Cutting out submanifolds with "orthogonal" functions

Question

Let $Z$ be an embedded manifold in some $\mathbb{R}^M$. Then locally $Z$ is cut out by independent functions $(g_1, \ldots, g_l): \mathbb{R}^M \to \mathbb{R}^l$, where $l = \operatorname{codim} Z$. That is, for every $z \in Z$, there exists a neighborhood $\tilde{U}$ of $z$ in $\mathbb{R}^M$ such that $\tilde{U} \cap Z = \bigcap_{i=1}^l g_i^{-1}(0)$. (See for example page 24 in Guillemin and Pollack.) Here, independent means that $d(g_1)_z, \ldots, d(g_l)_z$ are linearly independent in a neighborhood of $z$. My question is:

Can the functions $g_1, \ldots, g_l$ be chosen so that $d(g_1)_z, \ldots, d(g_l)_z$ are actually orthogonal?

My idea is that we can first apply Gram-Schmidt to $\{d(g_i)_z\}$ first to obtain orthogonal vectors $\{\widetilde{d(g_i)_z}\}$, and then take $\tilde{g}_1, \ldots, \tilde{g}_l$ to be the corresponding linear combinations of $\{g_i\}$ with differentials $\{\widetilde{d(g_i)_z}\}$. Then I believe $\{\tilde{g}_i\}$ still cuts out $Z$. However, being orthogonal at one point $z$ generally does not mean that the differentials are orthogonal in a neighborhood of $z$.

In fact, I'm interested in the special case where $Z$ is actually a submanifold of some other embedded manifold $Y \subseteq \mathbb{R}^M$. Then there exists independent functions $(g_1,\ldots,g_l)$ locally cutting out $Z$ such that $(g_{k+1},\ldots,g_l)$ cut out $Y$ locally. In this case, is it possible to make $\{d(g_1)_z, \ldots, d(g_k)_z\}$ orthogonal to $\{d(g_{k+1})_z, \ldots, d(g_l)_z\}$?

Answer 1

What a great question! As you realized, you can easily arrange this at any given particular point. But to achieve what you desire locally will require examining the proof of the Tubular Neighborhood Theorem (see Guillemin and Pollack, p. 76, #16). You want to impose some differential geometric conditions that the diffeomorphism from the normal bundle to a neighborhood of the submanifold preserve orthogonality along the submanifold. I believe that if you check the details of the proof Guillemin and Pollack outline, you'll end up with that condition. (Otherwise, it can be achieved by following geodesics in your ambient manifold $Y$.)