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The entries of the covariance matrix of a convex body $K$ are defined as \begin{equation} \label{last} (a_{ij}) = \frac{\int_K x_ix_j}{|K|} - \frac{\int_K x_i}{|K|}\frac{\int_K x_j}{|K|}. \end{equation} We define the isotropic constant of any convex body $K$ using \begin{equation} \label{last2} L^{2n}_{K} :=\frac{\text{Det}\text({Cov{K}})}{|K|^2}. \end{equation} Is $L_K$ scaling invariant? To me that determinant looks scaling invariant. If the isotropic constant is scaling invariant where I can find a proof?

Johan Aspegren
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  • I think I have the scaling invariance in two dimensions, but I cannot work it out in $n$ dimensions. – Johan Aspegren May 17 '21 at 20:08
  • Moreover, I know sources that says it's invariant under invertible affine transformations, so that's the reason I keep trying. – Johan Aspegren May 17 '21 at 20:11
  • Ok, I move this as soon as I know how. – Johan Aspegren May 17 '21 at 20:20
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    What step are you stuck on? You can do a change of variables when computing the constant for $\lambda K$ to see how the covariance matrix scales, then use properties of determinants. – Jose27 May 18 '21 at 07:24
  • @Jose27 has it right. First, assume that $$\int_K x^i,dx= 0$$ and figure out how $a_{ij}$ scales. In other words, how are $a_{ij}(K)$ and $a_{ij}(\lambda K)$, where $\lambda > 0$, related? I recommend that you write your integrals more carefully as, for example, $$\int_K x^ix^j,dx$$ and $$\int_{\lambda K} y^iy^j,dy$$ The $dx$ and $dy$ factors are important, because they remind you how to do the change of variables correctly. – Deane May 18 '21 at 15:41
  • I'm not stuck anymore, thanks. I worked this already for balls, now it's good to work this to $K$ directly. – Johan Aspegren May 18 '21 at 16:30

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