# Non-commutative distance formula

I am trying to find a distance formula or a method that can give the non-commutative distance between two points in a feature space.

Suppose there are two movies represented in an R^n feature space. Now I want that when I try to find the distance/similarity between these movies using the feature vectors, I get different values with respect to which movie is the reference point i.e.,

Dist(Mov1, Mov2) != Dist(Mov2, Mov1)

I know this is slightly vague, but I am trying to understand my thoughts and would like your help. I found that there is a field called non-commutative geometry, but that seems a little too much and the last resort to dive into. Kindly direct my thoughts.

• What other properties would you want your distance to have? Without more information, we're unlikely to help you much. Here's a thought anyway - for directed graphs, a sensible distance between two points is the shortest path along directed edges. This can easily not be commutative depending on the particulars of the graph. Commented Apr 7, 2021 at 18:36
• The Kullback-Leibler divergence is a non-symmetric (this is the term you should use instead of 'non-commutative') function that is intuitively thought of a distance. It's a function on probability distributions, not points, but still might be of interest to and helpful for you as it shows that sometimes there are important reasons to have a non-symmetric 'distance'. Commented Jun 7, 2021 at 19:18

Take for example the Euclidean distance L2, defined by: $$L_2(x,y) = \left(\sum_{i=1}^{d} (x_i-y_i)^2\right)^{1/2}$$ where $$d$$ is the vector dimension. You can easily add a term $$\alpha \in (0,1)$$ and put more weight on the first term, for example: $$L_2(x,y) = \left(\sum_{i=1}^{d} (\alpha x_i-(1-\alpha)y_i)^2\right)^{1/2}$$ That will certainly be non-cummutative and basically it can be interpreted as putting more weight on the 'reference' vector.