This problem comes from the response of the author of papers.

Consider two convex bodies $A$ and $B$:

$$A= \{X\in \mathcal{S}^4 : \operatorname{tr}(X) = 1, X\succeq 0 \}$$
$$B = \operatorname{conv} SO(3)$$

  1. $\mathcal{S}^4$ is the set of symmetric $4\times 4$ matrices.
  2. $A$ is a $9$ dimensional convex body. ($A$ is symmetric, so $10$ dimensional body, and $\operatorname{tr}(X) = 1$ will decrease one degree of freedom in the diagonal. Just imagine the vectorization of a matrix.) Of course, the extreme points of $A$ are those matrices in $A$ with rank one.
  3. $B$ is also a $9$ dimensional convex body (convex hull of rotation matrices).

I have two questions:

  1. What is the difference between "affinely isomorphic" and "isomorphic"? I try to search some lectures; however, I cannot fully understand it still. Hope for a plain explanation.
  2. How to prove that both convex bodies are affinely isomorphic?
  • 1
    $\begingroup$ Two convex bodies in $\mathbb{R}^n$ are affinely isomorphic, if there exists an affine isomorphism ($x \mapsto Ax + v$, where $A \in \mathrm{GL}(n)$) that maps one convex body onto the other. $\endgroup$ – Deane Yang Sep 12 '16 at 17:54

Consider a 3d-rotation with respect to the axis generated by the unit vector $(a,b,c)$ to the angle $\theta$. Its matrix is $$ M=\pmatrix{\cos \theta+a^2(1-\cos\theta)&ab(1-\cos\theta)-c\sin\theta& ac(1-\cos\theta)+b\sin\theta \\ab(1-\cos\theta)+c\sin\theta&\cos \theta+b^2(1-\cos\theta)&bc(1-\cos\theta)-a\sin\theta\\ac(1-\cos\theta)-b\sin\theta&bc(1-\cos\theta)+a\sin\theta&\cos \theta+b^2(1-\cos\theta)}. $$ Now take a 4d-projection of rank 1 to the line generated by the unit vector $(a\sin \frac{\theta}2,b\sin \frac{\theta}2,c\sin \frac{\theta}2,\cos \frac{\theta}2)$. Its matrix elements are $\cos^2\frac{\theta}2,a^2\sin^2 \frac{\theta}2$, $ab\sin^2 \frac\theta2$, $a\sin \frac\theta2 \cos \frac\theta2$ and so on. All of them are affine functions of the matrix elements of the above matrix $M$ and viceversa. It follows from the formulae $\cos\theta=2\cos^2\frac{\theta}2-1$, $1-\cos\theta=2\sin^2 \frac\theta2$, $\sin\theta=2\sin \frac\theta2\cos \frac\theta2$. These affine functions define the affine map which maps $A$ onto $B$.

  • $\begingroup$ What does "ace" mean here? $\endgroup$ – sleeve chen Sep 12 '16 at 18:12
  • $\begingroup$ Ah, axis, of course. Invariant line of a rotation. $\endgroup$ – Fedor Petrov Sep 12 '16 at 18:23
  • $\begingroup$ Just ask one more trivial question. The affine maps in your answer hold for extreme points in $A$ and $B$. It also holds for other points in $A$ and $B$ just because the affine map preserves convexity? $\endgroup$ – sleeve chen Sep 12 '16 at 20:34
  • $\begingroup$ Yes, of course, affine image of the convex hull of a set is the convex hull of its image. $\endgroup$ – Fedor Petrov Sep 12 '16 at 20:56

Here is a definition of affinely isomorphic for convex polytopes:


Ziegler, Günter M. Lectures on polytopes. Vol. 152. Springer Science & Business Media, 1995. p5.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.