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I'm trying to understand the connection between the notions of linear independence and general position. I have no background in geometry, so first I'll start with what I know and then I'll pose specific questions, please bear with me and correct me at any point.

Let $q$ be a prime power, $d$ be a nonnegative integer, and $V$ be a $(d+1)$-dimensional vector space over the finite field $F_q$ with $q$ elements. For $v \in V$ denote $$[v] = \left\{ cv \mid c \in F_q, c\neq 0 \right\}.$$ Then the collection of symbols $[v]$ can be seen as the points of the $d$-dimensional projective space PG(d,q). Furthermore, for a $(k+1)$-dimensional subspace $S$ of $V$, the set $$\left\{ [s] \mid s \in S \right\}$$ is a $k$-flat of $PG(d,q)$.

From what I read in pg. 19 of these notes I assume that this definition of the notion of "general position" is correct:

We say that $m$ points in $PG(d,q)$ are in general position if they are not contained in any $(m-2)$-flat.

So, to my understanding, the following statement is correct:

The points $[v_1], \ldots, [v_m]$ of $PG(d,q)$ are in general position if and only if the vectors $v_1, \ldots, v_m$ are linearly independent.


My proof. The points $[v_1], \ldots, [v_m]$ are in general position iff they are not contained in any $(m-2)$-flat, which is true iff $v_1, \ldots, v_m$ are not contained in any $(m-1)$-dimensional subspace in $V$. This is the same as linear independence of $v_1, \ldots , v_m$.

My questions are: Is the above statement correct? Are the preceding definitions accurate?

PS. The reason for my confusion is that I've read different definitions for "general position" that I don't understand well, as well as discussions were people explain that general position is not equivalent to linear independence (which I thought my statement above implies). While I'm trying to understand and digest things, it would be very helpful to know if I got the above correctly.

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The definition from the notes is not exactly correct. In a projective space of dimension $d$ (projective dimension), a set of points is usually considered to be in general position if no $d+1$ are contained in a hyperplane.

So for example if we consider the Fano plane $\mathrm{PG}(2,2)$, the four points associated with the vectors $$(1,0,0),\ (0,1,0),\ (0,0,1),\ (1,1,1)$$ are in general position, because no $3$ are contained in any line.

The definition Massimo gives is rather for affine independence; I'm guessing he referred to Wikipedia for the definition where the writing is slightly confusing. Note that according to Wikipedia, a set of at most $d+1$ points in general position is affinely independent (ie vectors linearly independent); however we can have sets of more than $d+1$ points that are in general position, in which case these two concepts are different.

Note that a collection of linearly independent vectors will define points that are in general position; but the concepts are not equivalent, as shown be the above example (though note that of those four vectors that I gave, any 3 of them form a linearly independent set).

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  • $\begingroup$ Thanks a lot. I also saw the definition in wikipedia as well as the definition with the hyperplanes which in fact didn't appear equivalent, and I've been banging my head hard over this, thanks for making it clear. Btw, my bounty expired so I started a new one, I'm allowed to assign it to the answer in a day or so. $\endgroup$ – geo909 Nov 11 '16 at 16:21
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    $\begingroup$ @geo909 I'm glad I was able to make it clear, it is definitely a weird definition. In trying to answer your question, I realized I intuitively knew it (from lots of working with it) but struggled to find words to explain :) $\endgroup$ – Morgan Rodgers Nov 12 '16 at 3:09

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