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Jun 29, 2017 at 17:29 comment added Majid @BenMcKay Your answer is definitely an answer to the previous version of my question. Thanks.
Jun 29, 2017 at 17:25 vote accept Majid
Jun 29, 2017 at 17:25
Jun 29, 2017 at 17:24 vote accept Majid
Jun 29, 2017 at 17:24
Jun 20, 2017 at 19:00 history edited Ben McKay CC BY-SA 3.0
explained why this is not the answer anymore
Jun 20, 2017 at 18:46 comment added Majid @RobertBryant Is that ok now?
Jun 20, 2017 at 18:43 comment added Majid @RobertBryant Now I understood what you mean. Let me see how to correct the question.
Jun 20, 2017 at 18:36 comment added Robert Bryant @Majid: Yes, to avoid confusion; by your definition, $\Sigma_2$ is not a sphere, even if $\Sigma_1$ is.
Jun 20, 2017 at 18:29 comment added Majid @Robert Bayant unfortunately I did not get what your comment says. You mean I should use the word "unitary ball"?
Jun 20, 2017 at 18:23 comment added Robert Bryant @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.
Jun 20, 2017 at 18:22 comment added Ben McKay 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.
Jun 20, 2017 at 18:11 comment added Majid What is the problem by containing both 1 and 2?
Jun 20, 2017 at 16:22 comment added Ben McKay If $\pi$ is not an isomorphism, then $\Sigma_2$ contains zero.
Jun 20, 2017 at 16:20 history answered Ben McKay CC BY-SA 3.0