What is a projective space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:33:56Z http://mathoverflow.net/feeds/question/7407 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7407/what-is-a-projective-space What is a projective space? Andrew Stacey 2009-12-01T09:31:27Z 2009-12-01T19:49:59Z <ol> <li>Is there a "recognition principle" for projective spaces?</li> <li>What categories are there with projective spaces for objects?</li> </ol> <p><strong>Background:</strong> Although the title is a nod to <a href="http://mathoverflow.net/questions/5957/what-is-a-metric-space" rel="nofollow">What is a metric space?</a>, this question is really built on top of my questions about Hilbert spaces (<a href="http://mathoverflow.net/questions/6552/what-is-an-intuitive-view-of-adjoints-version-2-functional-analysis" rel="nofollow">What is an intuitive view of adjoints? (version 2: functional analysis)</a>, <a href="http://mathoverflow.net/questions/5086/can-one-do-without-riesz-representation" rel="nofollow">Can one do without Riesz Representation?</a>, <a href="http://mathoverflow.net/questions/123/linearity-of-the-inner-product-using-the-parallelogram-law" rel="nofollow">Linearity of the inner product using the parallelogram law</a>). The answers to these questions have set me pondering about Hilbert spaces. In short, I'm wondering whether or not it is possible to change the slogan of Hilbert spaces from "Hilbert spaces are great because they are self-dual" to "Hilbert spaces are great because you don't have to mention the dual". So I'm looking at ways to remove any explicit mention of the dual from the basic constructions in Hilbert space theory. My current favourite replacement is with statements about complementary subspaces.</p> <p>This leads to projective spaces since they are the first non-trivial family of subspaces of some Hilbert space. The subspace analogue of Riesz representation says that there is an isomorphism between $\mathbb{P}H$ and $\operatorname{Gr}_{\infty -1}(H)$ (the space of codimension one subspaces). I want to be able to treat the latter space as if it were a projective space <em>in its own right</em> and say that this is an "isomorphism of projective spaces". However, I don't want to explicitly construct a vector space of which $\operatorname{Gr}_{\infty - 1}(H)$ is the projective space since that vector space would be $H^*$. So I want a recognition principle for projective spaces and a suitable notion of "isomorphism".</p> <p>An example of what the recognition principle might be like would be an expression of the structure of a projective space as an <a href="http://ncatlab.org/nlab/show/algebraic+theory" rel="nofollow">algebraic theory</a>, perhaps a many-sorted one. That would also make it obvious what the morphisms were.</p> <p>Obviously there are many versions of what the morphisms could be. What I would like is for the obvious functor from Hilbert spaces to projective spaces to be <a href="http://ncatlab.org/nlab/show/full+and+faithful+functor" rel="nofollow">full and faithful</a> but not <em>by construction</em>. That is, I don't simply want to take the category of Hilbert spaces and replace each Hilbert space by it's projective space because then to work out the morphisms between two projective spaces I first need to choose representing Hilbert spaces and I don't want to do that.</p> <p>To give this a connection to something a little deeper than just a different way of working with Hilbert spaces, recall that in twisted K-theory the starting place is a bundle of projective spaces that <em>cannot</em> be globally lifted to a bundle of Hilbert spaces. Nonetheless, some constructions related to a Hilbert space still work on such a bundle because they don't depend on the actual choice of Hilbert space so you can make a local choice and then prove it independent of that choice (for example, the space of Fredholm operators). Is there a way to go directly to that construction without making local choices?</p> http://mathoverflow.net/questions/7407/what-is-a-projective-space/7456#7456 Answer by Greg Kuperberg for What is a projective space? Greg Kuperberg 2009-12-01T16:29:54Z 2009-12-01T18:14:55Z <p>I have never seen finite-dimensional projective spaces defined by axioms, only by constructions of some kind from something else related to axioms. For example in algebraic geometry, you can define a projective <em>variety</em> as the Proj of a graded algebra, which could be adequately axiomatic. However, projective space in that setting comes from the relatively unsatisfying "axiom" that the algebra is freely generated from degree 1.</p> <p>I was about to say that it would be very difficult to make axioms for a Hilbert-projective space that are very different from the axioms for a Hilbert space. But then I thought of a way to do it. A von Neumann algebra is a Banach *-algebra over $\mathbb{C}$ that satisfies the <code>$C^*$</code> axiom and also has a predual as a Banach space. If $\mathcal{M}$ is a von Neumann algebra, it has a space <code>$\mathcal{M}^\diamondsuit$</code> of pure normal states, by definition the extremal normal, normalized, positive dual vectors. This is a generalized projective space. In particular if $\mathcal{M}$ is a Type I factor &mdash; the conditions for which need no direct mention of Hilbert spaces &mdash; then von Neumann's theorem identifies <code>$\mathcal{M}^\diamondsuit$</code> with the space of lines in a Hilbert space $\mathcal{H}$. (And of course $\mathcal{M}$ itself with $B(\mathcal{H})$.) No global phases are ever chosen in the definition.</p> <p>Does this meet your requirements? My motivation is the fact that quantum probability is the correct probability theory for quantum mechanics, and that in quantum mechanics global phases are always irrelevant. Reflecting that, the global phase doesn't exist in the von Neumann definition of a state.</p> <p><hr /></p> <p>Qiaochu in the comments makes the point that there is another very important set of axioms for a projective space, namely the classical incidence axioms for a projective geometry. My favorite version is that a projective geometry is a spherical type $A_n$ building. These axioms are different in that they don't even pick a field beforehand. Indeed, there are projective planes that are not the standard projective plane over a field.</p> <p>It would be interesting to make axioms for a topological type $A_\infty$ building corresponding to the projective space of a Hilbert space. It seems plausible, and it could be a very different model from the von Neumann algebra model. But maybe von Neumann, the person, is still there in this idea, because the incidence geometry of a Hilbert space is also known as <a href="http://en.wikipedia.org/wiki/Quantum%5Flogic" rel="nofollow">quantum logic</a>.</p> http://mathoverflow.net/questions/7407/what-is-a-projective-space/7489#7489 Answer by Gonçalo Marques for What is a projective space? Gonçalo Marques 2009-12-01T19:36:52Z 2009-12-01T19:49:59Z <p>The book "Modern Projective Geometry" has a system of axioms for projective spaces in a set $X$ using a function $f\colon X\times X\to P(X)$ (see page 30). Also there's a lot of projective geometry that can be done in the context of lattices. I thing this is related to Greg's answer through Von Neumann's "<a href="http://books.google.pt/books?id=onE5HncE-HgC&amp;dq=Continuous+geometry&amp;printsec=frontcover&amp;source=bl&amp;ots=gbfVKzAAbj&amp;sig=qYzkgkbW6ZJu7v4subUL2jrOieI&amp;hl=pt-PT&amp;ei=S2QVS8jBLIWy4QbX3aHiBg&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=3&amp;ved=0CBEQ6AEwAg#v=onepage&amp;q=&amp;f=false" rel="nofollow">Continuous geometry</a>". (see <a href="http://planetmath.org/encyclopedia/VonNeumannLattice.html" rel="nofollow">here</a>). Another book to look might be Baer's classic "<a href="http://books.google.pt/books?id=Ol8YFnZRJfIC&amp;dq=baer+linear+algebra&amp;printsec=frontcover&amp;source=bn&amp;hl=pt-PT&amp;ei=5mQVS4qvJoSl4QauoN3cBg&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=4&amp;ved=0CBgQ6AEwAw#v=onepage&amp;q=&amp;f=false" rel="nofollow">Linear algebra and projective geometry</a>". </p> <p>On Greg's comment to Qiaochu's comment: let's not forget that for every projective space $X$ we can always find a division ring $D$ such that every $X=P^n(X)$ for some n (in the the plane case you need that $X$ be a Desargue's plane).See <a href="http://www.cs.elte.hu/geometry/csikos/proj/pr5.dvi" rel="nofollow">this</a>. (But, as far as I know, these are all finite dimensional results so they might not be that interesting to Andrew...)</p>