If I have two homogeneous vectors say $v_1 = (x_1, y_1, z_1, w_1)$ and $v_2 = (x_2, y_2, z_2, w_2)$, then is their addition defined if $w_1 \ne w_2$? If $w_1 = w_2 = w$, then I can simply do $v_1 + v_2 = (x_1+x_2, y_1+y_2, z_1+z_2, w)$, but I am having trouble with interpreting result when $w_1 \ne w_2$.
*Edit:*If v1{5, 1, 2, 1} and v2{3, 3, 6, 1} are two homogeneous vectors then correct addition is add{8, 4, 8, 1} and not {8, 4, 8, 2}, which will be equivalent to {4, 2, 4, 1} in 3D space. It makes sense to define such operations on homogeneous 3D vectors when w1=w2 but I guess when w1 != w2 such operations are undefined but I am not sure so I am wondering if anyone can shed some light on it and confirm if my assumption is correct
Best Answer
It cannot be done in a well-defined manner in general.
For, consider the two homogeneous vectors $(0:0:0:1)$ and $(1:0:0:0)$ in $\Bbb P^3(F)$.
Any well-defined manner of associating to these a sum should be invariant under changing representatives, which are $(0:0:0:\lambda)$ and $(\nu:0:0:0)$. The "obvious" option $(1:0:0:1)$ actually corresponds to only one possible element in the span $(\lambda:0:0:\nu): \lambda,\nu \in F$.
If we go back to our intuition about projective space, we see that a "homogeneous coordinate" is actually not but a one-dimensional subspace of the underlying vector space $\Bbb A^4(F)$. Any operation of "adding" these should be expected to result in a two-dimensional subspace: the span written down above.
The reason why the addition works if we have two vectors with the same component non-zero (say the last), is that we can effectively "pretend" we are in $\Bbb A^3(F)$ by the mapping $(x:y:z:w) \mapsto \left(\frac xw,\frac yw, \frac zw\right)$; this is called an affine chart.
For further information, see the Projective space lemma on Wikipedia; investigation of projective space in this context falls under the subject algebraic geometry.