## November 16, 2006

### Segal on QFT

I spent a delightful afternoon, yesterday, discussing quantum field theory, the renormalization group, and such matters with Graeme Segal. Earlier, he gave a nice talk in the Geometry and String Theory seminar on his approach to QFT.

As you might expect, his formulation involves a Riemannian $d$-manifold, $X$, which is a cobordism between $d-1$ manifolds, $Y_1$ and $Y_2$. From that data, one imagines that a QFT is supposed to manufacture an isomorphism

(1)
$U_X:\, \mathcal{H}_{Y_1}\to \mathcal{H}_{Y_2}$

For a free, massive scalar field, $\mathcal{H}_{Y}= L^2_q\left(\Omega^0(Y)\right)$ the space of “square-integrable” functions on the space of smooth 0-forms on $Y$. I put a subscript “$q$” to remind us that “integration” over the topological vector space, $V= \Omega^0(Y)$, depends on a choice 1 of quadratic form $q\in\mathrm{Sym}^2 V^*$. This quadratic form, in turn, depends on the germ of a metric, $g$, on $X$, restricted to $Y$. $L^2_q\left(\Omega^0(Y)\right)$ is the quantization of the infinite-dimensional symplectic vector space, $\Sigma_Y = \Omega^0(Y)\oplus \Omega^{d-1}(Y)$ of “Cauchy data”, $(\phi,(*d\phi)|_Y)\in \Sigma_Y$.

Schematically, for $\Psi\in L^2_q\left(\Omega^0(Y_1)\right)$, $(U_X \Psi)(f_2) = \int_\substack{\phi:X\to \mathbb{R}\\ \phi_{Y_2}= f_2} e^{-S(\phi)} \Psi(\phi|_{Y_1}) \mathcal{D}\phi$

Graeme’s story is in Euclidean signature, but we can continue to complex metrics on $X$, so long as the quadratic form in $S(\phi)$ has positive definite imaginary part. The Minkowski theory arises (for $X=Y\times I$) at the boundary of this region in the space of complex metrics. There are several subtleties

• $L^2_q(V)$ is isomorphic to $L^2_{\tilde q}(V)$ if and only if $\tilde q q^{-1} -1$ is trace-class.
• Even then, the isomorphism is only projective. On one side, $\left\Vert e^{-\frac{1}{4}q}\right\Vert =1$; on the other, $\left\Vert e^{-\frac{1}{4}\tilde{q}}\right\Vert =1$. The above isomorphism gives the latter, only up to a factor of ${\det\left(\tfrac{\tilde{q}q^{-1}}{2\pi}\right)}^{-1/2}$
• To fix this, we set $\left\Vert e^{-\frac{1}{4}q}\right\Vert = {\text{det}_\zeta\left(\tfrac{q}{2\pi}\right)}^{-1/2}$ because the $\zeta$-function regulated determinant has the desired property $\det(\tilde{q}q^{-1}) \text{det}_\zeta(q) = \text{det}_\zeta(\tilde{q})$

All of the above is relatively standard stuff. Segal wants to go further, and study the notion of locality in this framework. To this end, he chops $Y^{d-1}$ into two pieces, $U_1$ and $U_2$, glued together on their common boundary, $Z^{d-2}$.

Naïvely, one is tempted to write

(2)
$\mathcal{H}_Y = \mathcal{H}_{U_1}\otimes \mathcal{H}_{U_2}$

with $\mathcal{H}_{U_i}= L^2(\Omega^0(U_i))$. But this is false2.

Let $\mathcal{G}_Y=\text{Heis}(\Sigma_Y)$ be the Heisenberg group, which is the central extension of $\Sigma_Y=V\oplus V^*$, $0 \to U(1) \to \text{Heis}(\Sigma_Y) \stackrel{\alpha}{\to} \Sigma_Y\to 0$ where $\alpha(g_\xi) = \xi$ and the group multiplication law is $g_{\xi_1}g_{\xi_2} = e^{i\omega(\xi_1,\xi_2)} g_{\xi_1+\xi_2}$ and $\omega(\cdot,\cdot)$ is the symplectic form on $\Sigma_Y$.

$\mathcal{H}_Y= L^2\left(\Omega^0(Y)\right)$ is an irreducible representation of $\mathcal{G}_Y$. It is highly reducible, as a representation of $\mathcal{G}_{U_1}\subset \mathcal{G}_Y$. Still the Reeh-Schlieder theorem tells us that $e^{-\frac{1}{4}q}$ is a cyclic vector for $\mathcal{G}_{U_1}$. In fact, $\mathcal{H}_Y$ is an irreducible representation of $\mathcal{G}_{U_1}\times \mathcal{G}_{U_2}\subset \mathcal{G}_Y$, but is not decomposable into (sums of) tensor products of irreps of $\mathcal{G}_{U_1}$ and $\mathcal{G}_{U_2}$. (Infinite dimensional groups are funny, that way…)

Still, there is something that makes sense, and replaces (2). It’s called the Connes Fusion Tensor Product. This was the punchline of the whole lecture and, alas, it went by too fast for me to give a coherent rendition of it. The idea is to let $V_1\cup V_2$ be a tubular neighbourhood of $Z$ (where $V_i\subset U_i$), and then let $\mathcal{H}_{L}= L^2\left(\Omega^0(U_1\cup V_2)\right)$ and $\mathcal{H}_{R}= L^2\left(\Omega^0(U_2\cup V_1)\right)$ be “thickened” versions of the Hilbert spaces we wrote before. The Connes Tensor product $\mathcal{H}_{L} \underset{\mathcal{G}_{V_1}\times \mathcal{G}_{V_2}}{*} \mathcal{H}_{R}$ can be completed to a Hilbert space isomorphic to $\mathcal{H}_Y$.

1 That is, we know how to do Gaussian integrals of the form $\int e^{- \frac{1}{2}q(v)} P(v) dv$ for some polynomial function, $P\in \mathrm{Sym}^n V^*$.

2 A pet peeve of mine is people who begin a discussion of the blackhole information paradox by drawing a spacelike slice ($Y$), dividing it into regions inside and outside the horizon ($U_1$, $U_2$), and writing $\mathcal{H}_Y = \mathcal{H}_{\text{in}}\otimes \mathcal{H}_{\text{out}}$.

Posted by distler at November 16, 2006 4:37 PM

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### Re: Segal on QFT

Re your pet peeve:

As we know by now, the cosmological constant is positive. Now, you turn the holographic crank and conclude (à la Banks) that the relevant Hilbert spaces are finite dimensional and all these subtleties go away! ;-)

Posted by: Robert on November 17, 2006 1:35 AM | Permalink | Reply to this

### À bas l’analyse fonctionnelle!

Now, you turn the holographic crank and conclude (à la Banks) that the relevant Hilbert spaces are finite dimensional and all these subtleties go away! ;-)

Well, yes, that would save me the trouble of having to really understand those accursed Type III factors, which are the source of all these difficulties!

Posted by: Jacques Distler on November 17, 2006 1:49 AM | Permalink | PGP Sig | Reply to this

### Re: Segal on QFT

I went to a great talk by Segal in Oz a few years ago, on Langlands. By the way, Lambda is ZERO. Come on, string theorists don’t want a lambda. Get rid of it.

Posted by: Kea on November 17, 2006 2:29 AM | Permalink | Reply to this

### Re: Segal on QFT

Hi Jacques,

Sorry to post a somewhat off-topic question, but you guys here probably have the answer I have been seeking for some time now:

Question:

What is “elliptic cohomology” exactly? And why is it considered so central to current theory physics research?

Thanks,
Brian

Posted by: Peter on November 17, 2006 4:12 AM | Permalink | Reply to this

### Elliptic Cohomology

What is “elliptic cohomology” exactly? And why is it considered so central to current theory physics research?

Who said it’s “central” to current theoretical physics research? Is that some weird “New Yorker map of the world” view of theoretical physics research?

Elliptic Cohomology is the biggest and baddest of the Generalized cohomology theories. A generalized cohomology theory satisfies all of the Eilenberg-Steenrod axioms, except for the dimension axiom: $\mathrm{H}^n(\text{point})= 0,\quad\forall n\gt 0$

Some generalized cohomology theories have a multiplicative structure, so that $\mathrm{H}^\bullet(X)$ forms a ring. Elliptic cohomology is one of those.

One characterization of elliptic cohomology is in terms of the formal group law satisfied by $\mathrm{H}^\bullet({\mathbb{C}P}^\infty)$ where ${\mathbb{C}P}^\infty=BU(1)$ is the classifying space of $U(1)$ bundles. $\mathrm{H}^\bullet({\mathbb{C}P}^\infty)$ is always a formal power series $\mathrm{H}^\bullet({\mathbb{C}P}^\infty)= H^\bullet(\text{point})[[t]]$ with a single generator, $t$, in degree 2 and coefficients in the ring $H^\bullet(\text{point})$.

For ordinary cohomology, this satisfies the formal group law $F(x,y) = x+y$ (multiplication is the cup product). For K-theory, the formal group law $F(x,y) = x+y + x y$ (multiplication is the tensor product of line bundles).

For elliptic cohomology, the formal group law is defined by an elliptic curve.

I think Urs had a nice description of all of this here.

Posted by: Jacques Distler on November 17, 2006 8:25 AM | Permalink | PGP Sig | Reply to this
Read the post Categorical Trace and Sections of 2-Transport
Weblog: The n-Category Café
Excerpt: A general concept of extended QFT and its relation to the Kapranov-Ganter 2-character.
Tracked: November 17, 2006 10:43 AM

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