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Let there be an embedding $\phi:S\rightarrow M$ where $M$ is a d-dimensional manifold and $S$ is a codimension-k submanifold. the space of all tangent vectors to the embedded surface $\phi(S)$ is simply the image of the pushforward. If we call it $T\phi(S)$ we can express it as

\begin{equation} T\phi (S)=\{ \vec{t}\in TM | \vec{t}=d\phi \vec{v}\ \forall \vec{v}\in TS \} \end{equation}

where $d\phi$ is the usual pushforward defined from the map $\phi$. The space of 1-forms $\tilde{n}$ normal to $\phi(S)$ is given by

\begin{equation} \begin{aligned} \tilde{n}\cdot \vec{t}&=0\\ \tilde{n}\cdot \big[ d\phi \vec{v}\big]&=0\\ \phi^* \tilde{n} \cdot \vec{v}&=0 \end{aligned} \end{equation}

Since this equation must hold for all $\vec{v}\in TS$, we get that the condition that defines the normal 1-forms is $\phi^* \tilde{n}=0$. Essentially, the cotangent space of normal 1-forms to $\phi(S)$ is the kernel of the pullback.

My problem

Let's use coordinates $y^a=(c^i,x^A)$ for $M$ where $i=1,\dots,k$ and $A=1,\dots,d-k$. The functions $c^i$ represent the constraints that determine the embedding, such that we can use a particular embedding $\phi_0 = y_0(0,\sigma^A)$. In these coordinates, we can use a 1-form $\tilde{\rho}\in T^*M$ and coordinates $\sigma^A$ on $S$ to obtain the condition for the normal 1-forms

\begin{equation} \phi^* \tilde{\rho}=\big(\rho_i \frac{\partial c^i}{\partial \sigma^A} +\rho_A \big)\tilde{d\sigma}^A \end{equation}

which gives us a set of $k$ normal 1-forms $\tilde{n}^i$ given by

\begin{equation} \tilde{n}^i = \tilde{dc}^i-a^i_A \tilde{dx}^A \end{equation}

where $a^i_A$ is only determined through it's pullback $\phi^* a^i_A=\frac{\partial c^i}{\partial \sigma^A}$.

Question

I'm familiar with the normal 1-forms being simply $\tilde{n}^i=\tilde{dc}^i$ which makes sense since $c^i=\text{constant}$ determines the foliation of $M$ by $S$. Why am I getting the extra term $a^i_A$? Is this related to the fact that I'd have to include a metric tensor to uniquely define the normal to $\phi(S)$?

If anyone has any good sources about hypersurfaces, foliations, etc. in the general context of not having a metric tensor, that would be great!

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  • $\begingroup$ I'm a bit confused what the question is - when the $c^i$ are constant on $S$, then $a^i_A = 0$, so this is perfectly compatible with your expectations, no? You just did a bit more general derivation here that includes the case where your surface is not the level set of the $c^i$ and there you get this extra term $\endgroup$
    – ACuriousMind
    Commented Dec 7, 2023 at 19:55
  • $\begingroup$ But it’s the pullback of $a^i_A$ which would be zero in that case, no? The actual $a^i_A$ as a function on M could be any function in the kernel of the pullback. $\endgroup$ Commented Dec 7, 2023 at 22:13

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