# Function for unique volume element

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).

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