Consider a symmetric smooth convex body $M$ in $R^4$, and a positive definite symmetric bilinear form $S$, there also exists a vector-valued symmetric linear form $Q$, such that if $S(v, w) = 0$, then $Q(v, w)$ is tangent to the convex body $M$ at $v$ and at $w$.

My question is that: Is this convex body an ellipsoid?

The background of this problem is that I try to generalize a lemma in my master thesis to a higher dimension. The answer is yes for convex body in $R^3$ and here is the proof:

If we consider a smooth convex body in $R^3$, for every 2-dimensional hyperplane that crosses the origin, there always exists a basis $\{e_1, e_2\}$, such that $S(e_1, e_1) = 1, S(e_1, e_2) = 0, S(e_2, e_2) = 1$. This means that $Q(e_1, e_2)$ is tangent to the convex body both at $e_1$ and $e_2$.

Now we consider a rotation with $x = cos\alpha, y = sin\alpha$, since $S(x e_1 - y e_2, y e_1 + x e_2) = 0$, $Q(x e_1 - y e_2, y e_1 + x e_2) = xy(Q(e_1, e_1) - Q(e_2, e_2)) +(x^2 - y^2)Q(e_1, e_2)$ is always tangent at $x e_1 - y e_2$ and $y e_1 + x e_2$ for rotation of every angle $\alpha$. Then $Q$ would map to $-Q$ when angle $\alpha$ goes from $0$ to $\frac{\pi}{2}$. So there exists a rotation such that $Q$ lies in the tangent plane spanned by $\{e_1, e_2\}$. Since for such a pair of vectors $(x_0 e_1 - y_0 e_2, y_0 e_1 + x_0 e_2)$, $Q$ has to be tangent at two points of the convex body restricted in 2-dimensional hyperplane. Since smooth convex body is symmetric, then $Q$ has to be zero.

So we change the basis of $\{e_1, e_2\}$ to the basis after rotation $\{x_0 e_1 - y_0 e_2, y_0 e_1 + x_0 e_2\}$ and denote it as {e_1', e_2'}. Then $Q(e_1', e_2') = 0$. So $Q(e_1', e_1') - Q(e_2', e_2')$ has to be tangent at every point of the 2-dimensional hyperplane. So it is a linear projector that maps the whole convex body to this hyperplane.

Then using Kakutani theorem, this convex body has to be an ellipsoid.

The difficulty of generalizing this proof to a higher dimension lies in that for every 2-dimensional hyperplane that crosses the origin, whether there always exists a basis of vectors $(v, w)$ such that the orbit of rotation of $Q(v, w)$ is a 1-dimensional linear subspace. It is possible that it is always 2-dimensional for arbitrary basis.

For sure, there might exist a different proof or there might exist a counterexample.