and more specifically what is the block decomposition of the curvature $\nabla$ on $E$ relative to the decomposition $E=L\oplus L’$? (Answer: Gauss-Codazzi equations, namely Gauss’ equation is the diagonal terms and Codazzi’s are the off-diagonal ones).

See Ricci Equation $(\overline R(U,V)X,Y) = (R^\nabla(U,V)X,Y) - (B_U X, B_V Y) + (B_V X, B_U Y)$ for slightly more details on this.

peek-a-boo
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  • Presumably instead of decomposing the covariant derivative with respect to a tangent vector of tangent and normal vectors, i.e. $\nabla_X Y$ and $\nabla_X \nu$, we can decompose it with respect to a normal vector, i.e. $\nabla_{\nu} Y$ and $\nabla_{\nu} \xi$?, for tangent vectors $X,Y$ and normal vectors $\nu,\xi$? – AlexInorbit Jun 17 '24 at 14:24
  • @AlexInorbit suppose $(M,g,\nabla)$ is a Riemannian manifold equipped with its Levi-Civita connection, and $S\subset M$ a submanifold. If $Y$ is merely a vector field tangent to $S$, then for a normal vector $\nu\in (T_pS)^{\perp}$, the quantity $\nabla_{\nu}Y$ is not even defined, nor is it possible to provide a well-definition. Just imagine you have a vector field defined on $\Bbb{R}^2\times{0}$; there are multiple ways to extend it (even locally) in the normal direction, so the normal derivative of such an object is not well-defined. – peek-a-boo Jun 17 '24 at 15:03
  • Similarly, asking about $\nabla_{\nu}\xi$ makes no sense, because $\xi$ is again a map $S\to (TS)^{\perp}$, so we can only differentiate along directions tangent to the domain, i.e we can only consider for each $p\in S$ and $X_p\in T_pS$, the covariant derivative $\nabla_{X_p}\xi$. – peek-a-boo Jun 17 '24 at 15:05
  • So, I think the ‘meta’ to keep in mind here is that the directions along which you can differentiate must always be tangent to the domain manifold (in this case $S$). The target space can be any vector bundle ($TS$ in the case of tangent vector fields, $(TS)^{\perp}$ in the case of normal vector fields, $S\times\Bbb{R}$ for ‘scalar functions’ etc) as long as that vector bundle carries a connection. If you look at the link, you’ll see throughout that I fixed the base manifold $M$ and considered various vector bundles over it. – peek-a-boo Jun 17 '24 at 15:07