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Consider a Finsler norm $\varphi$ 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 level surface of Finsler norm $\varphi$ at direction $v$ and $w$.

My question is that: If $Q$ is not identically zero, 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 the level surface of Finsler norm 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 ellipsoid theorem in convex geometry, 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.

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  • $\begingroup$ There's something missing in your description of your problem. For example, you don't assume any connection between $S$ and $M$ or put any condition on $Q$. If you let $S$ and $M$ be arbitrary and just take $Q(v,w) = 0$ for all $v$ and $w$, then your hypotheses are satisfied but $M$ need not be an ellipsoid. (This is a counterexample to your claim even in dimension $3$.) Moreover, since there is no connection between $S$ and $M$ assumed, for the basis $\{e_1,e_2\}$ that you first construct in your 'proof', there is no reason to suppose that $e_1$ and $e_2$ to belong to $M$. $\endgroup$
    – Robert Bryant
    Commented Dec 21, 2022 at 15:59
  • $\begingroup$ @RobertBryant Yes, you are correct. I try to reformulate it but it does not work. It should be below: If $S(v, w) = 0$, then $Q(v, w)$ is tangent to level surface of Finsler norm $\varphi$. It does not matter whether basis lies on convex body $M$. Secondly, yes, $Q$ cannot be identically zero. The point is that if it is zero for a zero measurable set, then by continuity, Kakutani still holds. $\endgroup$
    – threeautumn
    Commented Dec 22, 2022 at 11:51
  • $\begingroup$ @RobertBryant Also, this problem originates from my thesis of isometric embedding $f$ of second-order flat Finsler metric into Banach-Minkowski space. S(v, w) is an analog of second fundamental form, using projection of $d_p^2 f(v, w)$ into some fixed vector $\tau$. $Q(v, w) = d_p^2 f(v, w) - S(v, w) \tau $. First-order flatness implies that $d_p \varphi(d_p^2 f(v, w)) = 0$. So $d_p \varphi(Q(v, w)) = 0$ if $S(v, w) = 0$. We proved that 3-dim second-order flat Finsler manifold embedded into 4-dim Banach-Minkowski space has degenerate second fundamental form. But not sure for higher dim. $\endgroup$
    – threeautumn
    Commented Dec 22, 2022 at 11:57
  • $\begingroup$ Is the convex body $M$ the set of vectors that satisfy $\phi(v)\le 1$? Otherwise, you haven't said how $M$ is related to anything else, so I'm guess that this is what you mean. Let me know if you mean something different. $\endgroup$
    – Robert Bryant
    Commented Dec 22, 2022 at 14:39
  • $\begingroup$ @RobertBryant Yes, it is unit sphere of Finsler norm and its interior. $\endgroup$
    – threeautumn
    Commented Dec 22, 2022 at 20:56

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