Dimension of a linear space in $\mathbb{P}^{n}$ (Schubert Calculus)

Question

I am reading about Schubert calculus and have come across this definition of a linear space:

A linear space $L$ in $\mathbb{P}^{n}$ is defined as the set of points $P = (p(0), p(1), \ldots, p(n))$ of $\mathbb{P}^{n}$ whose coordinates $p(j)$ satisfy a system of linear equations $\sum_{j=0}^{n} b_{\alpha j}p(j) = 0$, where $\alpha = 1, \ldots, (n-d)$. [Here, I get confused: ] We say that $L$ is $d$-dimensional if these $(n-d)$ equations are independent, that is if the $(n-d)\times (n+1)$ matrix of coefficients $[b_{\alpha j}]$ has a nonzero $(n-d) \times (n-d)$ minor.

I would have thought that this space is of dimension $(n+1)$. Maybe because each equation is zero and the $p(j)$ satisfy such an equation, in each combination there is an "extra" vector by dependence. Thus, this leaves $(d+1)$ independent vectors, but not $d$ (this may have to do with the fact that we are in $\mathbb{P}^{n}$).

Any help is appreciated.

Answer 1

Think of how $\mathbb P^n$ is defined:

$$\mathbb P^n = \mathbb K^{n+1}\setminus \{0\} /\sim, $$

where $\vec x \sim \vec y$ if they are proportional. Now the matrix $B$ is nondegenerate, so the linear map $B : \mathbb K^{n+1} \to \mathbb K^{n+1}$ has a $d+1$ dimensional Kernel $\tilde L$.

However, when you take quotient $\sim$, then $L = \tilde L\setminus\{0\} /\sim$ has one less dimension (The reason is the same as why $\mathbb P^n$ is $n$-dimensional instead of $n+1$-dimensional).

Note that you might think that it is an (established) abuse of languages to call that $L$ linear: $\mathbb P^n$ is not a vector space, after all.

Answer 2

The reason for the "off-by-one" error you are seeing is that in projective space, points are really given by equivalence classes of coordinates. Therefore the point $$ (p_0,p_1,p_2,\ldots,p_n) $$ is in the same equivalence class as $$ (\lambda p_0,\lambda p_1,\lambda p_2,\ldots,\lambda p_n), $$ for any $\lambda\not=0$.

Answer 3

I figure I'll add my two cents as well, even though the other answers are also good. I agree that it is confusing that projective dimension is always one less than the vector space dimension. You have to be careful to specify which you mean. However there is a good reason for this double usage of the term, using geometric intuition.

We can view a vector space and its subspaces geometrically; the zero vector is included in every subspace. So one nonzero vector determines a line ($1$-dimensional), and two linearly independent vectors form a plane ($2$-dimensional). Moving to a projective setting, one nonzero vector now determines only a point ($0$-dimensional) while two linearly independent vectors form a line ($1$-dimensional). We need three linearly independent vectors to give a triangle/plane ($2$-dimensional) in this setting. So really the dimension can be interpreted in a geometric way.

Of course when we deal with a projective space in analytic geometry, we have the complication that we need to keep track of both systems simultaneously: we are representing geometric objects in the projective setting with geometric objects in a vector space setting. The same object has a different geometric dimension, depending on the point of view (i.e. a $(d+1)$-dimensional vector subspace corresponds to a $d$-dimensional projective subspace).