My ultimate goal is to learn some algebraic geometry with the more concrete immediate goal of understanding things like how $\mathbb R\mathrm P^2$ is embedded in $\mathbb C\mathrm P^2$ or how $\mathbb C\mathrm P^2$ can be realised as the symmetric product of $\mathbb C\mathrm P^1$.

It seems to me that a necessary prerequisite is a solid understanding of projective geometry.

My problem is that I find it hard to develop some intuition about it: I keep mentally translating complex lines into real planes and try to visualise it - but it is not getting me very far...

I have been reading these notes. While I have no troubles following them, I cannot really see things like two projective lines meet at a point.

So my question is: what would you recommend as far as the way of thinking about projective spaces and geometry is concerned? Should I first develop more confidence with computations and proofs and build intuition on that? Should I stop translating from complex to real coordinates?

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    $\begingroup$ Doing algebraic geometry over the complex numbers, one often draws real pictures just to improve the mental picture, halving the dimension in the process. For example the conic surface $x^2 + y^2 + z^2 + w^2$ in $\mathbb{C}P^3$ is 4 dimensional as a real manifold (and embedded in a 6-dimensional space!), so one uses the same polynomial and draws the surface in real (affine) space. When explaining projective complex geometry to non-mathematicians i sometimes joke that a 2-dimensional sphere ($\mathbb{C}P^1$) is a line to me. $\endgroup$ – Joachim Mar 28 '13 at 10:26
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    $\begingroup$ You can google "Kummer surface" for example to see an example.. $\endgroup$ – Joachim Mar 28 '13 at 10:29
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    $\begingroup$ Also, you might want to check out math.stackexchange.com/questions/85394/… .. Good luck with the studying! $\endgroup$ – Joachim Mar 28 '13 at 10:29
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    $\begingroup$ Thanks Joachim! When you joke about $\mathbb{C}P^1$ being a line, you mean a line in $\mathbb{C}P^n$, $n\geq 2$? $\endgroup$ – GFR Apr 2 '13 at 16:04
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    $\begingroup$ Dear @GFR, while i admit it is tempting to think that a line should be embedded in a bigger space, this is not necessary. In general in algebraic geometry, $kP^n$ with $k$ any field is called a linear variety (usually denoted $\mathbb{P}_k^n$ by the way). The special cases $kP^1$ and $kP^2$ are also called a (projective) line and plane respectively. This is just terminology. $\endgroup$ – Joachim Apr 3 '13 at 12:13

I feel that the thing to do is to start off by thinking about real projective spaces. There are also some tricks to help understand projective spaces in general.

One trick is to "lift" back to the underlying vector space. For example, to come to grips with the idea that any two projective lines in $\mathbb P^2(\mathbb R)$ meet in one point, we can think about what happens in the vector space $\mathbb R^3.$ A projective line is a linear collection of projective points, which themselves are described by lines through the origin in $\mathbb R^3.$ So, a projective line is described by a plane through the origin in $\mathbb R^3.$ Translating what "two lines in $\mathbb P^2(\mathbb R)$ meet in a point" means to their representatives in $\mathbb R^3,$ the statement becomes that any two planes through the origin intersect in a unique line. Can you visualize this?

Another trick is to remember that projective space decomposes (non-uniquely) into disjoint pieces: $\mathbb P^n = \mathbb A^n \sqcup\mathbb P^{n-1}.$ Often times, we can ignore the fact that we're working with projective space, if the behaviour we're interested in takes place entirely within the $\mathbb A^n$ piece.

For example, the projective plane is what we get by adding the projective line to the affine plane $\mathbb A^2$ (in our case $\mathbb A^2 = \mathbb R^2,$ but without the vector space "structure", i.e., we don't add points of affine space, or multiply them by scalars). Remember from your notes, the projective line $\mathbb P^1(\mathbb R)$ can be thought of as the circle $S^1$ with antipodal points identified. If we put $S^1$ inside $\mathbb R^2$ as the unit circle centred at the origin, then every line through the origin defines a unique point of $\mathbb P^1,$ by its intersection with $S^1.$ This means that the projective plane is what we get by adding to the affine plane one point for each direction (i.e. slope!) that a line can have in $\mathbb R^2.$ If we are interested in the intersection of two lines in $\mathbb R^2$ that we know have different slopes, then we can ignore projective geometry. But, if two lines are parallel, then they do not intersect in $\mathbb R^2,$ but they do define the same point of $\mathbb P^1,$ which means that they intersect in the $\mathbb P^1$ portion of $\mathbb P^2$ ("at infinity").

Regarding visualizing complex geometry as real geometry, well, it's limited, because visualizing geometry past 4 dimensions isn't really possible. So, as soon as we think about spaces of complex dimension two or more, we're stuck. I think it's better to understand the real picture where we can, and then try to accept that the complex numbers arise in a way that preserves (and simplifies) the essential algebraic properties of the real numbers. When you study algebraic geometry, you will learn that geometric properties (in the situations we can visualize) can be expressed by algebraic conditions (vanishing of polynomials, etc.). The algebraic conditions can make perfect sense over different fields and in any dimension, but you will still only be able to visualize the geometry in $\mathbb R^1,\mathbb R^2,\mathbb R^3$ (maybe also over finite fields...). This doesn't mean we cannot draw diagrams to help our understanding, but those diagrams will no longer be so closely linked to the literal solutions to our equations, if that makes sense. But it's surprising how far one can get with "meaningless" pictures, so I encourage you to draw them in the real case, and to keep them in mind over other fields!

Anyway, those notes you are reading look very nice, I would certainly recommend you to keep reading them, and things will become clearer the more you do.

  • $\begingroup$ Thanks Andrew, this was very useful. In particular I did not think of relating a projective line in $\mathbb{P}^2$ to a plane through the origin in $ \mathbb{R}^3$. Concerning the directions in which you can go to infinity, I have no problems with $\mathbb{R}^{n+1}\rightarrow S^n\rightarrow \mathbb{P}^n(\mathbb{R})$, but I cannot really "feel" $\mathbb{C}^{n+1}\rightarrow S^{2n+1}\rightarrow \mathbb{P}^n(\mathbb{C})$ even for n=1. I only see n=0: there is only one direction you can go to infinity in $\mathbb{C}$. $\endgroup$ – GFR Apr 2 '13 at 15:29
  • $\begingroup$ A question related to what you wrote is: can I substitute $\mathbb{R}$ with $\mathbb{C}$ and appropriately add the adjective complex in things like "any two projective lines in ℙ2(ℝ) meet in one point"? In this case, "any two (complex) projective lines in $\mathbb{P}^2(\mathbb{C}) meet in one point". I imagine there should be no problem since the relevant linear algebra should be the same whatever the field you are using. $\endgroup$ – GFR Apr 2 '13 at 15:33
  • $\begingroup$ Dear @GFR, that's right, this geometric property is certainly preserved. And in general, over the complex numbers things behave very nicely -- we can generalize this to: any two curves in the complex projective plane defined by the vanishing of polynomials $f$ and $g$ of degrees $d$ and $e$ meet in exactly $de$ points (counted with "multiplicity"). $\endgroup$ – Andrew Apr 2 '13 at 15:46

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