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

  • 1
    $\begingroup$ There is no general meaning to adding two 'homogeneous vectors', so you need to provide more context. What are you actually trying to do? $\endgroup$ – Rhys Apr 29 '13 at 16:08
  • $\begingroup$ I am working on writing a geometric kernel library in c++ and I was working on Vector classes. I realized that I need to treat 4D vectors different than 3D Vectors represented in their homogeneous form {x, y, z, w} and bumped into problem of defining addition of two homogeneous vectors with different w component. $\endgroup$ – Prasad Dixit Apr 29 '13 at 16:42

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.

| cite | improve this answer | |

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.