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?



How to show the two convex bodies are affinely isomorphic?
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 $B$.


