Musings

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

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

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 $\mathcal{H}_Y$.