So I have this one problem on trying to show two surfaces are diffeomorphic, and I cant seem to show yet. Could anyone help me show that the surface denotes by $\sigma = \{y \in \mathbb{R}^3 : y \bullet By = 1\}$ where $y$ is a column vector.

Question: Given that the matrix $B = A^T D A$, where $(A^TA=I)$ and $D$ is the $3 \times 3$ diagonal matrix with eigenvalues $\lambda_i , i=1,2,3$. How can we show that if $\lambda_i >0$ $;\forall i$, then the surface $\sigma$ is diffeomorphic to the unit sphere?

Further question: I also thought about it but what is the condition for $\lambda_i$ when it is diffeomorphic to say the cylinder with radius $1$ in $\mathbb{R}^3?$

I really need some good help with diffeomorphisms with surfaces as I get the concepts but do not have enough practice or exposure to show and solve problems with this concept included. I would appreciate any help.


What happens when you replace $D$ with $D(t) = \begin{pmatrix} \lambda_1 + (1-\lambda_1)t & 0 & 0 \\ 0 & \lambda_2 + (1-\lambda_2)t & 0 \\ 0 & 0 & \lambda_3 + (1-\lambda_3)t \end{pmatrix}$, especially at $t=0$ and $t=1$?

For your follow-on question, what happens when exactly one $\lambda_i$ is forced to be zero? Then the (1-dimenasional) subspace spanned by the corresponding eigenvector in $A$ is mapped to zero, so does not enter into the quadratic (distance) form $y^\mathrm{T}By$. (And if you go further, letting that one $\lambda_i < 0$, then you get a hyperboloid of one sheet, which is, of course, diffeomorphic to that infinite cylinder.)

In fact, it could be useful to you to recall that $A^\mathrm{T}DA$ is an eigendecomposition to eigenvalues $\lambda_i$ and eigenvectors (the columns of $A$), then think about $$ y^\mathrm{T}By = y^\mathrm{T}A^\mathrm{T}DAy = (Ay)^\mathrm{T}D(Ay) \text{.} $$ $A$ just changes the basis and $D$ scales the column vectors of $A$ to make them appear longer or shorter than they are. So $y^\mathrm{T}By$ is a distorted version of Euclidean distance in the basis given by the columns of $A$ and with the various components contributing to the "distance" unequally.

  • $\begingroup$ Thank you for your response. So why is it necessary we consider the form $D(t)?$ I am still not entirely grasping the entire picture of your help. $\endgroup$ – Aurora Borealis Feb 23 '18 at 4:48
  • $\begingroup$ It is not necessary, but it is useful. When struck, try deforming the problem. (Perhaps especially for diffeomorphisms.) $\endgroup$ – Eric Towers Feb 23 '18 at 4:49
  • $\begingroup$ From what I understand if I want to show say a $\Phi$ is a diffemorphism, then the inverse $\Phi^{-1}$ has to be smooth. With normal surfaces without the vector restriction I think I can by just finding some parametrisation and using the inverse function theorem, but in this case i still dont see what we are supposed to show? Is it possible we express the coordinates of $\mathbb{S}^2$ in terms of the vectors in $\sigma?$ Then find the $\det$ of the Jacobian of this mapping and apply IVT? $\endgroup$ – Aurora Borealis Feb 23 '18 at 4:52
  • $\begingroup$ Is there any chance that $D(t)$ gives you a bijection between points of the unit sphere and points of $\sigma$ as well as points of other deformations of the unit sphere? Maybe you can do something with that -- it is giving all of this from a linear function... $\endgroup$ – Eric Towers Feb 23 '18 at 4:54

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