Differential for surface of revolution

[Gene Naden]

O'Neill's Elementary Differential Geometry contains an argument for the following proposition:

"Let C be a curve in a plane P and let A be a line that does not meet C. When this *profile curve* C is revolved around the axis A, it sweeps out a surface of revolution M."

For simplicity, he assumes that P is the xy plane and A is the x axis. He says,
"If the profile curve is $C:f(x,y)=c$ we define a function g on $R^3$ by

$g(x,y,z)=f(x,\sqrt{y^2+z^2})$"

He says it is not hard to show, using the chain rule, that the differential dg is never zero on M. I could not show this except in particular cases:
IF $f=ax+by$ then $dg=adx+bd\rho$
If $f=xy$ then $dg=\rho dx+xd\rho$
If $f=x^2+(y-1)^2$ then $dg=2xdx+(pdy+qdz)$

where $\rho=\sqrt{y^2+z^2}$

How to show it generally?

 

[Orodruin]

He says it is not hard to show, using the chain rule, that the differential dg is never zero on M.

That sounds hard to prove as it seems wrong to me. In fact, the exact opposite should be true. If you have a submanifold that is the level curve of a function g (as in this case) then that function is constant on the submanifold and therefore dg=0 on the submanifold.

 

[Gene Naden]

Perhaps I didn't present the problem as clearly as I should have.

I think dg returns zero when acting on a vector that is tangent to the surface, but nonzero when acting on a vector that is directed away from the y axis..

I noticed a fairly general case: the curve C is defined by $y=h(x)$ then $f(x,y)=y-h(x)$ and $dg=d\rho-h^\prime dx \neq 0$ where $\rho=\sqrt{y^2+z^2}$

 

[Orodruin]

I suggest less thinking and restating the problem exactly as given in the source. To me the formulation ”... on M” suggests that we are only interested in the form dg on the submanifold, on which it is zero. This submanifold only contains its tangent vectors and by definition of the submanifold dg = 0 on it.

 

[Orodruin]

Even if dg is to be considered on all of $\mathbb R^3$, it is not true. Consider $f(x,y) = (y-1)^3$ for $c=0$. You need additional requirements on $f$.

 

[Gene Naden]

Well it is not a problem; it is a passage in the text, apparently intended to illustrate the application of a theorem. The theorem is: "Let g be a differentiable real-valued function on R3, and c be a number. The subset $M:g(x,y,z)=c$ of R3 is a surface if the differential dg is not zero at any point of M."

Here is more of the passage in the text:
"Surfaces of revolution. Let C be a curve in a plane C contained in R3, and let A be a line in P that does not meet C. When this profile curve is revolved around the axis A, it sweeps out a surface of revolution M in R3.

Let us check that M is really a surface. For simplicity, suppose that P is a coordinate plane and A is a coordinate axis - say, the xy plane and the x axis, respectively. Since C must not meet A, we put it in the upper half, y>0, of the xy plane. As C is revolved, each of its points (q1,q2,0) gives rise to a whole circle of points

$(q1,q2\: cos(\nu), q2\:sin(\nu))$ for $0<=\nu<=2\pi$

Thus a point $p=(p1,p2,p3)$ is in M if and only if the point $\bar{p}=(p_1,\sqrt{p2^2+p_3^2},0)$ is in C.

IF the profile curve is $C:f(x,y)=c$, we define a function g on R3 by $g(x,y,z)=f(x,\sqrt{y^2+z^2})$

Then the argument above shows that the resulting surface of revolution is exactly $M:g(x,y,z)=c$. Using the chain rule, it is not hard to show that dg is never zero on M, so M is a surface."

 

[Orodruin]

Then it is just wrong. I provided a counter-example in #5. More explicitly with coordinates $x$ and $\phi$ such that $y = r \cos(\phi)$ and $z = r \sin(\phi)$, for which $g = f(x,r)$:
$$
dg = f_x dx + f_r dr = 0\, dx + 3 (r-1)^2 dr = 0,
$$
since the surface is given by $(r-1)^3 = 0$ and therefore $r = 1$.

 

[Orodruin]

Well it is not a problem; it is a passage in the text, apparently intended to illustrate the application of a theorem. The theorem is: "Let g be a differentiable real-valued function on R3, and c be a number. The subset $M:g(x,y,z)=c$ of R3 is a surface if the differential dg is not zero at any point of M."

Also note that this implication goes in the other direction, i.e., the levels of $g$ are surfaces if $dg \neq 0$. It does not mean that $dg = 0$ if the levels are surfaces (as is the case in my example - $f = 0$ still defines a surface, but on that surface $dg = 0$.

 

[Gene Naden]

So you are saying that the theorem is wrong? The theorem is: "Let g be a differentiable real-valued function on R3, and c be a number. The subset $M:g(x,y,z)=c$ of R3 is a surface if the differential dg is not zero at any point of M."

 

[Orodruin]

No, the theorem is not wrong, but the example is. The "if" is not an "if and only if". In other words, that $dg \neq 0$ implies that $g = c$ is a surface, but that $g = c$ is a surface does not imply that $dg \neq 0$.

 

[Gene Naden]

Help me to understand what part of the example is wrong... that point sets formed by revolution of a curve are not necessarily surfaces? That $dg=0$ for some points? That the author invoked the converse of the theorem, which does not necessarily hold true?

 

[Orodruin]

It is perfectly possible to have $dg = 0$ on a level surface. Therefore the converse is not true and the example is seemingly saying that it is. The author is trying to say $dg \neq 0$ because the surface of revolution is a surface. This is not what the theorem is saying. In particular, it is easy to find an example of a level surface that has $dg = 0$, I presented one such example in #5. In fact, given any level surface $g = c$, we can always construct a new function $h = (g-c)^3$ for which $h = 0$ is the same surface as $g = c$ but has $dh = 0$ because $dh = 3(g-c)^2 dg$.

 

[Gene Naden]

Perhaps you could have another look at #6... he is saying that $dg\neq0$ because of the chain rule. That was my original question: how do you apply the chain rule to get this result. He then goes on to say that because $dg\neq 0$ everywhere, it is a surface.

This is a really important point in the chapter: that point sets formed by revoultion of a curve are necessarily surfaces. If that is not always true then it calls into question more of the material in the book.

 

[Orodruin]

Perhaps you could have another look at #6... he is saying that $dg\neq 0$ because of the chain rule. That was my original question: how do you apply the chain rule to get this result.

You don't, because it is wrong.

As long as you have a curve, the surface of revolution will be a surface. The problem comes when the function is constant in some finite region.

In general (in coordinates $x$ and $r$), you would find that
$$
dg = f_x dx + f_r dr = f_x dx + f_r (r_y dy + r_z dz).
$$
With $r = \sqrt{y^2 + z^2}$, $r_y = y/r$ and $r_z = z/r$ and so, by the chain rule,
$$
dg = f_x dx + f_r(y dy + z dz)/r.
$$
It is perfectly possible to have $f_x = f_r = 0$ and still have a level surface. In order for $dg \neq 0$, you need to have a non-zero gradient for $f$ everywhere on the level surface.

 

[Gene Naden]

I can show that if the curve is defined by $y=h(x)$ then $dg\neq 0$ everywhere

$f(x,y)=y-h(x)$
$g(x,y,z)=f(x,r)=y-h(r)$
$dg=dy-h^\prime dr\neq 0$

I think you said you had a counter-example. Does your counter-example involve a curve that does not contact the y axis? That is a given in the discussion.

 

[Orodruin]

I can show that if the curve is defined by $y=h(x)$ then $dg\neq 0$ everywhere

$f(x,y)=y-h(x)$
$g(x,y,z)=f(x,r)=y-h(r)$
$dg=dy-h^\prime dr\neq 0$

I think you said you had a counter-example. Does your counter-example involve a curve that does not contact the y axis? That is a given in the discussion.

Sorry, but this is wrong. It is only true because you chose your $f$ in a particular way. You would have the same curve if you defined $f = (y-h)^3 = 0$. It does not depend on the curve, it depends on the definition of $f$.

 

[Gene Naden]

You would have the same curve if you defined f=(y−h)3=0f=(y−h)3=0f = (y-h)^3 = 0

Now I am really confused. Which of my three equations is wrong? Or all three?

The way you have defined f, h is a function of x?

When you say "it does not depend on the curve", what is "it"?

Do you have a curve for which $dg=0$ somewhere on the revolution?

 

[Orodruin]

You cannot assume that because $y = h(x)$, your function $f$ is necessarily given by $f = y-h(x)$. For example, $f = (y-h)^3$ has the same level curves but still gives $dg = 0$ for the level surface $g = 0$.

I think you said you had a counter-example. Does your counter-example involve a curve that does not contact the y axis? That is a given in the discussion.

I not only said I had a counter-example. I explicitly gave you the counter-example, which is a cylinder of radius one.

 

[Gene Naden]

Yes I see that dg=0 for your example. Well, thank you for an interesting discussion.

It is unfortunate that we cannot inform O'Neill of his error. He passed away some years ago, according to Wikipedia.

 

[Gene Naden]

So, @Orodruin, how can we patch this up? I think you agree that if you rotate a curve about an axis, that makes a surface. You said my argument

$f(x,y)=y-h(x)$
$g(x,y,z)=f(x,r)=y-h(r)$
$dg=dy-h^\prime dr\neq 0$

is wrong. So how to prove the proposition?

 

[Orodruin]

The point is that you cannot start with $y = h(x)$ and uniquely identify $f$ from it. Letting $f = y - h$ is just one option, but that does not imply that $dg \neq 0$ for any function $f$ that gives $y = h(x)$. In fact, any monotonous function of $y - h$ will have the same level curves and if that function has zero derivative anywhere, then $dg = 0$ for that level surface. Therefore, the inference that $dg \neq 0$ for $y = h(x)$ cannot be made. It can only be made (or not) if you start by defining your function $f$, not if you start from $y = h(x)$.

The easiest way to show that the surface of revolution is a curve is to note that any curve can be parametrised by a single parameter $s$. The surface of revolution is then parametrised by $s$ and an angle $\phi$ and there is your surface. It should be pretty clear that the surface of revolution of a curve $\Gamma$ is just $\Gamma \times S_1$.

Edit: Let's be clear. If you have $y = h(x)$, then one possibility is $f = y - h(x)$. For this possibility, it is certainly true that $dg \neq 0$ and therefore the theorem guarantees that you have a surface. However, it is not true that any $f$ that gives a surface will give $dg\neq 0$.

 

[Gene Naden]

Have not yet come to formulas like $\Gamma \times S_1$ in my studies. What is it, please?

 

[George Jones]

Have not yet come to formulas like $\Gamma \times S_1$ in my studies. What is it, please?

If $A$ and $B$ are sets, what is $A \times B$?

Here, set $\Gamma$ is a set, i.e., a (n image of a) curve, and $S_1$ is a 1-dimensional sphere, i.e, a (perimeter) of a circle. For example, suppose point $p$ is on the curve $\Gamma$. As $p$ is revolved, it traces out a circle.

 

[Orodruin]

More generally, for any two sets $A$ and $B$, $A\times B$ is the set of pairs $(a,b)$ where $a\in A$ and $b \in B$. If $A$ and $B$ are manifolds, there is an obvious way of making $A\times B$ a manifold as well.

 

[mathwonk]

referring to posts # 12 and #21:
in an earlier chapter, (p.20), the author requires the function f defining a curve to have some non zero partial, so (x-1)^3 does not define a "curve" in his sense. he explicitly recalls this 8 lines above his discussion of surfaces of revolution on p. 129. so perhaps both the theorem and the example are ok in the sense intended by the author.

 

[Orodruin]

in an earlier chapter, (p.20), the author requires the function f defining a curve to have some non zero partial, so (x-1)^3 does not define a "curve" in his sense. he explicitly recalls this 8 lines bove his discussion of surfaces of revolution on p. 129.

Well, that is of course a sufficient requirement for leaving $dg \neq 0$. However, it is not a necessary requirement for a level curve to be a curve. It is also not something that has been clear from the statements in this thread.

 

[mathwonk]

agreed, and your examples indeed make clear just what is needed and why the author chose his restrictions on the word "curve" as he did.

 

[mathwonk]

the example of giving different functions that vanish on the same set, such as (x-1) or (x-1)^2 is treated carefully in algebraic geometry, where nowadays one gives more than just the zero locus when describing the "curve" defined by a given function. I.e. an algebraic variety is nowadays thought of as not just a set where some polynomials are zero, but also one gives the sheaf of functions that will be considered as regular on this set. Thus a plane curve now is essentially a pair, a point set, plus a function considered as defining the structure on the set. Thus if the set is the x-axis e.g., then the two curves (x axis, y) and (x axis, y^2) are considered different curves. As a result, the function y is considered as identically zero in the first case, but not in the second case. In the second case a polynomial say of x and y, is not considered to be zero on the fuzzy x-axis unless every term is divisible not just by y, but by y^2. This means the fuzzy x-axis has enough structure that we can not only evaluate functions on it at its points, but we can take their derivatives normal to the x-axis also. Throwing this extra information away is called taking the reduced form of the variety, so in our case the usual x-axis is the reduced form of the fuzzy x axis. For a plane polynomial, taking its reduced form means replacing its irreducible factorization f^r.g^s...h^t, by the reduced factorization f.g...h. so the reduced curve defined by (x-1)^3 would be the curve defined just by (x-1).

anyway, thanks to Orodruin for a very perceptive and nice example.

 

[Gene Naden]

I overlooked the condition on the function that implicitly defines the curve when I presented this problem. Perhaps if I had seen it much of this thread would have been simpler. It is still not clear to me what motivates the condition that the partials be nonzero. And it begs the question:

"If a point set is defined by the condition $f(x,y)=c_f$, where f is differentiable and the partials vanish at some point, is it always possible to define the same point set by $g(x,y)=c_g$ where the partials of g do not vanish?" If not then it seems the condition omits some "curves"

 

[mathwonk]

yes indeed, it omits all curves that cross themselves or that have kinks, such as the union of the x and y axes, or a figure eight or a "cusp". the idea is to include only those curves that have a unique "simple" tangent line at each point*, i.e. to include only one dimensional "manifolds". they are trying to exclude "singular" points. hence what is called a "curve" in o'neill would more generally be called a "smooth curve" or a "non singular curve".

* note that the tangent cone to an algebraic curve is defined by the lowest order term of its equation, so the tangent cone to the cusp y^2 = x^3 at x=0 is defined by y^2 -= 0, which is set theoretically a line, but is a "double line" and not a simple line. hence although the tangent cone to the cusp at (0,0) is only one set theoretic line, it is not algebraically a "simple" line, so the origin is not a non singular point.

The broader meaning of the word "curve" is to refer to any "one dimensional" algebraic set. so "curve" can be considered to refer only to the dimension of the set being one, but is being restricted further by o'neill to refer also to the absence of singular points.

https://www.encyclopediaofmath.org/index.php/Node

https://en.wikipedia.org/wiki/Cusp_(singularity)here is a free book on algebraic curves that requires some abstract algebra:
http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

and here is another highly recommended but cheap one that requires less background but is more old fashioned:

algebraic curves, by robert walker.
https://www.abebooks.com/servlet/Se...n=robert+walker&tn=algebraic+curves&kn=&isbn=

 

[Gene Naden]

Regarding post #30, yes I see that xy=0 gives you the union of the x and y axes, hardly a smooth curve.