\def\Ext{\mathop{\rm Ext}}
\def\Z{{\bf Z}}
\def\RP{{\bf R}P}
\def\CP{{\bf C}P}

Transcript of the qualifying examination of Dmitri Pavlov.

August 25, 2008, 14--17, room 959, Evans Hall.

This is a rough transcript of my qual, which occurred on August 25, 2008.
Since I wrote this transcript using my memory, not the actual recording,
the words ascribed to the participants do not coincide with the original phrases.

Participants: Dmitri Pavlov, Peter Teichner (advisor), Constantin Teleman (committee chair), Dan-Virgil Voiculescu, Raphael Bousso (Department of Physics).

\obeylines%
Teichner: What do you want to start with?
Me: Let's start with topology.
Teichner: How about the cohomology ring of $S^2\times S^2$.
[I compute group structure by K\"unneth formula.]
Teichner: What about ring structure?
[I use Poncar\'e duality.  It leaves two variants for ring structure.  I choose the wrong one.]
Teichner: How did you obtain this?
[I correct myself and point out the right ring structure.]
Teichner: Yes, and K\"unneth formula is valid for ring structure.  Can you write it down?
[I write down how to obtain this ring structure via K\"unneth formula.]
Teichner: But can give an example of manifold with the ring structure you originally wrote?
Me: Maybe connected sum of two-dimensional projective space with itself will work.
Teichner: Can you compute the cohomology of this manifold?
Me: One can use Mayer-Vietoris sequence or de Rham cohomology.
Teichner: You can use either.
[I write down Mayer-Vietoris sequence and compute cohomology groups.]
Teichner: But what about the ring structure?
Me: We can use de Rham cohomology.  We immediately obtain that $xy=0$, where $x$~and~$y$ are two generators of second cohomology group.
Teichner: What about $x^2$ and $y^2$?
Me: By Poincar\'e duality $x^2$ and $y^2$ are generators of~$H^4$.  Whether we have $x^2=y^2$ or $x^2=-y^2$ depends on orientation.
Teichner: Finally you came to the question of orientation of $\CP^2$.  First let's decide whether $\CP^2$ and $\CP^2$ are different manifolds.  In other words, can you tell us whether there is an orientation-reversing diffeomorphism of $\CP^2$?
Me: Such a diffeomorphism should induce a negative identity on top cohomology.
Teichner: You already computed ring structure on cohomology of $\CP^2$.
Me: Yes, from this structure one immediately obtains that there is no orientation-reversing diffeomorphism of $\CP^2$.
Me: The manifolds $\CP^2\#\CP^2$ and $\CP^2\#\overline{\CP^2}$ have cohomology rings $x^2=-y^2=z$ and $x^2=y^2=z$.
Teichner: Therefore these manifolds are not homeomorphic.
Teichner to Teleman: You want to ask some questions on topology?
Teleman: Can you compute the cohomology of $\RP^2\times\RP^2$.
Me: Yes, we can again use K\"unneth formula.  This time the cohomology has torsion
and we have to use the version with Tor functors.
[I write down the K\"unneth exact sequence.  I make a mistake and write $n-1$ instead of $n+1$ in the sum of Tor terms.]
Teleman: Is it $n-1$ or $n+1$ there?
Me: I think it is $n-1$.
Teleman: OK, go on and compute cohomology.
Me: First let's write down the left term of K\"unneth exact sequence.  For this we need cohomology of $\CP^2$.
[I write down {\it homology\/} of $\RP^2$.]
Teleman: Are we computing homology or cohomology?
Me: Oops, this is homology.
Teleman: How do you compute cohomology from homology?
Me: Universal coefficient theorem.
[I write down the universal coefficient theorem and apply it to $\RP^2$.  I say that $\Ext_\Z(\Z,\Z)=\Z$ but then correct myself.]
[I substitute the cohomology of $\RP^2$ into K\"unneth exact sequence.  The fifth cohomology group turns out to be nonzero.]
Me: Oops, we really should have $n+1$ in the K\"unneth formula.
Teleman: Signs should change when you pass from homology to cohomology.  That's how I remember all these formulas.
Teichner: There are all these fancy topics in the syllabus.  Let me choose one.  What's Dold-Thom theorem?
[I write down the Dold-Thom theorem.  I explains what symmetric product is and say that it converts Moore spaces to Eilenberg-Mac Lane spaces.]
Teichner: Is symmetric product a monoid?
Me: Yes, it is a commutative topological monoid.
Teichner: Is it an abelian group?
Me: No, the symmetric product is a free commutative topological monoid on a topological space, therefore never a group.
Teichner: But Eilenberg-Mac Lane space is a group.
Me: Yes.
Teichner: Can you tell us what will happen if we replace free commutative monoid by free commutative abelian group?
Me: We obtain zero singular chain group.  Reduced zero singular chain group.
Teichner: What I meant is connection with Eilenberg-Mac Lane spaces.
Me: One can obtain Eilenberg-Mac Lane spaces by iterating classifying space construction.  In this way one immediately obtains that Eilenberg-Mac Lane space is an abelian group.
Teichner: Let's make a five-minute break.
[Break.]
Teleman: Tell us about the relation between divisors and line bundles.
[I define Weil divisor group as the free abelian group generated by hypersurfaces.]
Teleman: Can you define hypersurfaces?
[I write down the definition in local charts.]
Teleman: These are smooth hypersurfaces.
Me: Yes, generally in a local chart one should have a zero set of a holomorphic function.
Teleman: Yes.  OK, what about line bundles?
Me: There is a canonical map from the group of divisors to the Picard group of line bundles.  For each hypersurface there is a unique line bundle and a global section of this bundle such that its zero set coincides with hypersurface.
Teichner: Is it well-defined?  I think the map goes the other way.
Me: Yes, it is well-defined.  The map cannot go the other way because there is no canonical way to attach a global holomorphic section to a line bundle.
Teichner: Yes, but is it well-defined?
Me: There is another way to construct this map.  Consider short exact sequence of sheaves $0\to O^*\to K^*\to K^*/O^*\to0$.  The boundary map $H^0(K^*/O^*)\to H^1(O^*)$ is exactly the map we need.
Teleman: Sheaves of what?
Me: Sheaves of abelian groups.
Teleman: With what group structure?
Me: Multiplicative.
Teleman: Why $H^0(K^*/O^*)$ and $H^1(O^*)$ are the groups we need?
[I explain the isomorphism for the Picard group via local trivializations.]
Teleman: What is $K^*$?
Me: Sheaf of meromorphic functions.
Teleman: But you mentioned only holomorphic functions and their sets of zeroes.
Me: For meromorphic functions one should subtract the divisor of poles from the divisor of zeroes.
Teleman: The map from divisors to line bundles, is it bijective, injective or surjective?
Me: No.  The long exact sequence I wrote down before tells us it isn't.  The kernel is the group of principal divisors, $H^0(K^*)$.  They are precisely the divisors defined by global meromorphic functions.
Teleman: And the cokernel?
Me: The cokernel is $H^1(K^*)$.  It can also be nonzero.
Teleman: Do you know when it is zero?
Me: It is zero for projective manifolds.
Teichner: What is Serre duality?
[I write it down as an isomorphism.]
Teleman: What about the pairing?
[I write Serre duality as a nondegenerate pairing.]
Teleman: What is the isomorphism from $H^{n,n}$ to complex numbers?
Me: Integration.
Teleman: I would like to ask about the Hodge conjecture, but it isn't on the syllabus.
Teichner: It is.
Teleman: Then tell us about the Hodge conjecture.
[I write down the Hodge conjecture.]
Teleman: This description suffices as the first approximation.  But if we want to make sense of this definition, we need to say something about the cohomology group in your statement.
Me: We can embed sheaf cohomology with complex coefficients into Dolbeault cohomology.
Teleman: Yes, we can do this, but this doesn't bring us any closer.  We need something else.  It is on your syllabus.
Me: I don't know.
Teleman: Can you tell us what Hodge decomposition is?
[I write down Hodge decomposition.]
Teleman: So now we can identify de Rham cohomology with direct sum of Dolbeault cohomology.
Teleman: Can you tell us what do you mean by the rational 1,1-cohomology?
Me: It is the intersection of Dolbeault cohomology and rational singular cohomology.
Teichner: Let's have another five-minute break.
[Break.]
Voiculescu: What is a von Neumann algebra?
Me: It is a weakly closed $*$-subalgebra of the algebra of bounded operators.
Voiculescu: Can you give another definition?
Me: It is a $*$-subalgebra of the algebra of bounded operators coinciding with its double commutant.
Voiculescu: What I want to do is to ask some questions on commutative von Neumann algebras and then go to type II$_1$ factors.  What can you say about commutative von Neumann algebras?
Me: Every von Neumann algebra is isomorphic to $L^\infty$ of some measurable space.
Voiculescu: In your definition a von Neumann algebra acts on some Hilbert space. 
Me: $L^\infty$ acts on $L^2$ by multiplication.
Voiculescu: What is an isomorphism of von Neumann algebras?
[I say that it is a norm-preserving bijection.  By bijection I meant an isomorphism of $*$-algebras, but forgot to say this.  Voiculescu and Teichner were uncontent by such wording and eventually made write down all the properties of $*$-algebra isomorphism.  Only then I noticed my terminological mistake.]
Voiculescu: So basically you want to say that an isomorphism of von Neumann algebras is a C*-algebra isomorphism.
Me: Yes.
Voiculescu: This is correct, although usually another definition is used and then this statement is derived as a corollary.  Do you know another definition of isomorphism?
Me: No, I don't.
Voiculescu: You can replace norm-preservation by ultraweak continuity.
Teichner: Can you say what ultraweak topology is?
[I write down the definition.]
Voiculescu: Can you define what a type II$_1$ factor is?
Teichner: By the way, what is a factor?
Me: A factor is a von Neumann algebra with a trivial center.
Me: First we define an order on the set of all projections of a factor.
[I write down the definition.  I use non-standard notation for two orders on projections.]
Voiculescu: You really want to use another notation for projections.
[I rewrite the definitions using another notation.]
Me: Now a type II$_1$ factor is a factor whose ordering of projections is isomorphic to~$[0,1]$ and the unit projection is finite.
Voiculescu: What is a finite projection?
Me: A finite projection is a projection that is not isomorphic to any of its subprojections.
Voiculescu: Can you define type II$_1$ factors in another way?
Me: Yes.  A type II$_1$ factor is a factor with faithful ultraweak continuous normalized trace.
Voiculescu: Yes, this definition is correct, even though you can omit some of the conditions.
Teichner: So you can get rid of faithfulness? 
Voiculescu: Yes.  You can get rid of faithfulness, you can get rid of ultraweak continuity, you can get rid of almost anything.
Voiculescu: Can you say what is a general condition that guarantees that group algebra of a group is a type II$_1$ factor?
Me: All conjugacy classes are infinite except for the unit.
[Voiculescu then asked a question on a topic not on a syllabus.  I did not know this topic.  Voiculescu did not insist, because this topic was not on my syllabus.]
[I leave the room and wait about five minutes.  Then Teleman opens the door and congratulates me.  Other committee members follow.]
\bye