User aaron bergman - MathOverflow

Answer by Aaron Bergman for Motivation of Virasoro algebra

2012-12-27T16:19:18Z

1 is standard (it's the correspondence between analytic complex functions and conformal maps). For 2, in physics, one really deals with projective representations, not just ordinary representations (this is because a quantum state is really a ray in Hilbert space rather than a vector). A projective representation of an algebra without central charge is the same as an ordinary representation of the algebra with a (potentially) non-zero central charge. It's easier to work with ordinary representations, so people use the centrally extended algebra.

Answer by Aaron Bergman for Derived category of varieties and derived category of quiver algebras

2012-11-25T22:34:02Z

Any variety with a strong exceptional collection that generates the derived category of coherent sheaves will work. This goes back in some form to Bondal and is a consequence of Rickard's derived Morita equivalence. (This is sufficient, but not necessary. For non-compact varieties, for example, you can get derived equivalences with the path algebra of quivers with loops that are not given by the endomorphism of a strong exceptional collection, but are closely related. See Bridgeland's 0502050 for an example, and more in my paper with Nick Proudfoot 0512166.)

In general, given a generator of the derived category (or any triangulated category or stable infinity-category I suppose), the (generalized) theory of derived Morita equivalence says that there is an equivalence of categories between the derived category of modules over the endomorphisms of this object and the original category. If the generator only has endomorphisms of degree zero (ie, $Ext^i(E,E)$ is zero for non-zero i), the endomorphisms form an algebra, and you get an equivalence between the derived categories of this algebra and the original category. The exceptional collection makes it easy to interpret this algebra as the path algebra of a quiver with relations.

If the generator does have endomorphisms of non-zero degree, we instead have to think of the endomorphisms as a dg- or A-infinity algebra (or some other spectrum thingie if you're not working over a field of characteristic zero I think). Since the quintic is a compact Calabi-Yau, Serre duality means that any object that has endomorphisms of degree zero also has endomorphisms of degree 3, so we can't represent its derived category of coherent sheaves as the derived category of the path algebra of a quiver with relations.

Answer by Aaron Bergman for What does Yang-Mills and mass gap problem has to do with mathematics?

2012-07-26T01:04:35Z

There is a long, long list of mathematical subjects that were either pioneered or significantly inspired by results in quantum field theory. However, while physicists may trust the manipulations they do in QFT, and the results of those manipulations have been spectacularly successful, for almost every interesting quantum field theory, there isn't even a rigorous definition or existence proof, much less a justification behind the manipulations that led to the invention of, for example, Seiberg-Witten invariants.

Solving the mass gap problem in Yang-Mills would represent the successful rigorous existence of a very non-trivial quantum field theory and the demonstration of a very nontrivial result about that field theory (that hasn't even been adequately demonstrated using physical techniques). While there probably aren't many direct mathematical consequences to the existence of a mass gap, the techniques involved would almost assuredly be applicable towards the large number of QFT inspired results in mathematics.

Classifying Space of a Group Extension

2010-01-17T21:18:38Z

Consider a short exact sequence of Abelian groups -- I'm happy to assume they're finite as a toy example: $$ 0 \to H \to G \to G/H \to 0\ . $$

I want to understand the classifying space of $G$. Since $BH \cong EG/H$, $G/H$ acts on $BH$ and we can write $BG \cong E(G/H) \times_{G/H} EG/H$. Thus, we have a fiber bundle (which I'll write horizontally) $$ BH \to BG \to B(G/H) $$

On the other hand, the central extension is classified by an element of the group cohomology $H^2(G/H,H)$ which is the same as $H^2(B(G/H),H)$. The latter is an element in the homotopy class of maps $[B(G/H),K(H,2)]$ and $K(H,2)\cong BBH$. This map looks like it classifies a principal $BH$ bundle over $B(G/H)$. I find it hard to imagine that this 'principal' $BH$ bundle is not 'the same' as the bundle above, so the question is, how do you see that? From this construction, it's not even obvious to me that the above bundle is a principal bundle.

I would guess (and being a poor physicist, I'm not so up on my homotopy theory), there's a sense that the classifying space of an Abelian group is an 'Abelian group', and taking classifying spaces of an exact sequence gives you back an 'exact sequence'. That gets you a 'principal bundle' (aren't quotation marks fun?), but even then I'm not sure how to see that the classifying map of this bundle is the same as the class in group cohomology.

Any references to the needed background would also be greatly appreciated.

Answer by Aaron Bergman for Reference for the derived category of $X$, $[X/G]$ and $X/G$

2011-11-16T00:17:54Z

Bridgeland, King and Reid, "The McKay correspondence as an equivalence of derived categories" has a nice discussion expanding on David Roberts's comment.

Answer by Aaron Bergman for Does the derived category of coherent sheaves determine the hodge theory?

2011-11-08T16:16:32Z

This isn't quite responsive, but you can show for nice dg-categories that the Hodge-de Rham spectral sequence degenerates in many situations. See Kaledin's notes, 0708.1574. This is a special case of Kontsevich's degeneration conjecture. For a lot of discussion along these lines, see Katzarkov, Kontsevich and Pantev on nc-Hodge structures.

Answer by Aaron Bergman for What exactly is the relation between string theory and conformal field theory?

2011-10-11T02:33:13Z

This is reiterating a lot of what Jeff said, but maybe I can explain from a different perspective.

There are two things going on here (as there always are in string perturbation theory.) The first is the string worldsheet, and the second is what is going on in spacetime.

The string worldsheet is a non-linear sigma model into spacetime. Here, spacetime is a Riemannian manifold (with plenty of other structure depending on the exact string model you're using.) The "nonlinear sigma model" on the string worldsheet (a surface potentially with multiple punctures/boundaries) has a metric (different from the metric on the target manifold) and a map from the worldsheet into the spacetime manifold -- there are other fields in fancier versions of string theory, but I'll neglect them. In string perturbation theory, you integrate over the moduli space of metrics and embeddings. The resulting theory is invariant under conformal transformations, and because metrics in two dimensions don't have a huge amount information in them, an essential part of the theory on the worldsheet ends up being a conformal theory. There are various other conditions which ensure that the CFT gives rise to a full theory of 2D quantum gravity, meaning that you really can integrate over the space of the metrics. If those conditions hold, using the CFT, you can compute string amplitudes corresponding to your punctured Riemann surface. The can be thought of as scattering string in spacetime.

The important thing is that the amplitudes computed above are supposed to be terms in an asymptotic expansion of, er, something. This is why it's called string perturbation theory: in analogy to quantum field theory, combining individual string amplitudes of higher and higher genus in a formal power series (where the parameter is called the "string coupling") is supposed to be an expansion arising from some "nonpertubative" theory. What this theory is in complete generality is still unknown (although we know a lot in various special cases).

We can try to ask what this all looks like from the point of view of spacetime. Now, a basic fact about perturbation theory is that it only really makes sense (or, at least, makes the best sense) when you're perturbing around a solution. All of this is a roundabout way of saying that the spacetime only makes sense when the target manifold and its various structures give rise to a good perturbation expansion which means that the two dimensional theory is a conformal field theory.

This is what people mean when they say that a 2D CFT is a solution to the equations of motion of string theory. In fact, you can drop the requirement that your 2D theory is a "non-linear sigma model", ie, that it has the structure of maps into a manifold. Then you get into the "moduli space" of two dimensional field theories. Which, as far as I know, is completely undefined. But, even in this case (the world of string field theory), the "classical solutions" are the ones where you can define a good perturbation expansion around, and those are the conformal field theories.

Added 10/12:

I wouldn't go too far with the entire "first quantization"/"second quantization" thing. You could imagine a free string field theory where the string isn't allowed to interact, but the nice thing about string perturbation theory is that the interactions and the propagation are different aspects of the same thing. This is in contrast to the perturbation theory of quantum field theories where the interactions and the propagators are different things (one is pointlike, the other is lifelike). The CFT (really, a theory of 2D quantum gravity) is what you start with, and you automatically get both the "free" string theory and the interactions. The question of "second quantization" (a term I really hate) is whether or not you can derive the formal power series resulting from adding the various amplitudes associated with Riemann surfaces of varying general as the perturbation expansion of another theory.

To answer your questions about how you go from fields on the worldsheet to fields in spacetime, you quantize the theory on the cylinder, and each vector in the Hilbert space corresponds to a spacetime field (because you can Fourier expand fields on the cylinder, this isn't as crazy as it sounds). However, because you really are doing 2D quantum gravity, you have to deal with the gauge invariances. The nice way to do this is using BRST quantization, and the actually physical fields are the cohomology of the BRST operator acting on the CFT Hilbert space.

This is pretty standard material in a first course on string theory. I don't have them on me, but I'd expect Eric D'Hoker's lectures in the IAS volumes on QFT and strings for mathematicians would do this.

General Equilibrium for Mathematicians

2010-04-18T23:30:47Z

I've been reading up a lot on the recent financial crisis, and central to the story is the existence of general equilibrium models in economics, say, as proven by Arrow and Debreu (and MacKenzie?). Regardless of the real-world validity of these models (they're not looking so hot these days), I'm interested in them as a purely mathematical exercise. So, that said, does anyone know of a good (preferably available online) exposition of the existence of a general equilibrium aimed at mathematicians?

Answer by Aaron Bergman for Why are operads so closely connected to mathematical physics?

2011-08-10T03:08:12Z

It's not really possible to give a precise answer to this question, so I apologize for being vague here. One answer is because a lot of multiplications in physics are associated with moving two things close to each other and looking at the result as a single object. The archetype for this is the operator product expansion in quantum field theory. This naturally leads to thinking about a little n-spheres operad with various decorations. Most examples that come to mind right now really reduce to that.

Answer by Aaron Bergman for If you were to axiomatize the notion of entropy .....

2011-07-01T15:41:29Z

From a more physical perspective, there's the work of Lieb and Yngvason:

http://arxiv.org/abs/math-ph/0204007

Answer by Aaron Bergman for topological actions

2011-04-27T11:46:11Z

They are trying to define the Chern-Simons action over a manifold $M$ by writing it as the integral of $\int F \wedge F$ over a bounding manifold $B$. When the bundle is nontrivial, they consider a more general cochain and show that there exists a $B$ over which the bundle extends such that $\partial B$ = $n$ copies of $M$. So, you can define

$$ n S = \int_B F \wedge F $$

But, because actions enter into imaginary exponentials in the path integral, this is really only defined mod 1 (once you reenter all the coefficients that I omitted). So, the action $S$ is only defined mod $1/n$.

They show how the second formula resolves the ambiguity in the text that follows, but it's probably best to think of it as a differential character or in terms of differential cohomology. A more rigorous presentation might be http://arxiv.org/abs/hep-th/9111004.pdf .

Update:

Let me try another explanation. We know from the above that $$ n S_{CS} = \int_B F \wedge F \quad \mbox{mod 1} $$

Thus, $$ S_{CS} = \frac{1}{n}\int_B F \wedge F + \frac{q(B,E)}{n} $$ with $q(B,E) \in \mathbb{Z}$. The simplest guess is that $q = 0$, but it's easy to see that the resulting action is not independent of the choice of $B$. In particular, we would want, for closed $B$, that $\frac{1}{n} \int_B F \wedge F \in \mathbb{Z}$, but it's only in $\frac{1}{n}\mathbb{Z}$.

So, the goal is to choose a $q(B,E)$ such that the action makes sense. Since you want something that is an integer when applied to a closed $B$, it's not too hard to guess something like DV's action.

Answer by Aaron Bergman for Heuristic behind $A_{\infty}$ - algebras

2011-03-06T21:54:43Z

The standard example is the based loop space $\Omega M$ thought of as the space of continuous maps from $[0,1]$ to $M$ where 0 and 1 are mapped to the base point. This has a product on it given by taking the loop you get from going around one loop and then the other. In other words, if you have loops $f$ and $g$, if $t\in [0,1/2]$, $(fg)(t) = g(2t)$, and if $t\in [1/2,1]$, $(fg)(t) = f(2t-1)$.

It's easy to see that this product is not associative, but it is associative up to a reparametrization of the circle. Thus, there's a homotopy between $f(gh)$ and $(fg)h$ which is a map

$$ [0,1] \times \Omega M^{{}\times 3} \to \Omega M $$

For four loops, you can draw a pentagon of homotopies: $$ f(g(hi)) \sim f((gh)i) \sim (f(gh))i \sim ((fg)h)i \sim (fg)(hi) \sim f(g(hi)) $$

Let $K_4$ be the pentagon. This is a map:

$$ \partial K_4 \times \Omega M^{{}\times 4} \to \Omega M $$

These homotopies are coherent which means that this extends to a map

$$ K_4 \times \Omega M^{{}\times 4} \to \Omega M $$

This pattern continues and gives Stasheff's Associahedra. A space, $H$, possessing a set of maps (and my memory's not so great here, so I'll assume $i>1$ which might not be correct)

$$ K_i \times H^{{}\times i} \to H $$

where $K_2 = pt$, $K_3 = [0,1]$, $K_4$ is as above, etc. is called an $A_\infty$-space. It's a theorem of Stasheff that any connected $A_\infty$-space is homotopic to a based loop space.

Now, pass to chains on the space and you get an $A_\infty$ algebra. The theorem of Stasheff is then the statement that the bar construction on an $A_\infty$ algebra is a dg-coalgebra.

This can all be thought of in terms of operads, of course, but I think this all came first. I don't know which introduction of Keller's you're using (he has several), but I believe this is all in this one.

Answer by Aaron Bergman for How are the classifying space of $E_8$ and $K(\mathbb{Z},4)$ related?

2011-01-17T05:04:20Z

For (1), after searching a bit, I think the original reference in the physics literature is Witten's "Topological Tools in Ten-Dimensional Physics", Int. J. Mod. Phys. A 1, 39 (1986). I think there's a reference to the fact about homotopy groups there, but I haven't read it in years.

Just to expand a bit on Jeff's answer for (2), M-theory contains a three-form with a four form curvature. Horava and Witten shower that one could associate the $E_8 \times E_8$ heterotic string with M-theory on $S^1/\mathbb{Z}_2$. The boundaries each have an $E_8$ gauge theory on them. Soon after, Witten used $E_8$ bundles to determine the quantization of the four form in M-theory in hep-th/9609122. This quantization is, interestingly, shifted from being integral. $E_8$ index theory was used spectacularly to compare the partition functions of IIA and M-theory in the ginormous paper of Diaconsecu, Moore and Witten, hep-th/0005090. As Jeff says, the paper of Diaconescu, Moore and Freed is the most modern way of looking at the subject using (shifted) differential cohomology. One of the conclusions of that paper is we just don't know whether the use of $E_8$ bundles for quantizing the M-theory four-form is real or just a convenient trick. But given the other ways $E_8$ seems to be hanging around M-theory (for example, the split real form of $E_8$ shows up when you compactify M-theory on $T^8$), I'd guess the former.

Topological Content of the Kakutani Fixed Point Theorem

2010-04-23T03:00:26Z

Reading about general equilibria, the Kakutani fixed point theorem seems to be a central tool. It states (following Wikipedia)

For $S \subset \mathbb{R}^n$, non-empty, compact and convex, and $\phi : S \to 2^S$, if $\phi$ has a closed graph and, for all $s \in S$, $\phi(s)$ is nonempty and convex, then there exists an $x \in S$ such that $x \in \phi(x)$.

It is often called a generalization of the Brouwer fixed point theorem, but I'm not sure I exactly see that as Brouwer holds for any subset homeomorphic to the closed ball, and plenty of those are nonconvex. Regardless, the proofs that I've found on the internet usually proceed by reducing to a simplex. Then, for every subdivision of that simplex into smaller simplices of side length $1/n$, you can generate a function to $S$ (not $2^S$) that agrees with $\phi$ on the vertices. Brouwer that thing, and you get a series of fixed points for each sized subdivision. Pass to a convergent subsequence, take the limit and you prove that you get a fixed point in the sense of the theorem by appealing to convexity of the target sets. Or something like that; I might have missed a detail or three.

Yuck. Maybe I'm betraying my anti-analysis prejudices, but compared to the proof of the Brouwer fixed point theorem using homology, I'm left a little unsatisfied. So, the question is, is there a way to think about the Kakutani fixed point theorem topologically? Perhaps not, given the role convexity seems to play in the proof, but then you could ask, like Brouwer for convex sets, is Kakutani a special case of a more topological theorem?

Answer by Aaron Bergman for A differentiable approximation to the minimum function

2010-08-11T05:47:02Z

If signs aren't a big deal, use the generalized mean formula

$$ \left(\frac{1}{n}\sum x_i^k\right)^{1/k} $$

for $k\to -\infty$.

Answer by Aaron Bergman for (How) is category theory actually useful in actual physics?

2010-08-08T06:18:54Z

Categories (and higher categories) seem to be a good way of expressing the locality of the path integral in physics. In particular, it is the idea of gluing of local structures that is important. This line of thought leads to the axiomatization of (parts of) various QFTs, with the most success in topological and conformal field theories. This idea has its origins with Atiyah, Segal, Baez-Dolan, Freed and probably a ton of other people I'm forgetting. Braided fusion categories as in the previous answer are an example of this in three dimensions. Most recently, there's Lurie's classification of TQFTs in all dimensions in terms of $(\infty,n)$ categories.

Answer by Aaron Bergman for Hochschild (co)homology of A and of Mod_A

2010-07-30T03:33:33Z

Basically this follows from the fact that the derived category of bimodules over two algebras is equivalent to the (suitably defined) functor category between the derived category of modules of each algebra. Say, Toen's paper on derived Morita equivalence. Then, the identity functor is given by the algebra itself interpreted as a bimodule, so the Hochschild cohomology is $\mathrm{Ext}^i_{A-A}(A,A)$. You can compute this using the bar resolution and a quick calculation gives you the usual definition of Hochschild cohomology.

Answer by Aaron Bergman for Are Calabi-Yau manifolds in dimension >= 3 algebraic?

2010-07-05T15:59:38Z

It depends a little bit on your definition of CY. If you're using a good one, it will imply that the Hodge numbers $h^{0,p} = 0$ for $p \neq 0,d$ (see, for example, Prop. 5.3 of Joyce's http://arxiv.org/abs/math/0108088). This implies that $H^2(X) \cong H^{1,1}(X)$. Since the Kaehler cone is an open set in $H^{1,1}(X)$, it contains an rational class, and we can scale that to be an integral class. So, by Kodaira and Chow, we're done.

Answer by Aaron Bergman for How should I think about B-fields?

2009-10-21T23:09:00Z

We like to do more than that, actually. The B-field is an element in the differential cohomology class $\check{H}^3(M)$, or, more geometrically, a connection on an abelian gerbe. Thus, there is a class $[H] \in H^3(M,Z)$ characterizing the gerbe. In the B-model, this twists the derived category. The connection is the part that changes the A-model, and when $[H] = 0$, you exactly get that the differential cohomology group is $H^2(X,U(1))$. In the geometric language, it's a flat connection on a trivial gerbe.

Answer by Aaron Bergman for Derived Physics

2010-06-11T16:27:53Z

The short but ahistorical answer is that topological string theories turn out to be examples of $(\infty,1)$-categories. The mathematical formulation of this statement is in Lurie's classification of topological field theories http://www.math.harvard.edu/~lurie/papers/cobordism.pdf (building on work of Atiyah, Segal, Getzler, Costello, Baez-Dolan, Kontsevich and probably a bunch more I'm forgetting.)

The content of this statement is that when you write down the axioms for a topological string theory, the collection of "boundary conditions" or "D-branes" look like the collection of objects in an $(\infty,1)$ category.

Of course, you can ask why the derived category of coherent sheaves. Historically, the answer to that is that it is very easy to write down a boundary condition for a holomorphic vector bundle in the topological B-model. It's not a huge leap from there to coherent sheaves, and if you start mumbling words like tachyon condensation, you can get to the derived category with a fair bit of hand waving.

That's from the physics side of things. On the math side, Kontsevich got there first, possibly by noting that the space of closed string states in the B-model ($H^\bullet(\wedge^\bullet TX)$) is exactly the Hochschild coohomology of the derived category of coherent sheaves. He then followed up by associating the (still not yet defined?) Fukaya category with the A-model and conjecturing that mirror symmetry is an equivalence of the two (with some Hodge structure goodies thrown in). Subsequently, it looks like you have to add in some things called coisotropic branes to cover all your bases, but the basic idea is right.

Kontsevich formulated all this in terms of $A_\infty$ categories which in the Lurie language turn into $(\infty,1)$ categories which are just TQFTs in disguise. So, Kontsevich's homological mirror symmetry is then the statement that two TQFTs are the same, just like mirror symmetry in string theory.

From the physics side of things, this was all a bit of a mess, but we now understand that the derived category really arises via Block's construction of the derived category (I'm being intentionally vague as to which version of the derived category) as arising from integrable super-connections of graded smooth vector bundles http://www.math.upenn.edu/~blockj/papers/BottVolume.pdf. You can see this explicitly in the physics from a few sources, particularly Kapustin, Rozansky and Saulina, and Herbst, Hori and Page, but I'm rather fond of my own contribution http://arxiv.org/abs/0808.0168.

Answer by Aaron Bergman for {transcendental numbers} \ {computable transcendental numbers}

2010-05-30T00:50:42Z

The other standard example is to order the Turing machines and take the binary number with the nth decimal being 1 if the nth Turing machine stops. The computability of this number is (obviously) equivalent to the Halting problem.

Here computable means that the digits are literally computable by a Turing machine. Thus, Sqrt(2) is certainly computable. The book you're referring to defines a notion of "real-time computable" which also puts restrictions on how many steps it takes to compute the digits.

Answer by Aaron Bergman for A book explaining power and limitations of Peano Axioms?

2010-03-21T18:17:26Z

There are people far more expert on this than I, but for the answer to (2), Goodstein's theorem is a famous example. There's a bit about it here

http://www.math.niu.edu/~rusin/known-math/95/independence

Answer by Aaron Bergman for "Homotopy-first" courses in algebraic topology

2010-03-21T02:06:00Z

This isn't quite what you mean, but I took Igor Frenkel's algebraic topology course as an undergrad. He taught out of Massey's book, A Basic Course in Algebraic Topology. It starts with the classification of 2-manifolds, does the fundamental group and the Seifert-von Kampen theorem, and then does singular homology and cohomology. De Rham cohomology is only there as an appendix. I think the fundamental group is a little bit easier to grasp early on in a first course than singular homology. For cohomology first, you could do something like Bott & Tu, I suppose, but I think this way is a bit more useful because de Rham cohomology is a little too nice for its own good.

Constructing Twisted K-theory

2009-11-29T21:51:18Z

There is a simple, intuitive "construction" of twisted K-theory if we are allowed to ignore that many things only hold up to homotopy. We know that maps to $K(Z,2)$ give line bundles on a space and that $K(Z,2)$ forms a group corresponding to the tensor product of line bundles. Line bundles also act as endomorphisms of K-theory given by the tensor product. Thus, there is an action of $K(Z,2)$ on $F$ (where $F$ is the classifying space for $K^0$). $K(Z,2)$ principal bundles are classified by maps to $BK(Z,2) \cong K(Z,3)$, ie, elements of $H^3$. Choosing such a map, we get a principal $K(Z,2)$ bundle, $E$, and we can form the associated bundle $E \times_{K(Z,2)} F$. Twisted K-theory is then the homotopy classes of sections of this bundle.

The usual constructions of twisted K-theory that I have seen make the above precise by choosing representatives of the relevant objects so that all the needed relations hold on the nose. My question is whether you can avoid doing that. In other words, can you define all the various notions up to homotopy and obtain a definition of twisted K-theory that way?

Answer by Aaron Bergman for Periods and commas in mathematical writing

2009-11-24T11:32:56Z

Yes, but I always use "\ ." or "\ ," to separate the punctuation from the formula.

Answer by Aaron Bergman for What is to Quantize something?

2009-11-20T16:45:10Z

Just to restate some facts already stated in other answers, quantization can mean a few different things. In deformation quantization, we start with a classical theory given by a Poisson manifold. Then, (by definition) the algebra of functions forms a Poisson algebra. A quantization of this algebra is a noncommutative algebra with operators $X_f$ for $f$ a function. There is also a formal parameter $\hbar$. This algebra satifies $$ X_f\ X_g = X_{fg} + \mathcal{O}(\hbar)\ . $$

The idea of quantization is that the Poisson bracket becomes a commutator, or

$$ [X_f,X_g] = \hbar X_{\lbrace f,g \rbrace} + \mathcal{O}(\hbar^2)\ . $$

Thus, we have a noncommutative version of classical mechanics. The existence of such an algebra is a theorem of Kontsevich (the case of a symplectic manifold was solved much earlier, but I forget by whom).

In mathematics, there are plenty of interesting analogous situations where you have a noncommutative thingie which is, in some sense, a formal deformation of a commutative thingie. You can see the other direction of the above as an example of the following general fact. Given a filtered algebra whose associated graded is commutative, there is a natural Poisson structure on the associated graded.

In physics, however, it's not enough to just deform the algebra of functions; we have to now represent things on a Hilbert space. This introduces a whole host of other problems. In geometric quantization, this is split into two steps. Let's say we have a symplectic manifold whose symplectic form is integral. Then we can construct a line bundle with connection whose curvature is that symplectic form. The Hilbert space is the space of $L^2$ sections of this bundle. This is much too large, however, so you have to cut it down (which is step 2). In various cases, well-defined procedures exist, but I don't believe this is well-understood in general. For example, I'm not sure it's possible to represent every function as an operator.

It's probably worth pointing out that, from the point of view of physics, quantization is backwards. It is the quantum theory that is fundamental, and the classical theory should arise as some limit of the quantum theory. There's some interesting mathematics there, and also a whole lot of philosophy too.

Answer by Aaron Bergman for ubiquity, importance of path algebras

2009-11-15T18:25:51Z

I want to say something like, "quivers (with relations) are finitely-generated unital algebras over $k^{\oplus n}$ ", but it's fairly vacuous, and I'm about to head off to the airport, so I don't have time to really think it through.

In physics, that is pretty much the situation once you throw in the word graded (it's easy to prove a version of Gabriel's theorem in this case). Quivers arise when you have a finite set of objects in a (pretriangulated dg-/A${}_\infty$/stable infinity/triangulated/whatever) category of D-branes, and the endomorphism algebra of the sum of these objects is presented as the path algebra of a quiver. If those objects form a nice generator, you get the usual equivalence between the original category and the derived category of quiver reps. The simple reps corresponding to the nodes of the quiver are called "fractional branes" in the physics literature, and the arrows in the quiver correspond to massless string states in the physics (as they are given by Ext^1s b/w the simple reps.)

Answer by Aaron Bergman for Wick rotation in mathematics

2009-11-14T04:22:38Z

Yes. It's called analytic continuation.

Comment by Aaron Bergman

2013-04-02T17:10:11Z

Was a little too quick there -- there are closed spacelike curves; just not the first things that come to mind.

Comment by Aaron Bergman

2013-04-02T16:56:27Z

This should probably be closed because it's pretty standard. First of all, you don't really mean closed geodesics (which would be examples of closed timelike curves, ie, time travel). You want a universe with some foliation of spacelike compact surfaces. The easiest example is a cylinder: S^1 x R. For the twin paradox to hold, you would need the Lorentz group to act isometrically on this space for a given flat metric. But, it's easy to see that the full Lorentz group won't act here. In fact, the choice of a metric defines for you a distinguished frame, solving the 'paradox'.

Comment by Aaron Bergman

2012-11-26T12:44:28Z

Those are the noncompact examples I was referring to in the parenthetical remark in the first paragraph. The last paragraph is evaded there because Serre duality doesn't hold for noncompact varieties.

Comment by Aaron Bergman

2012-07-28T21:52:37Z

I suspect he's thinking about the Maurer-Cartan equation that arises in deformation theory a la DGLAs with Kodaira-Spencer theory being the usual (but not only) example that arises in theoretical physics.

Comment by Aaron Bergman

2012-06-27T23:15:30Z

Hi Sandor, $\mathbb{C}^3$ is an example of what I was talking about above: there are two cones over $\mathbb{P}^2$ that are resolutions of Calabi-Yau cones: the total space of $\mathcal{O}(1)$ and $\mathcal{O}(3)$. The former, however, is not a crepant "resolution" of $\mathbb{C}^3$ while the latter is a crepant resolution of $\mathbb{C}^3/\mathbb{Z}_3$. I'm not sure if you can approach Yuji's question through the non-crepant resolution, but it seems plausible.

Comment by Aaron Bergman

2012-06-27T04:08:19Z

Or you might be able to approach things from the non-crepant resolution into a Fano orbifold surface (some power of the resulting line V-bundle is the canonical bundle for quasi-regular SE links). It's been a while and I've forgotten what little I knew of toric world, though. Maybe ask James Sparks?

Comment by Aaron Bergman

2012-06-27T04:06:38Z

Sure it is. It's the total space of the trivial $C^3$ bundle over a point. I think this should still work for most of the toric CY singularities (the ones in hep-th/0602041, for example). (Also, I should have said canonical bundle above not anti-canonical). The other possibility, I think, is that the resolution of the singularity gives something one dimensional with a $C^2$ bundle over it (like the conifold) -- there might still be vanishing theorem that will work. cont'd.

Comment by Aaron Bergman

2012-06-27T00:52:23Z

Hi Yuji, if I can speculate a bit, I suspect the situation you want is that Z is the total space of a line V-bundle over a Fano orbifold (the easiest example is the total space of the anti-canonical V-bundle over a Fano surface). Then you can write the tangent space to the total space in terms of the tangent space of the base and push-forward everything to the base. My guess is that the result will follow from the Kodaira vanishing theorem for orbifolds (where the 2 arises because the Fano orbifold is at most a surface), but I don't have the time to work it through.

Comment by Aaron Bergman

2012-01-28T22:11:16Z

Presumably this math.sunysb.edu/~mlyubich/Archive/Geometry/…

Comment by Aaron Bergman

2011-11-03T00:06:01Z

Shouldn't this be something like Pontryagin duality for the multiplicative group of positive reals (maybe with some analytic continuation thrown in?).

Comment by Aaron Bergman

2011-10-14T00:27:37Z

This is the magic of moving from one dimension to two dimension, from particles to strings. Instead of having to put in your interactions by hand, you get them for free. They're completely determined by your choice of theory of 2D quantum gravity. Admittedly this isn't too far from what you're saying, but I think it's a better way to think about it. The CFT comes first, and you get an interacting theory automagically.

Comment by Aaron Bergman

2011-10-14T00:13:30Z

I just don't like second quantization as a way of thinking about field theory, but that's really besides the point here which is the difference between one and two dimensional manifolds. When you are looking at quantum gravity in one dimension, there's just not all that much going on. In two dimensions, things are a lot more interesting and there is the connection to 2D CFTs. You don't only quantize on the cylinder and not get a CFT; you have a 2D theory of quantum gravity that makes sense on all 2D manifolds. cont'd

Comment by Aaron Bergman

2011-09-08T23:10:32Z

This is an exercise in Gelfand-Manin, right? Section V.2, maybe?

Comment by Aaron Bergman

2011-05-08T03:58:57Z

Try a constraint which fixes one of the x_i to something other than K/n.

Comment by Aaron Bergman

2011-04-27T21:45:52Z

The action they wrote down is completely unambiguous. You could have written down a different one like S = 1/n \int F /\ F, but that wouldn't have the correct properties.