### Crossed Products and Not All That

#### Posted by Guest

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[The following entry is by A. Rivero. He volunteered # to report from a lecture series currently given by Alain Connes, concerned with the application of spectral geometry to the physics of the standard model #.
This entry here is a warmup. - urs]
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Hi, this is A. Rivero speaking. I am invited to the guest account mainly to follow the topic of Connes’s spectral action in a couple of postings, but while we wait for the Connes-Marcolli-Chamseddine paper, I am doing a bit of effort to train myself on crossed products and Morita equivalence.

You should know there is a tradition of that when a book is title “and all that” - it aims for rigourous theorems. So, note that the title of this post is “not all that”.

(Also I hope you are seeing rightly this post, this Solaris cluster has the default mathml.css and it is all Greek to me).

As for C*-algebras, the core of crossed product theory is nicely packaged in appendix 2.C of the red book. I can not find more detail in other books; of course not Pedersen’s, but even Blackadar’s has only about one or two extra examples.

We start with an algebra $A$ and a group $G$ acting on it, call this action $\alpha_g$ . For my examples, I am going to take $A$ a finite abelian algebra; you can think of it as the algebra of functions on a discrete space of $n$ points.

The cross-product of $A$ and $G$ is a new algebraic structure on the space of functions from $G$ to $A$. So you can think we have pasted two spaces, do you? Not at all! It is true that you have more coordinates: first you fix $g \in G$ to get an element $f(g) \in A$ and then, in our examples, you could fix a $k$ from $1$ to $n$ to get an actual value $f_k(g)$. But the point is that the whole algebra $A \rtimes_\alpha G$ is actually Morita-equivalent to a smaller one, usually smaller than the original $A$: we use cross products to quotient two spaces, not to multiply them.

Take $A$ to be the algebra of functions on a three point space, $G$ to be the ${\mathbb{Z}_3} = \{e^0, e^1,e^2\}$ group, and $\alpha$ the obvious action $(\alpha_{e^i} b)_k= b_{k+i}$. We can built a covariant representation of this structure via a unitary representation of the group, namely we have

$\pi(\alpha_{e_1}(b))= \begin{pmatrix} b_2 & & \\ & b_3 & \\ & & b_1 \end{pmatrix}= \pi_G(e_1) \pi(b) \pi(e_1)^{-1}= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} b_1 & & \\ & b_2 & \\ & & b_3 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$

and so on… Now from this covariant representation we can obtain a involutive representation of the cross product algebra, via the formula from the appendix (let me dispose randomly of some $\pi$s, just remember we are playing representations):

$\tilde f = \sum_{g\in G} f(g) g$ where summation is my awful own notation for a Haar measure in the group, mostly because I do not know what a Haar measure is, and partly because we are speaking of discrete groups. In this concrete example we get: $\tilde f= \sum_{g\in \mathbb{Z}_3}\begin{pmatrix} f_1(g) & & \\ & f_2(g) & \\ & & f_3(g) \end{pmatrix} \pi(g)=\begin{pmatrix} f_1(e^0) & f_1(e^1) & f_1(e^2) \\ f_2(e^2) & f_2(e^0) & f_2(e^1) \\ f_3(e^1) & f_3(e^2) & f_3(e^0) \end{pmatrix}$ Not surprisingly, an element $f \in A \rtimes \mathbb{Z}_3$ is specified by nine values, corresponding to the 3 times 3 combinations of “coordinate values” in the product space. But the nine values do not have a diagonal arrangement anymore: two elements $s, t \in A \rtimes \mathbb{Z}_3$ are to be multiplied exactly as the representation tell us, matrix-wise.

Thus it happens that the resulting algebra is the algebra of $3\times 3$ matrices, which is now to be Morita equivalent to the algebra of $1\times1$ matrices, which is the algebra of functions over a single point. So yeah, we have quotiented out the three points space with an action of the $\mathbb{Z}_3$, and we have got a single point. Amusingly, we have got it in a non commutative representation that “remembers” the size of the space(s) we started from.

Okay, I have no idea what Morita equivalence is. I am trying to learn it, remember. But roughly, it seems to me that the statement about a pair of $n\times n$ and $m \times m$ matrix algebras to be Morita-equivalent amounts to the existence of a pair of families of $m \times n$ and $n \times m$ matrices (forming not a pair of algebras, but a pair of modules) whose combinations can be used to built the previous pair. Exercise 1: proof that the quaternion algebra is Morita equivalent to a single $1\times 1$ matrix algebra (is it?) Exercise 2: Do some bibliography quest to understand why the algebra of compact operators on a infinite hilbert space is also Morita equivalent to the single $1\times 1$ matrix algebra. Note 1: In each Morita equivalence class, there is at most a commutative algebra. Note 2: the whole topic of Moritaness is related to theory of representations of algebras, groups, etc.

Do we need to guess the unitary representation in order to get the involutive one? Amusingly, not! And this is the content of the really strong formula of the start of the appendix: $(r * s)(g') =\sum_{g\in G} r(g) \alpha_g(s(g^{-1} g')$

When we apply it to our particular example, it does not appear to be a typical rows-times-columns matrix product arrangement… $(r * s)_i(e^k)= r_i(e^0) s_{i+0}(e^0 e^k) + r_i(e^1) s_{i+1}(e^{-1} e^k) + r_i(e^2) s_{i+2}(e^{-2} e^k)$ …but we can cheat because we have already worked out the answer, and we know that while the row is the $k$ index, the column is actually $k+i$ (modulus 3 to get the index right inside the matrix). So,

$\tilde f^c_i = f_i (e^{c-i})$

Thus the above product can be written: $(\tilde r \tilde s)_i^{i+k}= \tilde r_i^i \tilde s_i^{i+k} +\tilde r_i^{i+1} \tilde s_{i+1}^{i+k} +\tilde r_i^{i+2} \tilde s_{i+2}^{i+k}$

The amusing point being that the dummy index reponsible of the matrix multiplication does not start always from 1 in the way we have built the product, but it is still a dummy index and it covers all the components of the multiplication. At this point I will suggest to print this webpage, clean a corner of the blackboard, and enjoy yourself doing the multiplication trick.

Next example. Take A to be the algebra of functions over 6 points, which we imagine to be the points of the group $\mathbb{Z}_6$ in order to introduce the obvious action of $G=\mathbb{Z}_3$ as subgroup of $\mathbb{Z}_6$. (It is actually a nuisance that $A$ does not carry the extra information of being the algebra of functions from some group; and I guess this is the motivation to take cross-products seriously in the scope of Hopf-algebras. Annotation: remember to check if the cross product structure in quantum groups textbooks is the same as here).

We turn the same machinery and we get a algebra of 6x6 matrices whose elements are non zero only if row minus column is an even number. Or, the column index is now $c=i+2k$. Of course this algebra can to be arranged to see it to be the algebra of a diagonal 2x2 matrix whose components are 3x3 matrices. Again the Morita equivalence reduces this a 2x2 matrix whose componets are 1x1 matrices. So we get the algebra of functions over $\mathbb{Z}_2$, ie we have got $\mathbb{Z}_2$ from $\mathbb{Z}_6/\mathbb{Z}_3$… but the algebra remembers us that we started from $\mathbb{Z}_6$.

Can we obtain $\mathbb{Z}_2$ as the quotient $\mathbb{Z}/2\mathbb{Z}$? We can, in the same way that in the previous example, but now the two diagonal boxes are composed of compact, infinite-dim, operators.

Can we distinghish $(\mathbb{Z}/2\mathbb{Z})^n$ from $\mathbb{Z}^n/2\mathbb{Z}^n$? Here things start to be interesting, if we can. Also, as intermediate exersice, we could consider $\mathbb{Z}^2/2\mathbb{Z}$ against $\mathbb{Z} \times \mathbb{Z}_2$.

Can we distinguish $(\mathbb{Z}/\mathbb{Z})^n$ from $\mathbb{Z}^n/\mathbb{Z}^n$? It sounds funny, but I am actually interested on this, because the quotient of a lattice from its dual has amusing 8-periodic gauss sums, and the case of self-dual lattices is particularly interesting.

## Re: Crossed products and not all that.

There are two characterisations of Morita equivalence (at least that come to my mind immediately): $A$ and $B$ are Morita equivalent if they are equivalent after tensoring with the compact operators (in which are in some sense the $N\to\infty$ limit of the $N\times N$ matrices and alternatively the existence of a Hilbert-bi-module. It depends on your preferences which you take as a definition and which as a theorem.

As far as philosophy goes my understanding is that you cannot “geometrically” distinguish Morita equivalent algebras. In that sense they encode the same (noncommutative) space. Especially, K-theory (which measures the topology of that space) cannot see the difference. Furthermore the two spaces have the same points (irreps of the algebra as Morita equivalent algebras have the same representation theory) and closely related: The same fields (projective modules) live on the two spaces (as seen most easily from the Hilbert-bi-module perspective). Thus the two spaces look the same, feel the same, smell the same,…

BTW, Connes’ book is not the best place to learn new things, for in introduction I would recommend Landi’s review.

And maybe you should also mention that the noncommutative torus is the crossed product of $S^1$ with $Z$ acting by translations by $\theta$ and that (as discussed earlier) T-duality is the crossed product with $R$.