0
$\begingroup$

Let $\alpha_1, \dots, \alpha_n$ be unit vectors in some vector space $V = R^d$. For any permutation $\pi: [n] \rightarrow [n]$, we can form the Gram-Schmidt orthogonal bases $\beta_{\pi,1}, \dots, \beta_{\pi,n}$ as $\beta_{\pi, 1} = \alpha_{\pi(1)}, \beta_{\pi,2} = \alpha_{\pi(2)} - (\alpha_{\pi(2)} . \beta_{\pi,1}) \beta_{\pi,1}$ etc.

Now, for any permutation $\pi$ define the set $X_{\pi} = \{ x \in V \mid (x.\beta_{\pi,i})^2 \geq \lambda \text{ for $i = 1, \dots, n$} \}$.

And define the set $X = \cup_{\pi} X_{\pi}$.

What does this set look like? Can one bound its volume?

Obviously, you can sum the volume of $X_{\pi}$ over $\pi \in S_n$. However, intuitively, it seems that the worst case for the volume of $X_{\pi}$ would be when the $\alpha$ vectors are orthonormal --- otherwise, the $\beta$ vectors become smaller. But in that case, $X_{\pi}$ does not depend on $\pi$ at all. So it seems like you should be able to avoid this extra factor of $n!$

$\endgroup$
12
  • $\begingroup$ What do you mean by $x.\beta^\pi_i$? The $i$th component of $x$ with respect to the orthogonal basis? $\endgroup$ Mar 5, 2014 at 22:42
  • $\begingroup$ @Andres, I mean the inner product. $\endgroup$ Mar 5, 2014 at 22:50
  • $\begingroup$ You probably meant $\leq$ rather than $\geq$ in the definition of $X_\pi$. $\endgroup$ Mar 5, 2014 at 22:54
  • $\begingroup$ @Mariano, no I mean $\geq$. It's not as nice geometrically I know. $\endgroup$ Mar 5, 2014 at 22:55
  • $\begingroup$ But then $X_\pi$ is the complement of a parallelopiped, which has infinite volume. $\endgroup$ Mar 5, 2014 at 22:56

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.