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Suppose that $\pi:(V_1,F_1)\to V_2$ is a linear surjective map, where $V_1$ and $V_2$ are vector spaces and $F_1$ is a Minkowski norm on $V_1$. Let $B_1$ be the unitary ball on $V_1$. Define $B_2:=\pi(B_1)$ and let $\Sigma_2$ be the board of $B_2$.

If $\Sigma_2$ is an indicatrix with respect to some Mnikowski norm on $V_2$?

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Yes, it is. The projection of the convex body enclosed by $\Sigma_1$ is a convex body in the space $V_2$. If you require, as in Finsler geometry, the quadratic convexity, then that is also preserved. In normed spaces this is just the construction of the quotient norm in the quotient of $V_1$ by the kernel of the projection (you can then identify the quotient with $V_2$). This construction comes up in the geometry of isometric submersions in Finsler geometry. See my paper with Duran.

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  • $\begingroup$ Thanks a lot for your answer. I am reading exactly this paper and this is why I asked this question here. Which pert of the paper is mentioning this construction? $\endgroup$
    – Majid
    Commented Jun 20, 2017 at 18:18
  • $\begingroup$ the quotient norm on $V_2$ is $F_2$ then? $\endgroup$
    – Majid
    Commented Jun 20, 2017 at 18:22
  • $\begingroup$ Propositions 2.1 and 2.2. Yes, it's just que quotient norm. $\endgroup$
    – alvarezpaiva
    Commented Jun 20, 2017 at 20:29
  • $\begingroup$ Trying to understand your paper, I have difficulties and some doubts about the parameterization that you are suggesting before Definition 2.2. I mean the vector-valued function $\alpha$. Suppose that $V=R^3$ and the Minkowski norm $F$ is the canonical metric, i.e. $F=||.||$. Define $\alpha: R^2\mapsto S$, where $S$ is the unit sphere, as $\alpha(x,y)=(x,y,\sqrt{1-x^2-y^2})$. Then $F^2(\alpha)=1$ and so each partial derivative of it is zero. In particular $g_{\varphi}(0,0,1)=0$. I am wondering the definition provided for the fundamental tensor is well-defined? $\endgroup$
    – Majid
    Commented Jun 26, 2017 at 13:37
  • $\begingroup$ It's supposed to just be the hessian of $\varphi^2$ in the affine space $V$ evaluated at the point $v$. Let me check what we wrote. $\endgroup$
    – alvarezpaiva
    Commented Jun 26, 2017 at 19:02
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This answers a previous variant of the question above:

Suppose that $V_1=\mathbb{R}^2$ with standard Euclidean metric, and $V_2=\mathbb{R}$. Let $\pi(x,y)=x+2y$. Then $\pi(\Sigma_1)$ contains both $1$ and $2$.

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    $\begingroup$ If $\pi$ is not an isomorphism, then $\Sigma_2$ contains zero. $\endgroup$
    – Ben McKay
    Commented Jun 20, 2017 at 16:22
  • $\begingroup$ What is the problem by containing both 1 and 2? $\endgroup$
    – Majid
    Commented Jun 20, 2017 at 18:11
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    $\begingroup$ You said you wanted $\Sigma_1$ to be the unit sphere. But if $\Sigma_2$ contains both 1 and 2, it is not the boundary of a convex body containing the origin, i.e. not the unit sphere of any Minkowski norm. $\endgroup$
    – Ben McKay
    Commented Jun 20, 2017 at 18:22
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    $\begingroup$ @Majid: The confusion is caused by the use of the word 'indicatrix', which you described as 'the unitary sphere' instead of as the 'unitary ball'. The unitary sphere (which is the usual meaning of 'indicatrix') is not convex, while the 'unitary ball' (whose boundary is the 'unitary sphere') is. $\endgroup$
    – Robert Bryant
    Commented Jun 20, 2017 at 18:23
  • $\begingroup$ @Robert Bayant unfortunately I did not get what your comment says. You mean I should use the word "unitary ball"? $\endgroup$
    – Majid
    Commented Jun 20, 2017 at 18:29

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