## June 2, 2006

### Mathai on T-Duality, III: Algebraic Formulation

#### Posted by urs The second part of my transcript of V. Mathai’s talk. Some facts about ${C}^{*}$-algebras

Let $A$ be some ${C}^{*}$-algebra with the action of a (locally compact) group $H$ on it

(1)$\alpha :H\to \mathrm{Aut}\left(A\right)$

such that

(2)$a↦{\alpha }_{h}\left(a\right)$

is norm continuous for all $a\in A$.

In such a situation there is something called the crossed product algebra

(3)$A{⋊}_{\alpha }H\phantom{\rule{thinmathspace}{0ex}}.$

At least for $H$ a finite group, forming the crossed product represents forming the global orbifold quotient of an $H$ space by $H$ ($\to$).

Here, however, we will implement T-duality on a space by forming the crossed product of the ${C}^{*}$-algebra representing that space with an infinite (abelian) euclidean group.

(Robert Helling, too, was in the audience, and particularly emphasized that this is somewhat remarkable, since it seems to suggest that there is a way to think of T-duality as orbifolding by a continuous group, in some sense. We didn’t come up with a good answer to this issue.)

So here is how to define that crossed product algebra.

Let ${C}_{c}\left(H,A\right)$ be the space of compactly supported $A$-valued functions on $H$. (Think of this as an $H$-graded algebra).

For ${f}_{1}$, ${f}_{2}$ in ${C}_{c}\left(H,A\right)$, their product is given by

(4)$\left({f}_{1}*{f}_{2}\right)\left(h\right)={\int }_{H}\phantom{\rule{thickmathspace}{0ex}}{f}_{1}\left(g\right)\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{g}\left({f}_{2}\left({g}^{-1}h\right)\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{dg}\phantom{\rule{thinmathspace}{0ex}},$

where the integral runs over the group $H$.

There is also a $*$-operation on ${C}_{c}\left(H,A\right)$, given by

(5)${f}^{*}\left(g\right)=\Delta \left(g{\right)}^{-1}\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{g}\left(f\left({g}^{-1}{\right)}^{*}\right)\phantom{\rule{thinmathspace}{0ex}},$

where $\Delta :H\to {ℝ}^{+}$ is the modular function relating the left and right Haar measure on $H$.

The crossed product algebra

(6)$A{⋊}_{\alpha }H$

is defined to be the completion of ${C}_{c}\left(H,A\right)$ is some universal norm.

Given this, there are a couple of theorems that we shall need.

Green’s theorem (version 1).

Let $A$ be a ${C}^{*}$-algebra with action $\alpha$ of $H$, and $H$ a normal closed subgroup of a group $G$, then one can form the induced ${C}^{*}$-algebra

(7)$\begin{array}{rl}B& ={\mathrm{Ind}}_{H}^{G}\left(A,\alpha \right)\\ & \left\{f:G\to A\phantom{\rule{thickmathspace}{0ex}}\mid \phantom{\rule{thickmathspace}{0ex}}f\left(t+g\right)=\alpha \left(g\right)\left(f\left(t\right)\right)\right\}\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

This induced algebra $B$ has an action $\beta$ of $G$. The theorem says that the two crossed products we thus get are Morita equivalent:

(8)$A{⋊}_{\alpha }H\stackrel{\text{Morita}}{\simeq }B{⋊}_{\beta }G\phantom{\rule{thinmathspace}{0ex}}.$

Green’s theorem (version 2).

With assumptions as before, let now $H,K\subset G$ be two subgroups of $G$. Then we have an equivalence of the form

(9)$C\left(G/K\right)⋊H\phantom{\rule{thickmathspace}{0ex}}\stackrel{\text{Morita}}{\simeq }\phantom{\rule{thickmathspace}{0ex}}C\left(H\G\right)⋊K\phantom{\rule{thinmathspace}{0ex}}.$

The Connes-Thom isomorphism

Let $A$ be a ${C}^{*}$-algebra with an action $\alpha$ of $G={ℝ}^{d}$, then there is a natural isomorphism of algebraic K-theories

(10)${K}_{j}\left(A\right)\phantom{\rule{thickmathspace}{0ex}}\simeq \phantom{\rule{thickmathspace}{0ex}}{K}_{j+d}\left(A{⋊}_{\alpha }{ℝ}^{d}\right)\phantom{\rule{thinmathspace}{0ex}}.$

(This will realize T-duality on the algebraic level. ${ℝ}^{d}$ should be thought of as the universal cover of the $d$-dimensional torus ${T}^{d}$ which is being T-dualized.)

Takai duality.

Let $A$ be a ${C}^{*}$-algebra with action $\alpha$ of $G={ℝ}^{d}$. Then $A{⋊}_{\alpha }{ℝ}^{d}$ has a natural action $\stackrel{̂}{\alpha }$ of the Pontryagin dual group ${\stackrel{̂}{ℝ}}^{d}$ given by

(11)$\left({\stackrel{̂}{\alpha }}_{g\prime }\left(f\right)\right)\left(g\right)=〈g,g\prime 〉\phantom{\rule{thinmathspace}{0ex}}f\left(g\right)$

for all $f\in {C}_{c}\left({ℝ}^{d},A\right)$, all $g\in {ℝ}^{d}$ and all $g\prime \in {\stackrel{̂}{ℝ}}^{d}$.

We have the following equivalence

(12)$A\phantom{\rule{thickmathspace}{0ex}}\stackrel{\text{Morita}}{\simeq }\phantom{\rule{thickmathspace}{0ex}}A{⋊}_{\alpha }{ℝ}^{d}{⋊}_{\stackrel{̂}{\alpha }}{\stackrel{̂}{ℝ}}^{d}\phantom{\rule{thinmathspace}{0ex}}.$

(This will imply that applying T-dualiy twice gets us back, up to Morita equivalence, to the original background.)

Before proceeding, let’s note the following remarks.

First of all, recall that the higher K-groups ($\to$) are defined in the algebraic setup by the formula

(13)${K}_{j}\left(A\right)={K}_{0}\left(A\otimes C\left({ℝ}^{j}\right)\right)\phantom{\rule{thinmathspace}{0ex}},$

where ${K}_{0}\left(A\right)$ gives the Grothendieck group of finitely generated projective modules for $A$ (to be thought of as vector bundles), while $C\left(X\right)$ denotes simply algebra of the continuous functions on $X$.

Second, note that when $H$ acts trivially on $A$, then the crossed product turns into the tensor product

(14)$A{⋊}_{\text{trivial}}H\simeq A\otimes {C}^{*}\left(H\right)\simeq A\otimes {C}_{0}\left(\stackrel{̂}{H}\right)\phantom{\rule{thinmathspace}{0ex}},$

where, for $H={ℝ}^{d}$, the second equivalence is given by Fourier transformation.

(Did I explain all the notation? Given a space $X$, $C\left(X\right)$ is the algebra of continuous functions on it. Given a group $G$, ${C}^{*}\left(G\right)$ is the ${C}^{*}$ algebra with product the convolution product on the group.)

Given all this, we can now

rephrase T-duality in terms of noncommutative topology:

We use the above theorems to establish three facts.

1) the algebraic action of T-duality

Consider the algebra

(15)$C\left(M×{T}^{n}\right)=C\left(M×{ℝ}^{n}/{ℤ}^{n}\right)$

describing a spacetime being a trivial $n$-torus bundle.

The group ${ℝ}^{n}$ acts on this in the obvious way (by translating along the torus cycles). So we may algebraically T-dualize by passing to the crossed product algebra

(16)$C\left(M×{ℝ}^{n}/{ℤ}^{n}\right)⋊{ℝ}^{n}\phantom{\rule{thinmathspace}{0ex}}.$

Now, by the second version of Green’s theorem we know that the result of this operation is Morita equivalent to

(17)$\cdots \simeq C\left(M×{ℝ}^{n}\{ℝ}^{n}\right)⋊{ℤ}^{n}\phantom{\rule{thinmathspace}{0ex}}.$

But ${ℝ}^{n}\{ℝ}^{n}$ is the trivial group - a point. So we are left with

(18)$\cdots \simeq C\left(M×\right)⋊{ℤ}^{n}\phantom{\rule{thinmathspace}{0ex}}.$

But here ${ℤ}^{n}$ acts on a point, hence trivially. Therefore, by the above remark, the crossed product here is equivalent to the tensor product

(19)$\cdots \simeq C\left(M\right)\otimes {C}^{*}\left({ℤ}^{n}\right)\phantom{\rule{thinmathspace}{0ex}}.$

However, the convolution algebra ${C}^{*}\left({ℤ}^{n}\right)$ is nothing but the algebra on the dual torus

(20)$\cdots \simeq C\left(M\right)\otimes C\left({\stackrel{̂}{T}}^{n}\right)=C\left(M×{\stackrel{̂}{T}}^{n}\right)\phantom{\rule{thinmathspace}{0ex}}.$

All in all, we find that the ${C}^{*}$-algebras of two T-dual torus bundles are equivalent

(21)$C\left(M×{T}^{n}\right)\simeq C\left(M×{\stackrel{̂}{T}}^{n}\right)\phantom{\rule{thinmathspace}{0ex}}.$

(Given that a torus and a dual torus are topologically the same, this doesn’t really sound all that shocking. As in the same talk in Vienna before ($\to$) there have been questions this time concerning the issue whether or not the ${C}^{*}$ algebras can distinguish a torus of radius $R$ from that of a radius $1/R$. V. Mathai seemed to claim that they can, while other people wondered if this did not require a full spectral triple, hence a noncommutative notion of metric geometry. I guess that how the ${C}^{*}$-algebras themselves depend on the radius is a subtle issue of langue, because, after all, the above says that the torus and the dual torus have, in fact, Morita equivalent algebras. Probably the subtlety is precisely in that higher-order notion of equality here. They are equivalent, but not equal. )

2) the action of algebraic T-duality on K-classes

The Connes-Thom isomorphism implies that the K-theory of the algebra of a torus bundle is ismorphic to that of its T-dual bundle (up to a shift) in the following way

(22)${K}_{j}\left(C\left(M×{ℝ}^{n}/{ℤ}^{n}\right)⋊{ℝ}^{n}\right)\simeq {K}_{j+n}\left(C\left(M×{ℝ}^{n}/{ℤ}^{n}\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

2) the involution property of algebraic T-duality

The Takai duality implies that applying algebraic T-duality twice gets us back, up to Morita equivalence, to the original background

(23)$C\left(M×{ℝ}^{n}/{ℤ}^{n}\right)⋊{ℝ}^{n}⋊{\stackrel{̂}{ℝ}}^{n}\stackrel{\text{Morita}}{\simeq }C\left(M×{ℝ}^{n}/{ℤ}^{n}\right)\phantom{\rule{thinmathspace}{0ex}}.$

So that’s how topological T-duality (in the absence of background $H$-flux) works in the language of commutative ${C}^{*}$-algebras.

Before considering nonvanishing $H$-flux and noncommutative $*$-algebras, we conclude the present section by saying how this motivates the following very abstract definition of T-duality.

Abstract definition of T-duality

Let $A$ be an algebra in some category $C$, such that we can associate algebraic K-theory ${K}^{•}\left(A\right)$ to $A$. Then any functor

(24)$T:C\to C$

shall be called an abstract T-duality if

1) ${K}^{•}\left(A\right)\simeq {K}^{•+r}\left(T\left(A\right)\right)$

and

2) $A\phantom{\rule{thickmathspace}{0ex}}\stackrel{\text{Morita}}{\simeq }\phantom{\rule{thickmathspace}{0ex}}T\left(T\left(A\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

Posted at June 2, 2006 2:01 PM UTC

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