Suppose $G : \mathbb R^n \to \mathbb R$ is a differentiable function and the level set $L = \{x \in \mathbb R^n: G(x)=0 \}$ is an $n-1$ dimentional surface near the point $x_0 \in \mathbb R^n$. Then the gradient $\nabla G(x_0)$ determines the subspace of all vectors perpendicular to $L$ at $x_0$.
In general $L$ might be some $m<n$ dimensional surface near $x_0$. In that case $\nabla G(x_0)$ does not determine he subspace of all vectors perpendicular to $L$. For example $G(x,y,z) = x^2 + y^2$ has the $z$-axis as a level set. At each point the normal vectors are all vectors in the $xy$-plane. While $\nabla G = (x,y,0)$ is in that plane it does not determine it.
What is the formula for the set of perpendicular vectors at point of a level surface, if we impose no restrictions on the dimensions?
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