What is the motivation for a vertex algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:47:19Z http://mathoverflow.net/feeds/question/53988 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra What is the motivation for a vertex algebra? AH 2011-02-01T14:52:34Z 2011-02-02T14:46:13Z <p>The mathematical definition of a vertex algebra can be found here:</p> <p><a href="http://en.wikipedia.org/wiki/Vertex_operator_algebra" rel="nofollow">http://en.wikipedia.org/wiki/Vertex_operator_algebra</a></p> <p>Historically, this object arose as an axiomatization of "vertex operators" in "conformal field theory" from physics; I don't know what these phrases mean.</p> <p>To date, I haven't been able to gather together any kind of intuition for a vertex algebra, or even a precise justification as to why anyone should care about them <em>a priori</em> (i.e. not "they come from physics" nor "you can prove moonshine with them").</p> <p>As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:</p> <p>What is the basic physical phenomenon/problem/question that vertex operators model?</p> <p>What is the subsequent story about vertex operators and conformal field theory, and how can we see that this leads naturally to the axioms of a vertex algebra?</p> <p>Are there accessible physical examples ("consider two particles colliding in an infinite vacuum...", etc.) that illustrate the key ideas?</p> <p>Also, are there alternative, purely mathematical interpretations of vertex algebras which make them easier to think about intuitively?</p> <p>Perhaps people who played a role in their discovery could say a bit about the thinking process that led them to define these objects?</p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/53991#53991 Answer by J Bass for What is the motivation for a vertex algebra? J Bass 2011-02-01T15:18:34Z 2011-02-01T15:18:34Z <p>This may hold you off until a REAL answer comes along. I found something resembling intuition on this subject when I first studied Quantum Field Theory. I'd recommend looking there before trying to tackle the many-headed beast that is conformal field theory. Many (but not all) of the VOA axioms can be seen in the properties of a QFT. As far as a "mathematician-friendly" place to learn about QFT goes I remember liking the book by Ryder. </p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/53995#53995 Answer by Charles Matthews for What is the motivation for a vertex algebra? Charles Matthews 2011-02-01T16:09:02Z 2011-02-01T16:09:02Z <p>Name-dropping here, but I can claim that Richard Borcherds discussed vertex algebras with me, before anyone knew what they were (not that I was able to be of any help). At that stage they were purely algebraic objects, with some kind of "Jacobi identity". At some point he said he had a definition as Lie algebra in an internal sense in a category - this idea was binned as not in fact useful. Much later he said something about a relationship with the Wightman axioms, but I gather that isn't really watertight when it comes down to it. So I doubt whether the emergence of an axiomatic theory can be "rationally reconstructed" without doing some damage to the history.</p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54004#54004 Answer by Yuji Tachikawa for What is the motivation for a vertex algebra? Yuji Tachikawa 2011-02-01T17:42:25Z 2011-02-01T18:22:36Z <p>Speaking as a string theorist, I would say Victor Kac's book "Vertex Algebras for Beginners" followed fairly closely what physicists originally thought. So you might want to have a look at it.</p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54008#54008 Answer by David Ben-Zvi for What is the motivation for a vertex algebra? David Ben-Zvi 2011-02-01T18:07:07Z 2011-02-01T20:58:31Z <p>Vertex algebras precisely model the structure of "holomorphic one-dimensional algebra" -- in other words, the algebraic structure that you get if you try to formalize the idea of operators (elements of your algebra) living at points of a Riemann surface, and get multiplied when you collide.</p> <p>Our geometric understanding of how to formalize this idea has I think improved dramatically over the years with crucial steps being given by the point of view of "factorization algebras" by Beilinson and Drinfeld, which is explained (among other places :-) ) in the last chapter of my book with Edward Frenkel, "Vertex algebras and algebraic curves" (second edition only). This formalism gives a great way to understand the algebraic structure of local operators in general quantum field theories -- as is seen in the recent work of Kevin Costello -- or in topological field theory, where it appears eg in the work of Jacob Lurie (in particular the notion of "topological chiral homology"). </p> <p>In fact I now think the best way to understand a vertex algebra is to first really understand its topological analog, the structure of local operators in 2d topological field theory. If you check out any article about topological field theory it will explain that in a 2d TFT, we assign a vector space to the circle, it obtains a multiplication given by the pair of pants, and this multiplication is commutative and associative (and in fact a commutative Frobenius algebra, but I'll ignore that aspect here). It's very helpful to picture the pair of pants not traditionally but as a big disc with two small discs cut out -- that way you can see the commutativity easily, and also that if you think of those discs as small (after all everything is topologically invariant) you realize you're really describing operators labeled by points (local operators in physics, which we insert around a point) and the multiplication is given by their collision (ie zoom out the picture, the two small discs blend and look like one disc, so you've started with two operators and gotten a third).</p> <p>Now you say, come on, commutative algebras are SO much simpler than vertex algebras, how is this a useful toy model? well think about where else you've seen the same picture -- more precisely let's change the discs to squares. Then you realize this is precisely the picture given in any topology book as to why $\pi_2$ of a space is commutative (move the squares around each other). So you get a great model for a 2d TFT by thinking about some pointed topological space X.. to every disc I'll assign maps from that disc to X which send the boundary to the basepoint (ie the double based loops of X), and multiplication is composition of loops -- i.e. $\Omega^2 X$ has a multiplication which is homotopy commutative (on homotopy groups you get the abelian group structure of $\pi_2$). In homotopy theory this algebraic structure on two-fold loops is called an $E_2$ structure. </p> <p>My claim is thinking about $E_2$ algebras is a wonderful toy model for vertex algebras that captures all the key structures. If we think of just the mildest generalization of our TFT story, and assign a GRADED vector space to the circle, and keep track of homotopies (ie think of passing from $\Omega^2 X$ to its chains) we find not just a commutative multiplication of degree zero, but a Lie bracket of degree one, coming from $H^1$ of the space of pairs of discs inside a bigger disc (ie from taking a "residue" as one operator circles another). This is in fact what's called a Gerstenhaber algebra (aka $E_2$ graded vector space). Now all of a sudden you see why people say you can think of vertex algebras as analogs of either commutative or Lie algebra (they have a "Jacobi identity") -- -the same structure is there already in topological field theory, where we require everything in sight to depend only on the topology of our surfaces, not the more subtle conformal geometry.</p> <p>Anyway this is getting long - to summarize, a vertex algebra is the holomorphic refinement of an $E_2$ algebra, aka a "vector space with the algebraic structure inherent in a double loop space", where we allow holomorphic (rather than locally constant or up-to-homotopy) dependence on coordinates.</p> <p>AND we get perhaps the most important example of a vertex algebra--- take $X$ in the above story to be $BG$, the classifying space of a group $G$. Then $\Omega^2 X=\Omega G$ is the "affine Grassmannian" for $G$, which we now realize "is" a vertex algebra.. by linearizing this space (taking delta functions supported at the identity) we recover the Kac-Moody vertex algebra (as is explained again in my book with Frenkel). </p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54029#54029 Answer by J Verma for What is the motivation for a vertex algebra? J Verma 2011-02-01T22:28:16Z 2011-02-01T22:28:16Z <p>I'll take a more physical POV. Although, we don't have any mathematically concrete definition of QFT in general, but we can define 2d Conformal Field theory (this is attributed to infinite symmetries and exact solvability), other that CFTs we can also define Topological QFTs (Atiyah).Greame Segal proposed a geometric definition of CFT. In Conformal field theory we deal with vertex operators analog to operators in QFT, we can write a Taylor like expansion of two vertex operators, k.a Operator product expansions (OPEs) which gives the QFT analog of two fields interacting. All these notions are captured by the axiomatization based on Vertex Algebras. To see a better picture,look at the classic paper of Belavin, Polyakov, and Zamolodchikov, where an algebraic approach of CFT was proposed. </p> <p>Recently Kapustin and Orlov proposed a more general definition of Vertex algebras and they showed the relation between their algebraic definition and Segal's geometric one.</p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54035#54035 Answer by Jeff Harvey for What is the motivation for a vertex algebra? Jeff Harvey 2011-02-01T23:39:32Z 2011-02-01T23:39:32Z <p>The answers here have focused on the mathematical aspects of VOAs and the motivation coming from QFT, the specialization to Conformal Field Theory, and then the further specialization to two-dimensional holomorphic CFT. Two-dimensional CFT's did arise from string theory where the fields on the 2d world-sheet of the string define a CFT. However there is another important part of the story which has not been mentioned and is more directly tied to physical phenomenon and that is the theory of critical phenomenon. Many systems, such as water-ice, magnetic systems and so on undergo phase transitions as a thermodynamic parameter such as temperature is varied. Typically these are first order transitions, meaning that there is a latent heat associated to the transition. Sometimes one can vary an additional parameter and find a line of first order transitions which terminates at a second order transition. The second order transition is characterized by fluctuations on all scales: the theory becomes scale and conformally invariant at that point. It also turns out that the behavior of thermodynamic quantities as one approaches the critical point are characterized by numbers called critical exponents which are universal for systems with the same symmetry structure. These exponents are related to what are called the conformal dimensions of operators in CFT and they are directly measurable in the lab for a variety of systems. One important tool which was used in the study of critical phenomenon is the operator product expansion or operator algebra of K. Wilson and L. Kadanoff. There is a huge literature on this. Here is a reference to an early paper on the operator algebra for the Ising model: <a href="http://prb.aps.org/abstract/PRB/v3/i11/p3918_1" rel="nofollow">http://prb.aps.org/abstract/PRB/v3/i11/p3918_1</a> . VOA's are a rigorous mathematical formalization of this kind of algebraic structure. For someone who wants to learn about CFT starting from a particular physical system (or at least a mathematical idealization of a physical system) the Ising Model is a good place to start. </p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54096#54096 Answer by Tim van Beek for What is the motivation for a vertex algebra? Tim van Beek 2011-02-02T13:39:31Z 2011-02-02T13:39:31Z <p>Rex asked:</p> <p>"*As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:</p> <p>What is the basic physical phenomenon/problem/question that vertex operators model?*"</p> <p>Please have a look at the nLab at the pages that have already been mentioned:</p> <p>ncatlab.org/nlab/show/vertex+operator+algebra and </p> <p>ncatlab.org/nlab/show/conformal+field+theory</p> <p>There is also a lot of material about QFT there.</p> <p>Here is a quick and dirty and oversimplified introduction to some aspects of QFT:</p> <p>In the Wightman framework we can think of quantum fields as operator valued distributions, that map test functions living on a given spacetime to (maybe unbounded) operators. Essentially selfadjoint operators represent observables that can be measured, in principle, by some experiment or device. Physicsists are mostly interested in systems that have interactions. We can think of elementary particles as localized excitations of quantumf fields that are solutions to specified wave equations (in the distributional sense). In order to have any interactions, these wave equations should have non-linear terms (actually you can define the notion of "free field" (a field with no interactions whatsoever) this way: A free field in the sense of physicists is a quantum field that is a solution of a linear equation).</p> <p>Non-linear terms entail products of quantum fields, which are undefined, in general, because products of distributions are undefined, in general. The history of QFT is about the struggle of physicists to dodge this problem in one way or another. One way to dodge this problem is to introduce, as an axiom, so called "(associative) operator product expansions". This is a way to formulate, as an axiom, (handwaving:) that the "severity of the singularity of products of distributions" is under control.</p> <p><strong>operator product expansion</strong></p> <p>An operator product expansion (<strong>OPE</strong>) for a family of fields means that there is for all fields and all $z_1 \neq z_2$ a relation of the form $$\psi_i(z_1) \psi_j(z_2) \sim \sum_{k \in B_0} C_{ijk} (z_1 - z_2)^{h_k - h_i - h_j} \psi_k(z_2)$$</p> <p>Here $\sim$ means modulo regular functions. (See ncatlab.org/nlab/show/conformal+field+theory for an explanation of the notation.)</p> <p>In a handwaving way, this axiom says that we assume we know something about the kind of singularity of the product of two fields, as their support comes closer and closer, and it is not so bad.</p> <p>A rigorous interpretation of an OPE would interpret the given relation as a relation of e.g. matrix elements or vacuum expectation values.</p> <p>An OPE is called <strong>associative</strong> if the expansion of a product of more than two fields does not depend on the order of the expansion of the products of two factors. <strong>Warning</strong>: Since the OPE has no interpretation as defining products of operators, or more generally the product in a ring, the notion of associativity does not refer to the associativity of a product in a ring, as the term may suggest.</p> <p>The axioms of vertex operator algebras are an <strong>axiomatization of OPEs</strong>. In this sense there is reason to expect that vertex operator algebras will play a key role, on one way or another, in a rigorous construction of interacting QFTs in four dimensions.</p> http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54099#54099 Answer by Richard Borcherds for What is the motivation for a vertex algebra? Richard Borcherds 2011-02-02T14:46:13Z 2011-02-02T14:46:13Z <p>The original motivation for vertex algebras is explained briefly in the original paper <a href="http://www.jstor.org/stable/27441" rel="nofollow">http://www.jstor.org/stable/27441</a> as follows. For any even lattice one can construct a space $V$ acted on by vertex operators corresponding to lattice vectors. More generally one can write down a vertex operator for every element of $V$. These vertex operators satisfy some complicated relations, which are then used as the definition of a vertex algebra. In other words, the original example of a vertex algebra was the vertex algebra of an even lattice, and the definition of a vertex algebra was an axiomatization of this example. </p> <p>This was motivated by my attempt to understand Igor Frenkel's work on the Lie algebra with Dynkin diagram the Leech lattice. Frenkel had constructed a representation of this Lie algebra using vertex operators acting on the space $V$, and I was trying to use his work to understand its root multiplicities. I did not use any insights from conformal/quantum/topological field theory or operator product expansions when defining vertex algebras (as implied by some of the other answers), for the simple reason that I had barely heard of these concepts and had almost no idea what they were. </p> <p>This is not all that helpful for understanding what a vertex algebra really is. A better view is to regard them as something like a commutative ring with an action of the (formal) 1-dimensional additive group. In particular any such ring is canonically a vertex algebra. The difference between rings with group actions and vertex algebras is that the "multiplication" of a vertex algebra from $V\times V$ to $V$ is not defined everywhere: it can be thought of as a "rational map" rather than a "regular map" in some sense. More precisely if we write $u^g$ for the action of a group element $g$ on $u$, then the product $u^gv^h$ is not defined for all group elements $g$ and $h$, but behaves formally like a meromorphic function of $g$ and $h$, which in particular may have poles when $g$ and $h$ are trivial. Making sense of this informal idea produces the definition of a vertex algebra. (For more details see the unpublished paper <a href="http://math.berkeley.edu/~reb/papers/bonn/bonn.pdf" rel="nofollow">http://math.berkeley.edu/~reb/papers/bonn/bonn.pdf</a>.) This means that vertex algebras behave like commutative rings: for example, one should define modules over them, tensor products of modules, and so on. </p>