1
$\begingroup$

This is an issue that I'm am trying to solve for a fine-tuning measure in particle physics, but it is purely mathematical. Consider three vectors $\{v_1, v_2, v_3\}$ in $\mathbb{R}^3$. I would like a measure $\Delta$ that captures their volume. This is obviously the parallelepiped with volume $v_1 \cdot (v_2 \times v_3)$. However, I would like this measure to limit to the area spanned as $v_i \rightarrow v_j$ (I don't want to lose information about the structure just because two vectors are parallel). That is, I would like the measure to reduce by a dimension, as two vectors become degenerate. One option is $$\Delta = \Delta_3 + \Delta_2$$ where $$ \Delta_3 = v_1 \cdot (v_2 \times v_3)$$ and $$ \Delta_2 = (v_1 \times v_2) \cdot (v_1 \times v_3) + (v_1 \times v_2) \cdot (v_2 \times v_3) + (v_1 \times v_3) \cdot (v_2 \times v_3)$$ This limits nicely to a 3-volume for orthogonal vectors, and a 2-area for degenerate vectors.

However, I would like this to work for $n$ vectors. Consider four vectors. Applying the measure above doesn't give the correct limit for two unique vectors $v_1, \; v_2 = v_3 = v_4$. One can normalise by $$[(v_1 \times v_2) + (v_1 \times v_3) + (v_1 \times v_4) + (v_2 \times v_3) + (v_2 \times v_4) + (v_3 \times v_4)]^{-1}$$ but this doesn't work for the other choice of degeneracy $v_1 = v_2, \; v_3 = v_4$.

I sense this "total volume" concept is captured somehow by the exterior algebra, but I believe the coefficients of each k-blade overcount degenerate contributions to the volume. Is this measure a known object? Can it be consistently constructed?

(If you need more context, chapter 4 of this work will quickly fill you in).

$\endgroup$
  • $\begingroup$ So what do you want your measure to be in the case when you have $n$ vectors lying in one plane? $\endgroup$ – fedja May 31 '19 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.