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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 $v$ and at $w$.

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.

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.

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 $v$ and at $w$.

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 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 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.

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 $v$ and at $w$.

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.

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 $v$ 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 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.

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 $v$ and at $w$.

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 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 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.