Proving Irregularity of {(x, y, z) within R^3 : x^2 + y^2 - z^2 = 0}

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In summary: If the definition is that a surface with equation ##F(x,y,z)=0## is regular if the gradient of the function doesn't vanish, then there isn't much to...argue about?
  • #1
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TL;DR Summary: Here is the exercise: "Prove that {(x, y, z) within R^3 : x^2 + y^2 - z^2 = 0} is not a regular surface".

"Prove that {(x, y, z) within R^3 : x^2 + y^2 - z^2 = 0} is not a regular surface". Any help would be so greatly appreciated!!! Soo confused. We are trying to show that it is impossible to create a diffeomorphism from a neighborhood to an open set (subsequently disproving the conditions of the definition), but we are struggling of course:) Thank you!!
 
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  • #2
labercrombie36 said:
TL;DR Summary: Here is the exercise: "Prove that {(x, y, z) within R^3 : x^2 + y^2 - z^2 = 0} is not a regular surface".

"Prove that {(x, y, z) within R^3 : x^2 + y^2 - z^2 = 0} is not a regular surface". Any help would be so greatly appreciated!!! Soo confused. We are trying to show that it is impossible to create a diffeomorphism from a neighborhood to an open set (subsequently disproving the conditions of the definition), but we are struggling of course:) Thank you!!
Choose a parameterization avoiding ##(0,0,0)## and show that you cannot perform a smooth transition in the third coordinate ##z## from ##z>0## to ##z<0## by using the mean value theorem.
 
  • #3
fresh_42 said:
Choose a parameterization avoiding ##(0,0,0)## and show that you cannot perform a smooth transition in the third coordinate ##z## from ##z>0## to ##z<0## by using the mean value theorem.
Ohhh interesting! Thank you so much for your help. I'm having some trouble connecting your usage of the mean value theorem here--are you picking some (0-#) and (0+#) as z and showing that there is no point between them that matches the slope of the secant line between these points? How does this show there is no "smooth transition" between them? Also, so sorry to ask so many questions--if we show that there is not a smooth function that can be drawn between points Z in the upper cone and points Z in the lower cone, how does this show the shape is not regular? Thank you again!!!
 
  • #4
##(0,0,0)## is the only point where the surface isn't regular.

Assume ##S:=\ldots ## is a regular surface. Then there are open sets ##U\subseteq \mathbb{R}^2\, , \,p:=(0,0,0)\in V\subseteq \mathbb{R}^3## that allow a parameterization ##F:U\rightarrow S\cap V.## Set ##u_0:=F^{-1}(p).## Now choose a path in ##U## from a point ##u_1=F^{-1}(x_1,y_1,-\varepsilon )## to ##u_2=F^{-1}(x_2,y_2,+\varepsilon )## that does not go through ##u_0.##

Finally, conclude that such a path cannot exist. Why?
 
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  • #5
What is your definition of a regular surface? If it is that the gradient of the function, which gives you the equation, is not zero, then it is much simpler.
 
  • #6
martinbn said:
What is your definition of a regular surface? If it is that the gradient of the function, which gives you the equation, is not zero, then it is much simpler.
Yes okay thank you!! I was wondering that! Because this is a cone of two sheets, we know the gradient of the point in which the two cones meet in the origin is zero, thus the surface is irregular at that point. This feels too easy though--am I missing something? Thank you so much!
 
  • #7
labercrombie36 said:
Yes okay thank you!! I was wondering that! Because this is a cone of two sheets, we know the gradient of the point in which the two cones meet in the origin is zero, thus the surface is irregular at that point. This feels too easy though--am I missing something? Thank you so much!
Why does it have to be hard? Where is the exercise from?
 
  • #8
martinbn said:
Why does it have to be hard? Where is the exercise from?
It's from our differential geometry course of exercises, and honestly they've all felt next to impossible, so I feel as though I must be missing something.
 
  • #9
labercrombie36 said:
It's from our differential geometry course of exercises, and honestly they've all felt next to impossible, so I feel as though I must be missing something.
The sketch of proof I gave is a standard approach. You take a path that has to exist in the open sets involved and show that if mapped to the charts leads to contradictions, e.g. to the MVT.

It can't be much easier than that and still rigorous.
 
  • #10
labercrombie36 said:
It's from our differential geometry course of exercises, and honestly they've all felt next to impossible, so I feel as though I must be missing something.
How is a regular surface defined in this course?
 
  • #11
martinbn said:
How is a regular surface defined in this course?
Even if you argue by the Jacobi matrix, you still have to use an arbitrary smooth path. So differentiating alone is not enough.
 
  • #12
fresh_42 said:
Even if you argue by the Jacobi matrix, you still have to use an arbitrary smooth path. So differentiating alone is not enough.
If the definition is that a surface with equation ##F(x,y,z)=0## is regular if the gradient of the function doesn't vanish, then there isn't much to do.
 
  • #13
martinbn said:
If the definition is that a surface with equation ##F(x,y,z)=0## is regular if the gradient of the function doesn't vanish, then there isn't much to do.
This is the definition I found:

##S\subseteq \mathbb{R}^3## is regular, if at any point ##p\in S## there exists an open set ##p\in V\subseteq \mathbb{R}^3## and a open set ##U\subseteq \mathbb{R}^2## with a smooth mapping ##F\, : \,U\longrightarrow \mathbb{R}^3## such that
  1. ##F(U)=S\cap V## and ##F\, : \,U\longrightarrow S\cap V## is a homeopmorphism.
  2. ##\operatorname{rank}D_pF = 2## for any ##p\in U.##
I guess we require a diffeomorphism instead of a homeomorphism.
 
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  • #14
fresh_42 said:
This is the definition I found:

##S\subseteq \mathbb{R}^3## is regular, if at any point ##p\inS## there exists an open set ##p\in V\subseteq \mathbb{R}^3## and a open set ##U\subseteq \mathbb{R}^2## with a smooth mapping ##F\, : \,U\longrightarrow \mathbb{R}^3## such that
  1. ##F(U)=S\cap V## and ##F\, : \,U\longrightarrow S\cap V## is a homeopmorphism.
  2. ##\operatorname{rank}D_pF = 2## for any ##p\in U.##
I guess we require a diffeomorphism instead of a homeomorphism.
The OP has to say what the course covers. Even in this level of generality it is likely they proved that in the case of a surface defined as the level surface of a smooth function the regularity is equivalent to th nonvanishing of the gradient.
 

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