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Apologies - a better explanation than I started with - thanks to people for helping. It is obvious that there are many bad cases for rank - the problem is are there a reasonable number of good cases?

A rank for fgp modules (fgp = finitely generated projective, as appropriate) over a unital C* algebra $A$ can be given by using a trace (from $A$ to complex numbers) of the matrix trace of the associated projection matrix. The problem is when this is multiplicative for bimodules - i.e. when the rank of $N \otimes_A M$ is the rank of $N$ times the rank of $M$. (It is likely easier to look at $N \otimes_A M$ in terms of dual bases rather than the original projection matrices.)

Classically this is just the rank of a vector bundle - and for locally trivial bundles there is no reason why this is multiplicative unless the topological space is connected (just have different local ranks on different components - a trace would somehow average this into a single number). We either need some condition on the bundle over different components, or an assumption that the space is connected.

The problem is to take a suitable unital algebra (with some form of connectedness condition) and show that there is a reasonable class of fgp modules with multiplicative ranks. I guess that $C^*$ algebras with normalised tracial states and some form of connectedness condition would be good. It is very likely that some regularity condition would be needed on the modules apart from being fgp.

Another way to look at the multiplicative problem: Take a dual basis $e^i\in E$ and $e_i\in E'$. Then the projection matrix is $P_{ij}=e_i(e^j)$. Given a dual basis for $F$, $f^i\in F$ and $f_i\in F'$, a projection matrix for $F\otimes_A E$ is $e_i(f_j(f^k).e^r)$. Now the map $\mathcal{P}_E:A\to M_n(A)$ given by $\mathcal{P}_E(a)=e_i(a.e^j)$ is an algebra map, sending 1 to $P$. Given the matrix trace and $\tau$ a trace on $A$, we have a trace $\tau\circ\mathrm{trc}$ on $M_n(A)$. Now if we had $\tau\circ\mathrm{trc}\circ\mathcal{P}_E:A\to\mathbb{C}$ being a fixed multiple of $\tau$ I guess we would have multiplicativity... [I seem to be editing my own question again - should I put this in an answer to my own question - I am unused to protocols hereabouts...]

[Another edit.] As far as I can see for multiplicativity, the key is that $\tau\circ\mathrm{trc}(P)$ should be independent of the algebra normalised trace $\tau$ (at least for a large number of traces). Then $\tau\circ\mathrm{trc}\circ\mathcal{P}_E:A\to\mathbb{C}$ is another (not normalised) trace, and applying this to the projection matrix for the other module should give the result. But what could give $\tau\circ\mathrm{trc}(P)$ independent of $\tau$? Is that the same as saying that just using the relation $\tau(xy)=\tau(yx)$ is enough to get $\tau$ applied to a multiple of the identity?

Remarks (1) there is such a rank for objects in a rigid braided monoidal ribbon category (see Majid's book), but that is too much machinery for the present case.

(2) A useful measure of `nice' modules from Morita theory is that the evaluation map $:E'\otimes_A E\to A$ is onto - this translates into the existence of suitably sized matrices $U,V$ so that the matrix trace of $UPV$ is the identity in the algebra.

(3) If there is a projection matrix (equivalently a dual basis) so that the matrix trace of the projection is a multiple of the identity, then (a) the rank is independent of the algebra trace, and (b) I suspect that the multiplicative result is true. However, this condition is likely to be too strong.

(4) By Blackadar's book, for a $C^*$ algebra $P$ can be chosen to be Hermitian, so for a tracial state the rank is positive. When is the matrix trace of $P$ (a positive algebra element) invertible?

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2 Answers 2

up vote 4 down vote accepted

Here is an answer to the second question (item (4) of the list in the question): The set of possible "matrix traces" will contain an invertible element if and only if the projection (or the module) is full. That is, the entries of the projection span the algebra as a two sided ideal in $A$. (This is also equivalent to the condition mentioned in item (2) of the list.)

Let me assume for simplicity that the projection $p\in A$ (rather than $M_n(A)$). If $$ p=\sum_{i=1}^n x_i^*x_i $$ and $$ a=\sum_{i=1}^n x_ix_i^* $$ then $a$ is a possible matrix trace for $pA$. Indeed, with $X=(x_1,x_2,\dots,x_n)$ we get $p=X^*X$ and $XX^*$ has matrix trace equal to $a$.

Now suppose that $p$ spans $A$ as a two-sided ideal. This entails the existence of $d_i\in A$ such that

$$ d_1^*pd_1+d_2^*pd_2+\cdots d_n^*pd_n=1 $$

Let $d=\sum_{i=1}^n d^*_id_i$, $x_i=\frac{1}{ \|d\|^{1/2}}d_ip$, and $y=(1-\frac{d}{\|d\|})^{1/2}p$. Then $$ \sum_{i=1}^n x_i^*x_i+y^*y=p(\frac{d}{\|d\|})p+p(1-\frac{d}{\|d\|})p=p. $$ On the other hand, $$ \sum_{i=1}^n x_ix_i^*+yy^*=1/\|d\|+(1-d/\|d\|)^{1/2}p(1-d/\|d\|)^{1/2}. $$ The right hand side is invertible, since we are adding a positive multiple of the identity to a positive element.

This proves the "harder" of the two implications. One can see that the possible matrix traces of a projection are always inside the 2-sided ideal generated by the projection (or its entries in the matrix case), which proves the other implication.

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One remark regarding the question on bimodules: If $M$ is an $A$-bimodule such that for another $A$-bimodule $N$ one has that $M\oplus N\cong A^n$ (as bimodules), then I believe that the projection $P$ from $A^n$ to $M$ will have all its entries in the center of the algebra $A$ (following from the fact that $P$ is a bimodule morphism). So this projection is really a projection over the center. Then, the multiplicative property for the trace will be true (since the trace restricts to a trace on the center), if, say, the center has connected spectrum. But not for a good reason. –  Leonel Robert Dec 8 '12 at 21:44
    
Thanks for that - it ties up several loose ends! –  Edwin Beggs Dec 10 '12 at 4:21
    
If I accept an answer, does it take the original question out of consideration? I would rather not do that, as I would really like to find out about the multiplicativity (paragraph added in question about another way to look at that). So much e=protocol to learn... Anyway, I am now happier - thanks. –  Edwin Beggs Dec 10 '12 at 5:05
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If the trace on $A$ is multiplicative then I think the answer to the first question is yes. That should be an easy computation. On the other hand, let $p$ be a central projection in $A$ and take $M = pA$ and $N = (1-p)A$. Then $M \otimes_A N = 0$ but the ranks of $M$ and $N$ are $\tau(p)$ and $1 - \tau(p)$, respectively. This suggests that unless $\tau$ is multiplicative (which would force $\tau(p) = 0$ or $1$) the answer to the main question is probably no.

I'm not sure how to parse the auxiliary question. My best guess is that you mean to ask whether there exist modules $M$ such that however we realize $M$ as $PM_n(A)$ for $P$ a projection in $M_n(A)$ we find that $tr(P)$ is a multiple of the identity. If so, I suppose the answer is yes, this happens if $M$ is free, but probably not otherwise.

But this is kind of a simplistic answer, so probably I have misunderstood the entire question ...

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Hi! Thanks for that. I am hoping that the trace being multiplicative is not needed. Here is an observation from Morita contexts: Modules and their duals are in some sense "well behaved" if the evaluation map from $E'\otimes E$ to $A$ is onto. In the case of a unital algebra this means that for a fin gen proj module with projection P we have matrices U,W of appropriate size so that the algebra valued matrix trace UPW is one. I suspect that this may be relevant... For the last bit, I suspect that the answer depends on a choice of P. –  Edwin Beggs Dec 6 '12 at 20:41
    
Hmm. I don't think the evaluation map from $E' \otimes E$ to $A$ being onto helps here. E.g. take $M = pA \oplus A$ and $N = (1-p)A \oplus A$. These modules satisfy your condition; their ranks are now $\tau(p) + 1$ and $2 - \tau(p)$ (assuming $\tau$ is normalized), but the tensor product is $A \oplus pA \oplus (1-p)A \cong A \oplus A$, with rank $2$. So you'd still need $\tau(p) = 0$ or $1$. –  Nik Weaver Dec 7 '12 at 0:59
    
I guess I still don't understand the second question. You mean you think $PM_n(A) \cong A^k$ does not imply that $tr(P) = k\cdot I$? Seems to me it does ... just checking the case $k = 1$ should suffice. –  Nik Weaver Dec 7 '12 at 1:01
    
The question would benefit from being expanded (e.g., what seems like a crucial hypothesis for the first part was relegated to a parenthetical remark). @Nik: For the second question, are you also assuming that P is central here? Otherwise I can't see how this is true. P could be in A and such that PA is isomorphic to A (e.g., with P in O_2). –  Leonel Robert Dec 7 '12 at 1:53
    
@Leonel: no, $P$ need not be central, however we are talking about isomorphism of $A$-modules, not isomorphism of C*-algebras. Does that make more sense? –  Nik Weaver Dec 7 '12 at 3:24
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