# Confusion about Surfaces, Normal Vectors in $R^n$

I think I'm having some confusions about some elementary concepts.

I'm studying the book "Calculus of Several Variables" by Lang. I realize the book is not completely rigorous but the following left me questioning some things, while I was trying to understand the author's argument about Lagrange multipliers.

See the screenshots below. The author argues that the gradient vector is the unique normal/orthogonal vector at a surface because it's normal to every differentiable curve passing through that surface.

But what if your "surface" was a line in $R^3$ - say the x axis? Then you could have two different orthogonal vectors that are normal to the surface (i.e. y and z axes). But I realize the x-axis can not be put in the form $g(x, y, z) = c$.

So what's going on here? Do points satisfying a constraint such as $g(x, y, z) = c$ always have one unique orthogonal vector at every point unlike a line in $R^3$ (in the sense the vector is orthogonal to any curve containing those points)? Can this be proven?

Is a line not a "surface" in $R^3$?

I hope what I'm saying makes sense.

• Is it true that both the $\;y,\,z\;$ axis are orthogonal to any curve through the origin and, thus, to the $\;x\;$ axis there? Because that's what has to be true. Commented Feb 12, 2018 at 11:24
• I don't get your question. Through the x-axis, there's only one curve - namely the x-axis. And the y- and z- axes are orthogonal at every point to this curve. Commented Feb 12, 2018 at 11:28
• I read that part of Lang and I understand now what you meant with the definition of the gradient. Commented Feb 12, 2018 at 11:34

The word "surface" in this context is defined in the first sentence of your screenshot. It is a result of differential geometry that such a set of points has exactly the $2$-dimensional shape that you have in mind when you hear the word "surface". It cannot be a line. See this question: in your case, since the target space is $\mathbb{R}$, $\text{grad}\ f(X)$ being non-zero is equivalent to d$_Xf$ being surjective, hence the preimage is locally a $3-1=2$-manifold.

Note also the gradient cannot be uniquely defined by the property you mention. There are infinitely many normal vectors to a surface at a given point (namely all multiples of the gradient).

I am quite surprised by your sentence "I realize the book is not completely rigorous". What do you mean? What is not rigorous?

Edit: A linear form is what you call a linear transformation in the special case where the codomain has dimension $1$. Say that the domain has dimension $n$. By the rank-nullity theorem, the kernel has to have dimension either $n$ (if the map is trivial) or $n-1$. Note that there is only one direction that is orthogonal to an $n-1$-dimensional subspace - and there is one unique orthogonal vector $v_0$ such that your linear form is the inner product $v\mapsto \left< v, v_0\right>$. That vector is nothing but the $1\times n$ matrix that represents the linear map.

In your case, when the linear form is the differential of a smooth function at a given point $p$, and provided that $\operatorname{d}_pf$ is surjective (hence has $n-1$-dimensional kernel), this one unique vector is what you call $\operatorname{grad} f$.

By smoothness, $\operatorname{d}f$ will stay surjective in a neighbourhood of $p$ and the gradient will therefore exist. In such a neighbourhood, $f$ is close to a linear map and its level sets $f^{-1}(c)$ look like $n-1$ dimensional affine subspaces, in the sense that they are affine subspaces up to a smooth change of variables.

This is known as the submersion theorem - see for instance these lecture notes, Theorem 1.

• Thanks for answering my question. Regarding the book's rigor, Lang himself says the book is not too concerned with theory - it wasn't meant as a slight. As far as vectors go, I meant that vectors that are scalar multiples of each other are equivalent. Can you prove or provide a reference for the claim that grad f is the only unique normal vector for every point (in the sense of being orthogonal to every curve) for a surface of the form $f(\vec x) = c$? The question you link to is above my level. If you feel the answer requires significant diff geo, I may pass. Commented Feb 12, 2018 at 20:22
• And as far as rigor goes, I don't think the author answers this question - but again I suspect the answer may take us too far afield. Commented Feb 12, 2018 at 20:28
• @AnlamK Here is rough explanation: by definition the gradient is the vector that represents the linear form $df$ at a given point. Along a curve lying in $f^{-1}(c)$, $f(\gamma(t))$ is constant and its derivative is $0$, therefore the tangent vector of such a curve at the given point is orthogonal to the gradient. Commented Feb 13, 2018 at 19:52
• Yes, the fact that any curve going through the surface is orthogonal to the grad follows easily from the chain rule; as you point out. My question is whether grad is the only unique such vector (ignoring scalar multiples) and why. Your original answer has some clues as to why but it's opaque to me. Commented Feb 14, 2018 at 12:10
• Thanks - if you can explain what a linear form is, what a kernel of a liinear form is (kernel for me only means null space of a linear transformation), why it has dim n-1 and expand on your answer and/or perhaps point to some additional references (is this the domain of differential geometry?), I'll accept your answer. Commented Feb 14, 2018 at 14:41