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A convex body $K$ in $\mathbb{R}^n$ is in isotropic position if, for all vectors $x \in \mathbb{R}^n$, we have

$$\frac{1}{\mathrm{vol}(K)}\int_K \langle x, y \rangle^2 dy = \|x\|^2.$$

My question: The geometric explanation of isotropic position of a convex body?

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    $\begingroup$ It is also assumed that the center of mass is at the origin. Isotropic position implies that the uniform probability distribution on $K$ has mean $0$ and covariance matrix equal to the identity. Any body can be put into isotropic position by a translation composed with a linear transformation. The integral on the left side is also related to the moment of inertia (see section on the inertia tensor here: en.wikipedia.org/wiki/Moment_of_inertia $\endgroup$
    – Deane Yang
    Commented Mar 11, 2017 at 3:56

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This definition is related to the moments of mass of a convex body.

The zero-th moment in this case is the volume, the first moment is the centre of volume, and the second moment, given by:

$$\int_K \langle x, y \rangle^2 dy$$

represents the amount of effort required to rotate the body, or the rotational interia.

A body is in an isotropic position if any vector that it is to be rotated about requires as much effort for the rotation as proportional to its magnitude.

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  • $\begingroup$ Can you give some concrete examples about convex bodies and their isotropic positions? Such as, a sphere, a ellipsoid or a prism? Thank you! $\endgroup$
    – Epsilon
    Commented Mar 13, 2017 at 1:57

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