# Connection between the $\Bbb R^3$ and $\Bbb R^2$ gradients of a level curve of a hill and the level surface in $\Bbb R^4$ represented by the same hill

I understand (or at least I think I understand) intuitively why the gradient of a surface in $$\Bbb R^3$$ (like a hill) must be perpendicular to the level sets that cut through the hill at various heights and that this vector represents the magnitude and direction of most rapid ascent. I think I also understand why the gradient is normal to a plane tangent to a surface at a particular point. It took me a while to reconcile these two things because I wondered how the gradient could be both of those things at once and the conclusion that I came to was that one gradient was a gradient in $$\Bbb R^2$$ describing the direction of greatest ascent for the hill in $$\Bbb R^3$$ and the gradient that describes the vector normal to a plane tangent to a surface (in this case, the hill) at a particular point is the gradient of the level surface in $$\Bbb R^3$$, the hill being now being considered a level surface to a hyper-surface in $$\Bbb R^4$$ which is described by a different function.

I guess firstly, I would like to know if my intuition is correct so far and then secondly, if it is, I would like to know if the gradient of the hyper-surface (the normal of a plane that's tangent to a particular point on the hill with the hill being considered a level surface to the hyper-surface) has any connection to the two dimensional gradient which lives in the $$xy$$-plane and shows the direction and magnitude of greatest ascent up the hill and is orthogonal to the level set that lives in the $$xy$$-plane.

What my brain was conjuring up was that maybe the $$\Bbb R^3$$ gradient was connected to the $$\Bbb R^2$$ gradient in that the $$\Bbb R^2$$ gradient was the projection of the $$\Bbb R^3$$ gradient/normal to the tangent plane, or perhaps the normal to the tangent plane's "underside" when considering a particular point on the hill. Essentially that the $$\Bbb R^3$$ gradient (or negative $$\Bbb R^3$$ gradient) could be considered to produce a "shadow" projection beneath it on the $$xy$$-plane and that projective shadow would correspond to the $$\Bbb R^2$$ gradient at that particular point on the hill's projection.

I am struggling to explain this in words so I tried to illustrate what I mean on the attached image. I feel like I am not understanding something here, but I don't exactly know what it is. Any input would be greatly appreciated. Thanks.

• This is an excellent question, but unfortunately it is off-topic for MO, which has a deliberately narrow focus on questions arising in original research. Would you like this migrated to math.stackexchange? Commented Dec 14, 2022 at 1:17
• Sure, thanks. Sorry, I didn't realize I wasn't in math.stackexchange to begin with Commented Dec 15, 2022 at 15:26

There's no such thing as the gradient of a surface; rather, it is a function that has a gradient.

A surface (in any dimension, really) can be represented by a function in two different ways.

• As a graph of a function $$g$$ (say, from a domain in $$\mathbb{R}^2$$ to $$\mathbb{R}$$). This is your "height function" representation. This way, to each surface in $$\mathbb{R}^3$$ and each (suitable) choice of coordinates corresponds, locally, a unique function, and to it, its gradient. Then, indeed, a gradient is an $$\mathbb{R}^2$$ vector pointing "uphill" and it has magnitude proportional to the slope rate. Note, however, that the "uphill" direction, and thus the gradient, depends both on a surface and on the choice of coordinates.
• As a level line of a function $$f$$ from $$\mathbb{R}^3$$ to $$\mathbb{R}$$, i.e., the set of points $$x\in\mathbb{R}^3$$ such that $$f(x)=c.$$ In that case, the gradient of that function will be an $$\mathbb{R}^3$$ vector orthogonal to the surface. A simple mind model for that is, indeed, a level line of the function $$g$$ given by an equation $$g(x)=c$$. You can view this level line as a section of a 3D graph, but there's little point in that - it's a flat line living in the plane $$\{x:x_3=c\}$$, one can just forget about ambient space. Similarly, in your question, one can just forget the $$\mathbb{R}^4$$.

A feature of the second approach is that the representation is highly non-unique. Indeed, consider the case $$c=0$$. Then, for any function $$h$$ that does not vanish anywhere, the level surfaces $$\{f(x)h(x)=0\}$$ and $$\{x:f(x)=0\}$$ are the same. Thus, $$fh$$ and $$f$$ represent (via the zero level line) the same surface. But their gradients may be very different! The only restruction is that they are collinear, as they are both orthogonal to the surface. So, in fact, you notion of "3D gradient of a surface", once you define it properly as "3D gradient of a function whose level line is the surface", can refer to any vector field normal to the surface, depending on the choice of the function.