## October 16, 2006

### Crossed Products and Not All That

#### Posted by Guest

[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]

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.

Posted at October 16, 2006 1:13 PM UTC

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### 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 $R$.

Posted by: Robert on October 16, 2006 5:47 PM | Permalink | Reply to this

### Re: Crossed products and not all that.

Thanks, I forget! I was thinking to title this thread “non-interesting crossed products”. Why? Because an action of a discrete group $\mathbb{Z}_n$ in the algebra of functions of another discrete set is always closing the periods. And things begin to be interesting when we act on [the algebra of funcions of] the circle with rotations, so that a irrational rotation never closes, and voila, we enter the non-commutative torus fair.

Posted by: Alejandro Rivero on October 16, 2006 6:02 PM | Permalink | Reply to this

### Morita equivalence

In general, two rings/algebras $A$ and $B$ are Morita equivalent if

- their categories of (left) modules are equivalent

(1)${}_A \mathrm{Mod} \simeq {}_B\mathrm{Mod}$

- which happens precisely if there is an $A$-$B$-bimodule $N$ and a weakly inverse $B$-$A$-bimodule $M$, meaning that

(2)$N \otimes_B M \simeq A$

and

(3)$M \otimes_A N \simeq B \,.$

Hence Morita equivalence of algebras is just ordinary equivalence of objects in the bicategory $\mathrm{Bim}$, whose objects are rings/algebras, whose morphisms are bimodules with composition being the tensor product over the respective algebra, and whose 2-morphisms are bimodule homomorphisms.

$A$ and $B$ are Moriate equivalent if they are equivalent after tensoring with the compact operators

This is a special case of the above for $C^*$-algebras, under suitable conditions.

Posted by: urs on October 16, 2006 6:04 PM | Permalink | Reply to this

### Re: Morita equivalence

This second point is, of course, the one I retorted into the rectangular matrices argument. Hoping that even some undergrad could be tempted to try the game.

Posted by: Alejandro Rivero on October 16, 2006 6:35 PM | Permalink | Reply to this

### T-duality as a crossed product

[…] (as discussed earlier) T-duality is the crossed product with $\mathbb{R}$.

For those reader who don’t know what “as discussed earlier” refers to:

this somewhat surprising statement - that T-duality is, algebraically, nothing but orbifolding by a continuous group (the additive $\mathbb{R}^n$, usually) - is a theorem that plays a central role in an approach that is known as topological T-duality, which we talked about in a series of SCT posts #.

Posted by: urs on October 16, 2006 6:47 PM | Permalink | Reply to this

### Re: T-duality as a crossed product

In the general case, topologists consider the crossed product with $\mathbb{R}$ to be the algebraic equivalent of the suspension map, if I recall correctly. About $\rtimes \, \mathbb{R}^n$ I can not tell.

Posted by: Alejandro Rivero on October 16, 2006 10:54 PM | Permalink | Reply to this

### Re: Crossed products and not all that.

Oh, I forgot the most important property: The crossed product is what cloesed strings do in an orbifold: The elements of the crossed product can be thought of as functions with two slots: The first is a point upstairs and the otherone is a group element. This denotes the field of a closed string with beginning at that point where the end of the string is at that point up to a group transformation denoted by the group element in the second slot.

The step of Morita equivalence shrinking $|\Gamma|\times|\Gamma|$ matrices to a point corresponds to projecting to only the invariant states.

The multiplication law in the crossed product is of course nothing but string concatenation on the orbifold.

Posted by: Robert on October 16, 2006 5:52 PM | Permalink | Reply to this

### C*-algebra of a global orbifold

The elements of the crossed product can be thought of as functions with two slots: The first is a point upstairs and the otherone is a group element.

So these two “slots” determine a morphism of the action groupoid associated to a global orbifold. The crossed product $C^*$-algebra is the category algebra of that action groupoid.

We talked about that for instance here.

Posted by: urs on October 16, 2006 6:39 PM | Permalink | Reply to this

### Crossed products, groupoids or strings?

Hmm, I think that the examples coming from strings should be presented, as the ones starting this post, as particular cases of the general one. A string to be a string needs to have some extra requeriments. Say, to be the -constrained if you wish- solution of a dynamical equation coming variationally from a Lagrangian.

Posted by: Alejandro Rivero on October 16, 2006 10:57 PM | Permalink | Reply to this

### Re: Crossed products, groupoids or strings?

Of course this is not full blown string theory. What I tried to say was something along the lines of “if you put strings on orbifolds you typically get low energy effective actions where the interactions can be coded as crossed products in the Lagrangian”.

Posted by: Robert on October 17, 2006 12:37 PM | Permalink | Reply to this

### twisted sectors and paths in groupoids

I think that the examples coming from strings should be presented, as the ones starting this post […] A string to be a string needs to have some extra requeriments

What Robert alluded to # are known as “twisted sectors” in orbifold string theory, related to paths in a space $X$ with a global group $G$ acting on it, which don’t close in $X$ but do close in $X/G$.

For the present discussion the details of the string model that features such a (global) orbifold setup are immaterial. What matters is just

• the concept of the action groupoid accociated to the action of $G$ on $X$.

This is the groupoid whose objects are the points $x$ of $X$ and which has a morphism $x \stackrel{(x,g)}{\to} y$ if and only if $y$ is obtained by acting with $g \in G$ on $x$.

• the concept of a category of paths in that groupoid.

Here, morphisms are ordinary paths in $X$ freely combined with “jumps” along the morphisms described above.

For instance a typical path with one jump might look like

(1)$x \stackrel{\gamma_1}{\to} y \stackrel{(y,g)}{\to} z \stackrel{\gamma_2}{\to} w \,.$

A path $x \stackrel{\gamma}{\to}y$ may not be closed in $X$ (meaning $x \neq y$), but it may be closable as a path in the action groupoid as

(2)$x\stackrel{\gamma}{\to} y \stackrel{(y,g)}{\to} x \,,$

if $x$ is the result of acting with $g \in G$ on $y$.

You can find more details on this idea of paths in groupoids here.

The category algebra of the action groupoid (described here) is the algebra characterizing the quotient space $X/G$ (which is in general no longer a manifold but just a (global) orbifold).

The category algebra of paths in the action groupoid is something that in principle plays a big role in string (field) theory (see for instance the recent discussion here), but which is much more subtle to define precisely, because it involves an infinite-dimensional space of morphisms.

Posted by: urs on October 17, 2006 12:38 PM | Permalink | Reply to this

### Re: Crossed products and not all that.

You should know there is a tradition of that when a book is title “and all that” it aims for rigourous theorems

I believe the original one is “1066 and all that”, which I’m pretty sure did not.

Posted by: Allen Knutson on October 16, 2006 6:09 PM | Permalink | Reply to this

### Re: Crossed products and not all that.

See the Wikipedia article:
http://en.wikipedia.org/wiki/1066_and_All_That

Posted by: Maarten Bergvelt on October 16, 2006 8:43 PM | Permalink | Reply to this

### all that literature

I assume Alejandro was referring to texts like

Posted by: urs on October 17, 2006 9:28 AM | Permalink | Reply to this

### Re: all that literature

Indeed. But it is true that the oldest of them, “PCT, Spin and Statistics”, refers to the aforemtioned 1066 about the idea of having only “memorable results”, and then extends it to “do not contain theorems that do not exist”.

Posted by: Alejandro Rivero on October 17, 2006 5:57 PM | Permalink | Reply to this

### Re: Crossed products and not all that.

http://en.wikipedia.org/wiki/1066_and_All_That

Appallingly, that article doesn’t even mention Streater & Wightman!

But now it does, of course … and it even links to Urs’s helpful list above!

Posted by: Toby Bartels on October 17, 2006 9:10 PM | Permalink | Reply to this

### Re: Crossed Products and Not All That

A. Rivero writes:

Okay, I have no idea what Morita equivalence is. I am trying to learn it, remember.

Robert writes:

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.

It takes a while to love Morita equivalence - but after one understands a few definitions and how they’re equivalent, it starts seeming quite beautiful.

The above definitions both involve some analysis. I prefer a more algebraic definition, which works for algebras over any commutative ring, not just algebras over the real or complex numbers. Here’s how it goes - taken from week209:

Higher categories show up quite naturally in the study of commutative rings and associative algebras over commutative rings. I’d heard of things called “Brauer groups” and “Picard groups” of rings, and something called “Morita equivalence”, but I only understood how these fit together when I learned they were part of a marvelous thing: a weak 3-groupoid!

Here’s how it goes. You don’t need to know much about higher categories for this to make some sense… at least, I hope not. Starting with a commutative ring R, we can form a weak 2-category Alg(R) where:

• an object A is an associative algebra over R,
• a 1-morphism M: A → B is an (A,B)-bimodule,
• a 2-morphism f: M → N is a homomorphism between (A,B)-bimodules.

This has all the structure you need to get a 2-category. In particular, we can “compose” an (A,B)-bimodule and a (B,C)-bimodule by tensoring them over B, getting an (A,C) bimodule. But since tensor products are only associative up to isomorphism, we only get a weak 2-category, or “bicategory” - not a strict 2-category.

This weak 2-category has a tensor product, since we can tensor two associative algebras over R and get another one. And, all the stuff listed above gets along with this process! When an n-category has a well-behaved tensor product we call it “monoidal”, so Alg(R) is a weak monoidal 2-category. But using a standard trick we can reinterpret this as a weak 3-category with one object, as follows:

• there’s only one object, R
• a 1-morphism A: R → R is an associative algebra over R
• a 2-morphism M: A → B is an (A,B)-bimodule
• a 3-morphism f: M → N is a homomorphism between (A,B)-bimodules.

Note how all the morphisms have shifted up a notch. What used to be called objects, the associative algebras over R, are now called 1-morphisms. We “compose” them by tensoring them over R.

Next, recall a bit of n-category theory from “week35”. In an n-category we define a j-morphism to be an “equivalence” iff it’s invertible… up to equivalence! This definition may sound circular, but really just recursive. To start it off we just need to add that an n-morphism is an equivalence iff it’s invertible.

What does equivalence amount to in the 3-category Alg(R)? It’s easiest to figure this out from the top down:

• A 3-morphism f: M → N is an equivalence iff it’s invertible, so it’s an isomorphism between (A,B)-bimodules.
• A 2-morphism M: A → B is an equivalence iff it’s invertible up to isomorphism, meaning there exists N: B → A such that:
• M ⊗B N is isomorphic to A as an (A,A)-bimodule,
• N ⊗A M is isomorphic to B as a (B,B)-bimodule.
In this situation people say M is a Morita equivalence from A to B.
• A 1-morphism A: R → R is an equivalence iff it’s invertible up to Morita equivalence, meaning there exists a 1-morphism B: x → x such that:
• A ⊗R B is Morita equivalent to R as an associative algebra over R,
• B ⊗R A is Morita equivalent to R as an associative algebra over R.
In this situation people say A is an Azumaya algebra.

In week209 I look at the biggest 3-group sitting inside the monoidal 2-category Alg(R), and turn that 3-group into a space with only its first three homotopy groups nontrivial. Here’s the upshot:

• The 1st homotopy group consists of Morita equivalence classes of Azumaya algebras over R. This is also called the Brauer group of R.
• The 2nd homotopy group consists of isomorphism classes of Morita equivalences from R to R. This is also called the Picard group of R.
• The 3rd homotopy group consists of invertible elements of R. This is also called the unit group of R.
Posted by: John Baez on October 18, 2006 6:07 AM | Permalink | Reply to this

### Re: Crossed Products and Not All That

It takes a while to love Morita equivalence - but after one understands a few definitions and how they’re equivalent, it starts seeming quite beautiful.

Relating this to our discussion on perfected understanding (& subsequent comments including this one), in situations like this do you believe that time will show that there a best way of organising different manifestions of the same entity? It’s clear that heuristically speaking it’s generally good to believe that different presentations are consequences of one base account. If I point out to you three manifestations of the normal distribution - central limit theorem; maximum entropy distribution with fixed first two moments; approached by distribution which is the projection onto 1 dimension of a uniform distribution over the $n$-sphere of radius $\sqrt{n}$ as $n$ increases - it’s hard not to imagine that there’s a unified story behind the scenes.

Where in the discussion mentioned above, you took the side of a plurality of understandings, I recall elsewhere your suggestion that there might be, in essence, just one proof of a result such as that primes equal to 1 mod 4 are the sum of two squares.

Posted by: David Corfield on October 19, 2006 10:29 AM | Permalink | Reply to this

### Re: Crossed Products and Not All That

Amusingly we have got some layers to discuss about equivalency. On this level we have the issue of how different views of a [active] concept -different legs of a theorem- are the same entity, “Morita equivalence”. And in at least one of these views, Morita equivalence itself is about how different algebras are representations of the same topological space!

Posted by: A. Rivero on October 19, 2006 3:21 PM | Permalink | Reply to this

### Re: Crossed Products and Not All That

Exercise 1: proof that the quaternion algebra is Morita equivalent to a single 1×1 matrix algebra (is it?)

Let me answer this exercise, because it brings more questions. We all know that quaternions can be build from pairs of complex numbers, and the standard construction is to multiply a column vector $\begin{pmatrix}\Phi \\ \bar \Phi\end{pmatrix}$ times a row vector $(\Psi, \bar \Psi)$ so we get the cuaternion $\begin{pmatrix} \Phi \\ \bar \Phi\end{pmatrix} \times (\Psi, \bar \Psi)= \begin{pmatrix} \Phi \Psi & \Phi \bar \Psi \\ \bar \Phi \Psi & \bar \Phi \bar \Psi \end{pmatrix} \in \mathbb{H}$

Now, notice: the set of column vectors of this kind is a left $\mathbb{H}$-module, the set of row vectors is a right $\mathbb{H}$-module, its product $\otimes_\mathbb{H}$ gives a number $(\Phi \Psi + \bar \Phi \bar \Psi) \in \mathbb{R}$ and in fact both sets are also $\mathbb{R}$-modules (but no $\mathbb{C}$-modules).

So the sets of column and row vectors as above (ie composed of a complex number and its conjugate) are respectively $\mathbb{H}-\mathbb{R}$ and $\mathbb{R}-\mathbb{H}$ left-right modules. Their $\otimes_\mathbb{R}$ product gives the quaternions, their $\otimes_\mathbb{H}$ product gives the reals. Conclusion: The algebra of quaternions is Morita equivalent to the algebra of real numbers.

But here comes the doubt: the algebra of real numbers is not Morita equivalent to the algebra of complex numbers; there is only a commutative algebra in each equivalency class. So, can we think that both algebras describe the same geometry or not? Time to check for some lecture notes on Gelfand (or Stone) duality.

Posted by: Alejandro Rivero on October 18, 2006 1:19 PM | Permalink | Reply to this

### Re: Crossed Products and Not All That

Time to check for some lecture notes on Gelfand (or Stone) duality

I am finding Rickart’s “General Theory of Banach Algebras” extremely soul-healing. It touches the complexification at two levels, first in the general case, then next chapter in the *-involutive case. It is a pity it is too old to include Morita equivalence.

It is misterious why the standard model needs the quaternions instead of other matrix algebra. I wonder if there is some explicit cross product getting the algebra of cuaternions as a result of quotienting some algebra of $n \times n$ complex matrices.

Posted by: Alejandro Rivero on October 22, 2006 4:49 PM | Permalink | Reply to this

### Ex. 1, answer is wrong, be careful

I just noticed that I missed a sign, so the 2x2 matrix does not represent a quaternion. I wonder in what dream I got the “We all know…” part :-(

Of course this solves a doubt I got afterwards, which is that Baez in some “weeks” invokes Morita equivalence classes of reals and quaternions, and they happen to be different. So Ok, they are.

Still, the example stands in the sense it builds an algebra, call it, ahem, $\bar H$, morita equivalent to the reals. It is sort of a mock algebra, because it has a imaginary unit $i^2=-1$ and two extra generators $j,k$ with $j^2=k^2=1$. Note here the sign mistake.

Posted by: Alejandro Rivero on November 6, 2006 5:37 PM | Permalink | Reply to this
Read the post Knowledge of the Reasoned Fact
Weblog: The n-Category Café
Excerpt: In a comment I raised the question of what to make of our expectation that behind different manifestations of an entity there is one base account, of which these manifestations are consequences. If I point out to you three...
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