2
$\begingroup$

In my lecture notes we have the following:

We consider $(K^3)^{\star}=K^3 \setminus \{(0, 0, 0)\}$ and we define the relation $$(a_1, b_1 , c_1) \sim (a_2, b_2, c_2) \Leftrightarrow (\exists \lambda\in K^{\star}, a_2 = \lambda a_1, b_2 = \lambda b_1 , c_2 = \lambda c_1)$$

This relation is an equivalence relation that divides the set $(K^3)^{\star}$ into equivalence classes.

The classes are lines of $K^3$ that passes through $(0, 0, 0)$.

The equivalence class of $(x, y, z) \in (K^3)^{\star}$ is $[x, y, z]$, that means that $$[x,y, z]=\{(x', y', z') \in (K^3)^{\star} | (x', y', z', ) \sim (x, y, z)\}$$

The set $$\mathbb{P}^2(K)=\{[x, y, z] | (x, y, z) \in (K^3)^{\star}\}$$ is called projective plane over $K$.

Can you explain why the part "The classes are lines of $K^3$ that passes through $(0, 0, 0)$." stand? Why are the equivalence classes lines?

$\endgroup$
  • 1
    $\begingroup$ Strictly speaking, they're lines through the origin with the origin deleted. $\endgroup$ – Hoot Jan 22 '15 at 21:54
0
$\begingroup$

You might want to read the answer I just posted to your similar question:

Lines in the projective plane

In short, you are identifying points in $(K^3)^*$ via the equivalence relation you wrote. This means that points which are equivalent are identified as one equivalence class in projective space. The relation you wrote says that two non-zero points will be considered the same if they are on the same line through the origin. Therefore under this relation, punctured lines are the equivalence classes, and correspond to points in $\mathbb{P}^2(K)$ which is the space of all equivalence classes.

If you are having trouble with equivalence relations in general, you might want to read up on some of the basics to help you understand the projective space construction.

$\endgroup$

Your Answer

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