The String Coffee Table

Ordinary representations of finite dimensional Lie groups can be multiplied using ordinary tensor products. For representations of loop groups however, the ordinary tensor product is ill suited, in particular because it does not preserve the level of the representation.

The right notion of product between loop group representations is that which is induced by the ‘fusion’ of primary fields in the corresponding conformal field theory. Hence the trick is to use the ‘operator/state correspondence’: translate a vector in a Hilbert space into an operator which generates that vector from the ‘vacuum vector’ and work with this operator instead of the original vector.

More precisely, this works as follows:

The representation of a loop group on some Hilbert $H$ space induces a von Neumann algebra $A$ of operators on that space. Hence we can regard $H$ as a left module for $A$. Since $H$ is also a Hilbert space, by assumption, this is called a Hilbert module. Many of the subtleties of Connes Fusion are due to the fact that we want to work with (bi)modules of algebras which at the same time are Hilbert spaces.

A special example of such a Hilbert space on which $A$ acts (from the left) can be obtained from the space of operators $A$ itself: fix any (faithful and normal) state $\varphi$ on $A$, i.e. a linear map

This induces an inner product $\langle \cdot ,\cdot ,{\rangle }_{\varphi }:A\times A\to ℂ$ on $A$ defined by

Here $x$ and $y$ are operators in $A$ and $y^{*}$ (or $y^{†}$, if you prefer) is the adjoint of $y$ (with respect to the inner product on $H$).

In general, $(A,\langle \cdot \cdot {\rangle }_{\varphi })$ will not be a Hilbert space itself, since it will not be complete with respect to the norm induced by that inner product. Hence one takes the completion of $M$ with respect to this norm and calls the result

The notation here is derived from the case where $A$ is the commutative von Neumann algebra $L^{\mathrm{\infty }}(M)$ of functions on some measure space $M$. Completing this with respect to some state yields the usual space $L^2(M)$ of square integrable functions.

The Hilbert space $L_{\varphi }^2(A)$ has a special vector, namely the identity operator $1\in A$, or rather its image $\hat{1}\in L^2(A)$. This is called the vacuum state

and $L_{\varphi }^2(A)$ is called a vacuum representation of $A$.

This now allows to define something like operator-state correspondence.

Let $K$ be any other (left) Hilbert module of $A$, i.e. a Hilbert space with a left action of $A$ on it. Consider the space

of linear operators from $L_{\varphi }^2(A)$ to $K$ that are compatible with the left $A$ action (i.e. which are left $A$-module homomorphisms). This space has a natural inclusion in $K$, obtained by sending every operator to the state obtained by applying it to the vacuum:

One nice thing about the space of operators $D(K,\varphi )$ is that, like every space of module homomorphisms, it is naturally a bimodule over $A$ itself. But bimodules have the nice property that we can form their tensor products over $A$.

The Connes fusion of two Hilbert $A$-modules $K$ and $L$ is defined to be the Hilbert space obtained by thus tensoring the respective operator spaces of $K$ and $L$ in the above sense:

In order for this to be not just a bimodule but a Hilbert bimodule we again need to specify an inner prodcut and take the completion with respect to that inner product. The natural inner product in the above space is similar to the 4-point function in CFT.

This is easily described once a certain obvious structure is made manifest: given two operators $x$ and $y$ in $D(K,\varphi )$, we get an element in $A$ (a linear operator from $H$ to $H$) simply by going from $H$ to $K$ using $x$ and then returning to $H$ using $y^{*}$. Let’s denote this as

and call it an $A$-valued inner product on $D(K,\varphi )$.

Given that, we may define an inner product on $D(K,\varphi ){\otimes }_{ℂ}D(K,\varphi )$ as

Here $(x_1,x_2)\cdot y_1$ denotes the left action of the operator $(x_1,x_2)\in A$ on $y_1$ regarded as an $A$-bimodule element.

This is like the ‘4-point’-function for the four operators $x_1,x_2,y_1,y_2$.

Finally, Connes fusion

of two Hilbert $A$-(bi)modules $K$ and $L$ is defined as the result of first forming the space $D(K,\varphi ){\otimes }_{ℂ}D(L,\varphi )$ and then completing with respect to the inner product given by the above ‘4-point function’.

The point of it all is that $K{\hat{\otimes }}_{A}L$ is (just) a slight variation of the naive tensor produc $K{\otimes }_{A}L$. The reason for the difference is a little twist which is introduced due to the fact that the inclusion of $D(K,\varphi )$ in $K$ given by the operator-state corespondence

respects the left action of $A$, but not the right action. In other words, it is not a bimodule homomorphism, just a one-sided module homomorphism. In fact, it is a bimodule homomorphism only up to a certain twist. This can be found in equation (4.3.7) of StolzTeichner.