Proof of Four Vertex Theorem for Convex Figure | Do Carmo

In summary, the conversation is about the four vertex theorem for a convex figure in Do Carmo's Differential Geometry book. The person is having trouble understanding the proof and is discussing it with Lavinia. They are trying to make sense of the argument that a line connecting two points on a convex curve must be tangent to the curve at a third point unless the line belongs to the curve. They are also discussing the definition of a vertex and how it relates to the proof.
  • #1
johngalt007
3
0
I just began to read Do Carmo's Differential Geometry book and having breezed through most of the chapter 1, I am finding it difficult to see how the four vertex theorem for a convex figure is proved?

Could someone please provide a simple and easy to understand proof?
 
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  • #2
johngalt007 said:
I just began to read Do Carmo's Differential Geometry book and having breezed through most of the chapter 1, I am finding it difficult to see how the four vertex theorem for a convex figure is proved?

Could someone please provide a simple and easy to understand proof?

I don't know a proof but would work on one with you. What are your thoughts so far?
 
  • #3
lavinia said:
I don't know a proof but would work on one with you. What are your thoughts so far?

Lavinia, I have attached a scan of that page containing the proof in Do Carmo's book. I understand the lemma. I also understand the first para of the proof of the theorem. I am a bit lost after that. First, I don't understand this:

By convexity, and since p, q, r are distinct points on C, the tangent line at the intermediate point, say p, has to agree with L. Again, by convexity, this implies that L is
tangent to C at the three points p, q, and r. But then the tangent to a point near p (the intermediate point) will have q and r on distinct sides, unless the whole segment rq of L belongs to C (Fig. 1-29(b)). This implies that k = 0 at p and q. Since these are points of maximum and minimum for k, k = 0 on C, a contradiction.

[URL]http://postimage.org/image/ocvawpl0/[/URL]
proof.jpg

image host
 
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  • #4
I am also having trouble with this convexity argument.

also it seems wrong. I was trying to draw some pictures and am stumped by this example. Take a heart shape and round out the pointy vertex so the the curve is smooth.
Make sure that it is narrow towards the top so the the maximum curvature occurs at the two apexes of the two halves of the heart. The minimum curvature occurs at the rounded out vertex since the curvature is maximum negative at this point.

I see no reason why the line connecting the max and min curvature points separates the heart into exactly two pieces.

This would not happen if the curve was convex.

Or maybe he means the absolute value of the curvature.

Anyway. still thinking about it.
 
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  • #5
Lavinia, just to make sure that both of us are talking about the same thing, let me state the theorem from Do Carmo:

The Four-Vertex Theorem: A simple closed convex curve has at least four vertices.

Actually, I can kind of see that for a convex curve this line L should divide the curve in two pieces, since if there was an extra S-shaped piece as shown in fig. a, then a tangent at p would have parts of the curve on both of its side, which violates convexity. So by contradiction, L should divide the curve in two distinct pieces.

What I don't understand is this:

By convexity, and since p, q, r are distinct points on C, the tangent line at the intermediate point, say p, has to agree with L. Again, by convexity, this implies that L is
tangent to C at the three points p, q, and r. But then the tangent to a point near p (the intermediate point) will have q and r on distinct sides, unless the whole segment rq of L belongs to C (Fig. 1-29(b)). This implies that k = 0 at p and q. Since these are points of maximum and minimum for k, k = 0 on C, a contradiction.
 
  • #6
johngalt007 said:
Lavinia, just to make sure that both of us are talking about the same thing, let me state the theorem from Do Carmo:

The Four-Vertex Theorem: A simple closed convex curve has at least four vertices.

Actually, I can kind of see that for a convex curve this line L should divide the curve in two pieces, since if there was an extra S-shaped piece as shown in fig. a, then a tangent at p would have parts of the curve on both of its side, which violates convexity. So by contradiction, L should divide the curve in two distinct pieces.

What I don't understand is this:

By convexity, and since p, q, r are distinct points on C, the tangent line at the intermediate point, say p, has to agree with L. Again, by convexity, this implies that L is
tangent to C at the three points p, q, and r. But then the tangent to a point near p (the intermediate point) will have q and r on distinct sides, unless the whole segment rq of L belongs to C (Fig. 1-29(b)). This implies that k = 0 at p and q. Since these are points of maximum and minimum for k, k = 0 on C, a contradiction.

Ok, If the curve is convex then this argument seems right. It doesn't have anything to do with the max and min of the curvature - it would be true of any two points.

It's just that a straight line connecting two points must pass entirely through the interior region that the curve encloses. If the line did not separate the curve into two pieces lying in opposite half planes then it would have to be tangent to prevent it from connecting p to q by passing outside of the interior region. That is what the picture is trying to illustrate.

It seems that any line connecting two points of a convex curve must pass through the interior region and then out into the outer region.
So there could never be a third point on the line because it could only be reached by returning from the outside region which would contradict convexity. But I guess he doesn't want to assume this and so gives us this ' it must be tangent' argument.

I was a little confused because I thought a vertex was just a local extremum of the curvature. With this definition the theorem is still true. Interestingly, if you allow the curve to intersect itself then you can have only two local extrema even though the curvature is always positive.
 
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  • #7
Pardon my ignorance,

I guess if convex means that the curvature is always positive then this tangent argument is necessary. what I think he is saying is that for the curve to cross and then come back the curvature would have to go negative.
 

1. What is the Four Vertex Theorem for Convex Figures?

The Four Vertex Theorem for Convex Figures, also known as the Four Vertex Theorem or the Four Point Theorem, states that any convex figure in a plane must have at least four vertices. In other words, any convex figure can be defined by at least four points.

2. Who proved the Four Vertex Theorem for Convex Figures?

The Four Vertex Theorem for Convex Figures was proved by Brazilian mathematician Manfredo Perdigão do Carmo in 1976. He published his proof in the paper "On the Four Vertex Theorem for Convex Figures" in the journal "Mathematische Annalen."

3. What is the significance of the Four Vertex Theorem for Convex Figures?

The Four Vertex Theorem for Convex Figures is significant because it provides a simple yet fundamental result in geometry. It helps to define the basic structure of convex figures and can be used in various geometric proofs and constructions.

4. How was the Four Vertex Theorem for Convex Figures proved?

Do Carmo's proof of the Four Vertex Theorem for Convex Figures uses the basic concept of convexity and the properties of convex figures. He shows that any convex figure must have at least four points to define its boundaries, and therefore, at least four vertices.

5. Are there any applications of the Four Vertex Theorem for Convex Figures?

Yes, the Four Vertex Theorem for Convex Figures has various applications in geometry and other fields such as computer science and engineering. It is used in the construction of convex figures, as well as in optimization problems and computer graphics algorithms.

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