Why Can't the Positive Cone Be a Submanifold in R^3?

  • Thread starter Thread starter Pietjuh
  • Start date Start date
  • Tags Tags
    Cone
Pietjuh
Messages
75
Reaction score
0
For a homework assignment i was asked to proof that the positive cone {x^2 + y^2 = z^2, z>= 0} cannot be a submanifold of any dimension of R^3.

It apparently goes wrong at the origin. I guess it's because you can't really speak of a tangent space at that point. So I tried to prove by contradiction you can't have a tangent space at that point. But I couldn't really arrive at a contradiction :confused:

Could someone give me a hint? :smile:
 
Physics news on Phys.org
Pietjuh said:
For a homework assignment i was asked to proof that the positive cone {x^2 + y^2 = z^2, z>= 0} cannot be a submanifold of any dimension of R^3.
It apparently goes wrong at the origin. I guess it's because you can't really speak of a tangent space at that point. So I tried to prove by contradiction you can't have a tangent space at that point. But I couldn't really arrive at a contradiction :confused:
Could someone give me a hint? :smile:
Try finding an equation for a tangent plane at the origin. That's the only thing I can think of (which is basically what you already had in mind!)

Alex
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top