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# projective basis

In the parent entry, we see how one may define dimension of a projective space inductively, from its subspaces starting with a point, then a line, and working its way up. Another way to define dimension start with defining dimensions of the empty set, a point, a line, and a plane to be $-1,0,1$, and $2$, and then use the fact that any other projective space is isomorphic to the projective space $P(V)$ associated with a vector space $V$, and then define the dimension to be the dimension of $V$, minus $1$. In this entry, we introduce a more natural way of defining dimensions, via the concept of a basis.

Throughout the discussion, $\mathbf{P}$ is a projective space (as in any model satisfying the axioms of projective geometry).

Given a subset $S$ of $\mathbf{P}$, the *span* of $S$, written $\langle S\rangle$, is the smallest subspace of $\mathbf{P}$ containing $S$. In other words, $\langle S\rangle$ is the intersection of all subspaces of $\mathbf{P}$ containing $S$. Thus, if $S$ is itself a subspace of $\mathbf{P}$, $\langle S\rangle=S$. We also say that $S$ spans $\langle S\rangle$.

One may think of $\langle\cdot\rangle$ as an operation on the powerset of $\mathbf{P}$. It is easy to verify that this operation is a closure operator. In addition, $\langle\cdot\rangle$ is *algebraic*, in the sense that any point in $\langle S\rangle$ is in the span of a finite subset of $S$. In other words,

$\langle S\rangle=\{P\mid P\in\langle F\rangle\mbox{ for some finite }F% \subseteq S\}.$ |

Another property of $\langle\cdot\rangle$ is the exchange property: for any subspace $U$, if $P\notin U$, then for any point $Q$, $\langle U\cup\{P\}\rangle=\langle U\cup\{Q\}\rangle$ iff $Q\in\langle U\cup\{P\}\rangle-U$.

A subset $S$ of $\mathbf{P}$ is said to be *projectively independent*, or simply *independent*, if, for any proper subset $S^{{\prime}}$ of $S$, the span of $S^{{\prime}}$ is a proper subset of the span of $S$: $\langle S^{{\prime}}\rangle\subset\langle S\rangle$. This is the same as saying that $S$ is a *minimal* spanning set for $\langle S\rangle$, in the sense that no proper subset of $S$ spans $\langle S\rangle$. Equivalently, $S$ is independent iff for any $x\in S$, $\langle S-\{x\}\rangle\neq\langle S\rangle$.

$S$ is called a *projective basis*, or simply *basis* for $\mathbf{P}$, if $S$ is independent and spans $\mathbf{P}$.

All of the properties about spanning sets, independent sets, and bases for vector spaces have their projective counterparts. We list some of them here:

1. Every projective space has a basis.

2. If $S_{1},S_{2}$ are independent, then $\langle S_{1}\cap S_{2}\rangle=\langle S_{1}\rangle\cap\langle S_{2}\rangle$.

3. If $S$ is independent and $P\in\langle S\rangle$, then there is $Q\in S$ such that $(\{P\}\cup S)-\{Q\}$ spans $\langle S\rangle$.

4. Let $B$ be a basis for $\mathbf{P}$. If $S$ spans $\mathbf{P}$, then $|B|\leq|S|$. If $S$ is independent, then $|S|\leq|B|$. As a result, all bases for $\mathbf{P}$ have the same cardinality.

5. Every independent subset in $\mathbf{P}$ may be extended to a basis for $\mathbf{P}$.

6. Every spanning set for $\mathbf{P}$ may be reduced to a basis for $\mathbf{P}$.

In light of items 1 and 4 above, we may define the *dimension* of $\mathbf{P}$ to be the cardinality of its basis.

One of the main result on dimension is the dimension formula: if $U,V$ are subspaces of $\mathbf{P}$, then

$\dim(U)+\dim(V)=\dim(U\cup V)+\dim(U\cap V),$ |

which is the counterpart of the same formula for vector subspaces of a vector space (see this entry).

# References

- 1 A. Beutelspacher, U. Rosenbaum Projective Geometry, From Foundations to Applications, Cambridge University Press (2000)

## Mathematics Subject Classification

05B35*no label found*06C10

*no label found*51A05

*no label found*

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